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 like:

${\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}}=1+2{\hat {N}}{\text{ where }}{\hat {N}}={\hat {a}}^{\dagger }{\hat {a}}$

since

${\hat {a}}{\hat {a}}^{\dagger }|n\rangle ={\hat {a}}(n+1)^{\frac {1}{2}}|n+1\rangle =(n+1)|n\rangle$

and ${\hat {N}}|n\rangle =n|n\rangle$

So, now we have that:

$\langle V\rangle ={\frac {k}{4\beta ^{2}}}\langle n|(1+2{\hat {N}}|n\rangle ={\frac {k}{4\beta ^{2}}}(2n+1)\langle n|n\rangle ={\frac {k}{2\beta ^{2}}}(n+{\frac {1}{2}})$

And, replacing $\beta ^{2}={\frac {m\omega _{0}}{\hbar }}$, we find that

$\langle V\rangle ={\frac {\hbar \omega _{0}}{2}}(n+{\frac {1}{2}})$

And can check that

$\langle T\rangle ={\frac {1}{2m}}\langle {\hat {p}}\rangle ={\frac {1}{2}}\langle E\rangle ={\frac {\hbar \omega _{0}}{2}}(n+{\frac {1}{2}})$

Which shows rather nicely that the Virial Theorem holds for the Quantum Harmonic Oscillator.

(See Liboff, Richard Introductory Quantum Mechanics, 4th Edition, Problem 7.10 for reference.)

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