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QM: Hamiltonians and energies of a system

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given known data
    Hi all.

    Lets I have a system, whose Hamiltonian is given by [itex]H = p_1 + p_2[/itex], where pi is the momentum of i, where this i can be whatever, e.g. momentum in some particular direction. When the Hamiltonian is given on this form, do I know for certain that the eigenenergies of the system are on the form:

    [tex]
    E_{m,n} = f(m) + g(n),
    [/tex]

    where f and g are two functions that depend on respectively m and n? I.e. I am wondering if the eigenenergies of the system follow the same form as the Hamiltonian.

    I hope you can help.

    Best regards,
    Niles.
     
  2. jcsd
  3. Feb 13, 2009 #2

    Tom Mattson

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    Shouldn't your Hamiltonian be quadratic in the momenta? Or are you doing relativistic QM in natural units?

    Yes, because Hamiltonians are linear operators. (Perhaps I should say that this can be done whenever the Hamiltonian is linear. I'm not sure if you can cook up a nonlinear one).
     
  4. Feb 13, 2009 #3
    Yes, it should. Sorry for that.


    Hmm, can this be seen from the following?

    [tex]
    H \psi = E\psi \quad \Rightarrow \quad H_1 \psi + H_2\psi = E_1\psi + E_2\psi = E\psi.
    [/tex]

    Thanks for taking the time to reply.
     
  5. Feb 13, 2009 #4

    Tom Mattson

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    Yes. So here [itex]E=E_1+E_2[/itex].
     
  6. Feb 16, 2009 #5
    I've thought about this for the last couple of days. What I have found out is that I believe the above only works if the wavefunction satisfies [itex]
    \psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2) [/itex], so it seperates Schrödingers time-independent equation. Do you agree?
     
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