Phy5645/HO problem2

In terms of the raising and lowering operators, the momentum operator is

${\hat {p}}=-i{\sqrt {\frac {m\hbar \omega }{2}}}({\hat {a}}-{\hat {a}}^{\dagger }).$

We now take its expectation value with respect to an arbitrary eigenstate of the harmonic oscillator:

$\langle n|{\hat {p}}|n\rangle =-i{\sqrt {\frac {m\hbar \omega }{2}}}\langle n|({\hat {a}}-{\hat {a}}^{\dagger })|n\rangle$

$=-i{\sqrt {\frac {m\hbar \omega }{2}}}(\langle n|{\hat {a}}|n\rangle -\langle n|{\hat {a}}^{\dagger }|n\rangle )$

$=-i{\sqrt {\frac {m\hbar \omega }{2}}}({\sqrt {n}}\langle n|n-1\rangle -{\sqrt {n+1}}\langle n|n+1\rangle )$

$=0\!$

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.

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