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  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 $\hat {\bf p}^2=-\hbar^2 \nabla^2$.

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

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