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In this pdf (p. 4) it is stated that the expectation value $\langle n | \hat{\pmb{E}}(\pmb{r}) | n \rangle = 0$. I don't see this immediately and I'm also not able to prove it. How can this be proved?

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  • $\begingroup$ Isn't it obvious as the ladder operator are traceless? $\endgroup$ – Young Kindaichi Nov 26 '20 at 13:38
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We can write it out using Eq. (4.12) to see

$$\langle n| \hat E(r) |n\rangle = i \omega \langle n| [A(r) \hat a - A^*(r) \hat a^\dagger] |n\rangle.$$

Since $\hat a |n\rangle = \sqrt{n} |n-1\rangle$ and $\hat a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle$, see Eq. (4.8), we have

$$\langle n| \hat E(r) |n\rangle = \alpha \langle n| n-1\rangle - \beta \langle n|n+1\rangle,$$

where $\alpha$ and $\beta$ are some numbers. Both brakets involve orthogonal states, and thus

$$\langle n| \hat E(r) |n\rangle = 0.$$

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  • $\begingroup$ Thanks, I thought r is an operator and I have to integrate over the space. $\endgroup$ – stonar96 Nov 26 '20 at 14:44

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