Phy5645/HO Virial Theorem

The average potential energy is given by

$\langle {\hat {V}}\rangle ={\tfrac {1}{2}}k\langle {\hat {x}}^{2}\rangle .$

Recall from a previous problem that

${\hat {x}}={\sqrt {\frac {\hbar }{2m\omega }}}({\hat {a}}+{\hat {a}}^{\dagger }),$

or

${\tfrac {1}{2}}k{\hat {x}}^{2}={\frac {\hbar k}{4m\omega }}({\hat {a}}+{\hat {a}}^{\dagger })^{2}.$

We can now write the average potential for the $n^{\text{th}}$ state of the harmonic oscillator as

$\langle V\rangle ={\frac {\hbar k}{4m\omega }}\langle n|({\hat {a}}+{\hat {a}}^{\dagger })^{2}|n\rangle$

$={\frac {\hbar k}{4m\omega }}\langle n|({\hat {a}}^{2}+{\hat {a}}^{\dagger 2}+{\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}})|n\rangle$

$={\frac {\hbar k}{4m\omega }}[\langle n|{\hat {a}}^{2}|n\rangle +\langle n|{\hat {a}}^{\dagger }|n\rangle +\langle n|({\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}})|n\rangle ]$

The first two terms are zero because

$\langle n|n-2\rangle =\langle n|n+2\rangle =0$

and the operator in the third term can be written as

${\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}}=2{\hat {n}}+1.$

Therefore,

$\langle {\hat {V}}\rangle ={\frac {\hbar k}{4m\omega }}(2n+1)\langle n|n\rangle ={\frac {\hbar k}{2m\omega }}\left(n+{\tfrac {1}{2}}\right),$

or, noting that $k=m\omega ^{2},\!$

$\langle {\hat {V}}\rangle ={\tfrac {1}{2}}\left(n+{\tfrac {1}{2}}\right)\hbar \omega .$

Similarly, using the fact that

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

we may show that

$\langle {\hat {T}}\rangle ={\frac {\langle {\hat {p}}^{2}\rangle }{2m}}={\tfrac {1}{2}}\left(n+{\tfrac {1}{2}}\right)\hbar \omega =\langle {\hat {V}}\rangle .$

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