Vibrational Motion

Key ideas

• Harmonic motion

• Parabolic potential

• Energy levels

• Wave functions

• Gaussian function

• Hermite polynominals

• The virial theorem

Vibrational Motion

Restoring force proportional to displacement = harmonic motion

F kx= −

k = force constant

F dVdx

= − and

21V kx2

=

“Parabolic potential energy”

Fig. 12-25

The Schrödinger equation is therefore

2 2

22

d 1- kx E2m dx 2

Ψ+ Ψ = Ψ

This is a standard equation with known solutions (later)...

For the boundary conditions that infinitely large extensions are

forbidden it can be shown that the permitted energies are

V1E (v )2

ω= + : 12k

mω =

v = 0, 1, 2, ...

And the separation between levels is

vv+1E E ω− = which is independant of v.

Fig. 12-26

Note also there is a zero point energy

01E2ω=

As with the particle in the box, the position is not completely

uncertain therefore the momentum cannot be zero.

Wave Function for Harmonic Motion

2 2 2

22

d 1- kx E2m dx 2

Ψ+ Ψ = Ψ

Has solutions of the form

(x) N (polynomial in x) (Gaussian function)Ψ = × ×

N = normalizing constant

Gaussian = bell shaped curve of the form 2xe−

Fig. 12-27

Polynomial = VH (y) - a Hermite polynomial

Solutions are

2y

2V VV N H (y)(x) e−=Ψ

1

2 4xy :α mk

α = =

Hermite’s satisfy the relationships

0H 1=

V V VH 2yH 2vH 0′′ ′− + =

And V+1 V V-1H 2yH 2vH−= (recursion)

v VH

0 1

1 2y

2 24y 2−

3 3y 12y8 −

4 4 216y 48y 12− +

5 35 y 120y16032y − +

And have the property

So the ground state V = 0 for the harmonic oscillator has

2y2

0 0(x) N e−

Ψ =2x22α

0N e−

=

12 4

mkα =

And therefore the probability density is

2

2x

2 2 α0 0(x) N e

−=Ψ

a bell shaped distribution.

Fig. 12-28

First vibrational level above ground state, v = 1

22

2xy

1 2α211

2N( N 2yα

x) e x e−−

= × × =Ψ

This has value 0 at x = 0 and the probability density 21 (x)Ψ has maxima at

x α=± 1

2 4

mkα =

Fig. 12-29

Fig. 12-31

Note as v→higher, the probability densities are higher at higher

potentials, consistent with a classical picture.

• Moving slowly near the turning points.

• Moving quickly near the vertical.

What are the νN ? - Normalization

2y

2V V(x) H (y)e

−Ψ = :

xyα

=

un-normalized

*V Vdx

∞

−∞

Ψ Ψ∫ *V Vα dy

∞

−∞

= Ψ Ψ∫22 -y

Vα H (y)e dy∞

−∞

= ∫ 1

V2α π 2 v!=

Therefore

1 12 2

V V

1(απ 2 v!)

N =

Note V f(v)N =

Calculating the Properties of Oscillators

• We have the wave functions.

• Need to calculate expectation values for the operator

corresponding to the property.

i.e. *V Vdx

∞ ∧

−∞

Ω = Ψ ΩΨ∫

E.g. for mean displacement, x we evaluate

*V Vx x dx

∞

−∞

= Ψ Ψ∫2 2y y

2 2V V

2V (H e )x(H e )dxN

∞− −

−∞

= ∫

Which looks horrendous

... but the Hermite polynomial properties help simplify the task...

2 2y y2 2

V V2

Vx (H e )x(H e )dxN∞

− −

−∞

= ∫ 2 2y y

2 2 2V V

2V (H e )y(H e )dyα N

∞− −

−∞

= ∫

22 yV V

2V H y H e dyα N

∞−

−∞

= ⋅ ⋅∫

Now we use the recursion relationship to write

V V-1 V+1yH vH + H12

=

So 2 2y yV-1 V V+1 Vx ν H H e dy+ H H e dy1

2

∞ ∞

−∞ −∞

= ∫ ∫

But from the properties of Hermite polynomial these are both equal to zero.

x (mean displacement) = 0 So the oscillator is equally likely to be on either side of x = 0.

For the mean square displacement we evaluate

V

2 * 2x x dx∞

−∞

= Ψ Ψ∫

In the same way... (by applying the recursion relation twice). In this case (∼2 pages of algebra!) we find that not all terms are zero

and

2

12

x (v+ )12 (mk)

=

Thus 2x increases with v corresponding to densities in Fig. 12-31

and to the classical increase in the swing of the oscillator with more

excitation (energy).

... with the result 212

x (v+ )12 (mk)

=

It’s easy to calculate the mean potential energy

122V (v )1 1 1 kkx ( )

2 2 2 m= = +

Which are wrote earlier as (v )1 1 ω2 2

= +

The total energy (see solution to Schrödinger equation)

E (v )1 ω2

= +

KE V (v )1 1 ω2 2

= = +

Which is a special case of the virial theorem:

“If the potential energy has the form bV ax= , then the potential energy and kinetic energy are related by

KE b(V)2 =

Here we have b = 2 so KE V=

Rotational Motion

• Rotation in 2D

• Rotation in 3D

• Spin

Rotation in 2D

Particle mass m on a circular path radius r in the (x,y) plane.

In classical terms

V = 0 everywhere so total KE E=

2

Kp2m

E E= =

Angular momentum around z axis

zJ pr= ±

2

z2

J2mr

E =

2mr moment of inertia I= =

2

zJ2I

E =

Now applying QM methods we will show that not all values of E are allowed.

i.e. both E and I are quantized.

Qualitative Argument

z prJ = ± : hλ

p =

zhrλ

J = ±

For constructive interference the only allowed λ are

l

2πrm

λ = , ml quantum number

Fig. 12-33

Thus zm hr m hhr

λ 2πr 2πJ l l= ± = =

z 0, 1, 2...J m , ml= = ± ±

And 2

zJE2I

=2 2

2Iml=

The corresponding wave functions are l

im

m 12

e( )(2π)

φ

φ =Ψ

(See next...)

More Rigorous Argument

The Hamiltonian (with V = 0) is

2 2 2

2 2H2m x y

∂ ∂= + ∂ ∂

And the wave function is a function of the angle φ .

Switching to spherical polar coordinates (recall rx cosφ= , ry sinφ= ) we derive

2

2 2 2 2

2 2 2 21 1r rrx y r φ

= + +∂ ∂ ∂ ∂ ∂+

∂∂ ∂ ∂ ∂

But, in this case, const.r = so we discard the r∂∂

terms (= 0)

2 2

2 2H -2mr

=Φ∂∂

(note 2I mr= )

H EΨ = Ψ

So we have 2

2

22IEΨ

=Φ∂∂

Ψ

Which can be solved....

2

2

22IEΨ

= ΨΦ∂∂ General solution is;

im

m 12

e( )(2π)

l

l

φ

φ =Ψ ; 2 2

Em

2Il= (any ml )

Acceptable solutions must be single valued.

i.e. Ψ must satisfy a cyclical boundary condition

( 2π) ( )φ φ+ =Ψ Ψ

So l

l

im ( 2π)

m 12

e

(2π)( 2π)

φ

φ+

=Ψ +l lim im 2π

12

e e

(2π)

φ

=

lim

12

e( )(2π)

φ

φ =Ψ

2imπ( )( 2π) eφφ =Ψ + Ψ

( )iπe 1= − so this is equivalent to ( ) l2m1 (0)( 2π)φ −Ψ + = Ψ

And m must be 0, 1, 2...± ±

We have 2 2

lEm

2I= , l 0, 1, 2...m = ± ±

All rotational (except l 0m = ) are doubly degenerate, i.e. the energy is independent of the sense of rotation.

Angular momentum

Recall xdp

i dx∧= similarly zi

i φ∧ ∂=

∂

zi∧

angular momentum operator

l

l

mz m

di

i dφ∧ ΨΨ = im

lim ei

φ=

im

12

e

(2π)

φ Ψ =

lmlm= Ψ

So the angular momentum is also quantized z zmll = , 0, 1, 2m = ± ±

Fig. 12-40

For given lm we know zl exactly however to locate the particle we

evaluate

l l

im im

m m 1 12 2

e e*(2π) (2π)

φ φ− + =

Ψ Ψ = 12π

Probability density is independent of φ and knowing the location on

the ring is impossible.

Angular momentum and angle are complementary observables.

Rotation in 3 Dimensions

Particle mass m, free to move anywhere on the surface of a sphere radius r.

Analogously to the particle on a ring, the wave functions must now satisfy two critical boundary conditions, giving rise to two quantum numbers.

The Schrödinger equation is

2

2H2m

= ∇ ( 22 2 2

2 2 2+ +x y z

∇ =∂ ∂ ∂∂ ∂ ∂

)

The wave functions are (switching to spherical polar coordinates)

(θ, )φΨ (since r = constant)

Which can be separated and solved.

Acceptable (cyclic) solutions have 2 quantum numbers

0, 1, 2...l = , -1, -2... -m l l l l l=

For each l there are (2l+1) values of lm .

The actual form of Ψ ’s are called the “spherical harmonics” l,mlY (θ, )φ

It follows from the solution that

2

E ( +1) 2Il l= , 0,1, 2l =

And that there are (2 +1)l Ψ ’s for each energy, so the levels are

(2 +1)l - fold degenerate (one for each ml ).

Since 2J

2IE = (classically)

12( +1)J l l= ; l = 0, 1, 2 ....

And we have already seen that

zJ ml= ; ml = l, l-1, ..., -l

Space Quantization and Vector Representation

Fig. 12-52

J is represented by a vector length 12( +1)l l , orientated so that the projection

on the z axis is length ml .

The idea that only certain orientations of angular momentum vector in space

are allowed is called “space quantization”

Note: Because we have specified zJ (using ml ) we can say nothing

about xJ or yJ (the 3 operators do not commute with each other). So

the vectors should be pictured as cones with the vectors sweeping

around the z-axis.

Fig. 12-59

Spin

Stern-Gerlach experiment

Ag atoms beamed through an inhomogeneous magnetic field – (2) bands of

Ag atoms emerge (non-classical result because of space quantization).

Fig 12-55

But if angular momentum l gives rise to (2l + 1) orientations then l=12

.

A non-integer ⇒ not rotation of the electrons around the nucleus (l must be

an integer).

⇒Must be the SPIN of the electron around its own axis.

Spin angular momentum = 12s(s+1) , Sm s, s-1, ... -s=