Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. First integral $$\int \Psi^*({\bf r},t)\hat {\bf p} \Psi({\bf r},t)\, d^3r,$$ where the $\Psi({\bf r},t)=e^{i({\bf k}\cdot{\bf r}-\omega t)}\,\,\,$ and $\hat {\bf p}=-i\hbar \nabla$.

  2. Second one $$\int \Psi^*({\bf r},t)\hat {\bf p}^2 \Psi({\bf r},t)\, d^3r,$$ where the $\Psi({\bf r},t)=e^{i({\bf k}\cdot{\bf r}-\omega t)}\,\,\,$ and $\hat {\bf p}^2=-\hbar^2 \nabla^2$.

share|cite|improve this question
@Nick Kidman these integrals are actually expectation value – nabla Jul 25 '12 at 10:32
@nabla: But the quantum state you give is an eigenstate of momentum operator.. There is no variation in momentum, then. – Siyuan Ren Jul 25 '12 at 10:37
@Karsus Ren how about uncertainty? – nabla Jul 25 '12 at 10:47
@nabla: It is momentum eigenstate, then there is no momentum uncertainty. – Siyuan Ren Jul 25 '12 at 11:08

You want

$$\langle\hat A\rangle:=\int \Psi^*(r,t)\hat A \Psi(r,t)\, d^3r,$$

and since $\Psi(r,t)=e^{i(kr-wt)}$ you have

$(\Psi(r,t))^*\hat p\Psi(r,t)=\\ =(e^{i(kr-wt)})^*(-i\hbar \nabla)e^{i(kr-wt)}\\ =e^{-i(kr-wt)}(-i\hbar (ik))e^{i(kr-wt)}\\ =\hbar k,$

and so

$\langle\hat p\rangle=\\ =\int \Psi^*(r,t)\hat p \Psi(r,t)\, d^3r\\ =\int \Psi^*(r,t)(\hbar k) \Psi(r,t)\, d^3r\\ =\hbar k \int \Psi^*(r,t)1\Psi(r,t)\, d^3r\\ =\hbar k \langle\hat 1\rangle,$

and similarly

$\langle\hat p^2\rangle=(\hbar k)^2 \langle\hat 1\rangle,$

which also implies

$\langle\hat p\rangle^2=\langle\hat p^2\rangle\langle\hat 1\rangle.$

The usual doctrine is that the wavefunction is a normed eigenstate, i.e. $$\int \Psi^*(r,t)\Psi(r,t)\, d^3r=1,$$ which means $\langle\hat 1\rangle$ is equal to the number 1. The problem is that the norm of the plane wave $\Psi(r,t)=e^{i(kr-wt)}$, for which the integrand is $\Psi^*(r,t)\Psi(r,t)=e^0=1$, diverges for an integral $\int d^3r$ over an infinite volume.

I'd be tempted to just say you should define

$$\langle\hat A\rangle:=\frac{\int \Psi^*(r,t)\hat A \Psi(r,t)\, d^3r}{\int \Psi^*(r,t)\Psi(r,t)\, d^3r},$$

as then in this case $\langle\hat p\rangle:=\tfrac{\hbar k\langle\hat 1\rangle}{\langle\hat 1\rangle}$ and so the object $\langle\hat 1\rangle$ formally factors out in the computation in this case. But in general, the problem is more complicated than that. Maybe you find this thread illuminating, and there are certainly others on physics.SE regarding such issues.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.