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Hamiltonian is commutable with momentum operator

  1. Jun 19, 2004 #1
    [H,P]=0 , where P is momentum operator.

    Hamiltonian is commutable with momentum operator. so H and p have
    wave function simultaniously, but in 1-dimensional potential well degeneracy
    not exist.

    what is the reason?
  2. jcsd
  3. Jun 21, 2004 #2
    Not so, e^{ip} and e^{-ip} have opposite momenta, but the same energy.
  4. Jun 21, 2004 #3


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    A particle in a well is not a free particle. The momentum eigenstates are not simple compex exponentials. Basically, a momentum eigenstate involves an integration over all space. Just because the particle has a zero probability of being outside the well does not allow you to ignore the well when considering the momentum eigenstates (otherwise, you would get a wavefunction that says the particle can be outside the well, i.e. periodic B.C.s).

    As far as I remember, momentum only commutes with the Hamiltonian for a free particle Hamiltonian (or in a configuration space orthogonal to all of the constraint degrees of freedom).
  5. Jun 22, 2004 #4
    thanks for your answer,but i don't get it well

    Is the Hamiltonian commutable with momentum operator in the 1- dimensional finite well?
    can you see [p^2/2m +v , P]=0 is to formed, for V=constant?
    why in this case, we can't find degeneracy?

    why eigenstate of momenta are not exsist when it is 1-dimentional infinite well?
    wave fuction of momenta which we could get through fourier transform for eigenfuntion of energy
    can not to be satify by wave funtion should be 0 in between potential boundary
    therefore si(0)=psi(L)=0 are not possible.

    we can know that though we solve making a eigenvalue eqeution of momenta, that is not satisfied with B.C.s

    How do I understand this?
  6. Jun 27, 2004 #5


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    I will appologize in advance for possibly not knowing what you are asking. Let me try this, though:

    For a 1-D single particle, consider H = f(X,P). That is to say, the Hamiltonian is formed by combining the position and momentum operators in some way.

    For a free particle, H = g(P). That is to say, the Hamiltonian does not depend on X, only on P. Now, P commutes with itself, so P commutes with g(P), and, therefore, [P,H] = [P,g(P)] = 0 for a free particle.

    For a particle in a well, you have H = g(P) + V(X). That is to say, the Hamiltonian depends on both X and P. Now, P does not commute with X, so P does not commute with V(X), and, therefore, [P,H] = [P,(g(P) + V(X))] = [P,g(P)] + [P,V(X)] = 0 + [P,V(X)] /= 0.

    Something I just noticed after rereading your post:
    I think you may be biasing yourself to the x-basis. Don't think of this in terms of a basis, think of it in terms of the entire space. If you just limit yourself to the range of x-values inside the well, that is like limiting yourself to consider only certain components of a vector when you consider the eigenvalue problem. This is a bad thing to do because, when you consider the same eigenvalue problem in a generally different basis, those particular components no longer have any meaning.
    Last edited: Jun 27, 2004
  7. Jun 28, 2004 #6

    V is not a constant. It is a pair of "step" functions. V changes at the boundaries of the well.

    Since P is like -id/dx,

    [P,V] = a pair of Dirac delta-functions ,

    which is certainly not zero.

    In this unphysical (because, V = infinity in some regions of space) "simplification" (which in some cases can be a good approximation), the "price" that we must "pay" is that the eigenfunctions of P just don't "fit into the picture".

    This makes perfect "physical" sense. Since an eigenfunction of P is just a free particle moving to the left or moving to the right, those eigenfunctions will not "fit into a picture" which forbids the wavefunctions from being anything but zero outside of the interval (a,b).
    Last edited: Jun 29, 2004
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