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: $$\langle \{x,p\}\rangle=~?$$
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$\begingroup$ Related. $\endgroup$– Cosmas ZachosNov 30, 2017 at 23:04
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$\begingroup$ If you ask for the expectation value you must mention in which state you want to know it, i.d. $<\psi|[x,p]|\psi>$. In any state? $\endgroup$– NikodemNov 13, 2022 at 0:21
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2 Answers
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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 $\{\hat{x},\hat{p}\}=\hat{x}\hat{p}+\hat{p}\hat{x}$.
answered Jun 16, 2012 at 2:54
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$\begingroup$ you means that $\langle -i\hbar\rangle=i\hbar$!? $\endgroup$– user8784Jun 16, 2012 at 11:32
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1$\begingroup$ $\langle -i\hbar\rangle= -i\hbar$. The expectation of any constant is just that constant. $\endgroup$ Jun 16, 2012 at 13:00
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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 $0$. $$\langle n|\{\hat x,\ \hat p\}|n \rangle = \langle\hat x \hat p + \hat p \hat x\rangle = \langle \hat x \hat p\rangle + \langle\hat p \hat x\rangle = \langle\frac{i\hbar}2\rangle + \langle\frac{-i\hbar}2\rangle = \frac{i\hbar}2 - \frac{i\hbar}2 = 0$$
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1$\begingroup$ Why is $\langle\hat x\hat p\rangle=\frac{i\hbar}{2}$? $\endgroup$ Nov 29, 2017 at 20:33
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2$\begingroup$ @JahanClaes It's not. This answer is incorrect. $\endgroup$– Chris ♦Nov 29, 2017 at 21:20
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$\begingroup$ OK, I've re-added the remark about the SHO. $\endgroup$ Nov 30, 2017 at 22:14
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$\begingroup$ JahanClaes, @Chris, can you please take a look at physics.stackexchange.com/questions/155089/… $\endgroup$ Dec 1, 2017 at 11:19
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1$\begingroup$ @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. $\endgroup$– Chris ♦Dec 1, 2017 at 11:30