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

Homework Help: Two particle harmonic oscillator

  1. Oct 21, 2007 #1
    Two particles are subjected to the same potential [itex]V=\frac{m \omega x^2}{2}[/itex].
    Particle one has with position x1 and momentum p1.
    Particle two has with position x2 and momentum p2.

    The problem asks how to show that the Hamiltonian is [tex]H=H_1+H_2[/tex]

    I am assuming that it is asking for a way to derive this relation but I'm not sure where to start from?
  2. jcsd
  3. Oct 21, 2007 #2
    I may be being pedantic- but it's not possible to say in QM that a particle has position x and momentum p. Is that really how the question is worded?
  4. Oct 21, 2007 #3
    this problem isn't unnecessarily a quantum mechanical problem, it's just in a QM book
  5. Oct 21, 2007 #4
    That's a good point.

    Well- in any case- write down the total hamiltonian for the system and see if it can be split into two separate hamiltonians.

    In the QM case you will need to write psi=psi(x1,x2) and then use a trial product wavefunction psi(x1,x2)=psi(x1)psi(x2)
    Last edited: Oct 21, 2007
  6. Oct 21, 2007 #5
    Do you mean I should start with the Hamiltonian:
    [tex] H=\frac{p_1^2}{2m}\frac{p_2^2}{2m}+\frac{m \omega^2 x_1^2}{2}\frac{m \omega^2 x_2^2}{2}[/tex]
  7. Oct 21, 2007 #6
    Yes- except you seem to have missed out the + signs. It's p1^2/2m1 + p2^2/2m2 and
  8. Oct 21, 2007 #7
    then isnt it trivial to show [tex]H=H_1+H_2[/tex]
  9. Oct 21, 2007 #8
    Pretty much. What's interesting though is that you can show that if you make the trial substitution psi(x1,x2)=psi(x1)psi(x2) that the total Hamiltonian separates nicely into two separate equations.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook