Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A few general questions

  1. Aug 25, 2009 #1
    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).
     
  2. jcsd
  3. Aug 26, 2009 #2

    olgranpappy

    User Avatar
    Homework Helper

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

    no. act on an eigenfuntion with d/dx... do you get the same function back again? no.
     
  4. Aug 26, 2009 #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.
     
  5. Aug 26, 2009 #4

    olgranpappy

    User Avatar
    Homework Helper

    Correct. And the particle in a box is not free (because there is a confining box/potential).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A few general questions
  1. A few questions (Replies: 8)

  2. A few questions (Replies: 2)

  3. A few QM questions (Replies: 2)

Loading...