Particle in an infinite potential box- expected values of energy

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SUMMARY

The discussion focuses on calculating the expected values of energy for a particle in an infinite potential box, specifically using the wavefunction ##\psi(x)=\sqrt{\frac{8}{3a}}sin^2(\frac{\pi x}{a})##. Participants calculated the expected kinetic energy as ##=\frac{\hbar^2}{2ma^2}## and compared it to the ground state energy levels. They also explored the probabilities of finding the particle in the ground and first excited states, emphasizing the need to calculate coefficients ##c_n## using integrals of the wavefunction and energy eigenstates. The discussion highlighted the importance of correctly interpreting the wavefunction and its relation to energy eigenstates.

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  • #31
Could anybody please respond?
 
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  • #32
Hello, Rorshach. I don't understand "by leveraging development coefficients" in part (c), so I don't think I can help much.

You've already calculated ##<\hat{p}^2>##, so once you determine ##<p>## you will easily be able to get ##\Delta p##.

Note that the wavefunction ##\psi## is an even function about ##x = a/2##. Since ##\hat{p}## is proportional to the derivative operator, you should be able to argue that ##\hat{p} \psi## is an odd function about ##x = a/2##. So, if you set up the integral for ##<\hat{p}>## you should be able to conclude what the answer is without actually carrying out the integration. But that doesn't appear to be the way you are expected to do it.

I will bring this problem to the attention of some of the other helpers.

I guess you don't have access to the person who made the question. It would be nice if you could get clarification on what he/she meant for you do do here.
 
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  • #33
Unfortunately, I have no contact with the person who made this question. I also don't want to integrate anything, since it was clearly stated in the problem to not to do it, however user tannerbk gave a formula a few replies above for expectated value for hamiltonian <H>, so maybe there is an analogic one for <p>?
 
  • #34
Maybe they want something like this. You know that you can expand your wave function in terms of the energy eigenstates:

##\psi(x) = \sum c_n \sin(k_n x)## where ##k_n = n\pi/a##.

Write each ##\sin(k_n x)## as ##\frac{1}{2i}(e^{ik_nx} - e^{-ik_nx})##.

Suppose you could show that each exponential ##e^{ik_nx}## is a momentum eigenstate corresponding to momentum ##\hbar k_n## while each exponential ##e^{-ik_nx}## is a momentum eigenstate corresponding to momentum ##-\hbar k_n##.

Can you then construct an argument showing ##<\hat{p}> = 0##?

Note, however, that this type of argument might not be sound. There are difficulties with defining a momentum operator for the particle in a box. The functions ##e^{ik_nx}## don't satisfy the boundary conditions of the particle in a box, so I'm not sure they qualify as legitimate states for the particle in a box.

See for example http://academic.reed.edu/physics/fa...ous Essays/Generalized Momentum Operators.pdf

or this thread https://www.physicsforums.com/showthread.php?t=285694
 
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