1

PH 3001: QUANTUM MECHANICS I (45L, 3C)

Dependencies: PH 1001 is recommended

Syllabus: Inadequacies of classical physics and evolution of quantum physics; Particles and wave packets; Heisenberg uncertainty principle and its consequences, some illustrations of uncertainty principle; Wave function and its interpretation, position probability density, superposition principle; Time-dependent Schrodinger equation; Conservation of probability, probability current density; Dirac bracket notation; Linear operators and their properties: eigenvalues and eigenfunctions of operators, Hermitian operators, adjoint operator; Expansions in eigenfunctions: Orthogonality, degeneracy, probability amplitudes, discrete and continuous spectra; Commutators, commuting observables, compatibility; Expectation values; Time-independent Schrodinger equation, stationary states; Energy quantisation; Properties of the energy eigenfunctions; General solution of the time-dependent Schrodinger equation; Solutions of the time-independent Schrodinger equation for a particle moving in a region of zero potential, step potential, barrier potential, finite square well potential, infinite square well potential, linear harmonic oscillator potential and square box potential; Symmetry and parity; One-electron atoms: separation of the time-independent Schrodinger equation in spherical polar co-ordinates, energy levels, quantum numbers, degeneracy, eigenfunctions of the bound states, probability densities; Orbital angular momentum and orbital magnetic dipole moment of electron; Stern-Gerlach experiment, existence of spatial quantisation, spin angular momentum and spin magnetic dipole moment of electron; Spin-orbit interaction; Total angular momentum; Spin-orbit interaction energy and the hydrogen energy levels; Transition rates and selection rules

Assessment: Mid & End of semester written examinations

References: Quantum Physics – R. Eisberg and R. Resnick

Quantum Physics – S. Gasiorowicz

Quantum Physics – M.S. Rogalski and S.B. Palmer

Introduction to Quantum Mechanics – B.H. Bransden and C.J. Joachain

Quantum Mechanics – R.L.Liboff

Introduction to Quantum Mechanics – D.J. Griffiths

Introduction: Transition from classical to quantum theory was required mainly due to experimental findings during

the period from late 1800s to early 1900, such as

• the interaction of electromagnetic radiation with matter was not entirely according to the laws

of electromagnetism as developed by Ampere, Laplace, Henry, Maxwell, etc

• the development of the theory of atomic structure of matter, as a result of the discovery of the

electron and the confirmation of the nuclear model of the atom

• the deviation of the experimentally measured motion of subatomic particles from what was

predicted under the assumptions of Newtonian mechanics.

2

To explain the new observations, what is known today as the "Quantum Theory" was developed. In

this course we will first briefly review the phenomena which forced physicists to introduce the new

concepts: the particle properties of radiation, the wave properties of matter and the quantization of

physical quantities, which led to the transition from classical to quantum theory. Simply speaking,

quantum theory (or quantum mechanics) is a physical science that deals with the behavior of matter

and energy on the scale of atoms and subatomic particles/waves. Blackbody radiation and Planck's hypothesis (Birth of Quantum Physics) Blackbody: A surface that totally absorbs all the radiation falling on it.

( )E ν - Monochromatic energy density (Spectral Distribution Function).

Max Planck (in 1900) derived the expression ( ) ( )3 3( ) 8 / 1 1h kTE hv c e νν π = − using the following

model for the radiating atoms.

* Each atom behaves as a harmonic oscillator oscillating with some frequency ν . * Each oscillator can absorb or emit radiation energy only in an amount E hν= .

(The classical e/m theory permits a continuous absorption of energy of an oscillator ∝ (amplitude)2).

The possible (quantized) values of the energy of an oscillator, nE nhν= , where n = 1, 2, 3, ….

Planck's expression agreed well with experimental values of ( )E ν at different temperatures. However, it was

not compelling enough for the conclusion of the quantum (particle) nature of radiation.

Photoelectric Emission - ejection of electrons from a surface by the action of light (Heinrich Hertz, 1887)

Irradiation of certain metals with visible and ultraviolet light showed that,

1. the photoelectron current, when exists, increases with an increase in the intensity of the incident

radiation.

2. for each substance, there is a minimum (threshold) frequency of the e/m radiation such that below

which no matter how intense the radiation, no photoelectrons are produced.

ν

E(ν) Rayleigh-Jeans formula

Wein formula

Planck formula

3

3. the kinetic energy of the photoelectrons is independent of the intensity of the light source but varies

linearly with the frequency of the incident radiation.

4. the time delay between the irradiation and the emission ~ 10−9 s (in classical theory) but it is <<<

10−9 s in actual observation. P.e. effect could not be explained using only classical theories. In 1905, Albert Einstein considered the radiation

to consist of a collection of quanta of energy (particle nature of radiation) and proposed that the energy absorbed

by an electron in a single process (in a time <<< 10−9 s) from e/m radiation of frequency ν is hν .

Then, if φ is the energy required by a photoelectron to escape from a metal by overcoming the

Coulomb potential barrier, the kinetic energy of the escaping electron is k = νE h φ−

The value of φ does not depend on the electron energy and its value is not the same for all electrons in the

metal. The minimum value of φ (= 0φ ) is called the work function of the metal. ( )k 0maxE hν φ⇒ = −

⇒ There is a minimum value of 0( )ν ν= such that 0 0 0hν φ− = ⇒ 0 0= hν φ

0ν is called the threshold frequency for photoemission.

• In the photon picture (particle nature) of radiation, the intensity is proportional to the number of photons

per unit area per unit time

⇒ The photo current is proportional to the source intensity (explains the 1st observation)

• When 0 <ν ν ⇒ ( )k max0E < ⇒ No photoelectric emission (explains the 2nd observation)

• k =E hν φ− ⇒ kE ∝ ν (explains the 3rd observation)

An important contribution to the acceptance of the quantum nature of radiation came from the theoretical

work of Einstein for explaining the photoelectric effect

The Compton Effect - Scattering of radiation by free electrons (Arthur Compton in early 1920’s) It provided the most direct evidence for the particle nature of radiation. When e/m radiation of wavelength 0λ

was scattered by free electrons through a given angle, the scattered beam consisted of two components, one with

the wavelength 0λ and the other with a wavelength λ′ shifted relative to 0λ by an amount that depends on the

scattering angle. This was not consistent with classical radiation theory, according to which the incident

radiation should be re-radiated by electrons and the intensity observed at an angle θ should vary as

2(1 cos )θ+ , independent of the wavelength of the incident radiation. Compton explained the modified

component by treating the incoming radiation as a beam of photons of energy 0hν , with individual photons

scattering elastically off individual electrons. (In an elastic collision, both momentum and energy are

conserved.)

By interpreting the above as a collision between the electron and a particle of zero rest mass called a photon

4

(energy, 0E hc λ= and momentum, 0p h λ= ), Compton showed that ( )( )0 1 coseh m cλ λ θ′ − = − ,

θ is the photon scattering angle, C eh m cλ = - Compton wavelength of the electron.

The measurements of the modified component agreed very well with the values predicted by the above theory.

The Balmer series In 1895, Balmer found an empirical formula which represents the wavelength of lines in the visible and near UV

regions in the line spectrum of atomic hydrogen.

The nine lines known at that time were found to satisfy the formula ( )2 2 4C n nλ = − , where C = 3646 Å,

n = 3, 4, 5, …….. for the Hα, Hβ, Hγ, ............ lines.

The existence of atomic line spectra which exhibit regularities could not be explained by the atomic models

based on classical physics. Then, N. Bohr explained this by introducing the quantum concept into the physics of

atoms.

The Bohr Model of the Hydrogen Atom Thomson model of the atom :- electrons are embedded in a continuous distribution of positive charge.

Scattering of α-particles by thin foils (Geiger and Marsden in 1908) showed occasional large angle scattering (1

in 10000).

Rutherford's nuclear model of the atom in 1911:- all of the positive charge and most of the mass of the atom are

concentrated in a region (the nucleus ~ 10−14 m) that is small compared with the dimensions of the atom (~ 10−10

m).

This model explained α-particle scattering quantitatively. However it could not account for the spectra of

radiation from atoms and the stabilising mechanism for atoms. Rutherford's model requires the electrons to

travel in planetary orbits about the nucleus. According to e/m theory, when a charged particle is accelerating in

a circular orbit its energy is lost as e/m radiation at a rate 2 2 306q a cπε . Then, in a time of the order of 10−10 s,

the electron should spiral into the nucleus resulting in a collapse of the atom.

θ Incident Photon

Electron

Recoiled electron

Scattered photon

For a photon,

E h hcν λ= =

p E c h λ= =

5

To overcome the above two difficulties, N. Bohr in 1913 assumed circular orbits for electrons and only a certain

set of stable orbits (stationary states) are allowed in contrast to the infinite number of orbits possible in classical

mechanics. His model was based on the two postulates.

(i) An electron in a stable orbit does not radiate e/m energy and radiation can only take place when a

transition is made between the allowed energy levels, b a abE E hν− = .

(ii) The magnitude of the orbital angular momentum of the electron moving in a circular orbit around the

nucleus can take on the values L n= , where n = 1, 2, 3, ...

Assuming circular orbits for electrons and if the nuclear charge of a one-electron (hydrogen-like) atom is Ze+ ,

and the radius of the orbit is r , by equating the Coulomb force with the centripetal force acting on the electron

⇒ 2 2 204 eZe r m v rπε = .

From Bohr's 2nd postulate, eL m vr n= = , we get 20(4 )v Ze nπε= and 2 2 2

0(4 ) er n Zm eπε= . The total energy of the electron 2 2 2 4 2 2 2

0 02 (4 ) 2 (4 )n e eE m v Ze r m Z e nπε πε= − = − .

Note: As n →∞ ⇒ 0nE → . n - principal quantum number. n = 1 - Ground state (K shell), n = 2, 3, 4, .... - Excited states (L, M, N, .... shells)

This discrete set of allowed electron energy levels explained the existence of regularities in atomic line spectra.

However, the Bohr model could not account for the lines in emission spectra of many-electron atoms, the

difference in their intensities and splitting in some of the lines. Correspondence Principle: -

The limiting cases of quantum mechanical results are the classical physics results. The limit is reached when the

quantum numbers are large. Ex: - Show that the above principle is satisfied by Bohr atomic model.

Consider the frequency emitted when an electron makes a transition from the orbit with quantum number n to

the orbit with quantum number 1n − , when n is large.

Classically, an electron moving in a circular orbit with velocity v would emit a radiation of frequency 23 2 3

02 ( 2 )( 4 ) (1 )cl er m Ze nν ν π π πε= = .

According to the Bohr model, the frequency of the emitted radiation 23 2 2 2

1 0( ) / ( 4 )( 4 ) 1 ( 1) 1n n eE E h m Ze n nν π πε− = − = − − .

When n ,1>> we have 2 2 31 ( 1) 1 2n n n − − ≈ and therefore clν ν→ .

Stationary States and their Experimental Evidence by Frank & Hertz Experiment (in 1914)

To explain the discrete (well-defined) frequency spectrum of atomic systems, Bohr assumed that the energy of

an atom can have only certain values (levels) 1 2 3, , .......E E E - quantization of energy. The states

corresponding to these energies are called stationary states.

6

( ) ( )i j i jh A E A Eν + → ( ) ( )j i jiA E A E hν→ +

Conservation of energy ⇒ ji j ih E Eν = − and ij j ih E Eν = − , and hence the frequencies which

constitute the absorption spectrum ( i jν ) are the same as those observed emission spectrum ( jiν ) of the system.

Consider the inelastic collision experiment (a part of the kinetic energy of the projectile is transferred as internal

energy to the target) between electrons and mercury atoms. *

1 2( ) ( )Gfast slowA E q A E q+ → + , (energy transferred to the target = 2 1E E− )

Let, 1E - ground state energy, 2E - first excited state energy, 212kE mv= - kinetic energy of the fast projectile

before collision, (Assume that, mass of the atom >> mass of the projectile ⇒ no recoiling of the atom.)

2 1kE E E< − ⇒ Collision is all elastic and the electron moves through the vapour, losing energy very slowly.

2 1~kE E E− ⇒ Collision is inelastic, the electron may lose ( 2 1E E− ) amount of energy in a single

encounter and the atom is excited to the first excited state. [k.e. of the projectile after the

collision, 2 1( )k kE E E E′ = − − ]. After exciting an atom, the energy is insufficient to

excite the other atoms and the electron may undergo elastic collisions.

12 EEEk −>> ⇒ It may suffer few more inelastic collisions, losing the energy ( 2 1E E− ) at each collision

and producing more excited atoms. (It is also possible to excite an atom to the second

excited state from the ground state)

In their experimental arrangement, a heated filament (F) emits electrons which are accelerated towards the grid

(G) by a variable potential V. The space between F and G is filled with mercury vapour. Between G and the

collecting plate P, a small retarding potential /V (~ 0.5 V) is applied so that those electrons which are left with

very little k.e. after one or more inelastic collisions can not reach the plate and are not registered by the

galvanometer. As V is increased the plate current I fluctuates with peaks occurring at a spacing of about 4.9 V.

E1

E2

E3

E4

E5

Absorption Emission

7

The first dip corresponds to electrons that lose all their k.e. after one inelastic collision with a Hg atom, which is

then left in an excited state. The second dip corresponds to those electrons that suffered two inelastic collisions

with two Hg atoms, losing all their k.e. and so on. The excited Hg atoms return to their ground state by emitting

a photon ( *Hg Hg hν→ + , with 2 1h E Eν = − ) of wavelength 2536 Å (corresponding to

4.86 eVhν = ), which has been observed spectroscopically as coming from the Hg vapour during the

passage of the electron beam. This experiment is one of the most striking proofs of the existence of stationary

states.

Particles and Fields The objects we touch and see have a well-defined shape and size and therefore localized in space. When the

above picture is used for fundamental particles (eg:- e, p, n, etc) as having shape and size (as well as mass and

charge) we get into a difficulty in explaining certain experimental results.

To describe the motion of fundamental particles, Louis de Broglie (in 1924) proposed to associate a ‘matter field

(wave)’ with each particle which describes the dynamical condition (momentum and energy) of the particle. [In

reverse, Planck had associated a particle (a photon) having precise momentum and energy with an e/m field.]

De Broglie assumed that the nature is symmetric and h pλ = can be applied to material particles as well as to

photons.

The wavelength λ and frequency ν of the matter wave associated with a particle of momentum p and energy

E are given by h pλ = and E hν = . λ - ‘de Broglie wavelength’

Wave number, 2k π λ= , angular frequency 2ω πν= ⇒ p k= and E ω= , where

342 1.0544 10 J.sh π −= = × .

If the above assumption is correct, whenever the motion of a particle is disturbed so that the matter wave

associated with it can not propagate freely, interference and diffraction phenomena should be observed as in the

case of e/m waves.

When the electrons are accelerated by a potential V , the gain in kinetic energy ⇒ 2 2 eeV p m= .

F

Galvanometer

/ ~ 0.5 VV

G P

V 5 10 15

4.9 V

V

8

De Broglie wavelength, 92 1.23 10 meh p h m eV Vλ −= = = ×

For 4 12~ 10 V ~ 10 mV λ −⇒ ~ wavelength of X-rays (e/m waves)

i.e. if one sends a beam of fast electrons through a crystal, he should obtain diffraction patterns resulting from

matter waves similar to those obtained for X-rays (by Bragg and others).

Davison and Germer carried out an experiment similar to one that Bragg did to observe X-ray diffraction, with

the X-ray beam replaced by an accelerated electron beam. They found that the electron current registered by the

detector was a maximum every time the Bragg condition, 2 sind nθ λ= , derived for X-rays was satisfied.

Particles and Wave Packets

Consider a free particle moving along the x-axis with a precisely known momentum p (≡ px) and

energy 2 2E p m= . As the matter wave of this particle, one can associate a harmonic wave with w.l.

h pλ = and freq. E hν = ⇒ i( )( , ) sin 2 ( ) sin( ) e k x tx t A x t A kx t A ωπ λ ν ω −Ψ = − = − = .

Ψ - Instantaneous value of the matter wave, A – amplitude, (k ≡ kx)

Prob. of finding the particle within a unit distance at a given place and time, 2 2 ConstantAΨ = = .

⇒ Position of the particle on the x-axis (from −∞ to +∞) is completely unknown, i.e. localization is not possible

for a free particle of well-defined (precisely known) momentum.

So, the harmonic wave represents the idealized situation of a particle of precisely known momentum moving

somewhere in a beam of infinite length.

The phase (or wave) velocity of the de Broglie wave, ( )( )pv k E h h p E pω νλ= = = =

If v is the non-relativistic velocity of the particle, 2 2E mv= , p mv= ⇒ 2pv v= ≠ particle velocity.

Now, consider a particle localized within a certain region x∆ of space. Then, the amplitude of its associated

matter wave should be large in that region and very small outside it. Such a wave can be formed by

superimposing (through interference) an infinite number of harmonic matter waves, each with infinitesimally

differing and kω (or and ν λ ) values. Then, the amplitude of the wave is modulated so that its value is non-

zero only over a finite region of space, in the vicinity of the particle. Such a superposition will result in a group

of waves or wave packet.

Detector

Incident electron beam

Diffracted electron beam

Crystal

9

The (group) velocity of propagation of the packet, d d d dgv k E p p m vω= = = = = particle velocity (since

p k= , E ω= and 2 2E p m= )

⇒ The group must move with the same velocity as the particle, along the x-axis.

Note: One cannot measure the phase velocity of a pure harmonic wave directly and can only measure the group

velocity of the waves.

Consider a superimposition of two harmonic waves (the simplest case) of equal amplitudes and slightly different

frequencies and wavelengths.

1( , ) sin( )x t A kx tωΨ = − ( )2 ( , ) sin ( d ) ( d )x t A k k x tω ωΨ = + − +

Their resultant, 1 2( , ) ( , ) ( , ) 2 cos[(d 2) (d 2) ]sin( )x t x t x t A k x t kx tω ωΨ = Ψ + Ψ = − −

(since d 2ω ω and d 2k k ) and its amplitude has been modulated to form a group.

The second term of Ψ is a wave of the same form as 1Ψ but Ψ is modulated by the first term so that its

oscillations fall within an envelope of periodically varying amplitude.

kdπ2

λ2π =k

),( txΨ t = 0 w

g

x 0

λ Vg

∆x Continuous wave train corresponding to an unlocalized particle

Wave packet of extension ∆x corresponding to a particle localized within the distance ∆x

10

Velocity of an individual wave is given by sin( )kx tω− as pv kω= = 2 (2 )πν π λ νλ=

Velocity of the ‘group’ is given by cos[(d 2) (d 2) ]k x tω− as (d 2) (d 2) d dgv k kω ω= =

For the group of matter waves associated with the particle, d dgv E p p m v= = = = particle velocity

Heisenberg's Uncertainty Principle for Position and Momentum Consider the wave packet (group of waves) associated with a particle, with an extension ~ x∆ , range

of wave numbers of interfering waves ~ xk∆ .

From Fourier analysis, it can be shown that 12. xx k∆ ∆ ≥ . [Derivation is not necessary]

x xp k= ⇒ . 2xx p∆ ∆ ≥ ← Heisenberg Uncertainty Principle

Similar relationships can be written for particle motion in y- and z- directions. Other acceptable forms of writing the H.U.P.

. xx p∆ ∆ . ~xx p∆ ∆ . xx p∆ ∆ h . ~xx p h∆ ∆

Physical interpretation

If a particle is within the region 12x x− ∆ and 1

2x x+ ∆ (where x∆ is the uncertainty in the position),

its associated wave packet is represented by superposing waves of momenta between 12x xp p− ∆ and

12x xp p+ ∆ (where xp∆ is the uncertainty in the momentum) so that . 2xx p∆ ∆ ≥ .

OR

The relationship . 2xx p∆ ∆ ≥ implies that, as 0 xx p∆ → ⇒ ∆ → ∞ and, as

0xp x∆ → ⇒ ∆ → ∞ . i.e., if we have a very precise knowledge of the position/the linear

momentum of the particle, then we have a very imprecise measurement of its linear momentum/the

position. Furthermore, as we cannot make 0, 0xx p∆ = ∆ = at the same time, it is not possible to

measure both position and momentum in that direction of a particle exactly at the same time.

The restriction by the u.p. applies only to ‘complementary pairs’ ( , )xx p , ( , )yy p and ( , )zz p . The

restriction does not apply to other pairs ( , )yx p , ( , )xy p , ( , )xz p etc.,

i.e. . 0yx p∆ ∆ = , . 0xy p∆ ∆ = , . 0xz p∆ ∆ = etc.

Note: If (or h) were zero, there would be no limitation in our simultaneous measurements of x and

xp -- Classical view.

Since h is very small, we do not notice the importance of H.U.P. in our ordinary experience.

11

Eg.: The speed of a bullet (m = 50 g) and the speed of an electron (m = 9.1 × 10−31 kg) are measured to

be the same, 300 ms−1 with uncertainty of 0.01%. With what fundamental accuracy could we

have located the position of each particle, if the position is measured simultaneously with the

speed in this experiment?

For the bullet (macroscopic object),

x xp mv= = 0.05 kg × 300 m s−1 = 15 kg m s−1

0.01100 0.01100

xx x

x

vv v

v∆

× = ⇒ ∆ = ×

1 3 10.01 0.01 0.0115 kg ms 1.5 10 kg ms100 100 100x x x xp m v m v mv − − −∆ = ∆ = × = × = × = ×

3432 17

3 16.6 10 J.s1HUP 3 10 m = 10 Diameter of a nucleus

2 4 1.5 10 kg m sxx

p π

−− −

− −

×⇒ ∆ ≥ × = = × ×

∆ × ×

So, the HUP sets no practical limit to our measurement. Uncertainties of this order are always masked

by experimental errors.

For the electron (microscopic object) x xp mv= = 9.1 × 10−31 kg × 300 ms−1 = 2.7 × 10−28 kg m s−1

0.01100 0.01100

xx x

x

vv v

v∆

× = ⇒ ∆ = ×

28 1 32 10.01 0.01 0.012.7 10 kg m s 2.7 10 kg m s100 100 100x x x xp m v m v mv − − − −∆ = ∆ = × = × = × × = ×

343 7

32 16.6 10 J.s1HUP 2 10 m = 0.2 cm = 10 Diameter of an atom

2 4 2.7 10 kg m sxx

p π

−−

− −

×⇒ ∆ ≥ × = = × ×

∆ × ×

So, the electron may be found anywhere between 1 – 107 atoms.

⇒ HUP sets a practical limit to our measurement.

If h ~ 6.6 × 10−4 J s, then for the bullet, ∆x = 3 ×10−2 m = 3 cm

⇒ HUP sets a practical limit to our measurement.

As we know, for the u.p. . 2xx p∆ ∆ ≥ to play a significant role in simultaneous measurements of x

and xp , the product . xx p∆ ∆ should take a value of the order of 34~ 10 J s− which is very small. In our ordinary experiences, we have to deal with macroscopic objects, which involve significantly large values of x and xp . Therefore, even with the best measuring techniques available, we get relatively large values for experimental uncertainties of x and xp in our measurements and they will obviously satisfy the u.p. On the other hand, when any of these values (say for x∆ ) is substituted in the above relation, the resulting value of xp∆ becomes so small. Since experimental uncertainties are very much larger than the values predicted by the u.p., it sets no practical limit (with regard to uncertainties) in simultaneous measurements of x and xp . So, the effect of the u.p. is hardly noticeable in our day-to-day life, which is not the case when dealing with microscopic objects.

12

The uncertainty relation for time and energy

Consider the wave packet associated with a particle.

Frequency width of waves = ∆ω

Time duration of the group of waves = ∆t

From Fourier analysis, it can be shown that, . 1 2t ω∆ ∆ ≥ (Derivation is not necessary).

E ω= ⇒ . 2E t∆ ∆ ≥ ← Time-energy uncertainty principle

Other equivalent forms of writing the u.p.

.t E∆ ∆ . ~t E∆ ∆ .t E∆ ∆ h . ~t E h∆ ∆

Here the interpretation is different.

Note: If an electron is excited from the ground state to an excited state, after a certain time, the electron

will make a radiative transition (emission of a photon) into another stationary state of lower energy.

The average length of time that the electron spends in an atomic state is called the lifetime of that

energy state

Physical Interpretation

If a stationary state of a system has a lifetime ∆t, then a measurement of energy of the state will be

uncertain by an amount E∆ so that . 2E t∆ ∆ ≥ . i.e. when ∆t is finite, the most probable value of

the energy of the state lies between 1 12 2andE E E E− ∆ + ∆ .

OR

The relationship . 2E t∆ ∆ ≥ implies that, if a state has a finite lifetime, then its energy can not be

measured exactly ( 0E∆ ≠ ) and, if the lifetime of the state is infinite, then only the energy of the state

can be measured exactly ( E∆ → 0).

For the ground state, ∆t → ∞ ⇒ ∆E → 0

t = 0

∆x

( , )x tΨ

x

x = 0

∆t

( , )x tΨ

t

Time duration

13

i.e. the energy of the ground state is known exactly.

For an excited state, ∆t is finite ⇒ 2

Et

∆ =∆ is finite.

i.e. the energy of the excited state is not known exactly.

∆E – ‘Natural width’ of the energy level. (nature’s limit, and not the uncertainty associated with the

measurement.)

Lifetimes of some levels of atomic hydrogen (in 10 − 8 s)

Level 2p 3s 3p 3d 4s 4p 4d 4f

Lifetime 0.16 16 0.54 1.56 23 1.24 3.65 7.3

Lifetime of 1s (ground) state = ∞, Lifetime of 2s (metastable) state = 1 7 s

For the 2p level, the ‘natural width’, 78~ 4.11 10 eV

0.16 10E −

−∆ = ×

×

Because of the finite energy widths in stationary states, the energy released or absorbed in a transition

is not well-defined. The range of energy of the photon (absorbed or released) is from 1 1

2 2 1 2 2 12 2( ) and ( )E E E E E E− ∆ − + ∆ − ⇒ Energy width of the photon (spectral line) =

22

Eτ

∆ = , where 2τ is the lifetime of the state with energy E2. (The state with energy E1 is assumed

to have an infinite lifetime.)

When both levels have finite lifetimes ( 1τ and 2τ ), energy width of the photon 1 2τ τ

= + (taking all

the possible ways of decaying of levels into account)

22

12 Δ EE +

0Δ 2 ≠E

22

12 Δ EE −

0Δ 1 =E 1E

2E

Excited state,

Lifetime, 2τ is finite

⇒

222 Δ

~Δτ =

tE is

finite.

Ground state,

Lifetime, 1τ is infinite

⇒ 1

1 1

Δ ~ 0Δ

Et τ= =

14

Illustration of HUP using ‘thought’ experiments (i) Diffraction by a slit Suppose one wants to determine the position x and momentum px of a particle simultaneously. If the

particle size is in the atomic scale, the act of measurement will appreciably disturb its state of motion.

To determine the x-cdt of the particle moving parallel to the y-axis with known energy, one can

observe whether or not the particle passes through a slit and leave a mark on a screen. The precision

will then be limited by the width of the slit. i.e. ∆x ~ b.

Before the particle went through the slit, its position x was not known but the momentum was known

perfectly. i.e. 0, 0y xp p≠ =

For the matter wave of the particle hp

λ= . Here the disturbance caused by the slit on the matter wave

of the particle results in a corresponding change in the motion of the particle (as seen by the diffraction

pattern produced).

The uncertainty in the particle’s momentum parallel to the x-axis ( xp∆ ) is determined by θ .

sin sinp pθ θ← → ⇒ 2 sinxp p θ∆ =

From the theory of the diffraction produced by a rectangular slit ⇒ sinb nθ λ λ= =

(n = 1 for the central maximum)

22 sin 2xh hp p

b bλθ

λ∆ = = × × ⇒ ~x b∆

⇒ . ~ ........... HUPxx p h∆ ∆

To determine the position of the particle very accurately (ie to x↓ ∆ ), use a narrow slit (ie to b↓ )

θ

x

y

p p

Slit

θ

xpΔ

Diffraction Pattern

Screen

15

⇒ θ ↑ ⇒ the central maximum in the pattern is widened ⇒ xp∆ ↑

Conversely, to reduce the uncertainty in the momentum (ie to xp∆ ↓ ), use a wide slit (ie to b↑ ) ⇒ x∆ ↑ .

So, the HUP is satisfied in this experiment. (ii) Heisenberg γ-ray microscope To determine the position of an electron placed under a microscope we have to illuminate it with light

of some wavelength λ.

x

y

Photons λhp =

θ

y

xΔ Electron

Microscope

Objective lens

xυ

α α Pph Pph

d

y For small α,

Sin α ≈ tan α = y

d2

For small θ,

Sin θ ≈ tan θ = y

x2

Δ

θ θ y

16

The momentum of scattered photons (by the electron) phhpλ

= and they must move within the cone

of angle 2α to pass through the objective lens.

sin sinph php pα α← →

The resulting uncertainty in the x-component of the momentum of the photon,

2 22 sin 2 sinx phh d h dhp p

y yα α

λ λ λ∆ = = × = × = .

Since some momentum is exchanged between the electron and the photon, the uncertainty in the x-

component of the momentum of the electron, xh dp

yλ∆ = ……….(1).

A microscope’s image of a point object is not just a point but a diffraction pattern. The diffraction of

light makes the position of the electron uncertain and this uncertainty is equal to the diameter of the

central maximum in the diffraction pattern.

2sin tan 2 sin

2x x x yy y

θ θ θ∆ ∆= = ⇒ ∆ =

Condition for diffraction, sind nθ λ=

For the central maximum, sin , ( 1)d nθ λ= =

22 yx yd dλ λ

∆ = × = …………..(2)

From (1) and (2), for the electron, 2. 2 ~ (or ~ )xhd yx p h h

y dλ

λ∆ ∆ = × =

⇒ HUP is satisfied in the detection of the electron.

To improve the accuracy of the measurement of x, we reduce ∆x.

This requires ↓ λ ⇒ .xp∆ ↑

To improve the accuracy of the measurement of xp , we reduce xp∆ .

This requires ↑ λ ⇒ .x∆ ↑

To see if the time-energy uncertainty principle is satisfied, consider the energy of the electron moving

in the x-direction. 2

2x x x

x x x xp p mv

E E p p v pm m m

= ⇒ ∆ = ∆ = × ∆ = ∆ ………… (3)

If ∆t is the time interval required for the position measurement, then the position uncertainty

xx

xx v t tv∆

∆ = ∆ ⇒ ∆ = ………………………………..……………… (4)

17

From (3) and (4), . . ~ (shown before)x x xx

xE t v p x p hv∆

∆ ∆ = ∆ × = ∆ ∆

. ~ (or )E t h⇒ ∆ ∆

We see that, at the atomic level, the act of measurement introduces a significant disturbance to the

system (i.e. the electron in experiment (ii)) due to the interaction between the measuring device (i.e.

the microscope with the light source) and the measured quantity (position of the electron).

U.P. implies that we can never define the path of a microscopic particle with absolute precision (i.e.

with zero uncertainty).

However, due to smallness of h, the uncertainties applied by the u.p. ( . ~xx p h∆ ∆ ) are much smaller

than the experimental errors in the measured values of x and xp for a macroscopic body.

e.g.:- Consider a cricket ball (m ~ 500 g) moving with speed v ~ 150 km/h. The value of v has been

measured with an uncertainty of 0.01 %.

31 3 1

0.01100 0.01100

150 100.010.5 kg ms 2.08 10 kg ms100 3600

v v vv

p mv p m v − − −

∆× = ⇒ ∆ = ×

×= ⇒ ∆ = ∆ = × × = ×

3431

3 16.63 10 J.sHUP . ~ ~ ~ 3 10 m

2.08 10 kg msxx p h x −−

×⇒ ∆ ∆ ⇒ ∆ ×

×

When measured with the best position measuring device available in the laboratory, 31~ 0.00001 m 3 ×10 mx −∆ .

Classical mechanics holds true for such ‘large’ bodies. The word ‘large’ is used in the sense that the

act of measurement does not disturb the state of motion of the body.

For particles of atomic dimensions (‘small’), paths can not be defined precisely.

Consider the path of a particle in the so called ‘phase space’ which is used to describe the dynamical

state (x and xp values) of the particle.

18

Example: Melons in quantum land

In quantum land, a strange land, where ħ = 10−3 J.s, melons with a very hard peel are grown. They have a diameter of approximately 20 cm and contain seeds with a mass around 0.1 g. Why do we have to be careful when cutting open melons grown in quantum land ? How big is the recoil velocity of a melon at the reflection of a visible photon of 628 nm wavelength ? Are such melons visible ? ∆x ~ 20 cm = 0.2 m, m = 0.1 g = 10−4 kg = 0.0001 kg Applying the H.u.p. . ~xx p∆ ∆ for melon seeds, their momentum uncertainty,

3

3 110~ 5 10 kg ms0.2

p−

− −∆ = ×

⇒ Velocity uncertainty of seeds, 3

14

5 10~ 50 ms10

pvm

−−

−

×∆∆ = =

The velocity (mean) with which seeds leave the melon when it is cut open, 1~ 180 km hv v −∆ = Momentum of a photon of wavelength 628 nm,

3 4 19

2 210 ~ 1 10 kg ms628 10ph

hp π πλ λ

− −−

= = = × × ×

x

xpΔ

xΔ

hpx x ~∆⋅∆

The quantum mechanical path of

a (‘small’) particle in the phase

space. Quantum mechanically x

and xp have uncertainties as

related by . ~xx p h∆ ∆ .

Therefore, the point representing

the dynamical state of the

particle lies in a cell of size

and xx p∆ ∆ resulting in a path

which is not well-defined.

px

( x, px )

x

px The classical path of a (‘large’) particle

in the phase space. Each point

represents the dynamical state of the

particle and the path is well-defined

(since 0 and 0)xx p∆ = ∆ = .

19

Radius of the melon, 1~ 10 cm = 10 mR −

Mass of the melon, 3 3 34~ Volume of melon Density of water ~ 10 kg m ~ 4 kg3

M Rπ −× ×

Taking the collision between the photon and the melon as elastic, the momentum of the melon after

collision, 4 1ph2 ~ 2 10 kg msMP p −= ×

The corresponding recoil velocity of the melon 4

3 1 12 10 5 10 ms 18,000 km h4

− −×= = × =

By the time the photon is seen, the melon is situated elsewhere.

Tutorial 1

1. The position and momentum of a 1 keV electron are simultaneously determined. If its position

is located to within 1 Ǻ, what is the percentage of uncertainty in its momentum ?

2. An electron microscope uses 40 keV electrons. Find its ultimate resolving power on the

assumption that this is equal to the wavelength of the electrons.

3. Wavelengths can be determined with accuracies of one part in 106. What is the uncertainty in

the position of a 1 Ǻ X-ray photon when its wavelength is simultaneously measured ?

4. (a) How much time is needed to measure the kinetic energy of an electron whose speed is 10

m s−1 with an uncertainty of no more than 0.1 percent ? How far will the electron have travelled

in this period of time ? (b) Make the same calculations for a 1 g insect whose speed is the

same. What do these sets of figures indicate ?

5. Find the line width and frequency spread for a 1-nanosecond (10−9 s) pulse from a ruby laser (λ =

6.3×10−7 m).

6. Suppose that a beam of electrons with a de Broglie wavelength of 10−5 m passes through a slit

10−4 m wide. What angular spread is introduced because of diffraction by the slit ?

7. A probe must always be smaller (at least by a factor of 10) than the object being studied; otherwise

there will be significant perturbation of the position and velocity of the object. Calculate the

minimum particle energy if (a) photons, (b) electrons, (c) neutrons are used to probe a nucleus

whose diameter is 10−14 m.

8. The velocity of a proton in the x-direction is measured to be an accuracy of 10−7 ms−1. Determine

the limit of accuracy with which the proton can be located simultaneously (a) along the x-axis (b)

along the y-axis. Repeat for a case in which the particle is an electron.

9. The position of an electron is determined with an uncertainty of 0.1 A. Find the uncertainty of its

momentum. If the electron's energy is of the order of 1 keV, estimate the uncertainty in its energy.

20

Repeat for a proton confined to a nuclear diameter (≈ 10−14 m) with an energy of the order of

2 MeV.

10. Verify that the uncertainty principle can be expressed in the form . 2L θ∆ ∆ ≥ , where L∆ is

the uncertainty in the angular momentum of a body and θ∆ is the uncertainty in its angular

position. [Hint: Consider a particle moving in a circle.]

11. Using the uncertainty principle arguments, show that electrons can not be present within nuclei.

You may take the nuclear diameter to be ≈ 10−14 m.

12. Use the uncertainty principle to determine the radius of the first Bohr orbit of the hydrogen atom.

13. An atom can radiate at any time after it is excited. It is found that in a typical case the average

excited atom has a lifetime of about 10−8 s. That is, during this period it emits a photon and is de-

excited. (a) What is the minimum uncertainty ∆ν in the frequency of the photon ? (b) Most

photons from sodium atoms are in two spectral lines at about λ = 5890 Å. What is the fractional

width of either line ∆ν/ν ? (c) Calculate the uncertainty ∆E in the energy of the excited state of the

atom. (d) From the previous results determine, to within an accuracy ∆E, the energy of the excited

state of a sodium atom, relative to its lowest energy state, that emits a photon whose wavelength is

centered at 5890 Å.

14. Use the uncertainty principle involving position and momentum to estimate the minimum energy

of a linear harmonic oscillator.

15. Consider a microscopic particle moving freely along the x-axis. Assume that at the instant t = 0

the position of the particle is uncertain by the amount ∆x0. Calculate the uncertainty in the

measured position of the particle at some later time t.

16. A boy on a top of a ladder of height H is dropping marbles of mass m to the floor and trying to hit

a crack in the floor. To aim, he is using equipment of the highest possible precision. (a) Show that

the marbles will miss the crack by an average distance of the order of ( )1 2m ( )1 4H g , where g

is the acceleration due to gravity. (b) Using reasonable values of H and m, evaluate this distance

(c) If the marbles are perfectly elastic and if they rebound again to the height H from the floor,

what is the expectation of the minimum horizontal distance of displacement from the point from

which it was dropped ?

17. The beam from a laser is usually extremely well-collimated. By applying the uncertainty principle

to estimate the transverse momentum of the photons, show that a helium-neon laser beam which

initially has diameter 1 mm can be collimated to an angle of about 33 10−× degrees. How small

would be the filament of an ordinary household torch have to be to achieve the same collimation?

21

Wave function We can talk about the path of a large object in the sense of classical mechanics. For such an object it is

possible to solve the classical equation of motion (Lagrange’s equation) and obtain an expression for

( )r t and hence describe the motion by defining a well-defined path.

But, we cannot talk about the trajectory of an atomic particle and, to describe its motion, we may use

the associated de Broglie’s matter field. In a de Broglie wave, the matter field is assumed to vary in

space and time (similar to e/m field varies in space and time in a light wave or pressure field varies

in space and time in a sound wave).and this variable quantity characterising de Broglie waves is

called the wave function. The value of the wave function associated with a moving particle at the

point x, y, z in space at the time t is related to the likelihood of finding the particle there at that time.

Wave function contains the maximum information that we have about the particle in a given

dynamical state. All observable properties of the particle can be obtained by Ψ( , )r t . In quantum

mechanics, to obtain Ψ( , )r t we solve the Schrodinger equation.

Eg: The electric field of a sinusoidal e/m wave of w.l. λ and freq ν can be written as

( , ) sin 2 ( )E x t A x tπ λ ν= − . The wave function of a particle moving in the x-direction with

a precise value of energy (i.e.ν) and linear momentum (i.e.λ) can be written as

( , ) sin 2 ( )x t A x tπ λ νΨ = − . The quantity E is a (radiation) wave associated with a photon

and Ψ is a (matter) wave associated with a material particle.

Suppose we have a particle which is confined to a small region of space (eg:- an electron confined to

an atom, diameter ~ 10−10 m). Its associated matter wave may be expressed in terms of standing waves

(the value of the wave is fixed in at each point of space) localized in the given region, with the

instantaneous value varying from point to point within the region and practically zero outside the

region. Let ( )xψ be the wave function associated with the particle.

22

2( ) * ( ). ( )x x xψ ψ ψ= is the intensity of the matter field or the ‘position probability density’.

Position probability density gives the probability per unit length of finding particle at a given point.

When the motion of the particle is in 3-d space, its wave function ( ) ( , , ).r x y zψ ψ= Wave

function can have a time dependence, ( . )r tΨ or ( , , , )x y z tΨ .

Particle is most likely to be found

x x

2( )xψ

Intensity

( )xψ Amplitude

21

Wave function We can talk about the path of a large object in the sense of classical mechanics. For such an object it is

possible to solve the classical equation of motion (Lagrange’s equation) and obtain an expression for

( )r t and hence describe the motion by defining a well-defined path.

But, we cannot talk about the trajectory of an atomic particle and, to describe its motion, we may use

the associated de Broglie’s matter field. In a de Broglie wave, the matter field is assumed to vary in

space and time (similar to e/m field varies in space and time in a light wave or pressure field varies

in space and time in a sound wave) and this variable quantity characterising de Broglie waves is

called the wave function. Wave function is a mathematical tool used in quantum mechanics to

describe any physical system and it is a function from a space that maps the possible states of the

system into the complex numbers. The value of the wave function associated with a moving

particle at the point x, y, z in space at the time t is related to the likelihood (probability) of finding

the particle there at that time. Wave function contains the maximum information that we have about

the particle in a given dynamical state. All observable properties of the particle can be obtained by

Ψ( , )r t and in quantum mechanics, it is found by solving the Schrodinger equation.

Eg: The electric field of a sinusoidal e/m wave of w.l. λ and freq ν can be written as

( , ) sin 2 ( )E x t A x tπ λ ν= − . The wave function of a particle moving in the x-direction with

a precise value of energy (i.e.ν) and linear momentum (i.e.λ) can be written as

( , ) sin 2 ( )x t A x tπ λ νΨ = − . The quantity E is a (radiation) wave associated with a photon

and Ψ is a (matter) wave associated with a material particle.

Suppose we have a particle which is confined to a small region of space (eg:- an electron confined to

an atom, diameter ~ 10−10 m). Its associated matter wave may be expressed in terms of standing waves

(the amplitude of the wave is fixed in at each point of space) localized in the given region, with the

amplitude varying from point to point within the region and practically zero outside the region. Let

( )xψ be the wave function associated with the particle.

Particle is most likely to be found

x x

2( )xψ

Intensity

( )xψ

22

2( ) * ( ). ( )x x xψ ψ ψ= is the intensity of the matter field or the ‘position probability density’.

Position probability density gives the probability per unit length of finding particle at a given point.

When the motion of the particle is in 3-d space, its wave function ( ) ( , , ).r x y zψ ψ= Wave

function can have a time dependence, ( . )r tΨ or ( , , , )x y z tΨ .

Wave function contains the maximum information that we have about the particle (system) in a

given dynamical state as allowed by the uncertainty principle. All observable (i.e. measurable)

properties of the particle can be obtained by Ψ( , )r t .

In classical mechanics, once we obtain ( )r t (by solving Lagrangian equation) we have solved the

problem.

In quantum mechanics, once we obtain Ψ( , )r t (by solving Schrodinger equation) we have solved

the problem.

In classical mechanics, position of a particle is given by ( )r r t= . Then we can find

Velocity ddrvt

=

Acceleration 2

2dd

rat

=

Force 2

2dd

rF mt

=

Linear momentum ddrp mt

=

Angular momentum L r p= ×

Total Energy k.e + p.eE = = 21 d ( , )

2 drm V r tt

+

In quantum mechanics, to get all the above information, what we have is the wave function.

Properties of the wave function Consider a particle moving in 3-d

(1) Ψ( , )r t is in general a complex function of ( , , )r x y z= and of time t and is sometimes called

the ‘probability amplitude’. Ψ( , )r t is an abstract quantity which is interpreted statistically.

Ψ( , )r t and Ψ( , )c r t , where c is a complex no, represent the same state of the particle.

(2) The probability of finding the particle in a state described by the wave function Ψ( , )r t within

the volume d d .d .dV x y z= about the point ( , , )r x y z= at time t = 2Ψ( , ) d ,r t V

3d d d = d .d .d (in cartesian co-ordinates)V r r x y z≡ ≡

23

e.g. i( )i( )( , ) e e x y z

y zk x k k tk r tr t A Aωω + + −⋅ −Ψ = =

i( )*Ψ ( , ) e k r t* r t A ω− ⋅ −=

2

2*P( , ) Ψ Ψ , if is complex.

= , if is real.

*r t A A A A

A A

= = =

(3) Probability of finding the particle within the volume 2Ψ( , ) dV

V r t r= ∫ (in 3-d)

Prob. of finding the particle within the points a and b on the x -axis 2Ψ( , ) db

ax t x= ∫

(in 1-d) -- Probability interpretation of the wave function

(4) The wave function Ψ( , )r t is said to be ‘square-integrable’, if the ‘normalization integral’

2

allspace

Ψ( , ) dr t r∫ is finite.

Note: In this course, we work with square-integrable wave functions only.

If 2Ψ( , ) ( ) ( ) ( ) ( )r t f x f y f z f t=

2

all allspace space

Ψ( , ) d ( ) ( ) ( ) ( ) d d d ( ) d ( ) d ( ) d . ( )r t r f x f y f z f t x y z f x x f y y f z z f t+∞ +∞ +∞

−∞ −∞ −∞

= =∫ ∫ ∫ ∫ ∫

(5) Since the probability of finding the particle somewhere must be unity, Ψ( , )r t must be

normalized to unity ⇒ 2

allspace

Ψ( , ) d 1r t r =∫ , for all t values.

dx dy

dz

r

z

y

x

2If P( , )d Ψ( , ) dr t r r t r=

2P( , ) Ψ( , ) Ψ ( , ).Ψ( , )*r t r t r t r t⇒ = =

P( , )r t is called the ‘position probability density’.

24

Since ψ and cψ represent the same state of a particle, it is convenient to work with square

integrable wave functions with the value of c chosen so that the wave function is normalized

to unity. 2

allspace

Ψ( , ) d finite (say )r t r a=∫

22

all allspace space

1 1Ψ( , ) d 1 Ψ( , ) d 1a

r t r r t ra

= ⇒ =∫ ∫

2

allspace

1Φ( , ) d 1, where Φ( , ) Ψ( , )r t r r t r ta

= =∫

Φ( , )r t is a wave function normalized to unity. Ψ and Φ represent the same state of the

system and it is more convenient to work with Φ.

Eg:

(i) Normalize the wave function i( )Ψ( , ) e k x tx t ω⋅ −= which represents a plane wave.

By writing i( )Ψ( , ) e k x tx t A ω⋅ −= , where A is the normalization constant,

* * i( )Ψ ( , ) e k x tx t A ω− ⋅ −=

Setting 2 2 2

allspace

Ψ( , ) d 1 d 1x t x A x A+∞

−∞

= ⇒ = ×∞ =∫ ∫

A can not be evaluated ⇒ plane wave functions can not be normalized.

(ii) Normalize the wave function 2 iΨ( , ) e a x btx t − −=

Writing 2 iΨ( , ) e a x btx t A − −= ⇒

2* * iΨ ( , ) e a x btx t A +−=

Setting2 22 2 22 2

0allspace

Ψ( , ) d 1 e d 2 e da x a xx t x A x A x+∞ +∞

−∞

− −= ⇒ = × ×∫ ∫ ∫

⇒ 21 4 1 4

2 i2 212 1 Ψ( , ) e2 2

a x bta aA A x taπ

π π− − = ⇒ = ⇒ =

We can generalize the above results for the case of a system of n particles. Eg: A many-

electron atom.

25

Wave function of particles, 1 2 3Ψ( , , , ........., , )nr r r r t , where 1 2 3, , , ........ , nr r r r are the

position vectors of particles.

Normalization integral = 21 2 3 1 2 3Ψ( , , , ........, . , ) d d .d ...........dn n

Vr r r r t r r r r∫

(6) The wave functions 1( )rψ and 2 ( )rψ are said to be ‘orthogonal’ if their scalar product,

2*1

allspace

( ) ( )d 0r Ψ r rψ =∫ or equivalently *

2 1allspace

( ) ( )d 0r r rψ ψ =∫ .

Dirac Bracket Notation

The scalar product of two functions 2*1

allspace

( ) ( )dr r rψ ψ∫ is denoted by 1 2ψ ψ Bracket.

2*

1 2 1allspace

( ) ( )dr r rψ ψ ψ ψ= ∫

In Dirac notation,

( )rψ ψ≡ - a ‘ket’ vector * ( )rψ ψ≡ - a ‘bra’ vector

Note: ( ) ( )* r rψ ψ ψ ψ≠ and only *( ) ( )dr r rψ ψ ψ ψ=∫

i. If 1 2 1 2c d c dψ ψ ψ ψ ψ ψ= + ⇒ = +

ii. If * * * * * * *1 2 1 2c d c dψ ψ ψ ψ ψ ψ= + ⇒ = +

iii. If c is a complex no ⇒ 1 2 1 2c cψ ψ ψ ψ=

⇒ 1 2 1 2*c cψ ψ ψ ψ=

iv. If the two wave functions 1ψ and 2ψ are orthogonal, ⇒ *1 2

allspace

( ) ( )d 0r r rψ ψ =∫

In Dirac notation, 1 2 0ψ ψ = or 2 1 0ψ ψ = .

v. For the scalar product of two wave functions 1( )rψ and 2 ( )rψ we have

*1 2 2 1ψ ψ ψ ψ= .

vi. A wave function ( )rψ is normalized, if allspace

( ) ( )d 1* r r rψ ψ =∫ .

26

In Dirac’s notation, 1ψ ψ = . vii. If 1( )rψ , 2 ( )rψ and 3 ( )rψ are wave functions 3 1 2 3 1 3 2ψ ψ ψ ψ ψ ψ ψ+ = +

Superposition Principle If 1 2 3( , ), ( , ), ( , ), ................ , ( , )Nr t r t r t r tΨ Ψ Ψ Ψ

are wave functions which represent different

states of a system (eg:- the ground and excited states of the electron in a hydrogen atom), then

1 1 2 21

( , ) ( , ) ( , ) .......... + ( , ) ( , ) N

N N n nn

r t c r t c r t c r t c r t=

Ψ = Ψ + Ψ + Ψ = Ψ∑

(N can be finite or infinite) where 1 2 3, , , ........., Nc c c c are constants (‘mixing coefficients’) such that

2 2 22 21 2 3

1......... 1

N

N nn

c c c c c=

+ + + + = =∑ , is also a wave function describing a

possible state of the system. 1

In Dirac notation, N

n nn

c=

Ψ = Ψ

∑

2nc is interpreted as the probability of the system in state ( , )r tΨ

being found in the state

( , )n r tΨ on measurement.

Eg: Hydrogen atom

Stationary State

Wave function

Probability of finding the electron on measurement

Nth excited state ( , )N r tΨ 2

Nc

…………………. ………………… ………………… …………………. ………………… ………………… …………………. ………………… …………………

3rd excited state 4 ( , )r tΨ 2

4c 2nd excited state 3 ( , )r tΨ

23c

1st excited state 2 ( , )r tΨ 2

2c Ground state 1( , )r tΨ

21c

When we have a set of wave functions 1 2 3, , , ............. , NΨ Ψ Ψ Ψ , we want each of them to be

normalized (to unity) and every pair to be orthogonal ----- An ‘Orthonormal set’ (basis).

i.e. * * * *1 1 2 2 2 2

all all all allspace space space space

d d d ............... d 1N Nr r r rΨ Ψ = Ψ Ψ = Ψ Ψ = = Ψ Ψ =∫ ∫ ∫ ∫

‘Mixture of states’

27

and * * * *1 2 2 3 1 3 3 4

all all all allspace space space space

d d d d ................. 0r r r rΨ Ψ = Ψ Ψ = Ψ Ψ = Ψ Ψ = =∫ ∫ ∫ ∫

Consider the ‘Kronecker Delta function’, defined as 1, if = 0, if

mn m nm n

δ == ≠

Then, the ‘condition of orthonormality’ of a set of wave functions is *

allspace

dm n mnr δΨ Ψ =∫

( = m n gives the ‘condition of normalization’ and m n≠ gives the ‘condition of orthogonality’.)

In Dirac’s notation, the ‘condition of orthonormality’ is m n mnδΨ Ψ = .

Consider,

1 1 2 2( , t) ( , t) ( , t) + .......... + ( , t) ................ ( , t)n n N Nr c r c r c r c rΨ = Ψ + Ψ Ψ + + Ψ

Multiplying throughout by *nΨ and integrating over all space,

* * * *1 1

all all all allspace space space space

d d ............ d ......... dn n n n n N n Nr c r c r c rΨ Ψ = Ψ Ψ + + Ψ Ψ + + Ψ Ψ∫ ∫ ∫ ∫

*

allspace

( , ) ( , )dn n nc r t r t r= Ψ Ψ = Ψ Ψ∫

1 2 3, , , ............, Nc c c c can be determined.

1

N

n nn

c=

Ψ = Ψ∑ ⇒ * * *

1

N

n nn

c=

Ψ = Ψ∑

* * * * *

1 1 1 1all all allspace space space

d d dN N N N

m m n n m n m nm n m n

r c c r c c r= = = =

Ψ Ψ = Ψ Ψ = Ψ Ψ∑ ∑ ∑ ∑∫ ∫ ∫

2* * *

1 1 1 1allspace

dN N N N

m n mn n n nm n n n

r c c c c cδ= = = =

Ψ Ψ = = =∑ ∑ ∑ ∑∫

Normalization of t),r(Ψ requires that *

allspace

d 1rΨ Ψ =∫ . This requires 2

11

N

nn

c=

=∑ .

Observables and Operators Operator + Wave function ⇒ Observable (l.m., a.m, k.e, p.e, posn, tot.energy, etc)

= 0 = 1 = 0

28

e.g.: To get the length of the table we measure it with a ruler.

In q.m., with each observable, there is an associated operator.

i. Position co-ordinate x, Operator is x x=

Position vector r , Operator is r r=

ii. A component of linear momentum xp , Operator is ˆixp

x∂

=∂

ix∂

= −∂

(in 1-d)

Linear momentum vector p , Operator is ˆ ii

p = ∇ = − ∇

(In 3-d)

In ˆ ip = − ∇

, ˆ ˆ ˆ ˆ( , , )x y zp p p p= and , ,x y z

∂ ∂ ∂∇ = ∂ ∂ ∂

.

Position and linear momentum operators have very simple forms. More complicated operators are

built up from their classical analogue.

If an observable F can be written in classical mechanics, as a function 1 2 3( , , , ........., )nf A A A A of

other observables 1 2 3, , , .........., nA A A A then, the operator F which represents the observable F

can be written in quantum mechanics, as 1 2 3ˆ ˆ ˆ ˆ( , , , ........, )nf A A A A , where 1 2 3

ˆ ˆ ˆ, , , .........., nA A A A are

the operators representing the observables 1 2 3, , , ........, nA A A A .

iii. Kinetic energy operator

2

2xp

Tm

= in classical mechanics (in 1-d)

2 2 2

2ˆˆ2 2

xpT

m m x− ∂

= =∂

in quantum mechanics (in 1-d)

2

2 22

ˆ ˆ ˆ: i ix x xNote p p px x x

∂ ∂ ∂ = ⋅ = − = − ∂ ∂ ∂

2

2ˆ2

Tm

−= ∇

(in 3-d)

iv. Orbital angular momentum operator

L r p= × in classical mechanics

ˆ ˆ ˆ iL r p r= × = × − ∇

in quantum mechanics

Note:

29

In classical machines r p p r× = − ×

In quantum machines ˆ ˆ ˆ ˆr p p r× ≠ − × ( )since ( i ) ( i )r r× − ∇ ≠ − − ∇ ×

⇒ i.e. the order of operation is important in quantum mechanics.

v. Total energy operator

( , )E T V r t= + in classical machines

ˆ ˆ ˆ( , )E T V r t= + in quantum machines

2 2

2ˆˆ ˆ( , ) ( , )

2 2p V r t V r tm m

−= + = ∇ +

ˆ ˆE H≡ ← Hamiltonian

Properties of Operators 1. Operators act on states (wave functions) to produce new states (new wave functions).

Let A be an arbitrary operator ⇒ 1 2Aψ ψ= (if 1ψ is normalized but 2ψ is not

normalized, we can normalize it so that, 2 3cψ ψ= )

In Dirac’s notation, 1 2A ψ ψ= .

Note: Unless otherwise stated, operators act on everything to the right of them.

( )1 2 1 2ˆ ˆA Aψ ψ ψ ψ=

e.g: 1 2i i1 2

ˆe , e ,k x k x Ax

ψ ψ ∂= = =

∂

1 2 1 2 1 2i i i i i( )1 2 2 2

ˆ e e i e e i ek x k x k x k x k k xA k kx

ψ ψ +∂= = =

∂

2. Two operators A and B are said to be equal, if for any wave function ψ , ˆ ˆA Bψ ψ= .

3. Operators can be added, subtracted or multiplied to one another to get new operators.

If B and C are operators, ˆ ˆˆA B C= ± and ˆ ˆˆA BC= are operators.

i.e. for any wave functionψ , we have ( )ˆ ˆˆA B Cψ ψ= ±

( )ˆ ˆˆA BCψ ψ=

4. Operator A is linear if it has the property

( ) ( ) ( )1 1 2 2 1 1 2 2

1 1 2 2

ˆ ˆ ˆ

ˆ ˆ =

A c c A c A c

c A c A

ψ ψ ψ ψ

ψ ψ

+ = +

+

30

where 1c and 2c are complex numbers.

We consider linear operators only in this course. The reason is that linear operators are

associated with physical observables (e.g: , , , , , )x p L T V E

• If A and B are linear operators, the sum ˆ ˆA B+ and the difference ˆ ˆA B− and the product

ˆ ˆA B⋅ are linear operators.

For any ψ , ( )ˆ ˆˆ ˆA B A Bψ ψ ψ± = ±

( ) ( )ˆ ˆ ˆˆ ˆ ˆA B A B A Bψ ψ ψ= =

• Any three operators A , B and C satisfy the associative property.

For any ψ , ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆA BC A BC A BC A B Cψ ψ ψ ψ= = =

⇒ ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆA BC A BC A B C= =

• Any three operators A , B and C satisfy the distributive property.

For any ψ ( )ˆ ˆ ˆ ˆ ˆˆ ˆA B C AC BCψ ψ ψ+ = +

⇒ ( )ˆ ˆ ˆ ˆ ˆˆ ˆA B C AC BC+ = +

Similarly ( )ˆ ˆ ˆ ˆ ˆˆ ˆA B C A B ACψ ψ ψ+ = +

⇒ ( )ˆ ˆ ˆ ˆ ˆˆ ˆA B C A B AC+ = +

• The square of an operator A is defined as 2ˆ ˆ ˆA A A=

⇒ ( ) ( )2ˆ ˆ ˆ ˆ ˆA A A A Aψ ψ ψ= =

In general, the order of multiplication of operators is important. i.e. A and B are not necessarily

commute i.e. ˆ ˆˆ ˆA B B A≠ , in general

5. Hermitian operators

Suppose A is an operator Ψ and Φ represent two states. Then, if

( ) ( )*all allspace space

ˆ ˆ( , t) ( , ) d Ψ( , ) ( , )d* r A r t r A r t r t rΨ Φ = Φ∫ ∫ - Condition of Hermiticity

is true for all Ψ and Φ , A is called a Hermitian operator.

In Dirac’s notation, * *ˆ ˆ ˆ ˆA A A AΨ Φ = Ψ Φ = Φ Ψ = Φ Ψ

31

* All the physical observables are represented by Hermitian operators.

6. If A is an operator representing an observable A and ψ is a wave function of a system and,

A aψ ψ= , where a is in general a complex number, then the system is said to be an

eigenstate of the operator A with an eigenvalue a. ψ is called an eigenfunction.

A aψ ψ= - Eigenvalue equation ( )ˆIn Dirac notation, A aψ ψ=

Physical significance: The eigenvalue equation represents a precise measurement of an

observable A which yields the value a.

Properties of eigenfunctions:

(i) An operator can have more than one eigenfunction.

(ii) Different eigenfunctions of the same operator can have the same eigenvalue or different

eigenvalues.

eg: Consider the operator ˆ ixpx∂

= −∂

and the wave function i( ) e k xxψ = .

i iˆ ( ) i e e ( )k x k xxp x k k x

xψ ψ∂

= − = =∂

⇒ ( )xψ is an eigenfunction of ˆ xp with eigenvalue k .

Consider the operator ˆ ixpx∂

= −∂

and the wave function

i( , , ) e ( , )k xx y z f y zψ = .

i iˆ ( , , ) i e ( , ) e ( , ) ( , , )k x k xxp x y z f y z k f y z k x y z

xψ ψ∂

= − = =∂

⇒ ( )xψ and ( , , )x y zψ are eigenfunctions of ˆ xp with the same eigenvalue k .

Consider the operator ˆ ixpx∂

= −∂

and the wave functions 1i1( ) e k xxψ = and

2i2 ( ) e k xxψ = .

⇒ 1 1 1 2 2 2ˆ ˆ( ) ( ) and ( ) ( ) x xp x k x p x k xψ ψ ψ ψ= =

⇒ 1( )xψ and 2 ( )xψ are eigenfunctions of ˆ xp with different eigenvalues

1k and 2k .

* The whole set of eigenvalues of an operator is called its spectrum and a spectrum may contain a

discrete or continuous set of values.

32

7. The eigenvalues of a Hermitian operator are real.

Pf: Let A be a Hermitian operator. Then, by definition of a Hermitain operator, for Ψ and Φ

( )**

all allspace space

ˆ ˆd dA r A rΨ Φ = Ψ Φ∫ ∫

When Ψ = Φ, ( )**

all allspace space

ˆ ˆd dA r A rΨ Ψ = Ψ Ψ∫ ∫ ……….. (1)

The eigenvalue equation ⇒ A aΨ = Ψ …………. (2)

Taking the complex conjugate ⇒ ( )* * *A aΨ = Ψ ……. (3)

By substituting in equn (1) from equns (2) and (3), * * *

all allspace space

d da r a rΨ Ψ = Ψ Ψ∫ ∫

Note: ( )* * *ˆ ˆA AΨ ≠ Ψ and ( )* * *a aΨ = Ψ .

( )* *

allspace

d 0a a r− Ψ Ψ =∫

*

allspace

d 0rΨ Ψ ≠∫ ⇒ * 0a a− = ⇒ *a a= ⇒ a is real.

Note: The inverse of the above statement is not necessarily true.

8. The eigenfunctions of a Hermitian operator corresponding to different eigenvalues are

orthogonal.

Pf: It is given that, ˆn n nA aΨ = Ψ and ˆ

m m mA aΨ = Ψ

It is required to prove that, if m na a≠ ⇒ *

allspace

d 0m n rΨ Ψ =∫ .

( )*all allspace space

ˆ ˆΨ d d*n m n mA r A rΨ = Ψ Ψ∫ ∫

Condition of Hermiticity

( )*all allspace space

Ψ d d*n n m n m ma r a rΨ = Ψ Ψ∫ ∫

33

* *

all allspace space

Ψ d d*n n m m n ma r a rΨ = Ψ Ψ∫ ∫

na is real ⇒ *n na a= ⇒ ( ) *

allspace

d 0n m n ma a r− Ψ Ψ =∫

n ma a≠ ⇒ *

allspace

d 0n m rΨ Ψ =∫ ⇒ nΨ is orthogonal to mΨ .

(Note: When n ma a= , we can not decide whether nΨ is orthogonal to mΨ or not.)

9. If two or more eigenfunctions have the same eigenvalue with an operator, then any linear

combination of these eigenfunctions will have the same eigenvalue.

Pf: Let A be the operator and, nΨ and mΨ be the eigenfunctions corresponding to the

eigenvalue a.

ˆm mA aΨ = Ψ , ˆ

n nA aΨ = Ψ (given)

Linear combination, nnmm cc Ψ±Ψ=Ψ , where mc and nc are complex numbers.

( )ˆ ˆ ˆ ˆm m n n m m n nA A c c c A c AΨ = Ψ ± Ψ = Ψ ± Ψ

. .m m n nc a c a= Ψ ± Ψ

( )m m n na c c a= Ψ ± Ψ = Ψ

⇒ Ψ is an eigenfunction of A with the eigenvalue a.

Degeneracy: An eigenvalue a is said to be degenerate if there is more than one linearly

independent eigenfunctions belonging to that eigenvalue.

The number of such eigenfunctions is the called the degree of degeneracy.

eg: Consider A with 1 2 3 4, , , , ..............., nΨ Ψ Ψ Ψ Ψ are all different eigenfunctions.

If 1 1 2 2 3 3ˆ ˆ ˆ ˆ, , , ................, n nA a A a A a A aΨ = Ψ Ψ = Ψ Ψ = Ψ Ψ = Ψ , then

a degeneracy exists and the degree of degeneracy = n

The eigenvalue a is said to be ‘n-fold degenerate’.

10. When two (normalized) eigenfunctions of a H. O. corresponding the same eigenvalue are not

known to be orthogonal, it is still possible to find two eigenfunctions of the operator having the

same eigenvalue and satisfy the normalization and orthogonality conditions.

Let A be a Hermitian operator and 1 2andΨ Ψ be two normalized eigenfunctions,

corresponding the same eigenvalue a and not known to be orthogonal.

34

In Dirac’s notation,

1 1A aΨ = Ψ 2 2A aΨ = Ψ

1 2andΨ Ψ are normalized ⇒ 1 1 2 21 , 1Ψ Ψ = Ψ Ψ =

Construct two linear combinations of 1 2andΨ Ψ as

1 1 1 2 2 ,c cΦ = Ψ + Ψ 2 3 1 4 2c cΦ = Ψ + Ψ ,

where 1 2 3 4, , andc c c c (assume as real) are 4 unknown constants to be evaluated.

Since it is difficult to evaluate all 4 constants, choose 1 1c = and 2 0c = .

⇒ 1 1Φ = Ψ , and it is therefore already normalized.

3 4andc c should be evaluated such that, 2Φ is normalized and orthogonal to 1Φ .

i.e. we require 2 2 1 21 , 0Φ Φ = Φ Φ = .

1 2 0Φ Φ = ⇒ ( )1 3 1 4 2 0c cΨ Ψ + Ψ =

⇒ 3 1 1 4 1 2 0c cΨ Ψ + Ψ Ψ = ⇒ 3 4 1 2c c= − Ψ Ψ

2 2 1Φ Φ = ⇒ ( ) ( )3 1 4 2 3 1 4 2 1c c c cΨ + Ψ Ψ + Ψ =

⇒ 2 23 1 1 3 4 1 2 3 4 2 1 4 2 2 1c c c c c cΨ Ψ + Ψ Ψ + Ψ Ψ + Ψ Ψ =

⇒ ( )2 23 4 3 4 1 2 2 1 1c c c c+ + Ψ Ψ + Ψ Ψ =

Substituting for 3c ,

⇒ ( ) ( )( )22 22 2 24 1 2 4 4 1 2 1 2 1c c cΨ Ψ + − Ψ Ψ + Ψ Ψ =

⇒ 24 1 21 1c = − Ψ Ψ ⇒ 2

3 1 2 1 21c = − Ψ Ψ − Ψ Ψ

When 1 2 0Ψ Ψ ≠ , 1 1Φ = Ψ and 1 2 1 22 2

1 21

− Ψ Ψ Ψ + ΨΦ =

− Ψ Ψ are

normalized and orthogonal to each other. In addition, according to result 9, they have the same

eigenvalue a as do 1 2andΨ Ψ .

The above procedure is known as the ‘Schmidt orthogonalization procedure’ and it can be

extended to any number of eigenfunctions.

In result 8, we proved that the eigenfunctions of H. O. corresponding to different eigenvalues as

orthogonal. So, if we consider results 8, 9 and 10 together, we can conclude that the set of

eigenfunctions of a H. O. can be made to satisfy the normalization and orthogonality conditions.

35

11. If A and B are two operators and ( , )r tΨ is some wave function,

Question: Is ˆ ˆˆ ˆ( , ) ( , )AB r t BA r tΨ = Ψ ?

Answer: Not in general.

Eg: ˆ ˆˆ ˆ, ixA x x B px∂

≡ = ≡ = −∂

Consider the state ( )xψ

( ) ( )ˆ ˆ ˆˆ ˆ ˆ( ) ( ) ( ) ix xAB x xp x x p x x xx

ψ ψ ψ ψ∂= = = −∂

( ) ( )( )

( )

ˆˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) i

i i ( )

x xBA x p x x p x x x xx

x x xx

ψ ψ ψ ψ

ψ

∂= = = −

∂∂

= − − Ψ∂

⇒ ˆ ˆ ˆ ˆ( ) ( )x xxp x p x xψ ψ≠ and ˆ ˆ ˆ( ) ( ) i ( )x xx p x p x x xψ ψ ψ− =

⇒ ( )ˆˆ ˆ ˆ ( ) i ( )x xxp p x x xψ ψ− = for any ( )xψ

⇒ ˆ ˆ ˆ ˆ ix xxp p x− =

Define the Commutator ˆ ˆ,A B of two operators A and B as ˆ ˆ ˆˆ ˆ ˆ,A B A B B A = − .

⇒ ˆ ˆˆ ˆ, ,A B B A = −

Note: Commutator ˆ ˆ,A B is also an operator.

If ˆ ˆ, ( ) 0A B rψ = , for any ( )rψ ⇒ ˆ ˆ ˆˆ ˆ ˆ, 0A B A B B A = ⇒ = ⇒ the two

operators A and B are said to commute.

Now consider, ˆ ˆ ˆ ˆ ix xxp p x− = ⇒ [ ]ˆ ˆ, ixx p =

x and xp are called canonically conjugate observables.

Since ˆ ˆ, iyy p = , [ ]ˆˆ, izz p = , the pairs , yy p and , zz p are also canonically conjugate

observables [ ] [ ]ˆ ˆ ˆ ˆ ˆˆ, , , ix y zx p y p z p = = = …………(A)

For all other pairs of ˆ ˆ ˆ, and x y z and ˆ ˆ ˆ, and x y zp p p we get zero commutators.

[ ] [ ] [ ]ˆ ˆ ˆ ˆˆ ˆ, , , 0x y y z z x= = =

[ ]ˆ ˆ ˆ ˆ ˆ ˆ, , , 0x y y z z xp p p p p p = = =

[ ] [ ] [ ] [ ]ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ, , , , , , 0y x z z x yx p y p x p y p z p z p = = = = = = …… (B)

36

Relationships (A) and (B) can be combined to be given in the form

ˆ ˆ, ii j ijx p δ = , 1i j x= → , , 2i j y= → , , 3i j z= →

12. Every operator commutes with itself or any function of itself

( )ˆ ˆ ˆ ˆ, 0, , 0A A A f A = = , where ( ) ˆ2ˆ ˆ , e , ........Af A A=

Commutator algebra

(i) ˆ ˆˆ ˆ, ,A B B A = −

(ii) ˆ ˆ ˆ ˆ ˆˆ ˆ, , ,A B C A B A C ± = ±

(iii) ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ, , ,A BC A B C B A C = +

(iv) ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ, , , , , , 0A B C B C A C A B + + =

Proof: (ii) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ,A B C A B C B C A ± = ± − ±

= ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆAB AC BA CA AB BA AC CA± − ± = − ± −

= ˆ ˆ ˆˆ, ,A B A C ±

Prove (i), (iii) and (iv) as exercises.

13. If two operators commute, then we can always find a set of eigenfunctions which are

simultaneous eigenfunctions of each operator.

Proof: It is given that, ˆ ˆ, 0A B = and for the set of functions ,nψ 1, 2, 3.........n =

ˆn nA aψ ψ= ………. (1) 1, 2, 3.........n =

ˆ ˆ ˆˆ ˆ ˆ, 0A B AB BA = ⇒ =

Operate with B on equation (1).

( ) ( )ˆˆ ˆn nB A B aψ ψ= ⇒ ( ) ( )ˆˆ ˆn nBA B aψ ψ= ⇒ ( )ˆ ˆ ˆn nAB aBψ ψ=

( ) ( )ˆ ˆ ˆn nA B a Bψ ψ= ⇒ A aφ φ= where ˆ nBφ ψ= .

Two cases are to be considered

(i) a is non-degenerate.(easy to prove) (ii) a is degenerate (difficult to prove)

We consider case (i) only in this course.

37

ˆn nA aψ ψ= ………. (1) A aφ φ= ………….. (2)

Since a is non-degenerate, nψ and φ should not be linearly independent.

i.e. nbφ ψ= (should be)

⇒ ˆ n nB bψ ψ= , n = 1, 2, 3……………………

Therefore , 1, 2, 3, ..........n nψ = are simultaneous eigenfunctions of both A and B .

Case (ii) can also be proven with some difficulty. Therefore, the result given under (13) can be

considered as always true.

i.e. when ˆ ˆ, 0A B = , the pair of observables represented by the operators A and B can be

measured simultaneously with perfect accuracy ( )0A B∆ = ∆ = .

However, if ˆ ˆ, 0A B ≠ , then the U.P applies to the observables represented by these

operators ( )A B∆ ⋅∆ ≥

The observables which can be measured simultaneously with perfect accuracy are called

compatible observables.

Eg: Components of r , i.e. , ,x y z [ ] [ ] [ ]( )ˆ ˆ ˆ ˆˆ ˆsince , , , 0x y y z z x= = =

Components of p , i.e. ˆ ˆ ˆ, ,x y zp p p [ ]( )ˆ ˆ ˆ ˆ ˆ ˆsince , , , 0x y y z z xp p p p p p = = =

Incompatible observables

x and ˆ xp since [ ]ˆ ˆ, i / 2x xx p x p= ⇒ ∆ ⋅∆ ≥

y and ˆ yp since ˆ ˆ, i / 2y yy p y p = ⇒ ∆ ⋅∆ ≥

z and ˆ zp since [ ]ˆˆ, i / 2z zz p z p= ⇒ ∆ ⋅∆ ≥

14. Expectation (mean, average) Values

If a series of measurements of an observable A (represented by A ) is made on a group

(ensemble) of systems in a state described by the wave function. ),( trΨ , then the expectation

value of A can be given by

*

allspace

avg *

allspace

ˆ( , ) ( , )d

( , ) ( , )d

r t A r t r

Ar t r t r

Ψ Ψ

=Ψ Ψ

∫

∫

38

If ( , )r tΨ is normalized, *

avgallspace

ˆ( , ) ( , )dA r t A r t r= Ψ Ψ∫

In Dirac’s notation, A

AΨ Ψ

=Ψ Ψ

and, if Ψ is normalized, ˆA A= Ψ Ψ

If A is representing a physical observable ⇒ A is Hermitian.

⇒ ( )**

all allspace space

ˆ ˆd dA r A rΨ Ψ = Ψ Ψ∫ ∫ - Condition of Hermiticity

Consider *avg

ˆ dA A r= Ψ Ψ∫ ………… (1) (assuming Ψ as normalized)

( )*avgˆ dA A r= Ψ Ψ∫

……….. (2) Using the condition of Hermiticity.

Taking complex conjugate of (1) ⇒ ( )**avg

ˆ d

A A r= Ψ Ψ∫

( )* avgˆ dA r A= Ψ Ψ =∫

(using (2))

⇒ avgA is a real value.

In Quantum mechanics, the uncertainty associated with measurements of an observable A is

defined (Statistically) as

( )1 22A A A ∆ = −

1 2 1 22 22 22 2A A A A A A A A A ∆ = − + = − +

1 222A A A ∆ = −

When Ψ is normalized,

*

allspace

ˆ ˆ dA A A r= Ψ Ψ = Ψ Ψ∫ , 2 *

allspace

ˆ ˆ ˆ ˆ dA A A A A r= Ψ Ψ = Ψ Ψ∫

15. The adoint of an operator In many situations, it is convenient to work with non-Hermitian operators. Then it is useful to introduce,

the adjoint or Hermitian conjugate of an operator A which is denoted by †A († - ‘dagger’)

†A satisfies the relation,

39

( )** †

all allspace space

ˆ ˆˆ ˆ ˆ ˆ( , ) ( , )d ( , ) ( , )dr t A r t r A r t r t rΨ Φ = Ψ Φ∫ ∫

for any functions ( , )r tΨ and ( , )r tΦ

If A is not a H. O.⇒ * *†ˆ ˆ ˆ ˆA A A AΨ Φ = Ψ Φ = Φ Ψ = Φ Ψ

If A is a H. O. ⇒ * *ˆ ˆ ˆ ˆA A A AΨ Φ = Ψ Φ = Φ Ψ = Φ Ψ

For any operator, if †ˆ ˆA A= , then A is said to be self-adjoint. A self-adjoint operator is

Hermitian.

In general † *ˆ ˆA A≠

Eg: Consider *ˆ ˆ ˆi ix xA p px x∂ ∂

= = − ⇒ =∂ ∂

Since ˆ xp is a H.O., †ˆ ˆ ix xp px∂

= = −∂

⇒ † *ˆ ˆx xp p≠

Also, in the matrix representation of operators;

If * *

** *

a b a bA A

c d c d

= ⇒ =

But ( )* * **† T *

* *

a c a cA A A

b d b d

= = = ≠

( unless * *b c= )

Eg: By using the definition of a Hermitain conjugate, show that

i. ( )† † †ˆ ˆˆ ˆA B A B± = ±

ii. ( )† † †ˆ ˆˆ ˆA B B A=

iii. ( )† * †ˆ ˆcB c B= , where c is some complex number.

iv. ( )††ˆ ˆA A=

For any operator A , if ˆiˆ e AQ = , then

ˆ(i )ˆ!

n

n

AQnψψ = ∑ , where ˆ nA in the expansion means

the operator A applied n times.

16. If Aψ ψ′ = , where A is some operator, and ψ ′ is the bra of the ket ψ ′ , then †Aψ ψ′ = .

Pf: For any φ and ψ , we have

40

* *† †ˆ ˆ ˆ ˆA A A Aψ φ ψ φ φ ψ φ ψ ψ φ ψ ψ′ ′ ′= = = = ⇒ =

17. Unit, Inverse, Unitary and Projection Operators The operator that leaves Ψ unchanged is called a unit operator.

i.e. I Ψ = Ψ In Dirac’s notation, I Ψ = Ψ

eg: IΦΨ = Φ Ψ

If for any operator A , there is B such that ˆ ˆˆ ˆ ˆBA AB I= = , then B is said to be the inverse of

A and denoted by 1A− .

An operator U , such that † †ˆ ˆ ˆ ˆ ˆU U U U I= = is called a unitary operator.

An operator Λ is said to be idempotent if 2ˆ ˆΛ = Λ . If, in addition, Λ is Hermitian, it is called a

projection operator.

Any function Ψ can be expressed in terms of two orthogonal functions Φ and Χ by means of

a projection operator.

Hint: writing Ψ = Φ + Χ with ˆΦ = ΛΨ and ( )ˆ1Χ = − Λ Ψ , show that 0Φ Χ = .

18. Closure Relation *

Consider the discrete orthonormal basis set nψ . If it is a complete set, every wave

function ψ can be expanded uniquely on the nψ s as

n nn

cψ ψ= ∑ .

As shown before, nc s are given by n nc ψ ψ= .

⇒ n n n n n n n nn n n n

cψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ

= = = =

∑ ∑ ∑ ∑

Since the above result is true for any ψ ⇒ ˆn nn

Iψ ψ =∑ - Closure Relation

Conversely, ˆ with .n n n n n nn n

I c cψ ψ ψ ψ ψ ψ ψ ψ= = = =∑ ∑

Note: The above results can be also shown to be true when the orthonormal set is a continuous

one. When mφ is a continuous orthonormal set, the closure relation takes the form

ˆdm m m Iφ φ =∫ .

Consider the discrete orthonormal basis set ( )n rψ . If it is a complete set, every wave function ( )rψ can be expanded uniquely on the ( )n rψ s as

41

*( ) ( ) ( ) ( )d ( )n n n nn n

r c r r r r rψ ψ ψ ψ ψ ′ ′ ′= = ∑ ∑ ∫ .

Interchanging the summation and the integration,

*( ) d ( ) ( ) ( )n nn

r r r r rψ ψ ψ ψ

′ ′ ′= ∑∫

From this it can be deduced that *( ) d ( ) ( ), where ( ) ( ) ( )n n

nr r r r r r r r rψ ψ δ ψ ψ δ′ ′ ′ ′ ′= − = −∑∫ and the Dirac’s Delta

function ( ) 1,

0,r r if r r

if r rδ ′ ′− = =

′= ≠

Note: The closure relation expresses the completeness of the set of functions nψ . Not all

Hermitian operators possess a complete set of eigenfunctions and those that do are called

observables.

Tutorial 2 Operators, Eigenvalues and Eigenfunctions 1. If A and B are two Hermitian operators, show that (a) A ± B is hermitian, (b) AB and BA may not

be hermtian (under what condition are they hermitian ?), but that (AB + BA) and i(AB − BA) are both hermitian, (c) (AB − BA) is not hermitian (d) the expectation value of A2 is always real and non-negative, (e) if ⟨A2⟩ = 2, find the eigenvalue.

2. If A and B are two operators with adjoints A† and B† respectively, show that (a) (AB)† = B† A†, (b)

(A ± B)† = A† ± B†, (c) A†† = A , where A†† = (A), (d) (cA)† = c*A†, where c is a complex number (e) (A + A†) and i(A − A†) are hermitian for any operator, as is AA†. Thus show that A can be put as a linear combination of hermitian operators (f) If A is not hermitian, is A†A hermitian ?

3. Given that the inverses A−1 and B−1 are known. (a) What is the inverse of S = AB ? (b) Does you

answer to part (a) depend on the hermiticity of A or B ? (c) Employ your answer to part (a) to establish that if A−1A = I, then AA−1 = I.

4. The operator Q satisfies the two equations Q†Q† = 0, Q†Q + QQ† = 1. The Hamiltonian for a

system is given by H = αQQ†, where α is a real constant. (a) Show that H is Hermitian. (b) Find an expression for H2, the square of H, in terms of H. (c) Find the eigenvalues of H as allowed by the result of part (b).

5. For the two operators: 3ˆ ( ) ( )P x x xψ ψ= and d ( )ˆ ( )d

xQ x xx

ψψ = , find the commutator ˆˆ,P Q .

6. Suppose the two operators, P and Q, satisfy the equations: P = Q†Q + 3, P = QQ† + 1.

(a) Show that P is self-adjoint (b) Find the commutator [Q†, Q]. (c) Suppose Ψ is an eigenfunction of P with eigenvalue p: PΨ = pΨ. Show that if QΨ ≠ 0, then QΨ is an eigenfunction of P and find the eigenvalue.

42

7. For any two operators A and B, it is given that BdtdA

dtdBAAB

dtd

+=)( . Is the following result

always true for an operator A: dtdAA

dtdA 2

2= . ? Show that .11

1−−

−−= A

dtdAA

dtdA

8. If xx

A +∂∂

= and xx

B −∂∂

= , find the products AB and BA and hence show that [A, B] = −2.

9. Show that the operator relation i iˆ ˆe exa xap p a− = + holds. 10. Write out fully the operators corresponding to the three rectangular components of the orbital

angular momentum prL ×= .

11. Show that the linear momentum operator −iħd/dx is hermitian. 12. Indicate which of the following functions are eigenfunctions of the operator d/dx: (a) eikx, (b) eαx,

(c) sin kx. Indicate in each case the eigenvalue. Repeat for the operator d2/dx2. 13. Show that the free-particle wave functions ψ(x) = e±ikx are eigenfunctions of the momentum

operator corresponding to the eigenvalues ±ħk, respectively. 14. Suppose one wants to find the eigenvalues of a Hermitian operator A. Show that if there exist

another operator X such that [A, X] = kX, then k is the spacing between the eigenvalues of A.

43

The Time-Dependent Schrodinger Equation (TDSE) This basic equation of non-relativistic quantum mechanics was proposed by E. Schrodinger in 1926 and is also called the wave equation.

In 3-d, ( , ) ˆi ( , )r t H r tt

∂Ψ= Ψ

∂

, where 2

2ˆ ˆ( , )2

H V r tm

= − ∇ +

In 1-d, ( , ) ˆi ( , )x t H x tt

∂Ψ= Ψ

∂ , where

2 2

2d ˆ( , )

2 dH V x t

m x= − +

This equation gives the time evolution of the wave function of a of a particle (quantum system) of mass m moving in a region of potential energy ( , )V r t . As solutions to the above equation one can obtain the wave function of the system.

In Dirac’s notation, the TDSE is given in the form, ˆ ˆi H Ht

∂Ψ= Ψ = Ψ

∂

Properties of the TDSE

• It is a linear equation (Ψ appears only in power in one). • It is a 2nd order differential equation in r and t. • If 1( , )r tΨ

and 2 ( , )r tΨ are solutions of the TDSE, then 1 2Ψ = Ψ + Ψ is also a

solution of the TDSE. This is a direct result of the linearity of the TDSE.

Probability Current Density Position probability Density, 2( , ) ( , ) * ( , ) ( , )P r t r t r t r t= Ψ = Ψ Ψ

⇒ all space

( , )d 1P r t r =∫ (true for all t values) ⇒

allspace

( , )d 0P r t rt∂

=∂ ∫

.

To show that the above relationship is satisfied in Schrodinger theory, Consider ( , ) * ( , ) ( , )P r t r t r t= Ψ Ψ

( , ) * ( , ) ( , ) * ( , ) ( , )t

P r t r t r t r t r tt t

∂ ∂ ∂= Ψ Ψ + Ψ Ψ

∂ ∂ ∂ ……………… (1)

Consider TDSE 2

2( , ) ˆi ( , ) ( , )2

r t V r t r tt m

∂Ψ= − ∇ + Ψ ∂

……………. (2)

TDSE* 2

2* ( , ) ˆi * ( , ) * ( , )2

r t V r t r tt m

∂Ψ− = − ∇ + Ψ ∂

……. (3)

ˆ ˆ ˆ( , ) * ( , ) if is realV r t V r t V= , and we do not consider complex potential energies in this course.

Now by substituting for t

∂Ψ∂

and *t

∂Ψ∂

from equations (2) and (3) in (1),

2 22 21 ( *) * ( ) * ( ) ( *)

i 2 2 2 iPt m m m

∂ = ∇ Ψ Ψ − Ψ ∇ Ψ = −∇ Ψ ∇Ψ − ∇Ψ Ψ ∂

44

Define the ‘Probability Current Density’ or ‘Probability Flux Density’ as

( , ) * ( ) ( *)2 i

j r tm

= Ψ ∇Ψ − ∇Ψ Ψ

Note: 0j ≠

only if Ψ is complex.

⇒ ( , ) . ( , ) 0P r t j r tt∂

+ ∇ =∂

……………. (4) in 3-d

The above equation is called the ‘Equation of Continuity’ and it describes the conservation of probability in a region. Note: In the above derivation, we initially considered a single particle, such as an electron. But,

experiments often involve beams of particles, for example, a beam of electrons from an accelerator fired at an atom. If the beam is not too intense (i.e. the particle density is not too high) each electron interacts with the atom independently of the presence of the other electrons, and the interactions between the electrons are negligible. This means that we may describe the problem involving the beam of electrons in terms of the same Schrodinger equation that describes a single electron moving in the potential produced by the atom.

Analogous to equation of continuity in quantum mechanics, in classical physics we have

. 0jtρ∂

+ ∇ =∂

, where j vρ=

and v is the velocity. For example,

• In Fluid mechanics, ρ ≡ mass density of a fluid and j vρ=

≡ mass flow density. ⇒ Equation of continuity describes the mass conservation for a volume element in a fluid (i.e.

mass can not be created or annihilated in a given volume.) • In Electrodynamics, ρ ≡ space charge density in a conducting medium and j vρ=

≡ electric current density. ⇒ Equation of continuity describes the charge conservation in a volume element in a

conducting medium (i.e. charge can not be created or annihilated in a given volume.)

Note: It can be written that ( , ) Re *i

j r tm

= Ψ ∇Ψ

and here i

p vm m∇

= =

and *Ψ Ψ is the

position probability density ⇒ position probabilty density velocity =j Pv= × .

As seen in the case of a classical fluid, prob. density × velocity should give probability (of finding particles) flux through a unit area per unit time in a beam of particles. This fact shows that it is reasonable to interpret ( , )j r t

as the probability current density.

Consider ( , ) . ( , ) 0P r t j r tt∂

+ ∇ =∂

.

The integral form of this equation over a finite volume V surrounded by a surface of area S

( , )d . ( , )d 0V V

P r t V j r t Vt∂

+ ∇ =∂∫ ∫

.

45

Green’s theorem, ( . )d .dV S

j V j S∇ =∫ ∫

⇒ ( , )d .d 0V S

P r t V j St∂

+ =∂ ∫ ∫

Physical Interpretation: In a region where there are no sinks or sources, any decrease (increase) in the probability with time in the volume V is accounted for by a probability current directed out of (into) it. Now set V → ∞

⇒ S → ∞ ⇒ r → ± ∞ ⇒ Ψ → 0 (as required by the square-integrability) ⇒ j

→ 0.

allspace

( , )d 0.P r t rt∂

⇒ =∂ ∫

Note: The above result has been obtained only by using the definition of position probability density and the TDSE (Schrodinger’s theory).

Solutions of the TDSE

22( , ) ˆi ( , ) ( , )

2r t V r t r tt m

∂Ψ= − ∇ + Ψ ∂

Assume V to be time-independent. ⇒ ˆ ˆ( , ) ( )V r t V r=

Then, the TDSE has a solution of the form ( , ) ( ). ( )r t r f tψΨ =

. Using the method of separation of variables, we get

22( ) ˆi ( ) ( ) ( ) ( ) ( ) ( )

2f tr f t r f t V r r

t mψ ψ ψ∂

= − ∇ +∂

Dividing throughout by ( ). ( )r f tψ 2

21 ( ) 1 1 ˆi ( ) ( ) ( ) constant ( , say)( ) 2 ( ) ( )

f t r V r r Ef t t m r r

ψ ψψ ψ

∂= − ∇ + =

∂

Function of t only Function of r only

1 d ( )i ( ) d

f t Ef t t

= ……. (1) 2

21 ˆ( ) ( ) ( )( ) 2

r V r r Er m

ψ ψψ

− ∇ + =

……. (2)

From (1), d ( )i ( )df t E f t

t= ⇒ d ( ) i ( )

df t E f t

t= −

Solution ⇒ i i( ) constant e e (choosing, constant = 1)Et h Et hf t − −= × =

From (2), 2

2 ˆ( ) ( ) ( )2

V r r E rm

ψ ψ − ∇ + =

or ˆ ( ) ( )H r E rψ ψ=

46

The above equation is called the ‘Time-Independent Schrodinger Equation (TISE)’. To get the solutions to TISE, one must know what ˆ( )V r is, and it is a characteristic of the problem under investigation Consider the TISE, ˆ ( ) ( )H r E rψ ψ=

Since H is the total energy operator, the above equation is also the energy eigenvalue equation.

( )rψ - energy eigenfunction, E - energy eigenvalue Since H is Hermitian, E must take real values only. Operate H on i( , ) ( ).e E tr t rψ −Ψ = , and by using the fact that H is time-independent,

( )i i iˆ ˆ ˆ( , ) ( ).e e ( ) e ( ) ( , )E t E t E tH r t H r H r E r E r tψ ψ ψ− − −Ψ = = = = Ψ .

ˆ ( , ) ( , )H r t E r tΨ = Ψ

← Another form of the energy eigenvalue equation. Position probability density, i i( , ) * ( , ) ( , ) * ( ).e ( ).e * ( ). ( ) ( )E t E tP r t r t r t r r r r P rψ ψ ψ ψ−= Ψ Ψ = × = = ⇒ Position probability density of i( , ) ( ).e E tr t rψ −Ψ = is constant in time. Such states

are called stationary states. Consider the TISE, ˆ ( ) ( ).H r E rψ ψ=

In general, TISE will have many energy eigenvalues

nE with the corresponding energy eigenfunctions ( )n rψ . i.e. 1 1 2 2 3 3( ), ( ), ( ), ................, etc,E r E r E rψ ψ ψ→ → →

so that, the TISE can be written as ˆ ( ) ( ) with 1, 2, 3, ...........n n nH r E r nψ ψ= =

. The corresponding solution to TDSE,

i( , ) ( ).e with 1, 2, 3, ...................nE tn nr t r nψ −Ψ = = ‘a particular solution’

The most general solution

i( , ) ( , ) ( ).e nE tn n n n

n nr t c r t c rψ −Ψ = Ψ =∑ ∑

Since we want ( , )r tΨ to be normalized to unity, the complex numbers cn are chosen so that

2 1nn

c =∑ .

Properties of Energy Eigenfunctions Basically, there are two kinds of energy eigenfunctions.

(i) Corresponding to bound states. ⇒ e.e.fns vanish at infinity. i.e. they can be normalized (square-integrable)

47

⇒ e.e.vas form a discrete set.

(ii) Corresponding to unbound (scattering) states. ⇒ e.e.fns are only finite at infinity. i.e. they can not be normalized (not square-

integrable) ⇒ e.e.vas form a continuous set (continuum).

Note: In case (ii), the difficulty of normalization can be avoided by enclosing the system in a very

large (but finite) box with impenetrable walls.

* dxψ ψ∞

−∞

= ∞∫ ⇒ ψ can not be normalized to unity.

* dL

L

x aψ ψ−

=∫ ⇒ 1aψ is normalized to unity.

(L is chosen to be very large and hence, a may be large but finite) Under the above ‘box normalization’ procedure, the energy spectrum will become descrete. As shown before, all the e.e.fns can be assumed to be normalized to unity. Since H is a Hermitian operator, e.e.fns corresponding to different e.e.vs (non-degenerate) are orthogonal. When an e.e.v. is degenerate, it is still possible to make the e.e.fns orthogonal by using some orhogonalization procedure. ⇒ Therefore, we can expect in all cases, the e.e.fns to form an orthonormal basis. i.e. they

satisfy the condition *

allspace

dm n m nrψ ψ δ=∫ .

Solutions to TISE for a Particle Moving in Various Potential Regions

(1) Free Particle (1-d) Consider a particle of mass m and total energy E moving along the x-axis in a region of constant potential energy.

0( ) ,V x V x= −∞ < < ∞

⇒ Force acting on the particle, d ( )( ) 0d

V xF xx

= − = .

Choose V0 = 0. (this choice will not affect the e.e.vas). Apply the TISE for the particle. 2 2

2d ( ) ˆ( ) ( ) ( ),

2 dx V x x E x x

m xψ ψ ψ− + = −∞ < < ∞

⇒ 2 2

2d ( ) ( ) .................. (1)

2 dx E x

m xψ ψ− =

Total energy, 2 2

( ) 02 2p pE V xm m

= + = +

x xp k p k p k= ⇒ = ⇒ =

(dropping x) 2 2

22

22

h k mEE km

= ⇒ =

48

2 22

2 2 2d ( ) 2 d ( )( ) 0 ( ) 0

d dx mE xx k x

x xψ ψψ ψ+ = ⇒ + =

Solutions (particular) of TISE, ⇒ i( ) e k xxψ ±= For physically acceptable solutions (i.e. ψ(x) → 0 as x → ± ∞), k has to be real ⇒ E ≥ 0. i.e. the energy spectrum of the particle is continuous and is extending from E = 0 to E = + ∞. Then, a general solution to TISE is i i( ) e + ek x k xx A Bψ + −= , where A and B are arbitrary constants. Solutions (particular) of TDSE are i i( , ) = e ek x E tx t ± −Ψ × . General solutions to TDSE is i i i i( , ) e e + e ek x E t k x E tx t A B− − −Ψ = .

i ( ) i ( )

1 2

( , ) e + e , where = . = ( , ) + ( , )

k x t k x tx t A B Ex t x t

ω ω ω− − +Ψ =Ψ Ψ

→ ← The solution is a sum of two plane waves traveling in +x and –x directions. Each wave represents a particle moving in +x-direction with a precisely known linear momentum p k= ( 0p∆ = ) and –x-direction with a precisely known linear momentum p k= − ( 0p∆ = ). Case (i): When the amplitude of one of the plane waves in Ψ(x, t) is zero, B = 0 (say).

⇒ i ( )1( , ) ( , ) e k x tx t x t A ω−Ψ = Ψ =

Consider normalization of Ψ(x, t).

( ) ( )2 2 2 21 , d , d d .x t x x t x A x A

∞ ∞ ∞

−∞ −∞ −∞

Ψ = Ψ = = ∞∫ ∫ ∫ ⇒ not normalizable.

⇒ Plane waves are not normalizable.

Consider the position probability density associated with Ψ(x, t).

( ) ( )2 2 21 ( , ) , , P x t x t x t A= Ψ = Ψ = = constant

⇒ Position of the particle moving in the +x direction is completely unknown on the x- axis. ⇒ ∆x = ∞

This is a direct consequence of the H.U.P, . 2xx p∆ ∆ ≥ (As ∆px → 0 ⇒ ∆x → ∞)

Consider the probability current density associated with Ψ(x, t)

( ) ( ) ( ) ( ) ( )* *d d, , , , ,2 i d d

j x t x t x t x t x tm x x

= Ψ Ψ − Ψ Ψ

= 2 2k pA Am m

= × = velocity × p.p.d.

Similar results can be obtained when A = 0 in the general solution of TDSE.

49

Case (ii): When the amplitudes of the two plane waves are equal, i.e. A = B.

( )i i i i( , ) e + e e = cos ek x k x t tx t A C k xω ω− − −Ψ =

A standing wave whose nodes fixed at values ( )2 , 0, 1, 2, ........nx n k nπ π= ± + =

( ) 2 2 2 ( , ) , cosP x t x t C k x= Ψ =

( ), 0j x t = (show this)

When the amplitudes of the two plane waves are equal, the probability flux associated with

one plane wave is cancelled by that of the other. This makes the prob. per unit time that the

particle will cross a point x is zero. Therefore, the resulting standing wave represents a

particle moving along the x-axis with precisely known momentum ( k ) but whose direction

is unknown. Since P = 0 at x = xn, the particle can not be found at these points. This is due

to an interference effect between the plane waves.

To obtain information about the position of a particle localized within a region ∆x, we must

superpose plane wave solutions of the form ( )i ( )( ) e k x k tA k ω− with infinitesimally different k

values to from a ‘wave packet’.

When k is continuous, Ψ(x, t) = ( ) ( )i ( )

allvalues

e dk x k t

k

A k kω−∫

When k is discrete, Ψ(x, t) = ( ) ( )i ( )e k x k t

kA k ω−∑

(2) Step Potential Consider a particle of mass m and total energy E approaching from the left of the potential

step which is defined as ( )0

0 00

xV x

V x<

= >

V0 Real potential situation

x < 0 x > 0

50

Since V(x) has a discontinuity at x = 0, we divide the region into two parts and solve TISE in regions I and II separately. We have to consider two cases,

(a) E < V0 (b) E > V0 Case (a): 0 < E < V0 In region I, the k.e of the particle = E − 0 = E In region II, the k.e. of the particle = E – V0 < 0 Classically, the particle should be reflected back to region I at x = 0. i.e. x > 0 is a classically

forbidden region.

In region I When ( ) 0V x =

( ) ( )2 2

I I2d

2 dx E x

m xψ ψ−

=

( ) ( )2

I I2 2d 2 0d

mEx xxψ ψ+ =

General solution to TISE

( ) 2I 2

2i ie e , where m Ek x k xx A B kψ −= + =

General solution to TDSE

( ) ( ) ( )I

i i, e ek x t k x tx t A Bω ω−− +Ψ = +

→ ← represents incident represents reflected particles particles

In region II When 0( )V x V= ,

( ) ( ) ( )2 2

II 0 II II2d

2 dx V x E x

m xψ ψ ψ−

+ =

( ) ( ) ( )2

II 0 II2 22d 0

dmx V E x

xψ ψ− − =

General solution to TISE

( ) ( )02II 2

2e e , where

m V Ex xx C Dα αψ α−−= + =

.

General solution to TDSE

51

( )IIi i, e ex t x tx t C Dα ω α ω− − −Ψ = + (Note: These are not plane waves.)

Since II ( )xψ has to be finite (small) in region II, e xα+ is not acceptable. So, D → 0 ⇒ ( )II e 0xx C xαψ −= > To determine A, B and C. we apply boundary conditions which govern the continuity of wave

function in regions I and II, because they represent the same particle.

Consider for smooth joining of a wave function

I II I II0 0 0 0d d( ) ( ) ( ) ( )d dx x x xx x x xx x

ψ ψ ψ ψ= = = == =

i ( )A B C k A B Cα+ = − = −

iikB Ak

αα

+= −

2ii

kC Ak α

= −

A, B and C are the amplitudes of the incident, reflected and transmitted waves respectively.

( ) i iI

ie e , 0ikk x k xx A xk

αψα

− += + < −

(a standing wave)

( )II2i e , 0

ik xx A x

kαψ

α −= > −

(an exponentially decaying function)

Intensity of the incident wave = 2 *A A A=

Intensity of the reflected wave = 2*B B B= 2

iik Ak

αα

+= −

2 2i ii ik k A Ak k

α αα α

+ − += = − − −

⇒ Reflection is total (100%).

smooth joining not joining not smooth joining

52

i.e., all the particles reaching the potential step with total energy 0 E V< bounce back.

This result is in agreement with the prediction of classical mechanics according to which all the

particles with total energy 0 E V< are reflected by the step.

Now consider II ( ) e 0xx C xαψ −= > ( )II ( ) 0xψ ≠

2II II

2*p. p. d. ( ). ( ) e xx x C αψ ψ −= = ⇒ There exist finite probabilities of finding the particle in the classically forbidding region

( 0x > ). This result is different from that is predicted by classical mechanics. This non- classical phenomenon is called the ‘penetration of classically forbidden region’ or ‘tunneling’.

Note:

Penetration doesn’t mean that the particle is stored in region 0x > permanently, because we see

that the incident particle is definitely reflected from the step.

Eg: Step potential is an approximation to the potential on a conducting electron moving near the

surface of a metal. Potential energy for an electron whose total energy is E increases at the

surface from a constant interior value (may be zero) to a higher constant exterior value Vr

and E < Vr. For Cu, Vr − E ~ 4 eV.

Flux of the incident beam of particles, 2 2i inc

kj v Am

ψ= × = ×

Flux of the reflected beam of particles, 2 2r ref

kj v Bm

ψ= × = ×

2r

2i

Reflected flux in region IReflection coefficient, 1Incident flux in region I

BjR

j A= = = =

⇒ Reflection is total (100%).

( )II e 0xx C xαψ −= >

( )II xψ decreases very rapidly as x increases. So the particle does not penetrate very far into the

classically forbidden region.

When 1 2, e 0xx αα

− → .

Therefore the probability is appreciable only over a ‘penetration depth’

( )0

12

xm V Eα

∆ = =−

.

53

A standing wave

incψ and refψ have equal amplitudes and velocities so they are in phase. Superposition of incψ and refψ will result in a standing wave. When V0 → ∞ ⇒ α → ∞ ⇒ II ( ) 0xψ →

⇒ No particle can penetrate into the classically forbidden region (i.e., an impenetrable wall)

Case (b): E > V0 In region I, the k.e. of the particle = E − 0 = E In region II, the k.e. of the particle = E – V0 > 0 Classically, the particle will transmit into region II and no reflection is possible. In region I, the solution of TISE has the same form as in case (a)

( ) 2I 2

2i ie e , where 0m Ek x k xx A B k xψ −= + = <

→ ← represents incident represents reflected particles particles In region II, TISE reads

( ) ( ) ( )2 2

II 0 II II2d 0

2 dx V x E x x

m xψ ψ ψ−

+ = >

( ) ( ) ( )2

II 0 II2 22d 0

dmx E V x

xψ ψ+ − =

General solution

( ) ( )02II 2

2i ie e , wherem E Vk x k xx E F kψ

−′ ′− ′= + =

→ ← represents transmitted represents reflected

particles particles

x

incψ

refψ

( )II , 0xx Ce xαψ −= >

54

In region II, the potential energy remains constant at V0

⇒ No reflection is possible. ⇒ F = 0

( )IIi 'e , 0k xx E xψ = >

To determine A, B and E, apply boundary conditions.

I II I II0 0 0 0d d( ) ( ) ( ) ( )d dx x x xx x x xx x

ψ ψ ψ ψ= = = == =

( )A B E k A B k E′+ = − =

k kB Ak k

′ −= ′+

2kE Ak k

= ′+

( ) i iI e e 0k kk x k xx A x

k kψ − ′ −

= + < ′+ (a standing wave)

( )II2 ie 0k k xx A x

k kψ

′= > ′+ (a pure traveling wave)

B ≠ 0 indicates that some particles are reflected at x = 0. This is again a result different from that is predicted classically.

Incident probability flux, 2i

kj Am

= ×

Reflected probability flux 2r

kj Bm

= ×

Transmitted probability flux 2t

kj Em′

= ×

Reflection coefficient,

Reflected flux in region IReflection coefficient, Incident flux in region I

R =

22 02

r2i 0

1 1

1 1

Vk Bj k k EmRkj k k VA

m E

− − ′ − = = = = ′ + + +

55

Transmitted flux in region IITransmission Coefficient, Incident flux in region I

T =

( )

2

t22i

4k Ej kkmTkj k kA

m

′′

= = =′+

1, 1 1R T R T< < ⇒ + = , as required by the conservation of the # of particles.

Incoming flux = Transmitted flux + Reflected flux Eg: This situation is an approximation to the potential acting on a neutron near the nuclear surface.

If the neutron gets energy so that E is a little greater than V0 there is a probability for it to be reflected back into the nucleus and thereby increasing the stability in it.

(3) Barrier Potential

Consider V(x) defined as 0 in the region 0 ( )

0 in regions < 0 and > V x a

V xx x a

≤ ≤=

A particle of mass m and total energy E is approaching the barrier from the left.

There are two cases.

(a) E < V0 ⇒ c.m. predicts a total reflection.

(b) E > V0 ⇒ c.m. predicts a total transmission.

For cases (a) and (b), the solutions to TISE in regions I and III are the same as those in the case of

free particles.

( ) 21 1I 1 2i i 2e e ......... (1), where 0k x k x m Ex A B k xψ

−= + = <

( ) 21 1III 1 2i i 2e e .......(2) , wherek x k x m Ex C D k x aψ

−= + = >

V(x) V0 I II III x

0 a

56

when E < V0

( ) ( )02II 2

2e e ........... (3), where , 0

m V Ex xx F G x aα αψ α−−= + = ≤ ≤

when E > V0

( ) ( )02II 2

2i ie e ........(4), where , 0m E Vx xx F G x aβ βψ β

−−′ ′= + = ≤ ≤

Since no reflection is possible in region III, D = 0

Note: We cannot set G = 0, because x is finite when 0 x a≤ ≤ . We can not set G′ = 0, because a reflection is possible at x = a due to change in potential energy.

Case (a) E < V0 Consider equations (1), (2) and (3). Appling boundary conditions at x = 0 and x = a,

I II II III0 0( ) ( ) ( ) ( )x x x a x ax x x xψ ψ ψ ψ= = = == =

I II II III0 0

d d d d( ) ( ) ( ) ( )d d d dx a x a

x xx x x x

x x x xψ ψ ψ ψ= =

= == =

Obtain four equations and solve them to determine B, C, F and G in terms of A (Do this as an

exercise.)

In region I, the wave function is not a pure standing wave (since A > B) and has a small traveling

wave component.

In region II, the wave function is a standing wave with decreasing amplitude

In region III, the wave function is a pure traveling wave.

Reflected flux in region IReflection coefficient, Incident flux in region I

R =

11

20 0I

2 2I

4 11

sinh

E EV VB v

RaA v α

−− − × = = + ×

Velocity in region I = Ikv

m=

57

Transmitted flux in region IIITransmission Coefficient, Incident flux in region I

T =

1

2 2III

2I

0 0

sinh14 1

C v aTE EA v

V V

α

−

× = = + × −

I IIIsince kv vm

= =

In this case (E < V0 ), T is more important than R, because it is not expected classically.

Classically we expect R = 1, T = 0.

But quantum mechanically, T ≠ 0 ⇒ The particle has a certain probability of penetrating the

barrier and appear in the x > a region even when E < V0.

This phenomenon is called the ‘Barrier penetration’ or ‘Tunnel effect’.

* As E → 0 ⇒ T → 0.

* 12

020

lim1

2

mV aT

E V

−

= + → . Here

202

mV a

is a measure of the ‘opacity’

of the barrier.

* In the classical limit, the opacity is very large. i.e. T = 0 for all 0 < E < V0 .

* When 1aα , 0 0

16 21 eE E aTV V

α −≅ −

⇒ T is very small.

Some consequences of barrier penetration (E < V0 )

The emission of α particles from radioactive nuclei (U238) through the potential barrier that they

experience in the vicinity of the nuclear surface.

Between A and B, an α- particle of total energy E ~ 4 MeV must penetrate the nuclear potential

barrier. (E < V0)

x

p.p.d Minimum ≠ 0 (A > B) a

I II III

*( ) ( )x xψ ψ

58

Case (b): E > V0

Consider equations (1), (2) and (4).

Apply boundary conditions at x = 0 and x = a to determine B, C, F ′and G′ in terms of A. (Do this

as an exercise) You will find that the wave function is oscillatory in all three regions.

Reflected flux in region IReflection coefficient, Incident flux in region I

R =

1

20 0

2 2

4 11

sinI

I

E EV VB v

RaA v β

−

− × = = + ×

Transmission coefficient,

1

2 2III

2I

0 0

sin14 1

C v aTE EA v

V V

β

−

× = = + × −

I IIIsince kv vm

= =

In this case (E > V0), R is more important than T, because classically we expect T = 1 and R = 0.

Since R ≠ 0, the particle has a certain probability of getting reflected back to x < 0 region.

(4) Infinite Square Potential Well (symmetric about at x = 0)

V(x)

V0 4 MeV x A B

+ ∞ V(x) + ∞ I II III a −a/2 0 +a/2

m

59

A particle of mass m is confined (trapped) to a potential well of width a which is defined as

0 in the region / 2 / 2( )

elsewherea x a

V x− ≤ ≤ +

= ∞

The walls of the well are impenetrable for the particle and therefore it is bound.

⇒ TISE will have solutions only for discrete values of energy.

ψ(x) must vanish in regions I and III, because no penetration is possible.

⇒ I III( ) ( ) 0x xψ ψ= =

we have to solve TISE only in region II

( ) ( ) ( )2 2

II2d , / 2 / 2 ( ) ( )

2 dx E x a x a x x

m xψ ψ ψ ψ−

= − ≤ ≤ + ≡

General solution, i i( ) e ekx kxx C Dψ −= + , where ( )2

22 / 2 / 2mEk a x a= − ≤ ≤ +

= cos sinA kx B kx+ ψ(x) must be continuous ⇒ requires that, ψ(x) = 0 at x = ± a/2

Since the potential makes an infinite jump at the boundaries d ( )d

xxψ need not be continuous.

At , cos sin 02 2 2a ka kax A B= − − =

At , cos sin 02 2 2a ka kax A B= + =

⇒ cos 02kaA = , sin 0

2kaB =

We can’t make both cos 02ka

= and sin 02ka

= for a given value of k. Also we can’t choose

both A = 0 and B = 0, because, then ψ(x) = 0. Therefore, we consider two classes of solutions.

1st Class

B = 0 and cos 02ka

=

⇒ , 1, 3, 5, .......2 2ka n nπ

= = ⇒ (say)nnk kaπ

= ≡

Wave function ( ) cos cos cosnn xx A kx A k x A

aπψ = = =

( ) cos ,n n nn xx A A A

aπψ = ≡

60

To find An, normalize ψn(x) ⇒ 2set ( ) d 1x xψ+∞

−∞

=∫ .

2 2 2I III

/ 2 / 2

/ 2 / 2( ) d ( ) d ( ) d 1n

a a

a ax x x x x xψ ψ ψ

− + +∞

−∞ − ++ + =∫ ∫ ∫

2 22/ 2 / 2

/ 2 / 2

1 2cos d 1 cos d2n n

a a

a a

n x n xA x A xa aπ π

− −

= + ∫ ∫ ⇒ 2 21

2n naA A

a= ⇒ = .

Wave function (energy eigenfunction), 2( ) cos , 1, 3, 5, .............nn xx n

a aπψ = =

Energy eigenvalue, 2 2 2

2 , 1, 3, 5, .........2

nnE n

maπ

= =

2nd class

A = 0 and sin 02ka

=

⇒ , 2, 4, 6, .......2 2ka n nπ

= = ⇒ (say)nnk kaπ

= ≡

Wavefunction (energy eigenfunction) 2( ) sin , 2, 4, 6, .............nn xx n

a aπψ = =

(n = 0 does not represent a physically meaningful solution.)

Energy eigenvalues 2 2 2

2 , 2, 4, 6, .........2

nnE n

maπ

= =

By combining the two classes of solutions

Energy eigenfunctions 2( ) cos , 1, 3, 5, .............nn xx n

a aπψ = =

2( ) sin , 2, 4, 6, .............nn xx n

a aπψ = =

Energy eigenvalues 2 2 2

2, 1, 2, 3, 4, 5, 6, .........

2n

nE nmaπ

= =

The above results are different from the classical prediction, according to which all the values of

energy are possible for the particle.

In the previous applications of TISE, there were solutions for every value of energy (a continuum).

• Energy spectrum consists of an infinite no. of discrete energy levels (quantized levels) ranging from 2 2 22maπ to ∞ corresponding to bound states.

• Energy levels are non-degenerate, i.e. for each energy eigenvalue there is only one energy eigenfunction.

• Energy eigenfunctions ψn(x) and ψm(x) corresponding to Energy eigenvalues En and Em (m ≠ n) are

orthogonal. i.e. / 2

/ 2( ) ( )d 0*

a

ax x xn mψ ψ

+

−

=∫ .

61

• Since each energy eigenfunction is normalized to unity, i.e./ 2

/ 2( ) ( )d 1*

a

na

x x xnψ ψ+

−

=∫ for any n, and we

have / 2

/ 2( ) ( )d*

a

ax x xn m nmψ ψ δ

+

−

=∫ ⇒ energy eigenfunctions from an orthonormal set.

• Lowest energy (zero-point energy) is not zero which is the case expected classically.

Consider 2 2 2

2 , 1, 2, 3, 4, 5, 6, .........2

nnE n

maπ

= =

2 2

1 min 21 02

n E Emaπ

= ⇒ = = ≠

(zero-point energy)

This is in agreement with the H.U.P, because

. xx p∆ ∆ and ~ ~ /xx a p a∆ ⇒ ∆ , ~ ~ /x xxp p p a∆ ⇒

2 2 2

2 2k.e. ~ ~2 2 2

xpE

m ma ma= ⇒

The total energy predicted by the H.U.P is of the order of the zero-point energy. 22 2 2

22 2 2

2 , 1, 2, 3, 4, ...........2

n nm anE n E n

m aπ

π= ⇒ = =

16 n = 4 9 n = 3 4 n = 2 1 n =1

n2

−a/2 0 +a/2 −a/2 0 +a/2

62

Mathematically, ( ) ( )f x f x− = - a symmetric (even) function

( ) ( )f x f x− = − - an anti-symmetric (odd) function

Parity: the ‘oddness’ or ‘evenness’ of wave function

Eigenfunctions of the 1st class, ( ) 2 cos , 1, 3, 5, ......nn xx n

a aπψ = =

⇒ ( ) ( )n nx xψ ψ− = ⇒ even functions of x or ‘even parity’ states

Eigenfunctions of the 2nd class, ( ) 2 sin , 2, 4, 6, .....nn xx n

a aπψ = =

⇒ ( ) ( )n nx xψ ψ− = − ⇒ odd functions of x or ‘odd parity’ states

• Classically, the particle is found with equal probability anywhere inside the well, but not so

quantum mechanically.

• As n → ∞ , we expect to have a constant probability of finding the particle inside the potential

well, which is the classical result, a result that confirms the correspondence principle. As a

increases, the spacing of the energy levels decreases. We get these energy eigenfunctions of

definite parity (even or odd) because the potential energy function is symmetric about x = 0,

i.e. ˆ ˆ( ) ( )V x V x= − - Even (symmetric) function of x

Ex: Show that, when ˆ ˆ( ) ( )V x V x= − , both ψ(x) and ψ(−x) are solutions to the TISE with the same energy

eigenvalue E.

Assume that ψ(x) is a solution to the TISE with the corresponding eigenvalue E.

2 2

2d ˆ( ) ( ) ( ) ( )

2 dx V x x E x

m xψ ψ ψ−

+ =

Changing x → −x and using the fact that, ˆ ˆ( ) ( )V x V x= − ,

⇒ 2 2

2d ˆ( ) ( ) ( ) ( )

2 dx V x x E x

m xψ ψ ψ−

− + − = −

⇒ ψ(−x) is a solution to the TISE with the same eigenvalue E.

Consider the two cases,

(i) the eigenvalue E is non-degenerate

i.e.ψ(x) and ψ(−x) then differ only by a multiplicative constant so that ( ) ( )x xψ αψ− = ⇒

eigenfunctions ψ(x) have a definite (i.e. odd or even) parity.

(ii) the eigenvalue E is degenerate

63

i.e. there can be more than one linearly independent eigenfunction corresponding to the eigenvalue

E. If these eigenfunctions ψ(x) and ψ(−x ) do not have a definite parity, then it is possible to

construct two linear combinations of them such that each has a definite parity. For example, 1 12 2( ) [ ( ) ( )] and ( ) [ ( ) ( )]e ox x x x x xψ ψ ψ ψ ψ ψ= + − = − − are even and odd parity states

having the same eigenvalue E as ( )xψ and ( )xψ − do. (Note: ( ) ( ) ( )e ox x xψ ψ ψ= + and

( ) ( ) ( )e ox x xψ ψ ψ− = − )

Parity Operator, ℘

Consider the operator ℘ which has the property of changing x to –x (or –x to x) in everything that it

acts on.

ˆ ( ) ( )x xψ ψ℘ = − ˆ ( ) ( )x xψ ψ℘ − = (in 1-d)

ˆ ( ) ( )r rψ ψ℘ = − , or ˆ ( , , ) ( , , )x y z x y zψ ψ℘ = − − − (in 3-d)

℘ is called the ‘parity operator’ and its operation is equivalent to a reflection through the origin.

eg: ℘( x2 + 2x) = (−x)2 + 2(−x) = x2 − 2x

Note: There is no algebraic form of ℘ and it is defined by what it does to a function.

For n particles, the definition of parity operator reads,

1 2 3 1 2 3ˆ ( , , , ........., ) ( , , , ........., )n nx x x x x x x xψ ψ℘ = − − − − (in 1-d)

1 2 3 1 2 3ˆ ( , , , ........., ) ( , , , ........., )n nr r r r r r r rψ ψ℘ = − − − − (in 3-d)

Properties of ℘

(i) 2 ˆˆ I℘ = - Unit operator

Pf: ˆ ( ) ( )x xψ ψ℘ = − ⇒ ( )ˆ ˆ ˆ( ) ( )x xψ ψ℘℘ = ℘ −

⇒ ( )ˆ ˆ ( ) ( )x xψ ψ℘℘ = ⇒ 2 ˆˆ ( ) ( ) ( )x x I xψ ψ ψ℘ = = ⇒ 2 ˆˆ I℘ = .

(i) The eigenvalues of ℘ are ± 1

Pf: ˆ ( ) ( )x xψ εψ℘ = - the eigenvalue equation

⇒ ( )ˆ ˆ ˆ( ) ( )x xψ ε ψ℘℘ = ℘ ⇒ 2ˆ ( ) ( )x xψ εεψ℘ =

⇒ 2ˆ ( ) ( )I x xψ ε ψ= ⇒ 2( ) ( )x xψ ε ψ=

⇒ 2 1ε = ⇒ 1ε = ± .

ˆ ( ) ( )x xψ εψ℘ = ⇒ ( ) ( )x xψ εψ− =

1 ( ) ( )x xε ψ ψ= ⇒ − = ⇒ even parity (symmetric) state, ( )e xψ

1 ( ) ( )x xε ψ ψ= − ⇒ − = − ⇒ odd parity (symmetric) state, ( )o xψ

64

Ex: Prove that,

(a) ℘ is a hermitian operator ( )†ˆ ˆ℘ = ℘

(b) ℘ anti-commutes with x and ˆ xp , commutes with H , when ˆ ˆ( ) ( )V x V x= − and

( )2ˆˆ ˆ

2xp

H V xm

= + ˆ ˆ ˆˆ ˆ ˆommutes with ˆ ˆ ˆˆ ˆ ˆanti commutes with

A c B AB BA

A B AB BA

⇒ = ⇒ = −

Note: Since ℘ commutes with H , i.e. ˆˆ , 0H ℘ = , we can find a set of simultaneous

eigenfunctions of ℘ and H .

(c) The odd and even parity states of ℘ are orthogonal to each other.

(5) Infinite Square Potential Well (not symmetric about x = 0) Consider a particle of mass m confined to a potential well of width a which is defined as

0 in the region 0 ( )

elsewherex a

V x≤ ≤

= ∞

Since the walls at x = 0 and x = a are impenetrable for the particle I III( ) ( ) 0x xψ ψ= = .

Solve TISE for the particle inside the potential well.

( ) ( ) ( )2 2

II2d , 0 ( ) ( )

2 dx E x x a x x

m xψ ψ ψ ψ−

= ≤ ≤ + ≡

2

2 22 2

d 2( ) ( ) 0,d

mEx k x kxψ ψ+ = =

General Solution, ( ) cos sin 0x A kx B kx x aψ = + ≤ ≤

Apply the boundary condition ( ) 0xψ = at x = 0 and x = a. We need not worry about the

continuity of d dxψ .

m

+ ∞ V(x) + ∞ I II III ψI(x) = 0 ψII(x) ψIII(x) = 0

0 a

65

At 0 0 ( ) sin , 0x A x B kx x aψ= ⇒ = ⇒ = ≤ ≤ At sin 0 (say) , 0, 1, 2, ........nx a B ka ka n k n a k nπ π= ⇒ = ⇒ = ⇒ = ≡ = n = 0 does not represent a physically meaningful solution.

(n = 0 ⇒ kn = 0 ⇒ ψII(x) = 0 for all x, 0 ≤ x ≤ a)

( ) sin , 0n nn xx B x a

aπψ = ≤ ≤

To find Bn, normalize ψn(x) ⇒ 2set ( ) d 1x xψ+∞

−∞

=∫ .

02 2 2

I III0

( ) d ( ) d ( ) 1n

a

ax x x x x dxψ ψ ψ

+∞

−∞

+ + =∫ ∫ ∫

2 22

0 0

1 2sin d 1 cos d2n n

a an x n xB x B xa aπ π = −

∫ ∫ ⇒ 2 212n naB B

a= ⇒ = .

Wave function (energy eigenfunction), 2( ) sin , 1, 2, 3, .............nn xx n

a aπψ = =

Energy eigenvalue, 2 2 2

2 , 1, 2, 3, .........2

nnE n

maπ

= =

n = 1, ψ1(x) → 2 2

1 22E

maπ

=

n = 2, ψ2(x) → E2 = 4E1

n = 3, ψ3(x) → E3 = 9E1

• We get infinite no of discrete e.e values ranging from 2 2

22maπ to ∞. They are all non-

degenerate. (i.e. for each E value, there is only one ψ(x) value )

• We don’t get two classes of solutions as V(x) is not symmetric about x = 0.

• Zero-point energy, 2 2

1 2 02

Emaπ

= ≠ , as required by H.U.P.

Existence of zero-point energy is typical in all systems in which a particle is confined to move in a

limited region.

22 2 22

2 2 22 , 1, 2, 3, 4, ...........

2n n

m anE n E nm aπ

π= ⇒ = =

• We can not talk of odd and even parity states as in the symmetric potential well case.

• Classically the particle should be found with equal probability anywhere inside the well but

not so in quantum mechanics.

• As n → ∞, the probability of finding the particle inside the well is almost constant.

66

• ψn(x) from an orthonormal basis

i.e. 0

* ( ) ( )da

x x xn m nmψ ψ δ=∫

• The particle can not have an arbitrary energy values as expected classically. Instead, the

energy values are quantized. This situation occurs whenever a particle is confined to move

in a limited region in space.

Recall: no quantization of energy is possible for

1. Free particle

2. Potential step

3. Potential barrier

*(6) Finite Square Potential Well

Consider a particle of mass m and total energy E bound to the potential well (i.e. E < V0, and we are

interested in bound states only) which is defined as

( )V x V

2a− 2a

I II III

0

9 n = 3 4 n = 2 1 n =1

n2

x

0 a 0 a

67

0 2 or 2( )

0 2 2V x a x a

V xa x a

< − >= − < <

When E < V0, classically the particle can be only within the well and E can have any value between

0 and V0. The above potential can be approximated to the following physical systems.

Eg; A neutron confined to the nucleus.

The potential acting on a conduction electron in a metal due to closely spaced positive ions.

Inside the well, the general solution to the TISE ⇒

( ) 2II 2

2i ie e , where 2 2

sin cos

m Ex xx A B a x a

A x B x

α αψ α

α α

−′ ′= + = − < <

= +

Outside the potential well, the general solution to the TISE ⇒

( ) ( )02I 2

2e e , where , 2

m V Ex xx C D x aβ βψ β−− += + = < −

( ) ( )02III 2

2e e , where , 2

m V Ex xx E F x aβ βψ β−− += + = >

For )(xψ to be finite outside the well (as ∞±→x ), C = 0, F = 0.

Apply boundary conditions: )(xψ and dx

xd )(ψ are continuous at 2ax ±= .

2ax −= , 2

2

sin ( 2) cos( 2) e

cos ( 2) sin( 2) e

a

a

A a B a D

A a B a D

β

β

α α

α α α α β

−

−

− + =

+ =

2x a= , 2

2

sin ( 2) cos( 2) e

cos ( 2) sin( 2) e

a

a

A a B a E

A a B a E

β

β

α α

α α α α β

−

−

+ =

− = −

We get,

one ion 3 ions in line m

nuclear diameter ~ 10 − 14 m

depth ~ 50 MeV

68 22 sin ( 2) ( ) e ....... (1)aA a D E βα −= − − , 22 cos ( 2) ( ) e .......... (2)aB a D E βα −= +

22 cos ( 2) ( ) e ........... (3)aA a D E βα α β −= − , 22 sin ( 2) ( ) e ......... (4)aB a D E βα α β −= +

If n0 and ( ) 0, Equ s (1) and (3) cot ( 2) ....... (5)A D E aα α β≠ − ≠ ⇒ = −

If n0 and ( ) 0, Equ s (2) and (4) tan ( 2) ......... (6)B D E aα α β≠ + ≠ ⇒ =

Since equations (5) and (6) can not be valid at once and we do not want all A, B, D and E to vanish,

we have to look for two classes of solutions. Note: In this problem, V(x) = V(−x)

For the 1st class: 0, and tan ( 2) ............. (6)A D E aα α β= = =

For the 2nd class: 0, and cot ( 2) ....... (5)B D E aα α β= = − = −

When A = 0 and D = E, from the expressions of I ( )xψ , II ( )xψ and III ( )xψ , we see that the

resulting eigenfunctions are even (symmetric) functions of x.

When B = 0 and D = −E, from the expressions of I ( )xψ , II ( )xψ and III ( )xψ , we see that the

resulting eigenfunctions are odd (anti-symmetric) functions of x.

To find energy values of even and odd parity states, we have to solve equns. (5) and (6) with the

definitions of α and β using a numerical or a graphical method. Here we use the graphical method.

cot ( 2)aα α β= − ⇒ ( 2) cot ( 2) 2a a aα α β= − ⇒ cotξ ξ η= − ,

tan ( 2)aα α β= ⇒ ( 2) tan ( 2) 2a a aα α β= ⇒ tanξ ξ η= ,

where 2aξ α= and 2aη β= .

22 22 2 2 2 0 0

2 22

( ) .4 4 2

mV mV aa aξ η α β+ = + = =

⇒ 2 2 2Rξ η+ = ,

which is the equation of a circle of radius ( )1 22 20 2R mV a= .

ξ and η are restricted to positive values. So, the energy values of even and odd parity states may

be found from the points of intersections in the first quadrant of each of the curves ηξξ =tan and

ηξξ −=cot plotted verses ξ with the above circle.

We see that the allowed values of E, when E < V0 are quantized. (when E > V0, allowed E values

form a continuum.)

69

2a

3ψ

2ψ

x

x

x

1ψ

2a− 0

The number of E values will depend on the radius

of the circle. Note that,

(1) The number of allowed energy values

depends on the product 20 aV and not on V0

and a separately.

(2) At least one even parity state is guaranteed

to exist. (No matter how small R is, the

circle intersects ηξξ =tan curve.)

(3) The ground state is an even parity state.

The 1st excited state gives an odd parity

state. The 2nd excited state gives an even parity state.

(4) To have the 1st odd parity state, it requires 2π≥R ⇒ m

aV2

222

0π

≥ .

(5) There exist only a finite number of discrete bound states.

For solutions of the 1st class, A = 0, D = E. To find the arbitrary constants, normalize )(xψ .

Set 2( ) d 1n x xψ+∞

−∞

=∫ ⇒ / 2

2 22 2 2 2

/ 2

/ 2

/ 2e d cos d e d 1

ax x

a

a

aD x B x x E xβ βα

−−

−

+∞

−∞+ + =∫ ∫ ∫

2

2 sine 12 2

a B aD aβ αβ α

− + + =

With this, use 2cos e2

aaB D βα − =

to get B and D. Similarly, find the normalization constants

for the other class of solutions.

E

E3 even

E2 odd

E1 even

0

1st class

2nd class

1st class

2a

− 2a

VO

η

ξ 0

2π π 3

2π

tanη ξ ξ=

cotη ξ ξ= −

1

2

m

70

Eigenfunctions extend into regions outside the well. Classically, the particle could never be found

in these regions when E < V0. However, quantum mechanically, it is possible to find the particle

outside the well and it moves further out as E increases.

(7) Linear Harmonic Oscillator Potential Consider the 1-d motion of a particle of mass m attached to a fixed center by a force proportional to

the displacement.

Eg: a mass attached to one end of a spring

The restoring force, αF x− ⇒ F kx= − , 2k mω= , k - force constant

d ( )d

V xF k xx

= − = − ⇒ 21( )2

V x kx= (potential energy of a L.H.O)

Classical mechanics predicts, the oscillating frequency 12

kmπ

Total energy, αE x , −∞ < x < ∞ ⇒ 0 < E < ∞ (continuous).

The importance of L.H.O. potential is that any arbitrary potential can be approximated to a L.H.O.

potential in the vicinity of a stable equilibrium position.

Consider any function ( )W x that has an equilibrium position at x = a.

x

V(x) = 21

kx2

0

x = 0 x = x

71

Taylor series expansion of ( )W x about x = a can be given by

( ) ( ) ( ) ( ) ( ) ( )21 ......2

W x W a x a W a x a W a′ ′′= + − + − +

For x values in the vicinity of x = a, ( ) ~ 0x a− ⇒ higher order terms can be neglected.

Since ( )W x has a minimum at x = a

d 0d x a

Wx =

= , 2

2d 0d x a

Wx =

> ⇒ ( ) ( ) ( ) ( )21 ......2

W x W a x a W a′′= + − +

Now by choosing a new set of axes such that X = x – a, ( ) ( ) ( )W X W x W a= −

⇒ ( ) 212

W X KX= , where ( )K W a′′=

L.H.O is a model for systems in which there are small vibrations about a point of stable

equilibrium. eg: Vibration motion of atoms in molecules and in solids.

TISE for a L.H.O reads

( ) ( ) ( )2 2

22

d 12 2d

x kx x E xm x

ψ ψ ψ−+ =

x−∞ < <∞

Solutions to this equation can be obtained with some difficulty. As it is going to be a time

consuming exercise, here we do not want to worry about it and we give only the final results.

The discrete set of energy eigenvalues are given by

( ) ( )1 12 2 , 0, 1, 2, 3, ............nE n n h nω ν= + = + =

The corresponding normalized energy eigenfunctions are given by

x

W(x)

a 0

Equilibrium position

72

( ) ( )2 2

1 22e , 0,1, 2, 3, ...........

2 !x

n nnαx H αx n

π nαψ −

= =

where 1 4 1 2

1 22

,m k m ma aω ωα = = = =

( )nH αx - Hermite polynomial of degree n, ( ) ( )2 2d1 e e

d

nn

n nx xH x

x−= −

Since ( ) ( )1 12 2 , 0, 1, 2, 3, ............nE n n h nω ν= + = + =

Quantum mechanical energy spectrum consists of an infinite sequence of discrete levels. 1 1

0 2 2 E hω ν= = , 3 31 2 2 E hω ν= = , 5 5

2 2 2 E hω ν= = , ………

Energy levels are equally spaced by the amount hω ν= .

Minimum (ground state) energy = 1 12 2 hω ν= .

The existence of zero-point energy is purely a quantum mechanical phenomenon (as required by

the H.U.P)

Classically, the minimum energy = 0 (when T = 0 and V = 0). This occurs when the particle is at

rest (no oscillation) at the equilibrium position. ⇒ x = 0, xp = 0.

This tells us that we know the position and the linear momentum of the particle exactly at the same

time ⇒ ∆x = 0, xp∆ = 0 ⇒ a contradiction to H.U.P

Total energy of the oscillator of amplitude A,

( )( )2 21 12 2

E m A A m Aω ω ω= = = ( )12 xA pω = 1 1. ~

2 2xx pω ω∆ ∆

Therefore, the energy of the lowest (ground) state is compatible with the H.U.P.

n ( )12nE n ω= + ( ) ( )

2 21 2

2e2 !

xn nn

αx H αxπ n

αψ − =

0 10 2E ω= ( )

2 21 2

20 e xx ααψ

π−

=

1 31 2E ω= ( )

2 21 2

21 2 e

2xx x ααψ α

π−

=

2 52 2E ω= ( ) ( ) 2 2

1 22 2 2

2 4 2 e8

xx x ααψ απ

− = −

3 73 2E ω= ( ) ( ) 2 2

1 23 3 2

3 8 12 e48

xx x x ααψ α απ

− = −

73

Since V(x) is symmetric (even) function of x, we get two classes of solutions as ( )n xψ for the

TISE.

( ) ( )n nx xψ ψ− = , when n = 0, 2, 4, ………. (even)

( ) ( )n nx xψ ψ− = − , when n = 1, 3, 5, ……….. (odd)

• The energy eigenvalues are non-degenerate. i.e. for a given value of n, there exists only one

( )n xψ corresponding to En,

i.e. 0 0 1 1 2 2, , , ..............E E Eψ ψ ψ→ → → etc.

• ( )n xψ from an orthonormal basis set. i.e. ( ) ( )* dn nmmx x xψ ψ δ+∞

−∞

=∫

For a given state ( )n xψ ,

ˆn nx xψ ψ= = ( ) ( )* d 0n nx x x xψ ψ+∞

−∞

=∫

ˆn nx xp pψ ψ= = ( ) ( )* d 0n nx i x xx

ψ ψ+∞

−∞

∂ − = ∂ ∫

The above results can be obtained only by considering the symmetry of the system.

Also, for a given state ( )n xψ

* 21 1 1

2 2 2

1ˆ ( ) ( )d 02n n n n nV V x kx x x n Eψ ψ ψ ψ ω

+∞

−∞

= = + = ≠

= ∫

( ) ( ) ( )*2 2

122

1 1

2 2

dˆ d 02 d

n n n n nT T x x x n Em x

ψ ψ ψ ψ ω+∞

−∞

= − = + = ≠

= ∫

⇒ On the average, the total energy is equally shared by the potential and the kinetic energy.

0

V(x)

n = 4 n = 3 n = 2 n = 1 n = 0

E4

E3

E2

E1

E0

x

74

Note: An easy way to determine T ( )V when V ( )T is known.

1 12 2

ˆ ˆ ˆ n n nH T V H T V E T E T E= + ⇒ = + ⇒ = + ⇒ =

Let’s treat a particle subjected to L.H.O motion classically.

The position of the particle ( ) sinx t A tω= , A - amplitude of the oscillation.

The speed of the particle, 2 2cosv A t A xω ω ω= = −

Total energy, 2 2 2 2 21 1 1 12 2 2 2E T V mv kx m A kAω= + = + = =

Classical turning points are given by ( )E V x= and they are located at 2EAk

= ± , 0 < E < ∞

Quantum mechanical value of the total energy ( )12 , 0, 1, 2, 3, ............nE n nω= + =

When E is chosen to be equal to nE in the expression of A, we can obtain the amplitude (or turning

points) of the L.H.O whose total energy is equal to the quantum mechanically allowed values.

i.e. 2

(say)nn

EA A

k= ± ≡

00

2EA

k kω

= ± = ± , 1

12 3E

Ak k

ω= ± = ±

, A2 = ……………….

Quantum mechanically, the position probability density, ( ) 2( )n nP x xψ=

Classical mechanically, the position probability density, ( )cP x is defined as, the fraction of the

total time spent by the particle in a unit length element on the x-axis.

i.e. 2 22 2

1 2 2 1( ) ~22c

vP xT T v A xA xπ ω

ω

= = =−× −

, 0 A< < ∞

Quantum mechanical value of the amplitude, 2 n

nE

Ak

= ±

To compare ( )cP x with ( )nP x , we replace A with nA in the expression of ( )cP x .

As it can be seen in the following figures, the eigenfunctions (or the corresponding probability

densities) do not fall sharply to zero at the classically turning points of the oscillator. However they

decrease very sharply. The amplitude of an oscillator is not sharply defined quantum mechanically

and as a result, there is a finite probability (i.e. non zero) of observing the particle outside the

classical limit. However, the particle is confined mainly to classically allowed region

75

As we can see, the general agreement between the classical and quantum results improves with

increasing n.

Algebraic Method for Obtaining the Energy Eigenvalues of a L.H.O

2 22 2 21 1

2 2ˆ ˆˆ ˆ ˆ ,2 2

x xp p kH kx m xm m m

ω ω= + = + =

Introduce the operators as done by Dirac. ˆiˆ ˆ

2 2xpma x

mω

ω± =

Using the facts that ˆ xp and x are hermitian operators and [ ]ˆ ˆ, ixx p = , it can be shown that †ˆ ˆa a+ −= , †ˆ ˆa a− += and [ ]ˆ ˆ, 1a a− + =

( ) ( ) ( )1 12 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2

H a a a a a a a aω ω ω− + + − − + + −= + = − = +

and also ˆ ˆ ˆ,H a aω± ± = ±

Ex: Verify the above results. If ψ is an eigenvector of H belonging to the eigenvalue E ⇒ H Eψ ψ=

ˆ ˆ ˆ,H a aω± ± = ± ⇒ ˆ ˆˆ ˆ ˆH a a H aω± ± ±= ±

Now, consider the state a ψ± .

⇒ ( ) ( )ˆ ˆˆ ˆ ˆ ˆH a a H a E aψ ω ψ ω ψ± ± ± ±= ± = ±

⇒ a ψ± is also an eigenvector of H corresponding to the eigenvalue

( )E ω± .

P3(x) = 23ψ

P2(x) = 22ψ

P1(x) = 21ψ

P0(x) = 20ψ

Pc(x)

Pc(x) n = 0

n = 3

n = 2

n = 1

x

76

Since a+ raises and a− lowers the value of E by the amount ω , they are called the ‘raising’ and ‘lowering’ operators respectively. Since H contains 2ˆ xp and 2x terms, H in any state can not be negative. Therefore, the eigenvalues of H must be non-negative. If 0E is the smallest eigenvalue and 0ψ is the corresponding eigenvector (i.e.

0 0 0H Eψ ψ= ), we must have 0ˆ 0a ψ− = . Otherwise, ( )0 0 0ˆ ˆ ˆH a E aψ ω ψ− −= −

would correspond to the eigenvalue ( )0E ω− which is a contradiction. Now, consider 0ˆ ˆa aω ψ+ − .

⇒ ( )10 02

ˆˆ ˆa a Hω ψ ω ψ+ − = −

0ˆ 0a ψ− = ⇒ ( )102

ˆ0 H ω ψ= − ⇒ 10 02H ψ ω ψ=

0 0 0H Eψ ψ= ⇒ 10 2E ω= ----------- Zero-point energy

To find the energy eigenfunction 0ψ , use 0ˆ 0a ψ− = with ˆiˆ ˆ

2 2xpma x

mω

ω− = +

.

⇒ 0 0di i 0

dm x

xψ ω ψ− − = ⇒ 0 0

dd

m xx

ωψ ψ= −

By integrating 2

20 0 e

m xA

ω

ψ−

= , where A0 is to be determined via normalization Consider the operation of a+ , n times on 0ψ .

( )0 0 0ˆ ˆ ˆH a E aψ ω ψ+ += + , ( )2 2

0 0 0ˆ ˆ ˆH a E aψ ω ω ψ+ += + + ,

( )3 30 0 0

ˆ ˆ ˆH a E aψ ω ω ω ψ+ += + + + , …………… ( )0 0 0ˆ ˆ ˆn nH a E n aψ ω ψ+ += + .

Taking 0ˆn

na ψ ψ+ ≡ and ( )0 nE n Eω+ = , we get ˆ n n nH Eψ ψ= .

0nE E n ω= + ⇒ ( )12nE n ω= + , where 0, 1, 2, 3, ..............n =

The eigenfunctions of excited states are obtained by applying the raising operator.

⇒ 2

2ˆ em x

nn nA a

ω

ψ−

+= As we have seen, the raising and lowering operators generate new solutions to the Schrodinger equation, but, these new solutions are not correctly normalized. Thus, ˆ na ψ+ is only proportional

to 1nψ + and a suitable normalization constant should be determined.

Let nψ be a normalized eigenvector corresponding to the eigenvalue nE and 1nψ + be a

normalized eigenvector corresponding to the eigenvalue 1nE + .

⇒ 1 1 ˆn n nC aψ ψ+ + += , where 1nC + is the normalizing coefficient.

77

⇒ * †1 1 ˆn n nC aψ ψ+ + +=

⇒ 2 †

1 1 1 ˆ ˆn n n n nC a aψ ψ ψ ψ+ + + + +=

Since 1 1 1n nψ ψ+ + = , †ˆ ˆa a+ −= and ˆ 1ˆ ˆ

2Ha aω− + = +

⇒ 2

1ˆ 11

2n n nHC ψ ψω+

= +

⇒ ( )2

11 1n n nC n ψ ψ+= +

⇒ 1 1 1nC n+ = + (assuming that 1nC + as real and positive)

Similarly, when 1 1 ˆn n nC aψ ψ− − −= , it can be shown that 1 1nC n− = . (verify this result)

Then, we have 1ˆ 1n na nψ ψ+ += + , 1ˆ n na nψ ψ− −= and hence

( ) ( )2 31 0 2 1 0 3 2 0

1 1 1 1ˆ ˆ ˆ ˆ ˆ, , , . . . . .2 2 3 3.2

a a a a aψ ψ ψ ψ ψ ψ ψ ψ+ + + + += = = = =

⇒ ( ) 01 ˆ

!n

n an

ψ ψ+=

The eigenvectors nψ with 0, 1, 2, 3, ..............n = form an orthonormal basis, nψ . i.e.

m n m nψ ψ δ= .

*(8) The Periodic Potential (1-d) Consider the motion of a particle of mass m in periodic potential of period l, which is defined as V(x + l) = V(x). Such a potential with rectangular sections is called the Kronig-Penney potential. It can be used as a model of the interaction to which an electron is subjected in a crystal lattice consisting of a regular array of single atoms separated by the distance l. Although a real crystal is of finite length, here we consider an idealized situation where x varies along the entire x-axis in the potential energy function. Since the kinetic energy term is unaltered by the change x x l→ + , with the above potential energy function, the whole Hamiltonian is invariant under displacements by l. For the case of zero potential, when the solution corresponding to a given energy 2 2 2E K m= is i( ) e K xxψ = , the displacement yields

78 i ( ) i( ) e e ( )K x l K lx l xψ ψ++ = = ,

which is the original solution multiplied by a phase factor, so that 2 2( ) ( )x l xψ ψ+ = . The observables will therefore be the same as at x as at x + l. Therefore, all physically acceptable solutions must satisfy the relation

i( ) e ( )nK lx nl xψ ψ+ = , with l K lπ π− ≤ ≤ and 0, 1, 2, ........n = ± ± - Bloch condition If we write ( )xψ in the form

i( ) e ( )KxKx u xψ = ⇒ ( ) ( )K Ku x l u x+ = - Bloch’s theorem

Bloch wave given by i( ) e ( )Kx

Kx u xψ = represents a traveling wave of wavelength 2 Kλ π= and amplitude ( )Ku x which is periodic with the same period l as that of the crystal lattice. Energy Bands With the zero of the energy scale sitting at the top of the wells, consider the two cases for which the electron energy E is such that 0 0 and 0V E E− < < > . Case 1: 0 0V E− < <

The TISE is 2

22

d ( ) ( ) 0d

x xxψ α ψ+ = , inside the wells

2

22

d ( ) ( ) 0d

x xxψ β ψ− = , between the wells

where ( )202

2m V Eα = +

, 22

2m Eβ = −

.

Solving the two equations for a cell b − l to b, we have

i i( ) e e , 0x xx A B b l xα αψ −= + − < < ( ) e e , 0x xx C D x bβ βψ −= + < <

Bloch’s theorem shows that, the electron does not belong to any one atom (ion) of the lattice, but has an equal probability of being found in the neighbourhood of any of the atom. In the cell (b − l , b), i( ) i( )( ) e e .........(1) , 0K x K x

Ku x A B b l xα α− − += + − < <

( i ) ( i )( ) e e ........(2) , 0K x K xKu x C D x bβ β− − += + < <

Since ( )Ku x is periodic, the above equations determine ( )Ku x (and hence ( )xψ ) for all values of x. In order to write ( )Ku x in the region b x l< < , using the fact that ( ) ( )K Ku x u x l= −

i( )( ) i( )( )( ) e e .........(3) ,K x l K x lKu x A B b x lα α− − − + −= + < <

For solutions to be physically acceptable, ( )xψ and d ( ) dx xψ (and hence ( )Ku x and d ( ) dKu x x ) must be continuous at both edges of the potential well (then it holds true for all wells) Equations (1) and (2) ⇒

A B C D+ = + ,

79

i( ) i( ) ( i ) ( i )K A K B K C K Dα α β β− − + = − − + ,

Equations (2) and (3) ⇒ i( ) i( ) ( i ) ( i )e e e eK c K c K b K bA B C Dα α β β− − + − − ++ = + ,

-i( ) i( ) ( i ) ( i )i( ) e i( ) e ( i ) e ( i ) eK c K c K b K bK A K B K C K Dα α β βα α β β− + − − +− − + = − − + , where c l b= − . From the above four equations, A, B, C and D can be determined. For a non-trivial solution, the determinant of the coefficients A, B, C and D must vanish. This result in the condition

2 2cos cosh sin sinh cos ............... (4)

2c b c b Klα βα β α β

αβ−

− =

from which energy E can be obtained. Case 2: 0E >

In this case, β becomes imaginary and this can be treated by setting β = ik with 22

2mk E=

.

Then for this case, the above condition can be rewritten as 2 2

cos cos sin sin cos ................ (5)2

kc kb c kb Klk

αα αα+

− =

Using the fact that the left-hand sides of the two conditions join smoothly at E = 0, one function of E is required for the full energy range 0E V> − . ⇒ ( ) cosF E Kl= . When the function F(E) is plotted for typical values of b, l and V0, it can be seen that Plot of F(E), which represents the left-hand sides of equations (4) and (5) as a function of the energy E. The heavy lines show the allowed values of E, corresponding to conduction bands; they are separated by forbidden energy gaps Since K is real ⇒ cos 1Kl ≤ ⇒ Values of E for which ( ) 1F E > are inaccessible.

⇒ Allowed values of E fall into bands satisfying the condition ( ) 1F E ≤ .

80 When the separation between the wells is increased, the energy bands for −V0 < E < 0 shrink and in the limit l → ∞ contract into discrete energy levels of an isolated potential well. If the separation between the wells ( l ) is increased, keeping c or V0 fixed

⇒ the energy bands for 0 0V E− < < become narrower and contract in the limit l → ∞ into discrete energy levels of an isolated potential well. Eg: a less affected electron in an atom in the presence of the other atoms in a crystal.

Bands corresponding to the lowest-lying levels are the narrowest. Eg: most tightly bound electrons

are less likely to be perturbed by the presence of the other atoms. Setting up of the boundary condition that ψ = 0 at both ends of the 1-d crystal, i.e. electron can not escape from the crystal, requires a standing wave solutions to TISE by the superposition of Bloch waves (traveling). - not easy to implement It is possible to apply periodic boundary conditions for a 1-d lattice forming a closed loop containing 2310N atoms. Single-valuedness of the wave function requires that,

( ) ( )x N l xψ ψ+ = i( ) e ( )nK lx nl xψ ψ+ = (shown before) ⇒ ie 1N K l =

⇒ = 2 , 0, 1, 2, .......K n Nl nπ = ± ±

*The Scattering Potential Consider the scattering of a particle by an arbitrary potential localized on a finite part of the x-axis

as shown below.

From the experience gained in previous applications, the general solutions to TISE in regions I and

III can be written straight away (consider unbound (scattering) states for which E > 0).

In region I ( ( ) 0V x = ) i i( ) e ek x k xx A Bψ −= + , where 2 22k mE=

In region III ( ( ) 0V x = ) i i( ) e ek x k xx F Gψ −= + , where 2 22k mE=

Since the potential is not specified in region II, we use the fact that TISE is a linear, second-order

differential equation.

In region II (V(x) is not specified) ( ) ( ) ( )x C f x D g xψ = + ,

where ( )f x and ( )g x are any two linearly independent particular solutions.

81

From the four boundary conditions (two joining regions I and II, and two joining regions II and III), two can

be used to eliminate C and D and the other two can be solved for B and F in terms of A and G.

⇒ 11 12B S A S G= + 21 22F S A S G= + .

The four coefficients i jS depend on k (and hence on E) and constitute a 2 × 2 matrix,

11 12

21 22

S SS

S S

=

called the scattering matrix or S-matrix. It gives the outgoing wave amplitudes

(B and F) in terms of incoming wave amplitudes (A and G).

B AS

F G

=

Note: For i jS , the following conditions hold.

2 2 2 2 * *11 12 12 22 11 12 21 221, 1, 0S S S S S S S S+ = + = + =

For particles incident from the left, G = 0, so the reflection and transmission coefficients are

22

112 ,lB

R SA

= = 2

2212 .l

FT S

A= =

For particles incident from the right, A = 0, so the reflection and transmission coefficients are 2

2222 ,r

FR S

G= =

22

122 .rB

T SG

= =

The S-matrix tells us everything about scattering from a localized potential.

If there are any bound states (E < 0) in the problem under consideration, then ( )xψ has the form

e (Region I),( ) ( ) ( ) (Region II),

e (R egion III),

x

x

Bx C f x Dg x

F

κ

κ

ψ−

= +

with 2 22mEκ = − .

In this case, boundary conditions are the same as before, so the S-matrix has the same structure

(with k replaced by iκ ). But, A and G are necessarily zero whereas B and F are not, and hence at

least two elements in the S-matrix must be infinite. Therefore, if one has got the S-matrix for E > 0

and want to locate the bound states, put in ik κ→ and look for energy values at which the

elements of S-matrix become infinite.

82

Motion in Three-Dimension

(9) Free Particle (3-d) Consider a particle of mass m moving in a region of potential energy ( ) ( ) 0, , 0V r V x y z V= = = , in the region , , x y z− ∞ ≤ ≤ ∞ − ∞ ≤ ≤ ∞ − ∞ ≤ ≤ ∞ .

TISE ⇒ ( ) ( ) ( ) ( )2

2

2r V r r E r

mψ ψ ψ− ∇ + =

Assume a solution of the form ( ) ( ) ( ) ( )x y zr x y zψ ψ ψ ψ=

and the corresponding energy

x y zE E E E= + + .

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 2

2 2 22

x y z y x z z x y

x y z x y z

x y z y x z z x ym x y z

E E E x y z

ψ ψ ψ ψ ψ ψ ψ ψ ψ

ψ ψ ψ

− ∂ ∂ ∂+ + ∂ ∂ ∂

= + +

Dividing throughout by ( ) ( ) ( ) ( )x y zr x y zψ ψ ψ ψ= and separating variables,

( ) ( )2 2

21

2 x xx

x Em x x

ψψ

− ∂= ⇒

∂ ( ) ( )

2

2 22

0,xx x

mEx x

xψ ψ∂

+ =∂

(1-d case)

Solution,

( ) i 22

2e ,xk x x

x xmE

x kψ ±= =

Similarly, in y- and z-directions,

( ) i 22

2e ,yk y y

y ymE

y kψ ±= =

( ) i 22

2e ,zk z z

z zmE

z kψ ±= =

The wave function ( ) i .e k rrψ ±=

with ( ), ,x y zk k k k=

and ( ), ,r x y z= describes a free

particle with energy 2 2

2kEm

= and linear momentum p k=

.

(10) Square Potential Box (3-d)

Consider a particle of mass m, confined to move in a rectangular potential box which is defined as

0 in the region 0 , 0 , 0 ( , , )

elsewherex a y b z c

V x y z≤ ≤ ≤ ≤ ≤ ≤

= ∞

83

Walls of the box are impenetrable for the particle ⇒ outside the box, ( ) 0rψ =

. Inside the box,

the wave function ( )rψ of the particle satisfies the TISE.

( )2

2 ( ) ( ) , since ( ) 0 inside the box2

r E r V rm

ψ ψ−∇ = ≡

Assume a solution of the form ( ) ( ) ( ) ( )x y zr x y zψ ψ ψ ψ= and the corresponding energy

x y zE E E E= + + .

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 2

2 2 22

x y z y x z z x y

x y z x y z

x y z y x z z x ym x y z

E E E x y z

ψ ψ ψ ψ ψ ψ ψ ψ ψ

ψ ψ ψ

− ∂ ∂ ∂+ + ∂ ∂ ∂

= + +

Dividing throughout by ( ) ( ) ( ) ( )x y zr x y zψ ψ ψ ψ= and separating variables,

2 2

21 d ( ) , 0

2 dx x

xx E x a

m xψ

ψ−

= ≤ ≤

2 2

21 d ( ) , 0

2 dyx y

yy E y b

mψ

ψ−

= ≤ ≤

2 2

21 d ( ) , 0

2 dzz z

zz E z c

mψ

ψ−

= ≤ ≤

We get three TISE’s for 1-d square potential wells placed in x-, y-, and z-directions.

Using the boundary conditions,

( ) 0 at 0, , ( ) 0 at 0, , ( ) 0 at 0,x y zx x a y y b z z cψ ψ ψ= = = = = = As in the1-d case, the solutions to the above TISE can be written as

2 2 2

22( ) sin ( ), , 1, 2, 3, .........

2x xx x

x n x n xn x n

x x E E na a ma

π πψ ψ= ≡ = ≡ =

2 2 2

22( ) sin ( ), , 1, 2, 3, .........

2y y

y yy n y n y

n y ny y E E n

b b mb

π πψ ψ= ≡ = ≡ =

z c b

a y

x

84

2 2 2

22( ) sin ( ), , 1, 2, 3, .........

2z zz z

z n z n zn z n

z z E E nc c mc

π πψ ψ= ≡ = ≡ =

The normalized energy eigenfunction,

8( , , ) ( ) ( ) ( ) sin sin sin ,x y z x y z

yx zn n n n n n

n yn x n zx y z x y z

abc a b cππ π

ψ ψ ψ ψ= =

0 , 0 0x a y b z c≤ ≤ ≤ ≤ ≤ ≤

Corresponding energy eigenvalues,

22 22 2

2 2 2 , , , 1, 2, 3, ........2x y z x y z

yx zn n n n n n x y z

nn nE E E E n n n

m a b cπ

= + + = + + =

Energy

spectrum is discrete. As the dimensions of the box increase, the spacing of the energy levels will

decrease ⇒ energy spectrum will become continuous.

A more interesting situation occurs when the box is cubical i.e. a = b = c

( )22 22 2 2 2 2 2 2

2 2 22 2 2 2 22 2 2x y z

yx zx y zn n n

nn n nE n n nm a a a ma ma

π π π = + + = + + =

n2

The values of energy are determined by the value of n.

38( , , ) sin sin sin ,

x y z

yx zn n n

n yn x n zx y z

a a aa

ππ πψ =

Ground level of the particle occurs when, nx = ny = nz = 1 = n2

( )2 2 2 2

2 2 2111 2 2

31 1 12 2

Ema maπ π

= + + =

( ) 31118, , sin sin sinx y zx y z

a a aaπ π πψ = ⇒ Ground level is non-degenerate

For higher energy levels, we see that the states characterized by the combination of integers nx, ny

and nz which give the same value for n, will have the same energy. However, we get different

eigenfunctions (states).

nx = ny = 1, nz = 2

nx = nz = 1, ny = 2 n2 = 6

nz = ny = 1, nx = 2

85

2 2

112 121 211 262

hE E Emaπ

= = =

( ) 31128 2, , sin sin sinx y zx y z

a a aaπ π πψ =

( ) 31218 2, , sin sin sinx y zx y z

a a aaπ π πψ =

( ) 32118 2, , sin sin sinx y zx y z

a a aaπ π πψ =

Since E112 = E121 = E211 but ψ112 ≠ ψ121 ≠ ψ211, a degeneracy exists in the 1st excited level. The

‘degree of degeneracy’ = 3 (a 3-fold degenerate level).

The 2nd excited level occurs when n2 = 9, i.e. nx, ny, nz, taking values 1, 2, 2 or 2,1, 2 or 2, 2,1 and

it is also a 3-fold degenerate level.

2 2 2

22x y zn n nnE

maπ

= ,

2 2

0 111 232

E Emaπ

= =

E/E0 = 1 for the ground level

= 2 for the 1st excited level

= 3 for the 2nd excited level

= 11/3 for the 3rd excited level

= 4 for the 4th excited level

Q. numbers (x, y, z) Deg. of degeneracy (2, 2, 2) 1 (1, 1, 3), (3, 1, 1), (1, 3, 1) 3 (2, 2, 1), (2, 1, 2), (1, 2, 2) 3 (1, 1, 2), (2, 1, 1), (1, 2, 1) 3 (1, 1, 1) 1

E/E0

4E0 11 30E 3E0 2E0 E0 E = 0

Small box ( a = b = c )

Large box

dE

86

The existence of degeneracy is a consequence of the symmetry of the potential

Special case: a < b < c

The ground level occurs when nx = ny = nz = 1 as before and the first excited level occurs when

nx = 1, ny = 1 and nz = 2 because, c > b and c > a.

The energy of the 1st excited level, 2 2

2 2 21121 1 4

2E

m a b cπ

= + +

, and no degeneracy exists

in the level.

Ex: When one of the dimensions of the box (say c) is zero, the rectangular potential box reduces to a

potential rectangle (in 2-d). Repeat the above derivation and look for degeneracies when a = b.

(11) Harmonic Oscillator (3-d) Consider a particle of mass m subjected to harmonic motion in 3-dimensional space

Anisotropic case: i.e. the spring constant k has 3 different values along x, y, z directions. (kx ≠ ky ≠

kz). Following the same procedure given in the potential box problem, it can be shown that

( ) ( ) ( )1 1 12 2 2x x y y z zz zx y x yn n n n n nE E E E n n nω ω ω= + + = + + + + + , where

2 2 2, ,x x y y z zk m k m k mω ω ω= = = .

( , , ) ( ) ( ) ( )z zx y x yn n n n n nx y z x y zψ ψ ψ ψ= , where

( ) ( )2 2

1 22e

2 !x

xxxxx

n xnx

n x H xn

ααψ α

π−

=

, with ( )1 2x xmα ω= and

nx, ny, nz = 0, 1, 2, …… . ( ) ( )andy zn ny zψ ψ take the same form.

Isotropic case: kx = ky = kz = k = mω2

⇒ ( ) ( )3 32 2x y zx y zn n nE n n n nω ω= + + + = + , where n = 0, 1, 2, …..

Ex: In this case, find the degeneracy in the five lowest energy levels.

87

Tutorial 3 Conservation of Probability, Probability Current Density 1. In obtaining the continuity equation we have assumed that the potential ),r( tV is a real quantity.

Prove that if ),r( tV is complex the continuity equation becomes

),r()],r([Im2),r(.),r( tPtVtjtPt

=∇+

∂∂ so that the addition of an imaginary part of the

potential describes the presence of sources if Im V > 0 or sinks (absorbers) if Im V < 0. Superposition Principle, Normalisation, Expectation values and Uncertainties 2. One thousand neutrons are in a one-dimensional box with walls at x = 0, x = a. At t = 0, the state

of each particle is ( , 0) ( )x Ax x aΨ = − . (a) Normalize Ψ and find the value of the constant A. (b) How many particles are in the interval (0, a/2) at t = 0 ? (c) How many particles have energy E5 at t = 0 ? (d) What is E at t = 0 ?

3. Let En denote the bound-state energy eigenvalues of a one-dimensional system and let ψn(x)

denote the corresponding energy eigenfunctions which satisfy the orthonormality condition. Let Ψ(x, t) be the wave function of the system, normalised to unity, and suppose that at t = 0 it is

given by .)(53)(5

2)0,( 21 xxtx ψψ +==Ψ Write down the wave function Ψ(x, t) at time

t. (a) Find the probability that at time t a measurement of the energy of the system gives the value E2 , (b) Does ⟨x⟩ vary with time ? Does ⟨px⟩ vary with time ? Does E = ⟨H ⟩ vary with time ?

4. Consider the function ψ(θ ) of the angular variable θ , restricted to the interval −π ≤ θ ≤ π . If the

wave function satisfy the condition ψ(π) = ψ(−π), show that the operator θdd

iL = has a real

expectation value. 5. Consider a particle whose normalised wave function is ψ(x) = 2α√αxe−αx, x > 0 and ψ(x) = 0, x <

0, where α is a constant. (a) For what value of x does P(x) = ψ(x)2 peak ? (b) Calculate ⟨x⟩ and ⟨x2⟩, (c) What is the probability that the particle is found between x = 0 and x = 1/α ?, (d) Calculate ⟨p⟩ and ⟨p2⟩, (e) Calculate ∆x and ∆p and show that the uncertainty principle is satisfied for the particle.

6. The wave function ψ(x) of a particle in the potential V(x) = ∞ for x < 0 and V(x) = Cx for x > 0 are

given by ψ(x) = 0 for x < 0 and ψ(x) = Ax exp(−bx) for x > 0, where A is a constant. (a) Sketch the potential distribution, (b) Use the normalisation condition to determine the value of A, (c) Find the expectation value of the total energy of the particle, (d) Find the value of b which gives rise to the ground state energy of the particle.

Tutorial 4 Schrödinger Equation and its applications 7. The wave function of a particle of mass m which moves in one dimension is i( )( , ) e k x tx t A ω−Ψ = , where

A, k, and ω are constants. Determine the potential V(x) in which the particle moves, in terms of m, , k and ω.

8. Consider an electron of energy E incident from right on the potential step defined by V(x) = 0 for x < 0

and V(x) = V0 for x > 0. (a) Find the probability current densities in both regions and obtain expressions for reflection and transmission coefficients, (b) Prove that the reflection and transmission coefficients are

88

the same as when the particle is incident on the potential from the left, (c) A uniform beam of electrons is incident normally on a metal. On the assumption that outside the metal the electrons have a kinetic energy of 1 eV, and that inside it their potential energy is 1 eV less than that of outside, calculate the fraction of electrons reflected from the metal surface.

9. A particle of total energy 9V0 is incident from the −x-axis on a potential given by V(x) = 8V0 for x < 0, V(x)

= 0 for 0 < x < a and V(x) = 5V0 for x > a. Find the probability that the particle will be transmitted on through to the positive side of the x-axis, x > a.

10. Consider a particle incident from the left on a potential step which is defined as V(x) = V1 for x < 0 and V(x)

= V2 for x > 0 with V1 < V2. Find the solution of the Schrödinger equation for the following cases: (i) V1 < E < V2 (ii) E > V2.

11. Set up the Schrödinger equation for a two-dimensional harmonic oscillator in the Cartesian co-ordinates

and obtain the energy eigenvalues and the eigenfunctions. 12. Estimate the zero-point energy of an electron confined inside a region of size 10−14 m, which is the order of

magnitude of nuclear dimensions. Compare this energy with both the gravitational potential energy and the Coulomb potential energy of an electron and a proton separated by the same distance. On the basis of this comparison, discuss the possibility that an electron can exist within a nucleus. Calculate the zero-point energy of a neutron which is confined within a nucleus.

13. Normalise the wave functions for a particle in a box showing that the normalisation constant N = (8/abc)1/2

= (8/V)1/2, where V is the volume of the box. Obtain an expression for the energy of the particle. Now consider a particle in a cubical potential box of side a that is very large. Find the number of quantum states (a) with energy between E and E + dE (b) with momentum between p and p + dp. Repeat your work in parts (a) and (b) for a particle in a large two-dimensional potential box with equal sides.

14. The ground state wave function of a particle of mass m subjected to a linear harmonic oscillatory motion in

the x-direction with a force constant k, is given by ψ0(x) = Nexp(−α2x2), where α = (mk/ħ2)1/4 and k = mω2 with ω being the angular frequency of the corresponding classical oscillator. (a) Evaluate N so that ψ0(x) is normalised to unity (b) Evaluate ⟨x2⟩, ⟨V⟩, ⟨T⟩ and ⟨p2⟩ for the oscillator in the ground state, where V and T are the potential and kinetic energies and p is the linear momentum (c) Evaluate the uncertainty product ∆x.∆p for the oscillator in the ground state, where ∆x = [⟨x2⟩ − ⟨x⟩2]1/2 and ∆p = [⟨p2⟩ − ⟨p⟩2]1/2 (d) Determine the position of the classical turning points for the oscillator in the ground state (e) Compute the probability that the particle is found in the classically forbidden region. (f) Obtain expressions for the probability density of the linear harmonic oscillator (quantum mechanical) in the ground state and that of the corresponding classical oscillator having the same energy. Then sketch roughly the variation of the quantum mechanical and classical mechanical probability densities as a function of position.

15. Find the eigenenergies and eigenfunctions for a particle moving under the potential

2 2 2 0( )

0m x xV x

xω >= ∞ ≤

.

16. A particle in an infinitely deep square potential well is found to be 70% of the time in its ground state

and the remainder in its first excited state. Write down the wave function of this particle. Show that this is normalised and calculate the expected result for the measurements (a) position, (b) momentum and (c) energy.

17. Show that, (a) ℘ is a Hermitian operator, (b) ℘ anti-commutes with x and px. (c) ℘ commutes with

px2, V(x) and H, (d) the eigenvalues of ℘ are +1 and −1 corresponding respectively to an even

eigenstate (ϕe(x) say) and an odd eigenstate (ϕo(x) say), (e) ϕe(x) and ϕo(x) are orthogonal to each other, in accordance with the fact that they belong to different eigenvalues of ℘.

83

Separation of the Schrodinger Equation in Spherical Polar Co-ordinates

Consider the motion of a particle of mass m in a central (radial) potential V(r) (i.e. V(r) depends only on r r=

). Since V(r) is then spherically symmetric, it is convenient to use spherical polar co-ordinates in this problem.

The Cartesian co-ordinates (x, y, z) of the point P and the spherical polar co-ordinates (r, θ, φ) are related by

sin cosx r θ φ= , sin siny r θ φ= , cosz r θ= , where 0 r≤ ≤ ∞ , 0 θ π≤ ≤ , 0 2φ π≤ ≤ .

22 2

2 2 2 2 2

In spherical polar co-ordinates ( , , ),

1 1 1sinsin sin

r

rr rr r r

θ φ

θθ θθ θ φ

∂ ∂ ∂ ∂ ∂ ∇ = + + ∂ ∂ ∂ ∂ ∂

3 2 2d d d sin d d d d , where d = sin d dr r r r r rθ θ φ θ θ φ≡ = = Ω Ω

Hamiltonian of the particle,

2 2 22 2

2 2 2

ˆ1ˆ ˆ ˆ( ) ( )2 2

LH V r r V rm m r rr r

∂ ∂ = − ∇ + = − − + ∂ ∂

,

where 2

2 22 2

1 1ˆ sinsin sin

L θθ θ θ θ φ

∂ ∂ ∂ = − + ∂ ∂ ∂

TISE for the particle,

2 22

2 2 2

ˆ1 ˆ( ) ( ) ( )2

Lr V r r E rm r rr r

ψ ψ ∂ ∂ − − + = ∂ ∂

Rearranging terms, assuming a solution of the form ( ) ( , , ) ( ) ( , )r r R r Yψ ψ θ φ θ φ≡ = and

separating variables

( )2

2 22 2

21 1ˆ ˆ( ) ( ) ( , ) (constant)( ) ( , )

m rr R r E V r L YR r r r Y

θ φ λθ φ

∂ ∂ + − = = ∂ ∂

( )22 2 2

21 ˆ( ) ( ) ( ) 0mr R r E V r R rr rr r

λ∂ ∂ + − − = ∂ ∂ …… (1) Radial Equation, to be

solved for R(r) 2 2ˆ ( , ) ( , )L Y Yθ φ λ θ φ= …………………… (2) Angular Equation, to be

solved for ( , )Y θ φ Let ( , ) ( ) ( )Y θ φ θ φ= Θ Φ , substitute in the above equation with the definition of 2L and separate variables.

y

x

P

r θ

φ

O

Q

z

84

22 2

2sin 1 ( )sin ( ) sin (constant)

( ) ( )mθ φθ θ λ θ

θ θ θ φ φ∂ ∂ ∂ Φ Θ + = − = Θ ∂ ∂ Φ ∂

⇒ 2

21 sin ( ) ( ) 0

sin sinmθ θ λ θ

θ θ θ θ

∂ ∂ Θ + − Θ = ∂ ∂ ……. (3)

⇒ 2

22

d ( ) ( ) 0d

mφ φφΦ

+ Φ = ………………………………(4)

Equation (4) has a particular solution, i( ) e mφφΦ = Using the requirement that eigenfunctions be single-valued,

( ) ( 2 )φ φ πΦ = Φ + (since φ and φ + 2π are the same angle) When φ = 0, ⇒ (0) (2 )πΦ = Φ ⇒ 1 cos 2 isin 2m mπ π= +

⇒ 0, 1, 2, 3, .......... or 0, 1, 2, 3, ........m m= = ± ± ±

Normalization of ( )φΦ requires that, when i( ) e mC φφΦ = , 2

2

0( ) d 1

πφ φΦ =∫ ⇒ 1 2C π= ⇒ i1( ) e , 0 2

2m

mφφ φ π

πΦ = ≤ ≤

In Equation (3), substitute cosw θ= with 1 1w− ≤ ≤

⇒ ( )2

22

d d1 ( ) ( ) 0, where ( ) ( )d d 1

mw P w P w P ww w w

λ θ − + − = Θ ≡ −

Since the wave function must be finite everywhere, ( )P w and ( )φΦ must be finite everywhere. All possible solutions those for which ( )P w is finite everywhere exist, only if (i) ( 1),l lλ = + where 0, 1, 2, 3, ..............l = (ii) m l≤ (i.e. l m l− ≤ ≤ and for a given value of l, there should be (2l + 1) values of m

such that , 1, 2, ..... , 0, ...., ......, 2, 1,l l l l l l− − + − + − − .) The angular part ( , )Y θ φ (normalized) of the wave function ( , , )rψ θ φ can be written as

( , ) ( ) ( )mml lm mlY C P wθ φ φ= Φ - known as ‘Spherical Harmonics’ ( )( ) ( )m

lP w P w≡

1 12 2(2 1)( )! (2 1)( )!1( 1) ( 1)

2( )! 4 ( )!2m m

lml l m l l mC

l m l mππ + − + −

= − × = − + +

⇒ 1

2i(2 1)( )!( , ) ( 1) (cos )e , 0

4 ( )!mm m m

l ll l mY P m

l mφθ φ θ

π + −

= − ≥ +

( )Note that, ( , ) ( , )mlm lY Yθ φ θ φ≡

In Dirac notation, the spherical harmonics, ( , )mlY l mθ φ ≡ * ( , )m

lY l mθ φ ≡ Spherical harmonics satisfy the orthonormality relation

( ) ( )2* * '

'0 0

( , ) ( , )d ( , ) ( , )d sin dm m m ml l l l l m mlY Y Y Y

π πθ φ θ φ θ φ θ φ θ θ φ δ δ′

′ ′′ Ω = =∫ ∫ ∫

85

In Dirac notation, the orthonormality condition, l l m ml m l m δ δ′ ′′ ′ = For spherical harmonics yielded by a given (fixed) value of l with

, 1, 2, ........., 1 , 0, 1, ........, 2, 1,m l l l l l l= − − + − + − − − , the orthonormality condition reduces to m ml m l m δ ′′ = . The low-order spherical harmonics

0 00 1

1 30 0 ( , ) , 1 0 ( , ) cos44

l m Y l m Yθ φ θ φ θππ

= = ≡ = = ≡

( )1 i 0 21 2

3 51 1 ( , ) sin e , 2 0 ( , ) 3cos 14 16

l m Y l m Yφθ φ θ θ φ θπ π

± ±= = ± ≡ = = ≡ −

………….. Some Important Properties of m

lY ’s (1) m

lY ‘s form a complete set in the variables θ and φ , i. e. any angular function ( , )F θ φ can be expanded in terms of them.

0( , ) ( , )

lm

l m ll m l

F a Yθ φ θ φ∞

= =−= ∑ ∑ ⇒ ( )*( , ) ( , )dm

l m la Y Fθ φ θ φ= Ω∫

(2) 2 2ˆ ( , ) ( 1) ( , )m m

l lL Y l l Yθ φ θ φ= + (from equation (2))

i.e., mlY ’s are eigenfunctions of 2L with the eigenvalue 2( 1)l l + .

(3) Behavior of m

lY ’s under the parity operation r r→ − ( )ˆi.e. ( ) ( )r rψ ψ℘ = −

. i.e. , ,r r θ π θ φ φ π→ → − → +

ˆ ( , ) ( , ) [cos( )] ( )mm m

l l mlY Y Pθ φ π θ φ π π θ φ π℘ = − + = − Φ +

[cos( )] [ cos ] ( 1) (cos )m m ml ml l lP P Pπ θ θ θ−− = − = −

( ) ( 1) ( )mm mφ π φΦ + = − Φ

⇒ ˆ ( , ) ( 1) (cos ) ( 1) ( )ml m mml mlY Pθ φ θ φ−℘ = − × − Φ

⇒ ˆ ( , ) ( 1) ( , )m l ml lY Yθ φ θ φ℘ = − i.e. m

lY has the parity of l (‘even’ for even l and ‘odd’ for odd l.)

θ

π φ+

φ

π θ−

z

y

x

86

Physical properties of 2L

Consider a particle of mass m, having linear momentum p and position vector r with respect to origin O. Then, the angular momentum of the particle with respect to O is defined as L r p= ×

. ( L

is perpendicular to the plane containing andr p . sinL L r p α= =

, where α is the

angle between andr p )

Taking r as a unit vector along r and t as a unit vector in the plane of andr p and perpendicular to r , ˆˆr tp p r p t= +

⇒ 2 2 2r tp p p= +

Since ˆ ip = − ∇

⇒ 2 2 2 2 2ˆ ˆ ˆr tp p p= − ∇ = +

As shown before, 2

2 22 2 2

ˆ1 Lrr rr r∂ ∂ ∇ = − ∂ ∂

⇒ 2 2

2 2 22 2

Lrr rr r∂ ∂ − ∇ = − + ∂ ∂

2 2

2 2 22 2

ˆˆ ˆr t

Lp p rr rr r∂ ∂ + = − + ∂ ∂

⇒ 2 2 2ˆ ˆ tL r p=

ˆr t× is a unit vector along the direction of the orbital angular momentum. Therefore, we

identify 2L as the operator corresponding to the ‘square of the orbital angular momentum’. In all isolated systems (with no external torques), the total angular momentum is conserved. In classical mechanics, angular momentum is an ordinary vector. But, in quantum mechanics, it is a vector operator and its three components do not commute with each other.

( , , )x y zL L L L=

, sinL L r p α= =

The cartesian components of L

are

( )x z yL y p z p= − ( )y x zL z p x p= − ( )z y xL x p y p= −

In q.m., the corresponding operators are

( )ˆ ˆ ˆ ˆˆ

i

x z yL y p z p

y zz y

= −

∂ ∂= − − ∂ ∂

( )ˆ ˆ ˆ ˆˆ

i

y x zL z p x p

z xx z

= −

∂ ∂ = − − ∂ ∂

( )ˆ ˆ ˆ ˆ

i

z y xL x p y p

x yy x

= −

∂ ∂= − − ∂ ∂

( )( )ˆ ˆThe above three relationships are represented by the vector operator iL r= − × ∇

It can be verified that L

is Hermitian. Consider the commutation relation ˆ ˆ,x yL L

p

y

x

z

m

α

rp

tp

t

r

r

87

( ) ( ) [ ] [ ]ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ, , , , ,z y x z z x z z y x y zy p z p z p x p y p z p y p x p z p z p z p x p = − − = − − + To evaluate each commutator individually, consider [ ]ˆ ˆ ˆˆ,z xy p z p

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, , , ,

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, , , ,ˆ ˆ = i

z x z x z x x z

z x z x x z x z

x

y p z p z y p p z y p p z p y p

z y p p y z p p z p y p z y p py p

= + = − −

= − − − −

−

Similarly, it can be shown that

[ ]ˆ ˆ ˆ ˆ, 0z zy p x p = , ˆ ˆˆ ˆ, 0y xz p z p = , ˆ ˆ ˆ ˆ ˆˆ , iy z yz p x p x p =

⇒ ( )ˆ ˆ ˆ ˆ ˆ ˆ,x y y xL L i x p y p = − ⇒ ˆ ˆ ˆ, ix y zL L L =

Similarly, it can be shown that ˆ ˆ ˆ, iy z xL L L = ˆ ˆ ˆ, iz x yL L L =

⇒ Operators representing any two components of the orbital angular momentum do not commute. i.e. they are not simultaneously measurable with absolute precision.

( )ˆ ˆ ˆ ˆ ˆThe above three relationships are equivalent to i , Note: 0L L L L L× = × ≠

Since 2 2 2 2ˆ ˆ ˆ ˆx y zL L L L= + + (Note:

22 2ˆ ˆL L L≡ ≡

)

2 2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , 0x x y z x y x z xL L L L L L L L L L = + + = + = (prove this)

Similarly, it can be shown that 2ˆ ˆ, 0yL L = , 2ˆ ˆ, 0zL L =

2 ˆˆThe above three relationships are equivalent to , 0L L =

Simultaneous eigenfunctions of 2L and one of L

’s components can be found. i.e. both the magnitude and one component of angular momentum can be measured simultaneously with absolute precision. In spherical polar co-ordinates, it can be shown that,

ˆ i sin cot cosxL φ θ φθ φ

∂ ∂= − − − ∂ ∂

, ˆ i cos cot sinyL φ θ φθ φ

∂ ∂= − − − ∂ ∂

, ˆ izLφ∂

= −∂

⇒ 2

2 22 2

1 1ˆ sinsin sin

L θθ θ θ θ φ

∂ ∂ ∂ = − + ∂ ∂ ∂

Note that, 2ˆ ˆ ˆ ˆ, , and x y zL L L L are purely angular operators. (i.e. independent of r)

⇒ 2ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, ( ) , ( ) , ( ) , ( ) 0x y zL f r L f r L f r L f r = = = =

Note that, i i1 1ˆ ( ) i e e ( )2 2

m mz m mL m mφ φφ φ

φ π π

∂Φ = − = = Φ ∂

⇒ m is an eigenvalue of zL with eigenfunction ( )m φΦ , where 0, 1, 2, 3, ........m = ± ± ± Therefore, a measurement of the z-component of the orbital angular momentum can yield the values 0, 1 , 2 , 3 , ........± ± ± Since the z-axis can be chosen arbitrarily, the component of the orbital angular momentum on any axis is quantized.

88

ˆ ˆ( ) ( ) ( ) ( ) ( ) ( )ˆ ( , ) ( , )

z m m z m mm m

z l l

L m L m

L Y m Y

φ φ φ θ φ θ

θ φ θ φ

Φ = Φ ⇒ Φ Θ = Φ Θ

⇒ =

The eigenvalues and eigenfunctions of 2L and ˆ

zL Since 2ˆ , 0zL L = , we can look for simultaneous eigenfunctions of 2L and ˆ

zL . We see that

( , )lmlY θ φ are simultaneous eigenfunctions of 2L and ˆ

zL , because

2 2ˆ ( , ) ( 1) ( , )l lm ml lL Y l l Yθ φ θ φ= + ˆ ( , ) ( , )l lm m

z ll lL Y m Yθ φ θ φ=

0, 1, 2, 3, ..............l = is called the ‘orbital angular momentum quantum number’.

ll m l− ≤ ≤ and 0, 1, 2, 3, ........lm = ± ± ± is called the ‘orbital magnetic quantum number’. Note: In order to show the connection to l, we have replaced m by lm . We designate the orbital angular momentum states corresponding to 0, 1, 2, 3, ..............l = by the symbols s, p, d, f, ………. (designation of the subshell system of atoms)

l 0 1 2 3

Designation

s

‘sharp’

p

‘principal’

d

‘diffuse’

f

‘fundamental’

[When there is more than one particle, we have 0, 1, 2, 3, ..............L = as the total orbital angular momentum quantum number and S, P, D, F, ……. are used to designate their states.] The angular part of the wave function ( , )lm

lY θ φ , is the same for all spherically symmetric (central) potentials. The actual shape of the potential, V(r), affects only the radial part of the wave function, R(r), which is determined by the radial equation

( )22 2 2

21 ( 1)ˆ( ) ( ) ( ) 0m l lr R r E V r R rr rr r∂ ∂ + + − − = ∂ ∂

…… (1)

By substituting ( ) ( )u r r R r= , Equation (1) can be simplified into the modified radial equation

2 2 2

2 2d ( ) ( 1)( ) ( ) ( )

2 d 2u r l lV r u r E u r

m r mr

+− + + =

…………………….(5)

It is identical to the one-dimensional TISE for a particle of mass m and total energy E moving in a

potential (effective), 2

eff 2( 1)( ) ( )2

l lV r V rmr+

= + . The term added to V(r) is called the centrifugal

term.

89

The Hydrogenic Atoms We consider the hydrogenic atom to consist of a heavy motionless nucleus (kept at the origin) of charge +Ze together with a much lighter electron of mass em and charge −e that circles around it, held in orbit by the mutual attraction of opposite charges. eg: H(Z = 1), He+(Z = 2), Li++(Z = 3), Be3+(Z = 4), ……

From Coulomb’s law, the potential energy (in SI units) is 2

0 0

( )( )( )(4 ) (4 )

ZeZe eV rr rπε πε

−−= = .

The Hamiltonian, 22 2

2

0

ˆˆ ˆ( )2 2 (4 )e e

ZepH V rm m rπε

−= + = − ∇ +

TISE for the electron reads, 22 2

22 2 2

0

ˆ1 ( ) ( )2 (4 )e

ZeLr r E rm r r rr r

ψ ψπε

−∂ ∂ − − + = ∂ ∂

Since the potential is central, the angular part of ( )rψ are spherical harmonics. Also, the radial equation can be shown to take the form of the modified radial equation (with ( ) ( )u r r R r= ).

22 2 2

2 20

d ( ) ( 1) ( ) ( )2 (4 )d 2e e

Zeu r l l u r E u rm rr m rπε

− +− + + =

(similar to equn (5))

Solutions to the above equation can be obtained for

(i) continuum states (with E > 0), describing electron-proton scattering

(ii) discrete bound states (with E < 0), representing the hydrogenic atom

Here we confine our attention to bound states only. It can be shown that, the physically acceptable solutions to this equation are obtained, when

22

2 20

142

em ZeEnπε

= −

where 1, 2, 3, ..............n = and n is called the principal quantum number. Note: For an arbitrary value of n, there are n number of possible values of l , as given by 0, 1, 2, .............., 1l n= − . (recall that, 0, 1, 2, 3, ..............l = ) Energy Levels The allowed bound state energy eigenvalues are given by

22

12 2 2

0

1 , 1, 2, 3, ..............42

en

m EZeE nn nπε

= − = =

This is the famous Bohr formula and Bohr obtained it in 1913 by using a mixture of classical physics and premature quantum theory. The Schrodinger equation did not come until 1924. The energy values can also be given in the form

90

2 2 22

2 20 0

1 ( )(4 ) 22n e

e Z ZE m ca n n

απε

= − = − ,

where the Bohr radius, 2

1000 2

(4 )0.529 10 m

ea

m eπε −= = ×

. ( = radius of the 1st Bohr orbit of H)

The radii of Bohr orbits are given by 2 22

0 02

4n

e

a nnaZ Zm e

πε = =

As n takes on values 1, 2, 3, ............., ∞ , bound state energy spectrum of hydrogenic atoms

contains an infinite number of discrete levels from 22

1 2042

em ZeEπε

= −

to zero. For atomic

hydrogen (Z = 1), 1 13.6 eVE = − . This negativeness in bound state energies provides stability to the atom. Spatial wave functions of different states in hydrogenic atoms are labeled by three quantum numbers n, l and lm as ( ) ( ) ( , )

l lnlm n l l mr R r Yψ θ φ=

. However, nE depends only on n and is independent of (i.e. degenerate with respect to) and ll m . Next, we can find the number of different states of this type possible for a bound level of energy nE , which is the degeneracy of that level. For each value of n, l may take on values, 0, 1, 2, .............., 1l n= − (i.e. n number of values) For each l value, lm may take on values in the range, , 1, ... 1, 0, 1, ........, 1,lm l l l l= − − + − − (i.e. 2l + 1 number of values)

* The total degeneracy of the bound state energy level nE = ( )1

2

0

2 ( 1)2 12

n

l

n nl n n−

=

−+ = + =∑

The hydrogenic energy levels depend on n and are independent of l and lm . The degeneracy with respect to lm is present for any central potential. But, the degeneracy with respect to l is a characteristic of the Coulomb potential. (i.e. ( ) 1V r r∝ ). The n-fold degeneracy with respect to l is removed when the dependence of V(r) is modified to become a non-Coulombic potential. Then, the energy of the electron becomes dependent on l and gives rise to n no. of distinct energy levels as nlE ( 0, 1, 2, .............., 1l n= − ). Eg: Alkali atoms The 2l + 1 - fold degeneracy w.r.t. lm is removed when an external magnetic field is applied. Then, the energy becomes dependent on ml.

Spectroscopically, atomic states are labeled as nl . For eg: 1s, 2s, 2p, 3s, 3p, 3d, ….. Ground level (n = 1) ⇒ l = 0 ⇒ lm = 0 ⇒ 1s state only ⇒ Total # of possible states = 1 (Degeneracy = n2 = 1)

91

First excited level (n = 2) ⇒ l = 0 ⇒ lm = 0 ⇒ 2s state ⇒ l = 1 ⇒ lm = −1, 0, +1 ⇒ 1 0 12p , 2p , 2p− + states ⇒ Total # of possible states = 4 (Degeneracy = n2 = 4) We see that the energy levels predicted by the Bohr theory and by Schrodinger theory agree with each other. But, the Schrodinger theory is more powerful, because, the eigenfunctions it yields enable us to calculate probability densities, expectation values of operators, transition rates, etc. The radial eigenfunctions of bound states It can be shown that, physically acceptable solutions to the radial equation are

2 12( ) e . ( )lln l n lR r N Lρ ρ ρ+−

+= ,

where N is a normalization constant and the associated Laguerre polynomials 1 2

2 1 1

0

[( )!]( ) ( 1)( 1 )! (2 1 )!

n l kl k

n lk

n lLn l k l k

ρρ− −

+ ++

=

+= −

− − − + +∑ with 0

2Z rna

ρ

=

Normalization of ( )nlR r ⇒ 2 2

0( ) d 1nlR r r r

∞=∫

⇒ 3

22 2 12 20

0

e ( ) d 12

lln l

naN L

Zρ ρ ρ ρ ρ

∞+−+

= ∫ ⇒

[ ]332 0 2 ( )!

12 ( 1)!

n n lnaN

Z n l+

= − −

Normalized radial eigenfunctions for the bound states of hydrogenic atoms are given as

1 232 12

30

2 ( 1)!( ) e . ( )2 [( )!]

lln l n l

Z n lR r Ln a n n l

ρ ρ ρ+−+

− − = +

Using the above expression, the first few radial eigenfunctions can be written as follows.

When n = 1 ⇒ l = 0 ⇒ ( ) ( )3 210 0 0( ) 2 expR r Z a Z r a= −

92

When n = 2 ⇒ l = 0 and 1 ⇒ ( ) ( ) ( )3 220 0 0 0( ) 2 2 1 2 exp 2R r Z a Z r a Z r a= − −

( ) ( ) ( )3 221 0 0 0

1( ) 2 exp 23

R r Z a Z r a Z r a= −

When n = 3 ⇒ l = 0, 1 and 2 ⇒ 30 ( ) .....R r = , 31( ) ......R r = , 31( ) ......R r = For the first few functions, ( )n lR r vs. r and 2 2 ( )n lr R r vs. r plots are as given below. We can use the known expressions of ( )n lR r and ( , )lm

lY θ φ in ( ) ( ) ( , )ll

mnlm n l lr R r Yψ θ φ=

to obtain the wave functions of the first few states of hydrogenic atoms as follows. n = 1 ⇒ l = 0 ⇒ 0lm = ⇒ 0

0 ( , ) 1 4Y θ φ π=

⇒ ( ) ( )3 2100 0 0

1( ) Z expr a Z r aψπ

= −

n = 2 ⇒ l = 0 ⇒ 0lm = ⇒ 00 ( , ) 1 4Y θ φ π=

⇒ ( ) ( ) ( )3 2200 0 0 0

1( ) Z 1 2 exp 22 2

r a Zr a Z r aψπ

= − −

⇒ l = 1 ⇒ 0lm = ⇒ 01

3( , ) cos4

Y θ φ θπ

≡

⇒ ( ) ( ) ( )3 2210 0 0 0

1( ) Z exp 2 cos4 2

r a Zr a Z r aψ θπ

= −

93

⇒ 1lm = ± ⇒ ( )11

3( , ) sin exp i4

Y θ φ θ φπ

± ≡ ±

⇒ ( ) ( ) ( ) ( )3 221 1 0 0 0

1( ) Z exp 2 sin exp i8

r a Z r a Z r aψ θ φπ± = − ±

In the above work on hydrogenic atoms, it was assumed the mass of the nucleus to be infinitely large when compared to the electron mass so that the nucleus remains fixed in space (at the origin). This is a good approximation even for hydrogen, because nuc 2000 em m≈ × . However, for a better agreement of the above theoretical results with accurate spectroscopic data, it requires the nuclear mass to be taken into account as finite. In such a situation, the electron and the nucleus revolve around their common centre of mass and it is possible to show that the electron moves around the nucleus as though the nucleus is fixed and the mass of the electron is slightly reduced to a value µ called the reduced mass of the system. The equations of motion of the electron-nucleus system are the same as those we have considered before, if we simply substitute µ for em

It can be shown that, e

e

m Mm M

µ =+

, where M is the mass of the nucleus. As

.eM mµ→ ∞ ⇒ → Under this change, the more accurate expressions of energy eigenvalues and energy eigenfunctions can be obtained by replacing em by µ in the relevant expressions. Eg:

22 2 2

2 2 20 0

14 (4 )2 2nZ e e ZE

an nµ

µπε πε

= − = −

and the modified Bohr radius, 2

002

(4 ) ema a

eµπε

µµ= =

.

The fraction by which the energy levels of a hydrogen atom is shifted = 1,836 0.999451,837e e

Mm m Mµ

= = =+

.

The more accurate expressions of radial wave functions and energy eigenfunctions can be obtained by replacing 0a by aµ in the relevant expressions.

Eg: ( ) ( )3 210 ( ) 2 expR r Z a Z r aµ µ= − ( ) ( )3 2

1001( ) Z expr a Z r aµ µψπ

= −

Consider an electron in a stationary state ( , , )

lnlm rψ θ φ .

Position Probability Density =22 2( , , ) ( ) ( , )

l lnlm n l l mr R r Yψ θ φ θ φ=

Probability of finding the electron in the volume element dr at the point r = 2( , , ) d

lnlm r rψ θ φ

94

Probability of finding the electron in the shell of inner and outer radii r and r + dr

= 2

2

0 0

( , , ) dlnlm r r

π π

θ φ

ψ θ φ= =∫ ∫

= 2

2 2

0 0

d sin d ( , , ) dlnlm r r r

π π

θ φ

θ θ φ ψ θ φ= =∫ ∫

= 222 2

0 0

( ) d d sin d ( , )lmn l lr R r r Y

π π

θ φ

θ θ φ θ φ= =∫ ∫

= 22 ( ) dn lr R r r

22

0 0

since d sin d ( , ) 1lmlY

π π

θ φ

θ θ φ θ φ= =

=

∫ ∫

= ( )dn lD r r ,

where the Radial Distribution Function, 2 2( ) ( )n l n lD r r R r= , 2 2( ) ( )n l n lR r R r=

2 ( )n lR r = Electron density along a given direction

2 2( ) ( )n l n lD r r R r= = Probability of finding the electron within a unit length at a distance r in any direction from the nucleus (i.e. within a shell of radius r and of unit thickness).

Consider the ground state of hydrogen (Z = 1), n = 1, l = 0, ml = 0.

( ) ( )3 2100 1 0 0

1( ) ( ) 1 expsr r a r aψ ψπ

≡ = −

As can be seen from the plots given below, 1 ( )s rψ and 2

1 ( )s rψ are exponentially decreasing functions of r with their maximum occurring at the origin To make a comparison of quantum mechanical results with those of Bohr theory, we look for the Radial Distribution Function of the ground state.

2 2

10 10( ) ( )D r r R r= , where ( ) ( )3 210 0 0( ) 2 1 expR r a r a= −

[Note: For states where ( )rψ is independent of θ and φ (eg: 1s, 2s, 3s, …..)

Probability distribution function = 224 rπ ψ .] The maximum in 10 ( )D r curve plotted versus r indicates that the probability of detecting the electron is maximum at some distance. This value of r is called the most probable distance and to find it

1 0d ( ) 0d

D rr

= ⇒ 0r a=

So, Q. theory is similar to the Bohr picture of the hydrogen atom. However, there is a finite probability of seeing the electron at other values of r between 0 and ∞ and therefore the size of the atom is not well-defined in Q. theory. The value r can also be calculated for the ground state of hydrogen, as follows.

2* 31 1 0 03

0 0 0 0

1 3( ) ( )d exp( 2 ) d sin d d m. p. d.2s sr r r r r r a r r a

a

π π

ψ ψ θ θ φπ

∞

= = − = >∫ ∫ ∫ ∫

95

As we see, the mean value is found to be larger than the most probable distance. This is because of the diffuseness incorporated by quantum theory to the electron orbit making it an ‘electron cloud’. According to quantum theory, the electron can be found at distances from r = 0 to r = ∞ from the nucleus with various probabilities (The maximum probability occurs at r aµ= ). So, their average value is found to be equal to1.5aµ . The Bohr theory predicts precisely aµ as the distance where the electron can be found. However, in quantum theory, that is the distance where the electron can be found with the maximum probability. To obtain a reasonable estimate for the size of the atom, we find a value (0.9)r such that there is a 90% probability that the electron is at a distance (0.9)r r≤ .

(0.9)2 2

100

Prob. ( )d 0.9r

r R r r= =∫ ⇒ (0.9)

203

0 0

2 exp( 2 ) d 0.9r

r a r ra

− =∫ ⇒ (0.9) 02.6r a= .

For most purposes, an effective atomic radius is taken as from 0 02.5a a− ~ 10−8 cm = 1 Å. Classically, a moving particle has a turning point when the total energy = potential energy. At this point, the kinetic energy (hence the velocity) = 0 and the particle is expected to be reflected by the potential barrier. For an electron in the ground state of hydrogen atom, classical turning point occurs, when

1( )V r E= ⇒ 2 2

0 0 0(4 ) 2(4 )e e

r aπε πε− −

= ⇒ 02r a= .

As shown by 10 ( )D r versus r curves, there is a finite probability of finding the electron for

02r a> . i.e. the electron has access to a region which is forbidden by classical theory. This effect of tunneling (penetration) through potential barrier is typical in q. theory results.

96

For values of 02r a> ⇒ V E> , and in order to satisfy the condition E T V= + , the kinetic energy T would have to be negative. This unrealistic situation can be explained by using H. u. p. as follows. In q.m., a particle is in a certain region means that its position measurement has been made (with some uncertainty). According to the u. p. uncertainty in position ⇒ uncert. in momentum ⇒ uncert. in kinetic energy It can be shown that, when the electron is in the classically forbidden region, the uncertainty introduced to its kin. energy is sufficiently large to compensate for the negative value required by conservation of energy. The energy eigenvalue equation (TISE) for a hydrogenic atom reads

ˆ ( ) ( )l lnlm n nlmH r E rψ ψ=

, where 12n

EE

n= , 1, 2, 3, ..........n =

Ionization potential in the nth shell, 2 2

20(4 ) 2P n

e ZI Ea nµπε

= =

ˆ ˆ( , , ) ( ) ( , ) ( )( 1) ( , )l l

l

m mlnlm n l n ll lr R r Y R r Yψ θ φ θ φ θ φ ℘ = ℘ = −

⇒ ˆ ( , , ) ( 1) ( , , )

l l

lnlm nlmr rψ θ φ ψ θ φ℘ = − ⇒

lnlmψ ’s have the parity of l.

In Dirac’s notation, ( )

lnlm lr n l mψ ≡

Eigenfunctions ( )lnlm rψ form an orthonormal basis as given by

* ( ) ( )dl ll ln l m n l m n n ll m mr r rψ ψ δ δ δ′ ′ ′ ′ ′ ′=∫

or equivalently l ll l n n ll m mn l m n l m δ δ δ′ ′ ′′ ′ ′ =

Once ( )

lnlm rψ functions are known, the solutions to TDSE are given by i( , ) ( ) e nl

E tn l mr t rψ −Ψ = .

Energy levels and spectrum of the hydrogen atom In his model, Bohr had postulated that transitions can take place between states with any two values of the q. no. n. This fact is justified by quantum mechanical calculations of transition probabilities. However, these calculations also suggests that for allowed transitions, the q. no. l changes by 1± . In addition, the q. no. lm changes by 0 or 1± . Therefore, the selection rules for the most common spectral transitions due to the interactions between the electron and the dipole electric field of the radiation are

1 21 2 1 2arbtrary, 1, 0, 1l l ln n n l l l m m m− = ∆ − = ∆ = ± − = ∆ = ± ,

where the symbols 11 1, , ln l m correspond to the initial state and the symbols

22 2, , ln l m correspond to the final state of the transitions. The states and allowed transitions for the hydrogen spectrum for n = 1, 2, 3, 4, are shown below. Shown below is only the gross structure of hydrogenic atom energy levels and possible transitions among them. Due to other interactions which exist within atoms, transitions do not occur exactly according to the dipole selection rules.

97

Note: Whenever an integration over whole space is performed on hydrogenic wave functions, even

if they are found to be independent of θ and φ, three separate integrations over r, θ and φ must be performed over the ranges, r = 0 to ∞, θ = 0 to π and φ = 0 to 2π with differentials,

2dr r , sin dθ θ and dφ , respectively.

When an integration is performed on a spherical harmonic function, ( , )lmlY θ φ , two

separate integrations over θ and φ must be performed over the ranges, θ = 0 to π and φ = 0 to 2π , with differentials sin dθ θ and dφ , respectively.

When an integration is performed on a Radial function, ( )n lR r , an integration over r must

be performed over the range r = 0 to ∞, with the differential 2dr r .

98

Orbital angular momentum and orbital magnetic Dipole moment of the electron Consider a hydrogen atom and assume the electron orbit as circular. Since this acts as a current

loop, there is a magnetic field in the direction perpendicular to the plane of the orbit.

At large distances from the loop, the B

field produced by the current loop is similar to that

produced by a magnetic dipole placed at the centre of the orbit.

The magnetic dipole moment due to a current loop enclosing a small area dA, dM I A=

M

is in a direction normal to the loop.

Current (magnitude) due to electron, 2 2

e e evT r v rπ π

= =

Area of the loop, 2dA rπ=

Orbital magnetic dipole moment (magnitude) of the electron, 22 2Lev evrM r

rπ

π= × =

Orbital angular momentum (magnitude) of the electron, eL m v r=

1 constant2 2e e

LM ev r eL m v r m

= × = =

Usually, LML

is written as, 2l l

B

e

LM eg gL m

µ= =

Orbital g-factor, lg = 1 Bohr magneton, 24 19.27 10 J T2B

e

em

µ − −= = ×

lB

LL

M gµ

=

⇒ lB

LL

M gµ

= −

( LM

is anti-parallel to L

)

( ) lB

L zzM g Lµ

= −

v

eme−

I Current

99

Quantum mechanically, it can be shown that ( )1L L l l≡ = +

, where l = 0, 1, 2, ….. is the

orbital angular momentum quantum number.

Note: This L is the average value of the magnitude of the orbital angular momentum.

Depending on the value of l, we can characterize the orbital angular momentum states of electrons

in atom as s, p, d and f sub shells (states).

Quantum mechanically, it can be shown that z lL m= , Lx = Ly = 0.

Note: These Lx, Ly, and Lz are the average values of the components of orbital angular momentum.

ml is the orbital magnetic quantum number. Quantum mechanically, it can be shown that, for a given value of l, lm varies in the range

ll m l− ≤ ≤ , in integer steps. ⇒ For a given values of l, there are (2l +1) number of ml values. Possible values of ml are − l, − l + 1, − l + 2, ……, 0, ..….. , l − 2, l − 1, l For s- states, l = 0 ⇒ 0 0lm≤ ≤ ⇒ ml = 0 one value

For p- states, l = 1 ⇒ 1 1lm− ≤ ≤ ⇒ ml = −1, 0, 1 three values

For d- states, l = 2 ⇒ 2 2lm− ≤ ≤ ⇒ ml = −2, −1, 0, 1, 2 five values

For f- states, l = 3 ⇒ 3 3lm− ≤ ≤ ⇒ ml = −3, −2, −1, 0, 1, 2, 3 seven values

Quantum mechanically, ( )

( )1

1ll l

BBBL L

g l lLM M g g l l

µµµ

+= = = = +

Quantum mechanically, ( ) l lB B

B ll lz

L zL

M g g m g mµµ

µ= = × =

For a given value of l, ( )L zM can have (2l + 1) number of discretely quantized values.

Note: These LM and ( )L zM are average values of the respective quantities.

Vector model of orbital angular momentum

( )1L L l l≡ = +

, where l = 0, 1, 2, …….

z lL m= , and for a given value of l, ll m l− ≤ ≤ . Lx = Ly = 0

The above facts suggest that the orbital angular momentum L

precess about the axis of

quantization (z-axis) so that its (2l + 1) # of projections on that axis are given by z lL m= (ml are

− l, − l + 1, − l + 2, ……, 0, ..….. , l − 2, l − 1, l). L

may be viewed as lying on the surface of a

100

cone with altitude of lm and z-axis as the axis of symmetry. All the orientations of L

on the

surface of this cone are equally probable.

Eg: When l = 1

l = 1, 2L = , Lz = − , 0, + cos 1 2θ =

Since q.m suggests the vector model of orbital angular momentum, it also suggests the vector

model of orbital magnetic moment.

( )L zM can have (2l + 1) # of discretely quantized values. However, classically L

(hence LM

) can have any # of orientations with respect to the z-axis. i.e.

( )L L LzM M M− ≤ ≤

An atom possesses magnetic dipole moment due to the orbital motion of the electron. When an

atom is in an external magnetic field of strength B

, the potential energy of the interaction

LV M B= − ⋅

The net force acting on the atom, ( ) ( )L LF V M B M B= −∇ = +∇ = ∇⋅ ⋅ ⋅

, , ,L L Lx y zB B B

F M F M F Mx y z

∂ ∂ ∂= = =

∂ ∂ ∂⋅ ⋅ ⋅

If B

is uniform, then Fx = Fy = Fz = 0.

Stern and Gerlach Experiment (1922) This experiment was carried out to measure the magnetic dipole moments of atoms by detecting the

deflection of an atomic beam by an inhomogeneous magnetic field.

z

θ

θ

2

2

2

+

−

LM

LM

LM

L

L

L

Axis of quantization

101

A collimated beam of Ag atoms entered the inhomogeneous magnetic field between the magnetic

poles and was detected after falling on a cooled collecting plate. (The whole setup was under

vacuum) Due to the geometry of the magnets, the force acting on the atoms was mainly in the z-

direction.

( )Lz zB

F Mz

∂=

∂⋅ ,

Bz

∂

∂ - field gradient

Along the x-axis,Bz

∂

∂ remains constant.

( )Lz zF M∝

Since the field gradient is a maximum at the center, Fz is maximum at the center and the deflection

should be maximum. If the magnetic moments of the atoms are randomly oriented as expected

classically, every value of ( )L zM such that ( )L L LzM M M− ≤ ≤ would occur and the deflected

beam would be spread into a continuous band (Fig. 2). Quantum mechanically only discretely

With no magnetic field

With magnetic field on (Classically expected)

With magnetic field on (Observed)

N

S

B field lines

Inhomogeneous magnetic field

Beam of silver atoms

N

S

Collecting plate

z

x y

Magnetic poles

102

quantized values of ( )L zM are possible ( ( ) B llL zM g mµ= ) and hence the deflected beam would

be split into several discrete components. (According to ( )Lz zB

F Mz

∂=

∂⋅ ). What Stern &

Gerlach observed was two distinct lines (Fig. 3). This is called the space quantization of the

components of the magnetic dipole moment along the direction defined by the magnetic field (i.e.

z-direction)

We know that,

l is an integer ⇒ # of possible ml values for a given l = (2l + 1 ) is an odd integer.

For any value of l, one of the ml values is zero. ⇒ There should be an undeflected

component of the Ag atomic beam (because ml = 0 result in a zero force)

Consider Ag atom

Ag (Z = 47): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d10

The net magnetic moment of an Ag atom is determined by the single 5s electron (all the other 46

electrons in completely filled sub shells contribute to nothing.)

For an s electron ⇒ l = 0 ⇒ ml = 0 ⇒ ( )L zM = 0 ⇒ Fz = 0

No deflection in the Ag beam is possible but, Stern & Gerlach observed two deflections!

An explanation came from Goudsmit and Uhlenbeck in 1925. i.e., in addition to the magnetic

moment produced by the orbital motion, electron may possess an intrinsic magnetic moment, sM

,

where the component of sM

in a given direction (z-axis) can take only two values. It can be

postulated that sM

is due to an intrinsic angular momentum ( )S

of the electron.

S

- Spin angular momentum of the electron

Similar to the relationship of LM

with L

, we have

sB

SSM g µ

= −

, sg – Spin g-factor (spin gyromagnetic ratio), sg ≈ 2

sM

- Spin magnetic moment of the electron.

Unlike the orbital angular momentum the spin angular momentum does not correspond to any

physical motion. (i.e., it is not possible to write a relationship equivalent to L r p= × for S

)

Quantum mechanically, S

and zS are defined to have average values

103

( )1S S s s≡ = +

, szS m=

For a given value of s, ss m s− ≤ ≤ ⇒

(2s + 1) # of sm values (−s, − s + 1, s + 2, ……. , s − 2, s − 1, s) are possible.

s – Spin angular momentum quantum number sm – Spin magnetic quantum number

Average values of 0xS = and 0yS = .

For electrons, 12s = ⇒ 1 1

2 2sm− ≤ ≤ ⇒ 1 1and2 2sm = − +

32

S = and and2 2zS = −

For other subatomic particles such as protons, neutrons, 12s = . They are commonly called ‘spin-

12 particles’ or ‘fermions’.

For photons, s = 1 ⇒ 1 1sm− ≤ ≤ ⇒ sm = −1, 0, +1

2S = and , 0 ,zS = − +

Quantum mechanically defined values of , , andx y zS S S S suggest the vector model of spin

angular momentum.

Eg: 12

s =

Since Sz values are quantized ⇒ ( )s zM

must also be quantized.

( ) ( )122 =s s s

B BB B B

zS s sz

SM g g m g m

µ µµ µ µ= − = − × = − = − × × ± ±

The force responsible for splitting the silver atomic beam in S & G experiment,

z

θ

θ S = 3 2

SM

SM

S

S

Axis of quantization

S = 3 2

“Spin up”

“Spin down”

2

− 2

104

( )z BzB B

F Ms z zµ

∂ ∂= = ±

∂ ∂⋅

Magnetic moment of the Ag atom

( ) ( ) ( ) ( )47 47

5 51 1

( 1)0 s

BL S L Si i s s

i i

s sM M M M M g

µ

= =

+= + = + = −∑ ∑

s - electron ⇒ l = 0 ⇒ ml = 0 s = 1

2 ⇒ ms = 12±

ml = 0 p - electrons ⇒ l = 1 ⇒ ml = −1, 0, +1 s = 1

2 ⇒ ms = 12±

ml = −1 ml = 0 ml = 1

This explains why it is possible to accommodate 2 electrons in an s-subshell and 6 electrons in a p-subshell,

etc.

Electron beams which have equal #s of ‘Spin up’ and ‘Spin down’ electrons are called ‘unpolarized

electron beams’ and those which have unequal #s of ‘Spin up’ and ‘Spin down’ electrons are called

‘polarized electron beams’.

The ‘degree of polarization’ = N NPN N↑ − ↓

=↑ + ↓

↑N and ↓N are the #s of spin up and spin down electrons in the beam.

For an unpolarized electron beam, N N↑ = ↓ ⇒ 0P =

For a polarized electron beam, N N↑ ≠ ↓ ⇒ 0P ≠

For a perfectly polarized beam, ↑N = 0 ( )or 0N ↓ = ⇒ 1P = ( )0 1P≤ ≤

S – G magnets are sometimes called spin filters.

The total angular momentum of a many-electron atom is i ii i

J S L= +∑ ∑

, and only those electrons in

incompletely filled subshells contribute to the above summations. Just as in the cases of orbital and spin

angular momenta, the (average) values of J

and its z-component zJ are given by ( 1)j j + and jm ,

ms = 12− ms = 1

2

105

respectively. Also, for a given value j (which can take integer or half integer values only), jm can take (2j +

1) number of values as given by − j, −j + 1, −j + 2, ..……, j + 2, j − 1, j. Since the total magnetic dipole

moment M

is determined by the values of J

and the z-component of the dipole moment zM is determined

by the values of zJ , one can expect (2j + 1) # of different orientations of M

w.r.t. the z-axis ⇒ (2j + 1) #

of zM values are possible.

Thus in a S-G experiment, if a beam of many-electron atoms (with some # of electrons in the unfilled

subshell) is found to produce x # of spots on the detection plate

⇒ x = 2j + 1 ⇒ ( )1 2j x= − .

The magnitude (average value) of the total angular momentum of the atom = ( 1)j j +

Tutorial 5 TISE in Spherical Polar Co-ordinates, Orbital Angular momentum, Hydrogenic Atoms 1. (i) If ( )

ln l m rψ describes the energy eigenfunction of the electron in the nth state of a hydrogen atom,

describe the quantum numbers which characterize energy eigenstates and give their limits. Sketch energy levels and indicate possible eigenstates (using the above notation) up to the 2nd excited level.

(ii) Explain briefly why the sub states (shells) 1p, 1d, 1f, 2d, 2f, 3f, … etc, can not occur in atoms.

(iii) Explain briefly why it is not possible to accommodate more than 02 electrons in a s-sub state, 06 electrons in a p-sub state, 10 electrons in a d-sub state, 14 electrons in a f-sub state, in an atom.

2. Find the energy eigenfunctions and the corresponding energy eigenvalues of a particle of mass m, that

rotates on a circle of radius R in a potential V(r), where V(R) = V0. 2

22 2

1 1In plane polar co-ordinates ( , ),r rr r r r

φφ

∂ ∂ ∂ ∇ = + ∂ ∂ ∂

3. Write down the time-independent Schrödinger equation for a particle of mass m that is constrained to

move on the surface of a sphere of radius R. Then, for the case in which the particle’s potential energy is zero, find the energy eigenvalues and energy eigenfunctions and check for any degeneracies.

2

2 22 2 2 2 21 1 1In spherical polar co-ordinates ( , , ), sin

sin sinr r

r rr r rθ φ θ

θ θθ θ φ

∂ ∂ ∂ ∂ ∂ ∇ = + + ∂ ∂ ∂ ∂ ∂

4. At a given instant of time, a quantum system is in the normalized orbital angular momentum state

( , ) sin sinNψ θ φ θ φ= , where N is the normalization constant.

(i) Determine the constant N. (ii) What possible values of zL will be found by measurement and with what probability will these

values be occurred ? (iii) What is 2L

for this state ?

(iv) What is zL for this state ?

Some useful spherical harmonics are 1 i1

3( , ) sin e8

Y φθ φ θπ

± ±= .

106

5. Consider an electron in a hydrogen atom whose wave function at t = 0 is the following superposition of

energy eigenfunctions ( )ln lm rψ

: 100 200 3221( , 0) = 2 ( ) 3 ( ) ( )14

r t r r rψ ψ ψ Ψ = − +

(i) Is the wave function an eigenfunction of the parity operator ? (ii) What is the probability of finding the system in the ground state (100) ? In the state (200) ? In the

state (322) ? In another energy eigenstate ? (iii) What is the expectation value of the energy (in terms of nE ): of the operator 2L

; of the operator

zL ? 6. Consider a hydrogen-like atom with atomic number Z, containing an electron in the ground state (1s).

The wave function of this electron is ( )1s ( ) expr N Z r aµψ = − , where N is the normalization

constant, r is the radial co-ordinate and aµ is the modified Bohr radius. (i) Determine the constant N. (ii) Determine the average value of the potential energy of the electron.

(iii) What is meant by the quantities 21sψ and 22

1s4 rπ ψ ? (iv) Evaluate the average value of the distance to the electron from the origin. (v) Evaluate the distance from the origin at which the probability of finding the electron is a

maximum. (vi) Comment on your results for parts (iv) and (v). (vii) Determine the classically forbidden region for the electron in the ground state of a hydrogen

atom, if its bound-state energy eigenvalue is −13.6 eV. (viii) What is the probability of finding the ground state electron in this region ?

It is given that, 10

!e d , where Real( ) 0 and is a positive integer.n a ynnI y y a n

a

∞−

+= = >∫

7. A system is found to be in an orbital angular momentum state which is a mixture of orthonormal basis

states corresponding to l = 1, as given by 1 1 1 0 1 1 ,α β γΨ = + + − where α, β and γ are real numbers. (i) What is the condition that needs to be satisfied by α, β and γ, for the normalization of Ψ ?

(ii) Find the expectation values of the operators 2ˆˆ and zL L

for the state Ψ of the system.

Electron's Spin Angular Momentum, Spin-Orbit Interaction 8. With regard to a Stern-Gerlach experiment, in each of the following cases, illustrate the recorded

configuration of the atomic beam that would be seen on the on the detecting plate giving reasons. (a) If the magnetic field was turned off (b) If the magnetic field was homogeneous (c) If the magnetic field was inhomogeneous and the space quantization did not exist (d) If the magnetic field was inhomogeneous and the space quantization did exist

9. A beam of hydrogen atoms, emitted from an oven running at a temperature 400 K, is sent through a Stern-

Gerlach magnet of length 1 m. The atoms experience a magnetic field with a gradient of 10 T/m. Calculate the transverse deflection of a typical atom in each component of the beam, due to the force exerted on its spin magnetic dipole moment, at the point where the beam leaves the magnet.

10. Consider a beam of particles of mass M and spin ½ , propagating in the +x-direction. The beam has a

cross section d. It interacts with an Stern-Gerlach apparatus whose field is in the z-direction. Employing relevant uncertainty relations, show that the smallest uncertainty in the normal displacement (∆z) grows large with decreasing mass.

Prepared by: Revised in September 2007 Dr. WMKP Wijayaratna Department of Physics

107 University of Colombo, Colombo 03.