0
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks alot for the explanation that makes alot of sense $\endgroup$ Commented Jan 2, 2021 at 22:42
1
$\begingroup$

$p$ can be positive or negative and when you average, they cancel. $\sqrt{<\hat p^2>}$ will give the sort of answer you would expect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.