Phy5645/HO problem2: Difference between revisions
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In terms of the raising and lowering operators, the momentum operator is | |||
<math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\langle\ | <math>\hat{p}=-i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}-\hat{a}^{\dagger}).</math> | ||
We now take its expectation value with respect to an arbitrary eigenstate of the harmonic oscillator: | |||
<math>\langle n|\hat{p}|n\rangle=-i\sqrt{\frac{m\hbar\omega}{2}}\langle n|(\hat{a}-\hat{a}^{\dagger})|n\rangle</math> | |||
<math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\langle n|\hat{a}|n\rangle-\langle n|\hat{a}^{\dagger}|n\rangle)</math> | |||
<math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\sqrt{n}\langle | <math>=-i\sqrt{\frac{m\hbar\omega}{2}}(\sqrt{n}\langle n|n-1\rangle-\sqrt{n+1}\langle n|n+1\rangle)</math> | ||
<math>=0\!</math> | |||
A similar intuitive argument as before would lead us to expect this result, due to the fact that the wave equation has a similar form in momentum space as it does in position space. | |||
Back to [[Harmonic Oscillator Spectrum and Eigenstates]] | Back to [[Harmonic Oscillator Spectrum and Eigenstates]] | ||
Revision as of 16:45, 8 August 2013
In terms of the raising and lowering operators, the momentum operator is
We now take its expectation value with respect to an arbitrary eigenstate of the harmonic oscillator:
A similar intuitive argument as before would lead us to expect this result, due to the fact that the wave equation has a similar form in momentum space as it does in position space.