Quantum Harmonic Oscillator ladder operator

[bobred]

Homework Statement

What is the effect of the sequence of ladder operators acting on the ground eigenfunction \psi_0

Homework Equations

\hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\psi_0

The Attempt at a Solution

I'm not sure if I'm right but wouldn't this sequence of opperators on the ground state result in zero?

 

[vela]

Yeah.

 

[bobred]

Thanks, I thought so. This is part of a larger question concerning the expectation value of \left\langle p^4_x \right\rangle. We are asked to show that

\left\langle p^4_x \right\rangle=\displaystyle\frac{ \hbar^4}{4 a^4}\left[ \displaystyle\int^\infty _{-\infty} \psi^*_0 \left(\hat{A}\hat{A}\hat{A}^\dagger\hat{A}^\dagger + \hat{A}\hat{A}^\dagger\hat{A}\hat{A}^\dagger + \hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\right) \psi_0 dx \right]=\displaystyle\frac{3 \hbar^4}{4 a^4}

Where \psi_0=\left( \displaystyle\frac{1}{\sqrt{\pi}a} \right)^\frac{1}{2}e^{-x^2/2a^2}

This is not what I get. Due to \hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\psi_0=0 I get \displaystyle\frac{2 \hbar^4}{4 a^4} am I missing something?
Thanks

 

[vela]

What are you getting for the other terms? It works out as it should for me. The first term should give you a 2 by itself.

 

[bobred]

As the first two sequences of operators preserve the ground eigenfunction and the last sequence zero, I get

\left\langle p^4_x \right\rangle=\displaystyle\frac{ \hbar^4}{4 a^4}\left[ 2 \displaystyle\int^\infty _{-\infty} \psi^*_0 \psi_0 dx \right]=\displaystyle\frac{2 \hbar^4}{4 a^4}

Thanks

 

[vela]

Remember that the raising and lowering operators don't simply change the state but introduce a multiplicative factor as well.

 

[bobred]

Right so I should be looking at \hat{A}\hat{A}^\dagger-\hat{A}^\dagger\hat{A}=1?

 

[bobred]

Hi

Thanks for your help what I needed was \sqrt{n}\psi_{n-1} and \sqrt{n+1}\psi_{n+1}. I now get the correct answer.