# The Simple Harmonic Oscillator

Michael Fowler

University of Virginia

### Einstein’s Solution of the Specific Heat Puzzle

The simple harmonic oscillator, a nonrelativistic particle in a potential ½Cx2, is an excellent model for a wide range of systems in nature.  Indeed, it was for this system that quantum mechanics was first formulated: the blackbody radiation formula of Planck.  A little later, Einstein demonstrated that the quantum simple harmonic oscillator resolved a long-standing puzzle in solid state physics—the mysterious drop in specific heat of all solids at low temperatures. Classical thermodynamics, a very successful theory in many ways, predicted no such drop—with the standard equipartition of energy,  ½kT in each mode, 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 an independent (quantum) simple harmonic oscillator, and, just as we discussed for black body radiation, these oscillators can 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.  Einstein’s picture was later somewhat refined—the 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 conclusion—the drop off in specific heat at low temperatures—was not affected.

### Schrödinger’s Equation and the Ground State Wave Function

The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass m attached to a spring having spring constant C is

.

The solution is

,

It is convenient to express the spring constant in terms of the oscillator frequency w, so the classical equation becomes

with the corresponding Schrödinger equation:

What will the solutions to this Schrödinger equation look like?  Since the potential ½mw2 x2 increases without limit on going away from x = 0, no matter how much kinetic energy we give the particle, if it gets far enough from the origin the potential energy dominates, and the (bound state) wavefunction will decay increasing rapidly as x increases further.  (Obviously, for a real physical oscillator there is a limit on the height of the potential—we will assume that limit is much greater than the energies in our problem.)

We know that when a particle penetrates a barrier of constant height V0  (greater than the particle’s kinetic energy) the wave function decreases exponentially into the barrier, as , where .   But, in contrast to this constant height barrier, the “height” of the simple harmonic oscillator potential continues to increase as the particle penetrates to larger x.  Obviously, in this situation the decay will be faster than exponential.  If we (rather naïvely) assume it is more or less locally exponential, but with a local a varying with V0, neglecting E relative V0  in the expression for a suggests that a itself is proportional to x, so maybe the wavefunction decays as ?

To check this idea, we insert  in the Schrödinger equation, and use

to find

The y(x) is just a factor here, and it is never zero, so can be cancelled out.  This leaves a quadratic expression which must have the same coefficients of x0, x2 on the two sides, that is, the coefficient of x2 on the left hand side must be zero:

.

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

.

So the conjectured form for the wave function is in fact the exact solution for the lowest energy state!  (It’s 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 lattice—helium 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    is the natural unit of length in this problem, and  that of energy, so to investigate higher energy states we reformulate in dimensionless variables,

.

Schrödinger’s equation becomes

.

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 be approximately of the form . (Of course, the equation itself will have the alternate diverging solution  except for special values of the energy.)

The standard approach to solving the general problem is to factor out the  term,

giving the differential equation for h(x):

We try solving this with a power series in x :    .   Inserting this in the differential equation, and requiring that the coefficient of each power x n  vanish identically, leads to a recurrence formula for the coefficients hn:

Evidently, the series of odd powers and that of even powers are independent solutions to Schrödinger’s equation.  For large n >>e, the recurrence relation simplifies to .  The series therefore tends to

.

Multiply this by the  factor to recover the full wavefunction, we find it diverges as .

Actually we should have expected this—for a general value of the energy, the Schrödinger equation has the solution  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 nondiverging solutions?  It is clear that the infinite power series must be stopped!  The key is in the recurrence relation: if the energy satisfies 2e = 2n + 1, with n an integer, hn+2 and all higher coefficients vanish. The remaining nth order polynomial is called a Hermite polynomial and written Hn(x).

The standard normalization of the Hermite polynomials Hn(x) is to take the coefficient of the highest power xn to be 2n .  The other coefficients are easy to find using the recurrence relation above, so:

So the bottom line is that the wavefunction for the nth excited state, having energy , is , where Cn is a normalization constant to be determined in the next section.

### Differential Operator Approach to the Simple Harmonic Oscillator

Actually, there is an easier way to find the energy levels of the quantum oscillator than the standard Schrödinger method given above.  Continuing with the same dimensionless variables, we introduce the differential operators

These operators only have meaning if they are operating on a function. It is crucial to note that if a product of these operators operates on a function, the  terms in each operator also operate on any ’s (in other operators) to their right. In particular, this means that the two operators above do not commute, in fact

Schrödinger’s equation

can be written in terms of these differential operators:

Exercise: check that this is true.

This leads to a surprisingly easy way to find all the solutions to the equation: suppose y(x) is a solution with energy e.

Multiplying the equation by ,

Now

Applying the operator gives the whole series of bound states, with energy increasing by one unit at each step, alternating between even and odd states.  The operator a, on the other hand, goes down the ladder: operating on a state of energy e, it generates a state having energy e - 1.  It must stop at the ground state, so that wave function is a solution of ay =0, or

This gives directly

for the ground state wavefunction, where we have added the normalization constant to give

.

Note that the appropriately normalized solution in standard (not dimensionless) units is

The nth excited state, having energy (n + ½) in dimensionless units, is

,

where Cn is a normalization constant.

To find the normalization constant Cn , which is set by requiring by , we first observe that .

Therefore

The last step can be established with an elementary integration by parts (or in vector space notation by observing that  is the adjoint of ).

The integral on the right is easily evaluated :

Therefore

,

taking the Cn’s real and positive.

In terms of the normalized wavefunctions yn(x), then, we have the important formula:

.

Exercise: find the corresponding formula for a.

The final point in normalizing the bound state is to express the function in terms of the nth order Hermite polynomial.  It is easy to check that the coefficient of x n  in   is 2n/2, and by definition that coefficient in the Hermite polynomial is 2n.

It follows that the correctly normalized (in x) wavefunction for the nth excited state is

### Time Dependent States of the Simple Harmonic Oscillator

Working with the time independent Schrödinger equation, as we have in the above, implies that we are restricting ourselves to solutions of the full Schrödinger equation which have a particularly simple time dependence, an overall phase factor , and are states of definite energy E.  However, the full time dependent Schrödinger equation is a linear equation, so if y1(x,t) and y2(x,t) are solutions, so is any linear combination Ay1+By2.  Assuming y1 and y2 are definite energy solutions for different energies E1 and E2, the combination will not correspond to a definite energy—a measurement of the energy will give either E1 or E2, with appropriate probabilities.  This is still a perfectly good, physically realizable wave function.

It is instructive to examine a combination state of this form a little more closely.  We know that for the ground state wave function,

and for the first excited state,

.

Suppose we simply add terms of this type together (neglecting the overall normalization constant for now), for example

.

Looking at this wave function for t = 0, we notice that the two terms have the same sign for x > 0, and opposite signs for x < 0.  Therefore, sketching the probability distribution for the particle’s position, it is heavily skewed to the right (positive x).  However, the two terms have different time-dependent phases, differing by a factor , so after time has elapsed, a factor of -1 has evolved between the terms.  If we now look at the probability distribution |y|2, it will be skewed to the left.  In other words, if the state is not of definite energy, the probability distribution can vary in time.  Of course, the total probability of finding the particle somewhere stays the same. Note that the probability distribution swings back and forth with the period of the oscillator.  This discussion also implies that an ordinary pendulum, which clearly swings back and forth, cannot be in a state of definite energy!