Find the possible values of momentum and the uncertainty of momentum for a given wave function

Question

I have the wave function $\psi (x) =A \sin (kx) $ for $- \infty <x< \infty$ and want to find all the possible values for momentum and $\Delta p$ . I have the solution but I can't understand it, my professor says that the possible values are $p= \pm \hbar k$,can someone explain how do we get to this? Also after finding $\langle p \rangle =0$ my professor finds $\langle p^2 \rangle = \frac{1}{2} \hbar^2 k^2 + \frac{1}{2} \hbar^2 k^2$, can you explain that?

Answer 1

I actually don't think this question makes much sense since your wave function $\psi(x)$ is not normalizable, but if you are willing to blindly accept (1) below then you can possibly understand the manipulations of your instructor.

If you know that $\phi(x)=e^{ikx}$ describes a state of momentum $+k$, then it's no surprise that $$ \psi(x)=A\sin(kx)=\frac{A}{2i} \left(e^{ikx}-e^{-ikx}\right) $$ describes a state that is a superposition of $+k$ and $-k$. The probability of getting $+k$ is the same as the probability of getting $-k$ since the coefficient $\vert A/2i\vert^2$ in front of each factor in the superposition is the same.

If the probability of obtaining $+k$ and $-k$ is the same, then the average value of $k$ will be $0$, irrespective of the constant $A$.

As to $\langle p^2\rangle$, note that $p^2\psi(x)=\hbar^2k^2\psi(x)$ so $$ \langle p^2\rangle =\frac{\int_{-\infty}^\infty dx\,\psi^*(x)\hbar^2k^2\psi(x)} {\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)}= \hbar^2k^2 \frac{\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)} {\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)}=\hbar^2k^2 $$ assuming (somewhat blindly) $$ \frac{\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)}{\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)}=1\, . \tag{1} $$

You are right in being suspicious about the convergence of the integral $\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)$.