[NOTES/QM-13003] Harmonic Oscillator ---- Eigenvalues and Eigenfucntions

We shall outline the steps for deriving energy levels and wave functions for harmonic oscillator in the coordinate representation. The eigenvalue equation $$ H\psi = E\psi \,\,\, $$ for the harmonic oscillator becomes the following differential equation in coordinate representation  \begin{equation} \left({-\hbar^2\over2m} {d^2\over dq^2} + {1\over2} m\omega^2q^2\right) \psi(q) = E\psi(q) \end{equation} The main steps in solution of the eigenvalue problem in coordinate representation are as follows.

1. In terms of dimensionless variables $\xi=\alpha q$, $\lambda=2E/\hbar\omega$, where $\alpha^2=m\omega/\hbar$, the Schr\"odinger equation (1) becomes $$ {d^2\psi\over d\xi^2} + (\lambda - \xi^2)\psi = 0\,\,\,. $$
2. It can be seen that for {\it large} $\xi$ solutions to the differential equation behave as a polynomial times $e^{\pm \xi^2/2}$.
3. Define $H(\xi)$ by means of the equation $$ \psi(\xi) = H(\xi) e^{-\xi^2/2} $$ then $H(\xi)$ satisfies equation. \begin{equation} H^{\prime\prime} - 2\xi H^\prime + (\lambda-1)H = 0\,\,\,. \end{equation}
4. The above equation is well known Hermite equation and can be solved by the method of series solution. To solve the Hermite equation we write a series expansion \begin{equation} H(\xi) = \xi^c(a_o+a_1\xi+a_2\xi^2+\cdots) \end{equation} The series (3) is substituted in (1), and coefficient of each power of $\xi$ coming from the L.H.S. of (2) must be set equal to zero. This gives value of $c$ $$ c(c-1) = 0~~\Rightarrow c= 0,1 $$ and recurrence relations for the coefficients an \begin{equation} a_{n+2} = {2n+2c+1-\lambda\over(n+c+1)(n+c+2)} ~ a_n\,\,\,. \end{equation}
5. {\it For} $c=0$ all the even coefficients are proportional to $a_o$ and all the odd coefficients are proportional to $a_1$, and $a_o$ and $a_1$ are arbitrary. Thus one gets \begin{equation} H(\xi) = a_1 y_1(\xi) + a_2 y_2(\xi) \end{equation} {\it For} $c=1$ the solution for $H(\xi)$ is proportional to $y_2(\xi)$ and is already contained in (5). Hence this case, $c=1$, need not be considered separately. Note the eqn.(2) is a second order differential equation and the most general solution is a linear combination of two independent solutions $y_1(\xi)$ and $y_2(\xi)$.
6. Next we must explore large $\xi$ behaviour of (5). The relation (4) for large $n$ takes the form $$ {a_{n+2}\over a_n} \sim {2\over n}\,\,\,. $$ which coincides with the ratio of the expansion coefficients, in the series for $\exp(\xi^2)$ $$ \exp(\xi^2) = \sum {\xi^{2n}\over n!} $$ Thus the two solutions $y_1(\xi)$ and $y_2(\xi)$ behave like $\exp(\xi^2)$ for large $\xi$ and \begin{eqnarray} \psi(\xi)&=& H(\xi) e^{\-xi^2/2}\\ \stackrel{\xi\to\infty}{} & \sim & e^{\xi^2}\times e^{-\xi^2/2} = e^{\xi^2/2} \end{eqnarray} This behaviour of $\psi(\xi)$ for large $\xi$ makes the solution unacceptable because $\psi(\xi)$ would not be square integrable.
7. The only way one can get a square integrable solution for $\psi(q)$ is that the solution $H(\xi)$ must reduce to a polynomial. If $H(\xi)$ is to contain a maximum power $n$ then we must demand the following conditions.
   • [(i)] $a_{n+2}=0$\\ $\Rightarrow 2n+2c+1-\lambda=0$\\ $\lambda = 2n+1$\\ and
   • [(ii]) $a_1=0$ if $n=$ even\\ $a_o=0$ if $n=$ odd.
8. The condition $\lambda=(2n+1)$ is equivalent to the energy quantization $$ E=(n+{1\over2})\hbar\omega\,\,. $$ The wave functions are obtained by using conditions, as in (i) and (ii) above, and the recurrence relations to solve for the coefficients $a_n$. The resulting solutions for $H_n$ are Hermite polynomials and the normalized eigenfunctions are given by $$ \psi_n(q) = \left({\alpha\over\sqrt{\pi}~2^nn!}\right)^{1/2} H_n(\alpha q) \exp(-\alpha^2q^2/2) $$ These coincide with the wave functions obtained from operator methods ${1\over n!}(a^{\dagger})^n\phi_0(q)$.