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

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

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.)

Back to Harmonic Oscillator Spectrum and Eigenstates