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?
2 Answers
$p$ can be positive or negative and when you average, they cancel. $\sqrt{<\hat p^2>}$ will give the sort of answer you would expect.
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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.
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Thanks alot for the explanation that makes alot of sense – goalgetter666 Jan 02 '21 at 22:42