*Michael Fowler*

*University of Virginia*

The simple harmonic oscillator, a nonrelativistic particle
in a potential ½*Cx*^{2}, 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, ½*kT* 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.

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 ½*m**w ^{2} *

We know that when a particle penetrates a barrier of
constant height *V*_{0} (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 *V*_{0},
neglecting *E* relative *V*_{0} 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 *x*^{0},
*x*^{2} on the two sides, that is, the coefficient of *x*^{2}
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.

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

_{}

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 2*e*
= 2*n* + 1, with *n* an integer, *h _{n}*

The standard normalization of the Hermite polynomials *H _{n}*(

_{}

So the bottom line is that the wavefunction for the *n*^{th}
excited state, having energy _{}, is _{}, where *C _{n}* is a normalization constant to
be determined in the next section.

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 *a**y* =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 *n*^{th} excited state, having energy (*n*
+ ½) in dimensionless units, is

_{},

where *C _{n}* is a normalization constant.

To find the normalization constant *C*_{n} ,
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 *C*_{n}’s real and positive.

In terms of the normalized wavefunctions *y*_{n}(*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 *n*^{th} order Hermite
polynomial. It is easy to check that
the coefficient of *x ^{ n}* in

It follows that the correctly normalized (in *x*) wavefunction
for the *n*^{th} excited state is

_{}

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 *y*_{1}(*x*,*t*)
and *y*_{2}(*x*,*t*)
are solutions, so is any linear combination *A**y*_{1}+*B**y*_{2}.
Assuming *y*_{1}
and *y*_{2}
are definite energy solutions for different energies *E*_{1} and *E*_{2},
the combination will not correspond to a definite energy—a measurement of the
energy will give either *E*_{1}
or *E*_{2}, 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!