Phy5645/HO Virial Theorem

Virial Theorem in the Case of the Quantum Harmonic Oscillator

Prove that the Virial Theorem holds for the Harmonic Oscillator. (Show that the average kinetic energy, $\langle T\rangle$ is equal to the average potential energy, $\langle V\rangle$.)

For the QHO, the average potential energy is written

$\langle V\rangle ={\frac {k}{2}}\langle {\hat {x}}^{2}\rangle$

It is convenient to re-write the position operator as

${\hat {x}}={\frac {1}{{\sqrt {2}}\beta }}({\hat {a}}+{\hat {a}}^{\dagger }){\text{ where }}\beta ^{2}={\frac {m\omega _{0}}{\hbar }}$

$\Rightarrow {\hat {x}}^{2}={\frac {1}{2\beta ^{2}}}({\hat {a}}+{\hat {a}}^{\dagger })^{2}$

Now, we can write the average potential for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n^{th} } state of the QHO like:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle V \rangle = \frac{k}{4\beta^2}\langle n|(\hat{a} + \hat{a}^\dagger)^2|n \rangle }

$={\frac {k}{4\beta ^{2}}}\langle n|({\hat {a}}^{2}+{\hat {a}}^{\dagger 2}+{\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}}|n\rangle$

$={\frac {k}{4\beta ^{2}}}\left[\langle n|{\hat {a}}^{2}|n\rangle +\langle n|{\hat {a}}^{\dagger }|n\rangle +\langle n|{\hat {a}}{\hat {a}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {a}}|n\rangle \right]$

Now, the first two terms disappear, as the raising and lowering operators act on the eigenkets:

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

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.)