# Expectation values of commutator and anti-commutator (momentum and position)

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: $$\langle \{x,p\}\rangle=~?$$

• Nov 30 '17 at 23:04

$i\hbar$ 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 $\{\hat{x},\hat{p}\}=\hat{x}\hat{p}+\hat{p}\hat{x}$.

• you means that $\langle -i\hbar\rangle=i\hbar$!?
– user8784
Jun 16 '12 at 11:32
• $\langle -i\hbar\rangle= -i\hbar$. The expectation of any constant is just that constant. Jun 16 '12 at 13:00

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 $0$. $$\langle n|\{\hat x,\ \hat p\}|n \rangle = \langle\hat x \hat p + \hat p \hat x\rangle = \langle \hat x \hat p\rangle + \langle\hat p \hat x\rangle = \langle\frac{i\hbar}2\rangle + \langle\frac{-i\hbar}2\rangle = \frac{i\hbar}2 - \frac{i\hbar}2 = 0$$

• Why is $\langle\hat x\hat p\rangle=\frac{i\hbar}{2}$? Nov 29 '17 at 20:33
• @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. Nov 30 '17 at 22:14
• JahanClaes, @Chris, can you please take a look at physics.stackexchange.com/questions/155089/… Dec 1 '17 at 11:19
• @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 1 '17 at 11:30