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

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

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

$$x\equiv\ell\xi \qquad E_x\equiv E_0\epsilon.$$

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:

$$- \frac{\partial^2\psi_x}{\partial \xi^2} + \xi^2 \psi_x = \epsilon \psi_x$$

Looks a lot cleaner, not?

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

$$\frac{\partial^2\psi_x}{\partial \xi^2} \approx \xi^2 \psi_x + \ldots$$

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

$$\psi_x = e^{\pm\frac12\xi^2+\ldots}$$

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

$$\psi_x \equiv e^{-\frac12\xi^2} h(\xi)$$

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

$$- \frac{\partial^2 h}{\partial \xi^2} + 2 \xi \frac{\partial h}{\partial \xi} + h = \epsilon h$$

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

$$h=\sum_p c_p \xi^p$$

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

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

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 $h$,

$$h = c_q \xi^q + c_{q+2}\xi^{q+2} + \ldots$$

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

$$h = c_0 + c_2 \xi^2 + c_4 \xi^4 + c_6 \xi^6 + \ldots$$

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

$$h = c_1 \xi + c_3 \xi^3 + c_5 \xi^5 + c_7 \xi^7 + \ldots$$

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:

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

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

$$c_{p+2} = \frac{2p+1-\epsilon}{(p+2)(p+1)} c_p$$

In par­tic­u­lar, for large $p$, by ap­prox­i­ma­tion

$$c_{p+2} \approx \frac{2}{p} c_p$$

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

$$0 = (2n+1-\epsilon) c_n \xi^n$$

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 [41, 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 [41, 29.15].

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

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

where the de­tails of the Her­mite poly­no­mi­als $H_n$ can be found in ta­ble books like [41, 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.64}.