# Question regarding the expected value of the momentum of a Gauß distribution

Given the position amplitude $$\psi(q) = \frac1{\sqrt{L \sqrt{\pi}}} e^{-\frac1{2}\left(\frac{q-q_0}{L}\right)^2}$$ I get the expected value for the position $$\langle \hat{q} \rangle = q_0$$ which makes sense to me but the expected value for momentum I get $$\langle \hat{p} \rangle = 0$$ which for me is kind of unintuitive. Is there a reasonable explanation of why this is the case? A Gauß distribution in my eyes should be one where the particle is the most localized (and the distribution with the lowest uncertainty) but shouldn't you still expect a non-zero momentum? To add to that, if u multiply a complex exponential function so u get $$\psi(q) = \frac1{\sqrt{L \sqrt{\pi}}} e^{-\frac1{2}\left(\frac{q-q_0}{L}\right)^2} e^{i k_0 q}$$ u also get a non-zero expected value for the momentum being $$\langle \hat{p} \rangle = \hbar k_0.$$ How can u interpret this physically?

Well of course you can take the Fourier transform of $$\psi(q)$$ to get $$\psi(p)$$, and you will find $$\langle p\rangle=0$$ but intuitively $$\psi(q)=\frac1{\sqrt{L \sqrt{\pi}}} e^{-\frac1{2}\left(\frac{q-q_0}{L}\right)^2}$$ is solution to the time-independent Schrödinger equation (in a harmonic potential) so that $$\langle x\rangle$$ will be independent of $$t$$. You can then think that $$\langle p\rangle = m\langle \dot{x}\rangle =m\frac{d}{dt}\langle x\rangle=0$$.
The second case, where you have shifted your Gaussian (in $$p$$-space) is no longer a solution to the TISE but in fact a coherent state and it's expectation values for $$\langle x\rangle$$ and $$\langle p\rangle$$ are discussed in this post.
$$p$$ can be positive or negative and when you average, they cancel. $$\sqrt{<\hat p^2>}$$ will give the sort of answer you would expect.