What does this notation mean in relation to quantum mechanics?


I know $[x,H]$ is just the commutator e.g $xH-Hx$, and the anti-commutator is $\{x,H\} = xH + Hx$ -- but what does the line above the commutator do?

Here is some context:

$$\frac{d \langle x_0\rangle}{dt} = \frac{1}{i \hbar}[\langle x\rangle \cos{\omega t}, H] -\omega\langle x\rangle \sin{\omega t} - \frac{1}{i \hbar}\bigg[\frac{\langle p\rangle }{m\omega}\sin{\omega t}, H\bigg] -\frac{\langle p\rangle}{m} \cos{\omega t}\qquad \tag{2} $$ $$= \frac{1}{i \hbar}\overline{[x,H]} \cos{\omega t} -\omega\langle x\rangle \sin{\omega t} - \frac{1}{m \omega}\frac{1}{i \hbar}\overline{[p,H]}\sin{\omega t} -\frac{\langle p\rangle}{m} \cos{\omega t} = 0\qquad \tag{3}$$

This is from: Problems and Solutions on Quantum Mechanics By Yung-kuo Lim Page 120 - Problem 1072 ISBN-10: 9810231334

  • $\begingroup$ Never seen this before. Maybe some reverse engineering is possible. What do the brackets in $<x>$ etc. mean. An expectation value? Then, maybe, the overbar also has this meaning. $\endgroup$
    – Urgje
    Nov 3, 2015 at 9:56

1 Answer 1


Comments to the question (v5):

  1. In this quantum case the overline/bar notation $\bar{A}=\langle A\rangle$ is borrowed from statistics and it denotes a quantum expectation value of a quantity $A$. See also Ehrenfest theorem.

  2. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator $$\tag{A} H~=~\frac{p^2}{2m} +\frac{m\omega^2}{2}x^2,$$ and defines an initial value position operator $$\tag{B} x_0~:=~x\cos\omega t - \frac{p}{m\omega}\sin\omega t, $$ and asks to show that the expectation value of the operator $$\tag{C} \frac{dx_0}{dt}$$ vanishes. Well, actually, it is straightforward to show with the help of Heisenberg's EOM that the operator (C) itself vanishes, and therefore its expectation value.

  3. The formula expression (3) with commutator inside the expectation value is correct, while the formula (2) with the expectation value inside a commutator seems to be an error.


  1. Yung-kuo Lim, Problems and Solutions on Quantum Mechanics; Problem 1072(a), p.120.
  • $\begingroup$ (This answer uses the Heisenberg picture rather than the Schroedinger picture.) $\endgroup$
    – Qmechanic
    Nov 4, 2015 at 10:52

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