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What are the expectation values of commutator and anti-commutator for momentum and position operators? In the case of commutator: $$\langle[x,p]\rangle=\langle i\hbar\rangle=~?$$ In the case of anti-commutator: $$\langle \{x,p\}\rangle=~?$$

Qmechanic
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$i\hbar$ is simply a number, or if you must regard it as an operator, a multiple of the identity. So $\langle i\hbar \rangle=i\hbar$, and so is $\langle -i\hbar \rangle$.

By the way, anticommutator of $\hat{x}$ and $\hat{p}$ is not $[\hat{p},\hat{x}]$, but $\{\hat{x},\hat{p}\}=\hat{x}\hat{p}+\hat{p}\hat{x}$.

Siyuan Ren
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Expectation values of constants or numbers are just those constants or numbers.

The expectation value of the anti commutator of $\hat x$ and $\hat p$, that is, $\langle\{\hat x,\ \hat p\}\rangle$, for the Harmonic Oscillator, or coherent states of the Harmonic Oscillator, is equal to $0$. $$\langle n|\{\hat x,\ \hat p\}|n \rangle = \langle\hat x \hat p + \hat p \hat x\rangle = \langle \hat x \hat p\rangle + \langle\hat p \hat x\rangle = \langle\frac{i\hbar}2\rangle + \langle\frac{-i\hbar}2\rangle = \frac{i\hbar}2 - \frac{i\hbar}2 = 0$$

Mauro Lacy
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    Why is $\langle\hat x\hat p\rangle=\frac{i\hbar}{2}$? – Jahan Claes Nov 29 '17 at 20:33
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    @JahanClaes It's not. This answer is incorrect. – Chris Nov 29 '17 at 21:20
  • OK, I've re-added the remark about the SHO. – Mauro Lacy Nov 30 '17 at 22:14
  • JahanClaes, @Chris, can you please take a look at https://physics.stackexchange.com/questions/155089/expected-value-of-xp-in-harmonic-oscillator – Mauro Lacy Dec 01 '17 at 11:19
  • @Chris, I've re-added the specific remark about the Harmonic Oscillator. You are right that it's not true in general, and I was wrong when I removed that remark. Thanks for pointing it out. The answer is now correct. – Mauro Lacy Dec 01 '17 at 11:26
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    @MauroLacy Is it true for coherent states? I believe it is only for stationary states. Which makes it pretty much useless as a relation. It certainly is misleading to write $\langle{\hat x,\ \hat p}\rangle = \langle\hat x \hat p + \hat p \hat x\rangle = \langle \hat x \hat p\rangle + \langle\hat p \hat x\rangle = \langle\frac{i\hbar}2\rangle + \langle\frac{-i\hbar}2\rangle = \frac{i\hbar}2 - \frac{i\hbar}2 = 0$: it should at least have something like $\langle n|{\hat x,\ \hat p} |n\rangle$ to show it is only for stationary states. – Chris Dec 01 '17 at 11:30
  • It's true for coherent states too, yes. – Mauro Lacy Dec 01 '17 at 11:30
  • @MauroLacy Regardless, that should be signified more clearly. – Chris Dec 01 '17 at 11:31
  • Chris is, *of course* right. In the phase space QM picture, this is self-evident; but, even in Hilbert space, displaced states, etc..., the bread-and-butter of squeezing, violate vanishing of the expectation of the anticommutator. Just because it can be made to hold for some states, exceptionally, does not permit using it as an example of an abstract relationship and conveying unsound impressions. – Cosmas Zachos Dec 01 '17 at 14:52