A question about terminology. I have seen both $\langle p\rangle$ and $\langle\hat{p}\rangle$ to calculate the expected value of momentum (same thing with position, energy etc.). The first one would suggest that we take the expected value of an observable while the second would suggest that we take the expected value of an operator. Which one is correct?


1 Answer 1


In terms of the notation: $\langle\hat{p}\rangle$ is the correct expression, since the mathematical operations are done with the operators, which are mathematical objects, rather than with the observables, which are the quantities measured experimentally.

More specifically, $\langle\hat{p}\rangle$ corresponds to \begin{equation} \langle\hat{p}\rangle = \langle\psi|\hat{p}|\psi\rangle = \int dx \psi^*(x)\hat{p}\psi(x), \end{equation} if the system is described by a wave function $\psi$, and to \begin{equation} \langle\hat{p}\rangle = \mathbf{Tr}[\hat{\rho}\hat{p}], \end{equation} if the system is described by densisty matrix $\hat{\rho}$.

Moreover, the observations will produce different values, corresponding to the eigenvalues of the operator, whereas the quantum mechanical average approximates the average of these observed values.

  • $\begingroup$ I see. So it is like in statistics where we take expected value of the stochastic variable and the stochastic variable here would be the operator. Would that be a correct interpretation? $\endgroup$ Mar 26, 2020 at 9:52
  • $\begingroup$ This is correct. The difference is that in usual probability/statistics you average functions over the probability distribution, whereas in quantum theory you average an operator over the wave functions. $\endgroup$ Mar 26, 2020 at 10:09
  • $\begingroup$ @user5744148 I added a couple of equation to my answer, to underscore that $\langle\hat{p}\rangle$ has very specific mathematical meaning. $\endgroup$ Mar 26, 2020 at 10:15
  • $\begingroup$ It seems strange then that Griffith always use the expectation value without a hat $\endgroup$ Mar 26, 2020 at 11:18
  • $\begingroup$ @user5744148 if it is clear what is the meaning, then it doesn't make much difference. In either case he is talking about the expectation of an operator. $\endgroup$ Mar 26, 2020 at 12:13

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