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The Simple Harmonic OscillatorMichael Fowler 11/13/06

Einsteins Solution of the Specific Heat Puzzle

The simple harmonic oscillator, a nonrelativistic particle in a potential 212

,kx is an

excellent model for a wide range of systems in nature. In fact, not long after Plancks

discovery that the black body radiation spectrum could be explained by assuming energy

to be exchanged in quanta, Einstein applied the same principle to the simple harmonic

oscillator, thereby solving a long-standing puzzle in solid state physicsthe mysterious

drop in specific heatof all solids at low temperatures. Classical thermodynamics, a very

successful theory in many ways, predicted no such dropwith the standard equipartition

of energy, kTin each mode (potential plus kinetic), the specific heat should remain more

or less constant as the temperature was lowered (assuming no phase change).

To explain the anomalous low temperature behavior, Einstein assumed each atom to be anindependent (quantum) simple harmonic oscillator, and, just as for black body radiation,

he assumed the oscillators could only absorb or emit energy in quanta. Consequently, at

low enough temperatures there is rarely sufficient energy in the ambient thermal

excitations to excite the oscillators, and they freeze out, just as blue oscillators do in low

temperature black body radiation. Einsteins picture was later somewhat refinedthe

basic set of oscillators was taken to be standing sound wave oscillations in the solid rather

than individual atoms (making the picture even more like black body radiation in a cavity)

but the main conclusionthe drop off in specific heat at low temperatureswas not

affected.

The Classical Simple Harmonic OscillatorThe classical equation of motion for a one-dimensional simple harmonic oscillator with a

particle of mass m attached to a spring having spring constant kis

2

2.

d xm kx

dt= -

The solution is

0sin( ), ,

k x x t

mw d w= + =

and the momentum p = mv has time dependence

0cos( ). p mx t w w d= +

The total energy2 2 2 2(1/ 2 )( )m p m x E w+ =

is clearly constant in time.


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It is often useful to picture the time-development of a system in phase space, in this case a

two-dimensional plot with position on the x-axis, momentum on the y-axis. Actually, to

have ( ),x y coordinates with the same dimensions, we use ( ), .m x pwIt is evident from the above expression for the total energy that in these variables the point

representing the system in phase space moves clockwise around a circle of radius 2mE

centered at the origin.

Note that in the classicalproblem we could choose any point ( ), ,m x pw place the systemthere and it would then move in a circle about the origin. In the quantum problem, on the

other hand, we cannot specify the initial coordinates ( ),m x pw precisely, because of theuncertainly principle. The best we can do is to place the system initially in a small cell in

phase space, of size / 2x pD D = h . In fact, we shall find that in quantum mechanics phase

space is always divided into cells of essentially this size for each pair of variables.

Schrdingers Equation and the Ground State Wave Function

From the classical expression for total energy given above, the Schrdinger equation for

the quantum oscillator follows in standard fashion:

2 22 2

2

( ) 1( ) ( )

2 2

d xm x x E x

m dx

yw y y- + =

h

What will the solutions to this Schrdinger equation look like? Since the potential2 21

2m xw increases without limit on going away from x = 0, it follows that no matter how

much kinetic energy the particle has, for sufficiently large x the potential energydominates, and the (bound state) wavefunction decays with increasing rapidity for further

increase in x. (Obviously, for a real physical oscillator there is a limit on the height of the

potentialwe will assume that limit is much greater than the energies of interest in our

problem.)

We know that when a particle penetrates a barrier of constant height V0 (greater than the

particles kinetic energy) the wave function decreases exponentially into the barrier, as

e x-a , where a= -2 02m V E( ) / h . But, in contrast to this constant height barrier, the

height of the simple harmonic oscillator potential continues to increase as the particle

penetrates to largerx. Obviously, in this situation the decay will be faster than

exponential. If we (rather navely) assume it is more or less locally exponential, but with a

local a varying with V0, neglecting Erelative to V0 in the expression fora suggests that

a itself is proportional to x (since the potential is proportional to x2, and Va ) so

maybe the wavefunction decays as ex-( )constant 2 ?

To check this idea, we insert2 2

/ 2( ) x bx ey -= in the Schrdinger equation, using


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2 2

2 2 4

1d x

dx b b

yy y= - +

to find

2 22 2

2 4

1 1( ) ( ) ( )

2 2

x x m x x E x

m b by w y y

- - + + =

h.

The ( )xy is just a factor here, and it is never zero, so can be cancelled out. This leaves aquadratic expression which must have the same coefficients ofx

0, x

2on the two sides, that

is, the coefficient ofx2

on the left hand side must be zero:

2 2

4, so

2 2

mb

mb m

w

w

= =h h

.

This fixes the wave function. Equating the constant terms fixes the energy:

2

2

1

2 2E

mbw= =

hh .

So the conjectured form for the wave function is in fact the exactsolution for the lowest

energy state! (Its the lowest state because it has no nodes.)

Also note that even in this ground state the energy is nonzero, just as it was for the square

well. The central part of the wave function must have some curvature to join together the

decreasing wave function on the left to that on the right. This zero point energy is

sufficient in one physical case to melt the latticehelium is liquid even down to absolute

zero temperature (checked down to microkelvins!) because the wave function spread

destabilizes the solid lattice that will form with sufficient external pressure.

Higher Energy States

It is clear from the above discussion of the ground state that bmw

=h

is the natural unit

of length in this problem, and wh that of energy, so to investigate higher energy states we

reformulate in dimensionless variables,

, x m E

xb

wx e

w= = =

h h.

Schrdingers equation becomes


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22

2

( )( 2 ) ( )

d

d

y xx e y x

x= - .

Deep in the barrier, the e term will become negligible, and just as for the ground state

wave function, higher bound state wave functions will have

2/ 2

e

x-

behavior, multiplied bysome more slowly varying factor (it turns out to be a polynomial).

Exercise: find the relative contributions to the second derivative from the two terms in2

/ 2 .n x

x e- For given n, when do the contributions involving the first term become small?Define small.

The standard approach to solving the general problem is to factor out the2

/ 2ex- term,

2/ 2( ) ( )h e xy x x -=

giving a differential equation for ( )h x :2

22 (2 1) 0

d h dhh

d dx e

x x- + - =

We try solving this with a power series in :x

2

0 1 2( ) ...h h h hx x x= + + = .

Inserting this in the differential equation, and requiring that the coefficient of each powernx vanish identically, leads to a recurrence formula for the coefficients hn:

2

(2 1 2 )

( 1)( 2)n n

nh h

n n

e+

+ -=

+ +.

Evidently, the series of odd powers and that of even powers are independent solutions to

Schrdingers equation. (Actually this isnt surprising: the potential is even in x, so the

parity operatorPcommutes with the Hamiltonian. Therefore, unless states are degenerate

in energy, the wave functions will be even or odd in x.) For large n, the recurrencerelation simplifies to

2

2, .

n nh h n

ne+ ?

The series therefore tends to


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( )

( )2

2 1222

2 .(2 2)(2 4)...2 1 !

nn n

en n n

xx xx

-

= =- - -

Multiply this by the2

/ 2e x- factor to recover the full wavefunction, we findy diverges for

large xas2

/ 2e x+ .

Actually we should have expected thisfor a general value of the energy, the Schrdinger

equation has the solution2 2

/ 2 / 2 Ae Bex x+ - + at large distances, and only at certain

energies does the coefficient A vanish to give a normalizable bound state wavefunction.

So how do we find the nondivergingsolutions? It is clear that the infinite power series

must be stopped! The key is in the recurrence relation.

If the energy satisfies

2 2 1, an integer,n ne= +

then hn+2 and all higher coefficients vanish.

This requirement in fact completely determines the polynomial(except for an overall

constant) because with 2 2 1ne= + the coefficients hm form < n are determined by

( )2

(2 1 2 1 )(2 1 2 )

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

m nmh h h

m m m m

e+

+ - ++ -= =

+ + + +.

This nth

order polynomial is called a Hermite polynomialand written ( ) .nH x The

standard normalization of the Hermite polynomials ( )nH x is to take the coefficient of thehighest power nx to be 2n

. The other coefficients then follow using the recurrence

relation above, giving:

2 3

0 1 2 3( ) 1, ( ) 2 , ( ) 4 2, ( ) 8 12 , etc. H H H H x x x x x x x x = = = - = -

So the bottom line is that the wavefunction for the nth

excited state, having energy

1

2

ne= + , is2 / 2

( ) ( )n n n

C H ex

y x x -= , where Cn is a normalization constant to be

determined in the next section.

It can be shown (see exercises at the end of this lecture) that ( ) ( )12n n H nH x x- = . Usingthis, beginning with the ground state, one can easily convince oneself that the successive

energy eigenstates each have one more nodethe nth

state has n nodes. This is also


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(Weve expressed a in terms of the original variables x, p for later use.)

From the commutation relation[ ], 1ip x = it follows that[ , ] 1.a a =

Therefore the Hamiltonian can be written:

1 1, where .

2 2 H a a N N a aw w

= + = + =

h h

Note that the operatorNcan only have non-negative eigenvalues, since

0.a a

N a ay y y y y y = =

Now

, ,N a a aa a a a a a a a = - = =

Suppose Nhas an eigenfunction n with eigenvalue n,

.Nn n n=

From the two equations above

( ) 1 Na a N a an n n n n = + = +so a n is an eigenfunction ofNwith eigenvalue 1.n+ Operating with a again and

again, we climb an infinite ladder of eigenstates equally spaced in energy.

a is often termed a creation operator, since the quantum of energy wh added each timeit operates is equivalent to an added photon in black body radiation (electromagnetic

oscillations in a cavity).

It is easy to check that the state a n is an eigenstate with eigenvalue 1,n- provided it is

nonzero, so the operatora takes us down the ladder. However, this cannot go on

indefinitelywe have established that Ncannot have negative eigenvalues. We must


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eventually reach a state for which 0,an n = the operatora annihilates the state. (At

each step down, a annihilates one quantum of energyso a is often called an annihilation

ordestruction operator.)

Since the norm squared of ,a n

2

,a a a N n n n n n n n n = = =

and since0n n > for any nonvanishing state, it must be that the lowest eigenstate (the

for which 0an n = ) has 0.n= It follows that the ns on the ladder are the positive

integers, so from this point on we relabel the eigenstates with n in place of .n

That is to say, we have proved that the only possible eigenvalues ofNare zero and the

positive integers: 0, 1, 2, 3 . Nis called the number operator: it measures the number

of quanta of energy in the oscillator above the irreducible ground state energy (that is,

above the zero-point energy arising from the wave-like nature of the particle).

Since from above the Hamiltonian

1 1

2 2 H a a N w w

= + = +

h h

the energy eigenvalues are

( )12 . H n n nw= + hIt is important to appreciate that Diracs factorization trick and very little effort has given

us allthe eigenvalues of the Hamiltonian

( )2 2 .2

Hw

p x= +h

Contrast the work needed in this section with that in the standard Schrdinger approach.

We have also established that the lowest energy state 0 , having energy 12

,wh must

satisfy the first-order differential equation 0 0,a = that is,

0( ) | 0 ( ) 0.d

i

d

x p x y x

x

+ > = + =

The solution, unnormalized, is2 / 2

0 ( ) .Cex

y x -=


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so

0 1 0 0 0 1 0 0

1 0 2 0 1 0 2 0

/ 2 , / 2 .0 2 0 3 0 2 0 3

0 0 3 0 0 0 3 0

x m p i mw w

-

- = = -

K K

K K

h hK K

K K

M M M M O M M M M O

These matrices are, of course, Hermitian (not forgetting the i factor in p).

To find the matrix elements between eigenstates of any product ofxs and ps, express all

the xs and ps in terms ofas and a s, to give a sum of products ofas and a s. Each

product in this sum can be evaluated sequentially from the right, because each a or a has

only one nonzero matrix element when the product operates on one eigenstate.

Normalizing the Eigenstates in x-space

The normalized ground state wave function is

14

2 2/ 2 / 2

0( ) ,

m xmCe e

x wwy x

p- - = =

h

h

where we have gone back to the x variable, and normalized using2

/ax

e dx ap

-

-= .

To find the normalized wave functions for the higher states, they are first constructed

formally by applying the creation operator a repeatedly on the ground state 0 . Next,

the result is translated into x-space (actually /x bx= ) by writing a as a differential

operator, acting on ( )0 .y xUsing

1 ,n a n n- =

( )1 0 .

!

n

aan n

n n

= - = =K

Now

( )( ) ( )( ) 1/ 2 1/ 2 / ,a i d d x p x x = - = -so


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( ) 14 2/ 21 1( ) 0 .

! ! 2

n n

n

a d me

dn n

xwy x xx p

- = = - h

We need to check that this expression is indeed the same as the Hermite polynomial wave

function derived earlier, and to do that we need some further properties of the Hermitepolynomials.

Some Properties of Hermite Polynomials

The mathematicians define the Hermite polynomials by:

2 2

( ) ( )n

n

n n

d H e e

d

x xx

x-= -

so2 3

0 1 2 3( ) 1, ( ) 2 , ( ) 4 2, ( ) 8 12 , etc. H H H H x x x x x x x x = = = - = -

It follows immediately from the definition that the coefficient of the leading power is 2n.

It is a straightforward exercise to check that Hn is a solution of the differential equation

2

22 2 ( ) 0,n

d dn H

d dx x

x x

- + =

so these are indeed the same polynomials we found by the series solution of Schrdingers

equation earlier (recall the equation for the polynomial component of the wave function

was

2

22 (2 1) 0

d h dhh

d dx e

x x- + - = ,

with 2 2 1ne= + .)

We have found ( )n

y x in the form

14

2/ 21 1

( ) ! 2

n

n

d m

edn

xw

y x x x p-

= - h .

We shall now prove that the polynomial component is exactly equivalent to the Hermite

polynomial as defined at the beginning of this section.

We begin with the operator identity:


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2 2/ 2 / 2d de e

d d

x xx

x x- - = -

Both sides of this expression are to be regarded as operators, that is, it is assumed that

both are operating on some function( )

f x .Now take the n

th power of both sides: on the right, we find, for example,

( )

( )

2 2 2 2 2 2 2 2

2 2

3

3/ 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2

33 / 2 / 2

3

d d d d e e e e e e e e

d d d d

de e

d

x x x x x x x x

x x

x x x x

x

- - - -

-

- = -

= -

since the intermediate exponential terms cancel against each other.

So:

2 2/ 2 / 2( )

nn

n

n

d de e

d d

x xxx x

- - = -

and substituting this into the expression for ( )n

y x above,

( ) ( )

( )

( )

2 2 2

2 2 2

2

1/ 4

/ 2 / 2 / 2

1/ 4

/ 2

1/ 4

/ 2

1

2 !

1

2 !

1, with .

2 !

nn

n nn

nn

nn

nn

d me e e

dn

m de e e

dn

m m H e x

n

x x x

x x x

x

wy x

x p

w

p x

w wx x

p

- -

- -

-

= -

= -

= =

h

h

h h

This established the equivalence of the two approaches to Schrdingers equation for the

simple harmonic oscillator, and provides us with the overall normalization constants

without doing integrals. (The expression for ( )ny x above satisfies 2| | 1n dxy = .)

Exercises:

Use ( ) ( ) 2 2nnn nd H e ed

x xxx

-= - to prove:

(a) the coefficient of nx is 2n.


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(b) ( ) ( )12n n H nH x x- =(c) ( ) ( ) ( )1 12 2n n n H H nH x x x x + -= -(d) ( )2 2 2 !nne H d nx x x p -

-=

(Hint: rewrite as ( )( ) 2nnn nd H e d d xx xx -- - , then integrate by parts n times, and use (a).)(e) ( ) ( )2 0, for .n me H H d m nx x x x -

-=

Its worth doing these exercises to become more familiar with the Hermite polynomials,

but in evaluating matrix elements (and indeed in establishing some of these results) it is

almost always far simpler to work with the creation and annihilation operators.

Exercise: use the creation and annihilation operators to find 4n x n . This matrix

element is useful in estimating the energy change arising on adding a small nonharmonic potential energy term to a harmonic oscillator.

Time-Dependent Wave Functions

The set of normalized eigenstates 0 , 1 , 2 , nK K discussed above are of course

solutions to the time-independentSchrdinger equation, or in ket notation eigenstates of

the Hamiltonian ( )12 . H n n nw= + h Putting in the time-dependence explicitly,( )12/, , 0 i n tiHtn t e n t e nw- +-= = =h . It is necessary to include the time dependence when

dealing with a state which is a superposition of states of different energies, such as

( ) ( )

1/ 2 0 1 ,+

which then becomes

( )( )/ 2 3 / 2

1/ 2 0 1 .

i t i t

e e

w w- -+ Expectation values

of combinations of position and/or momentum operators in such states are best evaluated

by expressing everything in terms of annihilation and creation operators.

Solving Schrdingers Equation in Momentum Space

In the lecture on Function Spaces, we established that the basis of x states (eigenstates

of the position operator) and that of k states (eigenstates of the momentum operator)

were both complete bases in Hilbert space (physicists definition) so we could work

equally well with either from a formal point of view. Why then do we almost always work

in x-space? Well, probably because we live in x-space, but theres another reason. The

momentum operator in the x-space representation is / p i d dx= - h , so Schrdingers

equation, written ( )( ) ( ) ( )2 / 2p m V x x E xy y+ = , with p in operator form, is a second-order differential equation. Now consider what happens to Schrdingers equation if we

work in p-space. Since the operator identity[ ], x p i= h is true regardless ofrepresentation, we must have / x i d dp= h . So for a particle in a potential ( )V x , writingSchrdingers equation in p-space we are confronted with the nasty looking operator


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( )/V i d dph ! This will produce a differential equation in general a lot harder to solve thanthe standard x-space equationso we stay in x-space.

Butthere are two potentials that can be handled in momentum space: first, for a linear

potential ( )V x Fx= - , the momentum space analysis is actually easierits just a first-order equation. Second, for a particle in a quadratic potentiala simple harmonic

oscillatorthe two approaches yield the same differential equation. That means that the

eigenfunctions in momentum space (scaled appropriately) must be identicalto those in

position spacethe simple harmonic eigenfunctions are their own Fourier transforms!

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