Phy5645/HO problem2: Difference between revisions

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(New page: Problem 2: The expectation value of p in eigenstate. <math>\langle\Psi_n|p|\Psi_n\rangle=-i\sqrt{\frac{m\hbar\omega}{2}}\langle\Psi_n|\hat{a}-\hat{a}^{\dagger}|\Psi_n\rangle</math> <ma...)
 
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<math>\langle\Psi_n|p|\Psi_n\rangle=-i\sqrt{\frac{m\hbar\omega}{2}}\langle\Psi_n|\hat{a}-\hat{a}^{\dagger}|\Psi_n\rangle</math>
<math>\langle\Psi_n|p|\Psi_n\rangle=-i\sqrt{\frac{m\hbar\omega}{2}}\langle\Psi_n|\hat{a}-\hat{a}^{\dagger}|\Psi_n\rangle</math>


<math>==-i\sqrt{\frac{m\hbar\omega}{2}}(\langle\Psi_n|\hat{a}\Psi_n\rangle-\langle\Psi_n|\hat{a}^{\dagger}\Psi_n\rangle)</math>
<math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\langle\Psi_n|\hat{a}\Psi_n\rangle-\langle\Psi_n|\hat{a}^{\dagger}\Psi_n\rangle)</math>
        
        
<math>==-i\sqrt{\frac{m\hbar\omega}{2}}(\sqrt{n}\langle\Psi_n|\Psi_n-1\rangle-\sqrt{n+1}\langle\Psi_n|\Psi_n+1\rangle)</math>
<math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\sqrt{n}\langle\Psi_n|\Psi_n-1\rangle-\sqrt{n+1}\langle\Psi_n|\Psi_n+1\rangle)</math>


<math>=0</math>
<math>=0</math>

Revision as of 01:37, 6 December 2009

Problem 2:

The expectation value of p in eigenstate.