Understanding the Infinite Square Well Problem | Bound States & Eigenfunctions

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In summary: This means that the Hamiltonian and momentum operators do not commute in this case, and the particle cannot be in an eigenstate of momentum. The probability of finding the particle is greater at the center of the well because the wavefunction is more spread out there, and thus the probability density is higher. This can be seen mathematically by looking at the probability density function, but it also makes sense physically.
  • #1
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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  • #2
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).
 
  • #3
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.
 
  • #4
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).
 

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