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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.

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  • $\begingroup$ Write down the probability density $|\psi(x)|^2$ and then try to perform a change of variable... $\endgroup$
    – Joe
    Commented Jan 30, 2013 at 13:25
  • $\begingroup$ For the momentum, "by inspection" maybe they mean that Fourier transform of a Gaussian is a Gaussian. $\endgroup$
    – twistor59
    Commented Jan 30, 2013 at 13:26
  • $\begingroup$ 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... $\endgroup$
    – Prook
    Commented Jan 30, 2013 at 14:00
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    $\begingroup$ 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. $\endgroup$
    – Martino
    Commented Jan 30, 2013 at 14:40

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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... http://en.wikipedia.org/wiki/Normal_distribution

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  • $\begingroup$ 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! $\endgroup$
    – Prook
    Commented Jan 30, 2013 at 15:09

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