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Asphyxion

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## Homework Statement

The wave function [tex]\psi_0 (x) = A e^{- \dfrac{x^2}{2L^2}}[/tex]

represents the ground-state of a harmonic oscillator. (a) Show that [tex]\psi_1 (x) = L \dfrac{d}{dx} \psi_0 (x)[/tex] is also a solution of Schrödinger's equation. (b) What is the energy of this new state? (c) From a look at the nodes of this wave function, how would you classify this excited state?

## Homework Equations

[tex]E_n = (n + 1/2) \hbar \omega[/tex]

SE:

[tex]-\frac{\hbar ^2}{2m} \dfrac{\partial}{\partial x} \psi + \frac{1}{2} m \omega ^2 \psi = E \psi[/tex]

## The Attempt at a Solution

[tex]\dfrac{\partial}{\partial x} \psi_0 = -\dfrac{xA}{L^2} e^{- \dfrac{x^2}{2L^2}}[/tex]

[tex]\dfrac{\partial}{\partial x} \psi_1 = - \dfrac{A}{L} e^{- \dfrac{x^2}{2L^2}} + \dfrac{x^2A}{L^3} e^{- \dfrac{x^2}{2L^2}}[/tex]

[tex]\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{xA}{L^3} \dfrac{x^2}{2L^2}} + \dfrac{2xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}[/tex]

[tex]\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{3xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}[/tex]

Put into SE this gives me:

[tex]-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 = E[/tex]

This is where I'm not getting any further. I just can't see how this is supposed to match the energy state of a harmonic oscillator. Am I thinking completely wrong, have I done the maths wrong. Or both? Any pointers would be greatly appreciated! (First time trying to post my LaTeX work on a forum, so be forgiving!)

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