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Say I have a particle with mass $m$, in a potential infinite well centered at $x=0$ with length $d$ which wave function at $t= 0$ is represented by: $$\Psi(x)=\begin{cases} \frac{1}{\sqrt{2}}\left[\sqrt{\frac{2}{d}}\cos\left(\frac{\pi x}{d}\right) +i\sqrt{\frac{2}{d}}\sin\left(\frac{\pi x}{d}\right)\right] &, |x|\le\frac{d}{2} \\ 0 &, |x|\ge\frac{d}{2} \end{cases}$$

We can clearly see from here that the wave function is represented by a superposition of 2 eigenfunctions of the Hamiltonian for this particular problem. And we can also see that we can get the first eigenfunction with probability of 0.5 and the second eigenfunction with a probability of 0.5 as well.

My question is, is the following way of calculating the expected value of momentum of this problem correct?

Since we know that the energy levels are given by: $$ E_n = n^2\frac{\pi^2\hbar^2}{2mL^2} $$

What we can deduce is the following: the particle is going to be with energy $E_1$ with probability 0.5, or with energy $E_2$, with probability 0.5.

Since we know the following relation:

relation between momentum and energy (when the particle is not relativistic, which we assume he is not)

We can deduce that, the particle is going to be with momentum P1, by probability of 0.5 or P2, by probability of 0.5, and therefore we can say that the expected value of the momentum is:

enter image description here

Is this correct?

PS: I know that the other way is to calculate it by definition which is: enter image description here

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    $\begingroup$ Please note that the preferred way to present equations here is MathJax, instead of images. I have partially replaced your images by MathJax. You can do the rest. $\endgroup$ – Thomas Fritsch Jul 27 '19 at 20:46
  • $\begingroup$ Hi and welcome to physics.SE! Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – ACuriousMind Jul 28 '19 at 10:12
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The relation you used, $p = \pm \sqrt{2m E}$, works iff the momentum operator and the (presumably time-independent) Hamiltonian operator are well-defined and share a set of common eigenstates. This is clearly not the case here, so you cannot use this formula for any average of momentum calculation. So only the integral calculation is valid. A kind piece of advice is not to use screenshots in writing here, but rather type the code (LaTex) by yourself.

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The expectation value of the momentum is zero. It is the expectation value of $\sqrt{p^2} $ that is not zero.

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  • $\begingroup$ what is wrong with my deduction in this case then? $\endgroup$ – Sammy Apsel Jul 27 '19 at 20:07
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I think I found the problem in my deduction:

The momentum-energy relation is not 100 percent correct, it should be this:

enter image description here

Which means that for every Energy E, there are 2 P's possible, one +, and one -. So yes, the final expected value will be 0.

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