Find the possible values of momentum and the uncertainty of momentum for a given wave function 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? 
 A: 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 $\int_{-\infty}^\infty dx\,\psi^*(x)\psi(x)$.  
