Solving the HO Differential Equation *

The differential equation for the 1D Harmonic Oscillator is.

$$\bgroup\color{black} {-\hbar^2\over 2m}{d^2u\over dx^2}+{1\over 2}m\omega^2x^2u = Eu .\egroup$$

By working with dimensionless variables and constants, we can see the basic equation and minimize the clutter. We use the energy in terms of $\hbar\omega$.

$$\bgroup\color{black} \epsilon={2E\over \hbar\omega} \egroup$$

We define a dimensionless coordinate.

$$\bgroup\color{black} y=\sqrt{m\omega\over\hbar}x \egroup$$

The equation becomes.

$$\begin{eqnarray*} {d^2u\over dx^2}+{2m\over\hbar^2}(E-{1\over 2}m\omega^2x^2)u=0 \\ {d^2u\over dy^2}+(\epsilon-y^2)u=0\\ \end{eqnarray*}$$

(Its probably easiest to just check the above equation by substituting as below.

$$\begin{eqnarray*} {\hbar\over m\omega}{d^2u\over dx^2}+\left({2E\over\hbar\omega... ...^2u\over dx^2}+{2m\over\hbar^2}(E-{1\over 2}m\omega^2x^2)u=0 \\ \end{eqnarray*}$$

It works.)

Now we want to find the solution for $y\rightarrow \infty$.

$$\bgroup\color{black} {d^2u\over dy^2}+(\epsilon-y^2)u=0\egroup$$

becomes

$$\bgroup\color{black} {d^2u\over dy^2}-y^2u=0\egroup$$

which has the solution (in the large $y$ limit)

$$\bgroup\color{black} u=e^{-y^2/2} .\egroup$$

This exponential will dominate a polynomial as $y\rightarrow \infty$ so we can write our general solution as

$$\bgroup\color{black} u(y)=h(y)e^{-y^2/2} \egroup$$

where $h(y)$ is a polynomial.

Take the differential equation

$$\bgroup\color{black} {d^2u\over dy^2}+(\epsilon-y^2)u=0\egroup$$

and plug

$$\bgroup\color{black} u(y)=h(y)e^{-y^2/2} \egroup$$

into it to get

$$\begin{eqnarray*} {d^2\over dy^2}h(y)e^{-y^2/2}+\epsilon h(y)e^{-y^2/2}-y^2h(y)e... ...\\ {d^2h(y)\over dy^2}-2y{dh(y)\over dy}+(\epsilon-1)h(y)=0 \\ \end{eqnarray*}$$

This is our differential equation for the polynomial $h(y)$.

Write $h(y)$ as a sum of terms.

$$\bgroup\color{black} h(y)=\sum\limits_{m=0}^\infty a_m y^m \egroup$$

Plug it into the differential equation.

$$\bgroup\color{black}\sum\limits_{m=0}^\infty[ a_m (m)(m-1)y^{m-2}-2a_m(m)y^m+(\epsilon-1)a_my^m ]=0\egroup$$

We now want ot shift terms in the sum so that we see the coefficient of $y^m$. To do this, we will shift the term $a_m (m)(m-1)y^{m-2}$ down two steps in the sum. It will now show up as $a_{m+2} (m+2)(m+1)y^m$.

$$\bgroup\color{black} \sum\limits_{m=0}^\infty[ a_{m+2} (m+2)(m+1)-2a_m(m)+(\epsilon-1)a_m ]y^m=0\egroup$$

(Note that in doing this shift the first term for $m=0$ and for $m=1$ get shifted out of the sum. This is OK since $a_m (m)(m-1)y^{m-2}$ is zero for $m=0$ or $m=1$.)

For the sum to be zero for all $y$, each coefficient of $y^m$ must be zero.

$$\bgroup\color{black} a_{m+2} (m+2)(m+1)+(\epsilon-1-2m)a_m=0 \egroup$$

Solve for $a_{m+2}$

$$\bgroup\color{black} a_{m+2}={2m+1-\epsilon\over (m+1)(m+2)}a_m \egroup$$

and we have a recursion relation giving us our polynomial.

But, lets see what we have. For large $m$,

$$\bgroup\color{black} a_{m+2}={2m+1-\epsilon\over (m+1)(m+2)}a_m\rightarrow {2\over m} a_m \egroup$$

The series for

$$\bgroup\color{black} y^2e^{y^2/2}=\sum{y^{2n+2}\over 2^nn!} \egroup$$

has the coefficient of $y^{2n+2}$ equal to ${1\over 2^nn!}$ and the coefficient of $y^{2n}$ equal to ${1\over 2^{n-1}(n-1)!}$. If $m=2n$,

$$\bgroup\color{black} a_{m+2}={1\over 2n}a_m={1\over m}a_m .\egroup$$

So our polynomial solution will approach $y^2e^{y^2/2}$ and our overall solution will not be normalizable. (Remember $u(y)=h(y)e^{-y^2/2}$.) We must avoid this.

We can avoid the problem if the series terminates and does not go on to infinite $m$.

$$\bgroup\color{black} a_{m+2}={2m+1-\epsilon\over (m+1)(m+2)}a_m \egroup$$

The series will terminate if

$$\bgroup\color{black} \epsilon=2n+1 \egroup$$

for some value of $n$. Then the last term in the series will be of order $n$.

$$\bgroup\color{black} a_{n+2}={0\over (n+1)(n+2)}a_n=0 \egroup$$

The acceptable solutions then satisfy the requirement

$$\begin{eqnarray*} \epsilon={2E\over \hbar\omega}=2n+1 \\ E={(2n+1)\over 2}\hbar\omega=\left(n+{1\over 2}\right)\hbar\omega \\ \end{eqnarray*}$$

Again, we get quantized energies when we satisfy the boundary conditions at infinity.

The ground state wavefunction is particularly simple, having only one term.

$$\bgroup\color{black} u_0(x)=a_0e^{-y^2\over 2}=a_0 e^{-m\omega x^2/2\hbar} \egroup$$

Lets find $a_0$ by normalizing the wavefunction.

$$\begin{eqnarray*} \int\limits_{-\infty}^\infty\vert a_0\vert^2 e^{-m\omega x^2/\... ...mega\over\pi\hbar}\right)^{1\over 4} e^{-m\omega x^2/2\hbar} \\ \end{eqnarray*}$$