Understanding the Infinite Square Well Problem | Bound States & Eigenfunctions

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quasar_4
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I need someone to tell me if I'm understanding things. :shy:

Let's say that we're studying the infinite square well problem, where the well extends from -L/2 to L/2 in 1 dimension. In this case, the energy of the system, E, is less than the potential at the barriers, so the eigenstates of the Hamiltonian (obviously) correspond to bound states.

Here is where I am confused - please tell me what I am thinking correctly and incorrectly:

- the Hamiltonian and momentum operators commute, so in general, they share a set of eigenfunctions. But the particle in this well can't be in an eigenstate of momentum, because it's in a bound state (and eigenstates of momentum correspond to scattering problems)?

- We know that the expectation value of momentum, <p>, must be zero for a particle in the well because bound states are stationary states, and a nonzero <p> would indicate that the particle was escaping the well (is this a good sort of physical reasoning)?

- The probability of finding the particle is greater at the center of the well then at the edges , but I can't really explain this physically (it seems to be more a mathematical result in my mind than a physical one, and I'm not sure how to describe the probability of finding the particle at some point, without thinking of probability density functions).
 
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quasar_4 said:
I need someone to tell me if I'm understanding things. :shy:

Let's say that we're studying the infinite square well problem, where the well extends from -L/2 to L/2 in 1 dimension. In this case, the energy of the system, E, is less than the potential at the barriers, so the eigenstates of the Hamiltonian (obviously) correspond to bound states.

Here is where I am confused - please tell me what I am thinking correctly and incorrectly:

- the Hamiltonian and momentum operators commute,
no, they do not. There is an external potential (the infinite barriers at +-L/2) which obviously makes the system not translational invariant.

so in general, they share a set of eigenfunctions.

no. act on an eigenfuntion with d/dx... do you get the same function back again? no.
But the particle in this well can't be in an eigenstate of momentum, because it's in a bound state (and eigenstates of momentum correspond to scattering problems)?

- We know that the expectation value of momentum, <p>, must be zero for a particle in the well because bound states are stationary states, and a nonzero <p> would indicate that the particle was escaping the well (is this a good sort of physical reasoning)?

- The probability of finding the particle is greater at the center of the well then at the edges , but I can't really explain this physically (it seems to be more a mathematical result in my mind than a physical one, and I'm not sure how to describe the probability of finding the particle at some point, without thinking of probability density functions).
 
Ah - so actually, H and p only (necessarily) commute for the case of the free particle. In any other situation I would have to check.
 
quasar_4 said:
Ah - so actually, H and p only (necessarily) commute for the case of the free particle. In any other situation I would have to check.

Correct. And the particle in a box is not free (because there is a confining box/potential).