D.12 The harmonic oscillator solution

If you really want to know how the harmonic oscillator wave function can be found, here it is. Read at your own risk.

The ODE (ordinary differential equation) to solve is

where the spring constant was rewritten as the equivalent expression .

Now the first thing you always want to do with this sort of problems
is to simplify it as much as possible. In particular, get rid of as
much dimensional constants as you can by rescaling the variables:
define a new scaled -coordinate and a scaled energy
by

If you make these replacements into the ODE above, you can make the coefficients of the two terms in the left hand side equal by choosing . In that case both terms will have the same net coefficient . Then if you cleverly choose , the right hand side will have that coefficient too, and you can divide it away and end up with no coefficients at all:

Looks a lot cleaner, not?

Now examine this equation for large values of (i.e. large ).
You get approximately

If you write the solution as an exponential, you can ballpark that it must take the form

where the dots indicate terms that are small compared to for large . The form of the solution is important, since becomes infinitely large at large . That is unacceptable: the probability of finding the particle cannot become infinitely large at large : the total probability of finding the particle must be one, not infinite. The only solutions that are acceptable are those that behave as for large .

Now split off the leading exponential part by defining a new unknown
by

Substituting this in the ODE and dividing out the exponential, you get:

Now try to solve this by writing as a power series, (say, a Taylor
series):

where the values of run over whatever the appropriate powers are and the are constants. If you plug this into the ODE, you get

For the two sides to be equal, they must have the same coefficient for every power of .

There must be a lowest value of for which there is a nonzero coefficient , for if took on arbitrarily large negative values, would blow up strongly at the origin, and the probability to find the particle near the origin would then be infinite. Denote the lowest value of by . This lowest power produces a power of in the left hand side of the equation above, but there is no corresponding power in the right hand side. So, the coefficient of will need to be zero, and that means either 0 or 1. So the power series for will need to start as either or . The constant or is allowed to have any nonzero value.

But note that the term normally produces a term
in the right hand side of the equation
above. For the left hand side to have a matching term, there
will need to be a further term in the power series
for ,

where will need to equal , so . This term in turn will normally produce a term in the right hand side which will have to be canceled in the left hand side by a term in the power series for . And so on.

So, if the power series starts with 0, the solution will take the
general form

while if it starts with 1 you will get

In the first case, you have a symmetric solution, one which remains the same when you flip over the sign of , and in the second case you have an antisymmetric solution, one which changes sign when you flip over the sign of .

You can find a general formula for the coefficients of the series by
making the change in notations in the left-hand-side sum:

Note that you can start summing at rather than , since the first term in the sum is zero anyway. Next note that you can again forget about the difference between and , because it is just a symbolic summation variable. The symbolic sum writes out to the exact same actual sum whether you call the symbolic summation variable or .

So for the powers in the two sides to be equal, you must have

In particular, for large , by approximation

Now if you check out the Taylor series of , (i.e. the Taylor series of with replaced by ,) you find it satisfies the exact same equation. So, normally the solution blows up something like at large . And since was , normally takes on the unacceptable form . (If you must have rigor here, estimate in terms of where is a number slightly less than one, plus a polynomial. That is enough to show unacceptability of such solutions.)

What are the options for acceptable solutions? The only possibility
is that the power series terminates. There must be a highest power
, call it , whose term in the right hand
side is zero

In that case, there is no need for a further term, the power series will remain a polynomial of degree . But note that all this requires the scaled energy to equal , and the actual energy is therefore 2. Different choices for the power at which the series terminates produce different energies and corresponding eigenfunctions. But they are discrete, since , as any power , must be a nonnegative integer.

With identified as , you can find the ODE for
listed in table books, like [40, 29.1], under the
name Hermite's differential equation.

They then
identify the polynomial solutions as the so-called “Hermite
polynomials,” except for a normalization factor. To find the
normalization factor, i.e. or , demand that the total
probability of finding the particle anywhere is one,
1. You should be able
to find the value for the appropriate integral in your table book,
like [40, 29.15].

Putting it all together, the generic expression for the eigenfunctions
can be found to be:

Hermite polynomialscan be found in table books like [40, pp. 167-168]. They are readily evaluated on a computer using the “recurrence relation” you can find there, for as far as computer round-off error allows (up to about 70.)

Quantum field theory allows a much neater way to find the eigenfunctions. It is explained in addendum {A.15.5} or equivalently in {D.65}.