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

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi_x}{\partial x^2}
+ {\textstyle\frac{1}{2}} m\omega^2 x^2 \psi_x
= E_x \psi_x
\end{displaymath}

where the spring constant $c$ was rewritten as the equivalent expression $m\omega^2$.

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 $x$-​coordinate $\xi$ and a scaled energy $\epsilon$ by

\begin{displaymath}
x\equiv\ell\xi \qquad E_x\equiv E_0\epsilon.
\end{displaymath}

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 $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{\hbar/m\omega}$. In that case both terms will have the same net coefficient $\frac12\hbar\omega$. Then if you cleverly choose $E_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar\omega$, the right hand side will have that coefficient too, and you can divide it away and end up with no coefficients at all:

\begin{displaymath}
- \frac{\partial^2\psi_x}{\partial \xi^2}
+ \xi^2 \psi_x
= \epsilon \psi_x
\end{displaymath}

Looks a lot cleaner, not?

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

\begin{displaymath}
\frac{\partial^2\psi_x}{\partial \xi^2}
\approx \xi^2 \psi_x + \ldots
\end{displaymath}

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

\begin{displaymath}
\psi_x = e^{\pm\frac12\xi^2+\ldots}
\end{displaymath}

where the dots indicate terms that are small compared to $\frac12\xi^2$ for large $\xi$. The form of the solution is important, since $e^{+\frac12\xi^2}$ becomes infinitely large at large $\xi$. That is unacceptable: the probability of finding the particle cannot become infinitely large at large $x$: the total probability of finding the particle must be one, not infinite. The only solutions that are acceptable are those that behave as $e^{-\frac12\xi^2+\ldots}$ for large $\xi$.

Now split off the leading exponential part by defining a new unknown $h(\xi)$ by

\begin{displaymath}
\psi_x \equiv e^{-\frac12\xi^2} h(\xi)
\end{displaymath}

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

\begin{displaymath}
- \frac{\partial^2 h}{\partial \xi^2}
+ 2 \xi \frac{\partial h}{\partial \xi}
+ h
= \epsilon h
\end{displaymath}

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

\begin{displaymath}
h=\sum_p c_p \xi^p
\end{displaymath}

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

\begin{displaymath}
\sum_p p (p-1) c_p \xi^{p-2} = \sum_p (2p+1-\epsilon) c_p \xi^p
\end{displaymath}

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

There must be a lowest value of $p$ for which there is a nonzero coefficient $c_p$, for if $p$ took on arbitrarily large negative values, $h$ 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 $p$ by $q$. This lowest power produces a power of $\xi^{q-2}$ in the left hand side of the equation above, but there is no corresponding power in the right hand side. So, the coefficient $q(q-1)c_q$ of $\xi^{q-2}$ will need to be zero, and that means either $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. So the power series for $h$ will need to start as either $c_0+\ldots$ or $c_1\xi+\ldots$. The constant $c_0$ or $c_1$ is allowed to have any nonzero value.

But note that the $c_q\xi^q$ term normally produces a term $(2q+1-\epsilon)c_q\xi^q$ in the right hand side of the equation above. For the left hand side to have a matching $\xi^q$ term, there will need to be a further $c_{q+2}\xi^{q+2}$ term in the power series for $h$,

\begin{displaymath}
h = c_q \xi^q + c_{q+2}\xi^{q+2} + \ldots
\end{displaymath}

where $(q+2)(q+1)c_{q+2}$ will need to equal $(2q+1-\epsilon)c_q$, so $c_{q+2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(2q+1-\epsilon)c_q$$\raisebox{.5pt}{$/$}$$(q+2)(q+1)$. This term in turn will normally produce a term $\Big(2(q+2)+1-\epsilon\Big)c_{q+2}\xi^{q+2}$ in the right hand side which will have to be canceled in the left hand side by a $c_{q+4}\xi^{q+4}$ term in the power series for $h$. And so on.

So, if the power series starts with $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the solution will take the general form

\begin{displaymath}
h = c_0 + c_2 \xi^2 + c_4 \xi^4 + c_6 \xi^6 + \ldots
\end{displaymath}

while if it starts with $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 you will get

\begin{displaymath}
h = c_1 \xi + c_3 \xi^3 + c_5 \xi^5 + c_7 \xi^7 + \ldots
\end{displaymath}

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

You can find a general formula for the coefficients of the series by making the change in notations $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2+\bar{p}$ in the left-hand-side sum:

\begin{displaymath}
\sum_{\bar p=q} (\bar p+2) (\bar p+1) c_{\bar p+2} \xi^{\bar p}
= \sum_{p=q} (2p+1-\epsilon) c_p \xi^p
\end{displaymath}

Note that you can start summing at $\bar{p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q$ rather than $q-2$, since the first term in the sum is zero anyway. Next note that you can again forget about the difference between $\bar{p}$ and $p$, 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 $p$ or $\bar{p}$.

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

\begin{displaymath}
c_{p+2} = \frac{2p+1-\epsilon}{(p+2)(p+1)} c_p
\end{displaymath}

In particular, for large $p$, by approximation

\begin{displaymath}
c_{p+2} \approx \frac{2}{p} c_p
\end{displaymath}

Now if you check out the Taylor series of $e^{\xi^2}$, (i.e. the Taylor series of $e^x$ with $x$ replaced by $\xi^2$,) you find it satisfies the exact same equation. So, normally the solution $h$ blows up something like $e^{\xi^2}$ at large $\xi$. And since $\psi_x$ was $e^{-\frac12\xi^2}h$, normally $\psi_x$ takes on the unacceptable form $e^{+\frac12\xi^2+\ldots}$. (If you must have rigor here, estimate $h$ in terms of $Ce^{\alpha\xi^2}$ where $\alpha$ 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 $p$, call it $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$, whose term in the right hand side is zero

\begin{displaymath}
0 = (2n+1-\epsilon) c_n \xi^n
\end{displaymath}

In that case, there is no need for a further $c_{n+2}\xi^{n+2}$ term, the power series will remain a polynomial of degree $n$. But note that all this requires the scaled energy $\epsilon$ to equal $2n+1$, and the actual energy $E_x$ is therefore $(2n+1)\hbar\omega$$\raisebox{.5pt}{$/$}$​2. Different choices for the power at which the series terminates produce different energies and corresponding eigenfunctions. But they are discrete, since $n$, as any power $p$, must be a nonnegative integer.

With $\epsilon$ identified as $2n+1$, you can find the ODE for $h$ 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. $c_0$ or $c_1$, demand that the total probability of finding the particle anywhere is one, $\int_{-\infty}^\infty \vert\psi_x\vert^2{\,\rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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:

\begin{displaymath}
h_n = {\displaystyle\frac{1}{\left(\pi\ell^2\right)^{1/4}}...
...sqrt{2^n n!}}}
\,e^{-\xi^2/2} \qquad n=0,1,2,3,4,5,\ldots %
\end{displaymath} (D.4)

where the details of the Hermite polynomials $H_n$ can 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 $n$ 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}.