Quantum Harmonic Oscillator Excited State

Relevant wiki: Schrödinger Equation

One must read off the eigenvalue corresponding to the operator:

$H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac12 m \omega^2 x^2.$

Call the given function $Cxe^{-\frac{m\omega x^2}{2\hbar}}$ . Then acting with the Hamiltonian, find:

$\begin{aligned} H\psi &= C \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac12 m \omega^2 x^2\right) xe^{-\frac{m\omega x^2}{2\hbar}} \\ &= C \frac12 m \omega^2 x^3 e^{-\frac{m\omega x^2}{2\hbar}} - C \frac{\hbar^2}{2m} \frac{d}{dx} \left(e^{-\frac{m\omega x^2}{2\hbar}} -\frac{m \omega}{\hbar} x^2 e^{-\frac{m\omega x^2}{2\hbar}} \right) \\ &=C \frac12 m \omega^2 x^3 e^{-\frac{m\omega x^2}{2\hbar}} - C \frac{\hbar^2}{2m} \left(-\frac{m\omega}{\hbar} x e^{-\frac{m\omega x^2}{2\hbar}} -\frac{m \omega}{\hbar} 2x e^{-\frac{m\omega x^2}{2\hbar}} +\left(\frac{m \omega}{\hbar}\right)^2 x^3 e^{-\frac{m\omega x^2}{2\hbar}}\right) \\ &= C \frac{\hbar^2}{2m} \frac{m \omega}{\hbar} 3x e^{-\frac{m\omega x^2}{2\hbar}} \\ &= \left(\frac{3}{2} \hbar \omega \right) \psi \end{aligned}$

The energy is thus the eigenvalue $\frac{3}{2} \hbar \omega$ .