Particle in a 1d Box Quantum Mechanics

7/15/2019 Particle in a 1d Box Quantum Mechanics

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Quantum MechanicsLecture-8

Concept of Modern Physics

by A. Beiser


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Exercise: A Particle limited to the x-axis has the wavefunction =ax between x=0 and x=1; =0 elsewhere.

(a)Find the probability that the particle can be found x=0.45and x=0.55.

(b)Find the expectation value < x > of the particles position.

(a) 0.0251 a2

(b) a2/4


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*

2 1

*

1 2

1 2

Condition for orthogonal wavefunction

d 0

or d 0

wavefunction and are mutually orthogonal

* *

1 1 2 2

1 2

Condition of normalized wavefunction

d 1 or d 1

and called normalized wavefunction.

*

i j

Condition of orthonormal function

d 0 if i j

= 1 if i = j


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A particle is confined to a one-dimensionalregion of space between two impenetrablewalls separated by distance L This is a one- dimensional box

The particle is bouncing elastically backand forth between the walls As long as the particle is inside the box, the

potential energy does not depend on itslocation. We can choose this energy value to be

zero V= 0, 0 < x < L, V, x 0 and x L

Particle in a one dimensional Box(infinite square well potential)

Particle in a one dimensional Box (infinite square well potential)


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Since the walls are impenetrable, there is zeroprobability of finding the particle outside the box.Zero probability means that

(x) = 0, forx< 0 andx>L

The wave function must also be 0 at the walls (x= 0and x = L), since the wavefunction must becontinuous

Mathematically,(0) = 0 and (L) = 0

In the region 0


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We can re-write it as

2

22 2 2

2

2

2

x 2mE 2mEx kx

xk x

x


The most general solution to this differential equation is

(x) =A sin kx + B cos kx

A and B are constants determined by the properties of thewavefunction as well as boundary and normalization conditions


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Since the particle cannot have infinite energy, it cannot exist outside the box.Therefore, the wave function must be zero outside the box. must be also zeroat the walls, that is, at x = 0 and x = L, for otherwise there would be discontinuities

at the walls.1. Sin(x) and Cos(x) are finite and single-valued functions

2. Continuity: (0) = (L) = 0

(0) =A sin(k0) +B cos(k0)=0B = 0

(x) =A sin(kx) (L) =A sin(kL) = 0 sin(kL) = 0kL = n, n =0, 1, 2

( n = 0 is not admissible because it yields zero everywhere which means that theparticle is no where).


2

2

22

2

2222

22

822

)(

2

n

mL

hn

mLm

nLE

mkEn

Lk

n

nnn

Not adm issible


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The particle can not have arbitrary energy , butcan have certain discrete energy corresponding

to n=1,2,3,

Each permitted energy is called eigenvalueof the

particle and constitute the energy level of thesystem, and the integer nthat specifies an energylevel En is called its principal quantum number.The wave function corresponding to each eigenvalue is called eigen functions.

Lowest level n = 1, energy notzero why ?


2 2

n 2

n hE

8mL


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2 2 2 2 2

n 2 2

n h nE

8mL 2mL

Thus the energy of the particle bounded in a box is quantized.

Another conclusion for the motion of particle in a box can also be drawn that the

particle can not have zero energy but has minimum energy and called as zero point

energy. The state corresponding to this energy is called ground state.

According to classical mechanics, when a particle is placed in a box, it can have

zero energy or continuous kinetic energy. Thus the quantization of energy is a

specific result derived from quantum mechanics.

If one assumes v = 0 for a particle in a box, the de-Broglie wave associated with it will

be =(h/mv) = , which ia an absurd result because there should be node at the

boundary for the bounded particle.

If the particle is bound in 1D box of width L. The particle can not have zero kinetic

energy, because from Heisenberg uncertainty principle, the uncertainty in position

of the trapped particle in a box is x = L, hence p as well as velocity of the particle

and their kinetic energy can not be zero.


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An e- is in a box 0.10 nm across, which is the order of magnitude ofatomic dimensions. Find its permitted energies.

A 10 g marble is in a box 10 cm across. Find its permitted energies.

2 2

n 2

2 34 218 2

n 31 10 2

2

n hE

8mL

n (6.63 10 J.s)E 6.0 10 n J

8(9.1 10 kg)(1.0 10 m)

38n eV

The minimum energy of the e- can have 38eV, corresponding to n=1.the sequence ofenergy levels continues with E2=152 eV, E3=342 eV, and so on. If such box existed, thequantization of a trapped e-s energy would be a prominent feature of the system. ( andindeed energy quatization is prominent in the case of an atomic election.)

The minimum energy of the marble can have 8.2810-65 J, corresponding to n=1. Amarble with this K.E. has a speed of only 1.310-31 m/s and therefore can not beexperimentally distinguished from a stationary marble.

Forv=1/3 m/s----corresponding energy level of quantum number n=1030.

Hence in the domain of everyday experience, quantum effects are unnoticeable, whichaccounts for the success of Newtonian Mechanics in this domain.

2 34 265 2

n 2 1 2

n (6.63 10 J.s)E 8.28 10 n J

8(1.0 10 kg)(1.0 10 m)


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2

2 2

2 2

2

n

2

2 nn n 2

2 2

n 2

x 2mEx

x

x k xx

(x) Asin(kx) Bcos(kx)

(0) Bcos(0) 0

(L) Asin(kL) 0

k L n

2mEnk n k

L Ln h

E8mL

n(x) Asin( x)

L



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Let us now consider the eigenfunctions of the particle.Substituting B=0 and k=n/Lin the equation of general solution.Solutions of the Schrdinger equation are

To find the value of A, we use the normalizationcondition

The normalized eignfunctions of the particle


n

n x(x) Asin

L

2

n

L

2 2

0

2

| (x) | dx 1

n xA sin dx 1

L

A L / 2 1

A 2 / L

n

2 n x(x) sin

L L

Initial wavefunctions and Energyfor the first four states in

a one-dimensional particle in a box

E4

E3

E2

E1

E4

E3

E2

E1

2 2

n 2

n hE

8mL

= L2

=2L3

=L

=2L

n

2Ln=1,2,3,...

n

2 2

n 2

n hE

8mL


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n

2 n xWavefunctions (x) sin

L L

n=1,2,3,...

*

n nProbability:

n=1,2,3,...

Note particle most likely

to be found in the middle

for n=1

2 2

2

Energy:

n hE= n=1,2,3,...

8mL

(a) The first two wavefunctions, (b) the correspondingprobability distributions, and (c) a representation of the

probability distribution in terms of the darkness ofshadin .


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No. of antinodes in eigenfunction = n, No. of nodes =No. of antinodes+1


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Electron in the 10nm Wide Well with Infinite Barriers

Calculate E1=?

Assume that a photon is absorbed, andthe electron is transferred from theground state (n = 1) to the secondexcited state (n = 3)

What was the wavelengths of the photon?

2 2

21 1 22

nE n E , where EmL

2 34 2

1 31 9 2

22

1

3.14 (1.05 10 )E

2 9.1 10 (10 10 )

E 6 10 J 0.00375 eV 3.75 meV


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Electron in the 10nm Wide Well with Infinite Barriers E1 = 3.75 meV

1 0 00375groundE E . eV

3

23 13 9 0 00375

Third excited state is E

E E . eV 0.0338 eV

3 1

34 8

19

0 00375 0 030

6 63 10 3 100 0300 030 1 6 10

124041333 41

0 03

(h ) E E 0.0338 . . eV

c .h . eV,. .

nm m

.


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2 22

2

2

1

2

2

th

n

n

n

For the n state

x

sin nL L

E nmL

E n E

Probability to Find particle in the Right Half of the Well

Particle in the Infinite Potential Well

L L

2 2

L/2 L/2

L

2

L/2

2| (x) | dx [ sin kx] dx

L2 2 L 1 1

sin (kx)dx [ ]L L 2 2 2


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Some trajectories of a particle in a box according toNewton's laws of classical mechanics (A), and accordingto the Schrdinger equation of quantum mechanics (B-D). In (B-D), the horizontal axis is position, and thevertical axis is the real part (blue) or imaginary part (red)of the wave function. The states (B,C,D) are energyeigenstates.


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Problems: Normalization.

1. Determine normalization constant

3

8:

22,cos)()

1

:2

2

exp)()

2:0,exp)()

2:0),/sin()()

2

2

NAnsxxNxiv

aNAnsxia

x

Nxiii

kNAnsxkxNxii

LNAnsLxLxnNxi


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Problems: Particle in a box.

1. An electron is confined to an one-dimensional infinity potential well of width

0.2 nm. It is found that when the energy of

the particle is 230 eV, its eigenfunction has 5

antinodes. Find the mass of the particle andshow that it can never have energy equal to

1 keV.

Answer m =9.1 x10E-31 Kg

n = 10.4


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Problems: Particle in a box.

1. Find the probability that a particle trapped in a box L wide canbe found between 0.45L and 0.55L for the ground and firstexcited state.

2. Write down Schrdinger equation for a one dimensional box,obtain the expression for eigen function and eigen values. If

length of the box ix 25, calculate the probability of findingthe particle with in an interval of 5 at the centre of the boxwhen it is in the state of least energy.

3. A particle is in a cubic box with infinitely hard walls whoseedges are L long. The wave function of the particle is given by

by

Find the normalization constant A.

3

8:sinsinsin),,(

LAAns

L

zn

L

yn

L

xnAzyx z

yx

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Particle in a 1d Box Quantum Mechanics