1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?