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, 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 T \rangle } is equal to the average potential energy, 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 } .)
For the QHO, the average potential energy is written
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}{2}\langle \hat{x}^2 \rangle }
It is convenient to re-write the position operator as
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 \hat{x} = \frac{1}{\sqrt{2}\beta}(\hat{a} + \hat{a}^\dagger) \text { where } \beta^2 = \frac{m\omega_0}{\hbar} }
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 \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 }
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 = \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 }
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 = \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:
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 n|n-2 \rangle = \langle n|n+2 \rangle = 0 }
and the operator in the third term can be written 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 \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} = 1 + 2\hat{N} \text{ where } \hat{N} = \hat{a}^\dagger\hat{a} }
since
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 \hat{a}\hat{a}^\dagger |n \rangle = \hat{a} (n+1)^{\frac{1}{2}}|n + 1 \rangle = (n+1)|n \rangle }
and 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 \hat{N}|n \rangle = n|n \rangle }
So, now we have that:
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|(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 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 \beta^2 = \frac{m\omega_0}{\hbar} } , we find that
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{\hbar\omega_0}{2}(n + \frac{1}{2}) }
And can check that
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 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.)