Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a bit puzzled about an excercise in which I have to find the expectation values for position and momentum. Normally this should be pretty easy but in this case I just don't get the point. Wavefunction is: $$ \psi(x) = \frac{1}{\sqrt{w_0 \sqrt{\pi}}} e^{\frac{-(x-x_0)^2}{2(w_{0})^{2}}+ik_0 x} $$

In a) you have to find the momentum rep. of this and in b) they ask you to find the expectation values of position and momentum.

Normally I would just compute the integral but in the solution they state, that "By inspection, it is easy to see that the expectation values for position and momentum are: $x_0 $ and $\hbar k_0 $" and I really don't know how to find these values. If anyone could briefly explain what they meant with "easy" I would be really happy.

share|cite|improve this question
Write down the probability density $|\psi(x)|^2$ and then try to perform a change of variable... – Joe Jan 30 '13 at 13:25
For the momentum, "by inspection" maybe they mean that Fourier transform of a Gaussian is a Gaussian. – twistor59 Jan 30 '13 at 13:26
I already computed the Fourier transformation of this to obtain the momentum representation... But I don't see how this could help... Especially because the resulting wave-function looks horrible. In which way is the probability density related to the expectation values? Sure, for the position value this might help to evaluate the integral but I think there has to be another way to "see" these values. Otherwise the solution would be another... – Prook Jan 30 '13 at 14:00
For the position, it is easy to see that when you take the mod-square, the function is a gaussian around $x_0$, so $x_0$ must be the mean value. – Bzazz Jan 30 '13 at 14:40
up vote 2 down vote accepted

If you build the square of the wave function, the result is a gaussian curve. If you compare your result with the general form of a normal distribution you can see that x0 is the expectation value...

share|cite|improve this answer
Thanks everybody. I think I know how to solve this now ;). That the Gaussian gives me the expectation value is clear but I thought I have to argue with the eigenvalue equations... Looks like this is not the case :) Thank you very much! – Prook Jan 30 '13 at 15:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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