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| ===Virial Theorem in the Case of the Quantum Harmonic Oscillator===
| | The average potential energy is given by |
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| Prove that the Virial Theorem holds for the Harmonic Oscillator. (Show that the average kinetic energy, <math> \langle T \rangle </math> is equal to the average potential energy, <math> \langle V \rangle </math>.)
| | <math> \langle \hat{V} \rangle = \tfrac{1}{2}k\langle \hat{x}^2 \rangle.</math> |
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| For the QHO, the average potential energy is written
| | Recall from a previous problem that |
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| <math> \langle V \rangle = \frac{k}{2}\langle \hat{x}^2 \rangle </math> | | <math>\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^{\dagger}),</math> |
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| It is convenient to re-write the position operator as
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| <math> \hat{x} = \frac{1}{\sqrt{2}\beta}(\hat{a} + \hat{a}^\dagger) \text { where } \beta^2 = \frac{m\omega_0}{\hbar} </math> | | <math>\tfrac{1}{2}k\hat{x}^2=\frac{\hbar k}{4m\omega}(\hat{a}+\hat{a}^\dagger)^2.</math> |
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| <math> \Rightarrow \hat{x}^2 = \frac{1}{2\beta^2}(\hat{a} + \hat{a}^\dagger)^2 </math> | | We can now write the average potential for the <math>n^{\text{th}}</math> state of the harmonic oscillator as |
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| Now, we can write the average potential for the <math> n^{th} </math> state of the QHO like:
| | <math> \langle V \rangle = \frac{\hbar k}{4m\omega}\langle n|(\hat{a} + \hat{a}^\dagger)^2|n \rangle </math> |
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| <math> \langle V \rangle = \frac{k}{4\beta^2}\langle n|(\hat{a} + \hat{a}^\dagger)^2|n \rangle </math> | | <math> = \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 </math> |
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| <math> = \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 </math> | | <math> = \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] </math> |
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| <math> = \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] </math>
| | The first two terms are zero because |
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| Now, the first two terms disappear, as the raising and lowering operators act on the eigenkets:
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| <math> \langle n|n-2 \rangle = \langle n|n+2 \rangle = 0 </math> | | <math> \langle n|n-2 \rangle = \langle n|n+2 \rangle = 0 </math> |
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| and the operator in the third term can be written like: | | and the operator in the third term can be written as |
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| <math> \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} = 1 + 2\hat{N} \text{ where } \hat{N} = \hat{a}^\dagger\hat{a} </math>
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| <math> \hat{a}\hat{a}^\dagger |n \rangle = \hat{a} (n+1)^{\frac{1}{2}}|n + 1 \rangle = (n+1)|n \rangle </math> | | <math> \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} = 2\hat{n}+1.</math> |
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| and <math> \hat{N}|n \rangle = n|n \rangle </math>
| | Therefore, |
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| So, now we have that:
| | <math> \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 ),</math> |
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| <math> \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}) </math> | | or, noting that <math>k=m\omega^2,\!</math> |
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| And, replacing <math> \beta^2 = \frac{m\omega_0}{\hbar} </math>, we find that
| | <math> \langle \hat{V} \rangle = \tfrac{1}{2}\left (n + \tfrac{1}{2}\right )\hbar\omega.</math> |
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| <math> \langle V \rangle = \frac{\hbar\omega_0}{2}(n + \frac{1}{2}) </math>
| | Similarly, using the fact that |
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| And can check that
| | <math>\hat{p}=-i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}-\hat{a}^{\dagger}),</math> |
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| <math> \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}) </math>
| | we may show that |
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| Which shows rather nicely that the Virial Theorem holds for the Quantum Harmonic Oscillator.
| | <math> \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.</math> |
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| (See Liboff, Richard ''Introductory Quantum Mechanics'', 4th Edition, Problem 7.10 for reference.)
| | Back to [[Harmonic Oscillator Spectrum and Eigenstates#Problems|Harmonic Oscillator Spectrum and Eigenstates]] |
The average potential energy is given by
Recall from a previous problem that
or
We can now write the average potential for the 
state of the harmonic oscillator as
![{\displaystyle ={\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 ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0646ed20c88df3868a5c31e21bb310181bf92675)
The first two terms are zero because
and the operator in the third term can be written as
Therefore,
or, noting that 
Similarly, using the fact that
we may show that
Back to Harmonic Oscillator Spectrum and Eigenstates