Homework Help: Two particle harmonic oscillator

1. Oct 21, 2007

indigojoker

Two particles are subjected to the same potential $V=\frac{m \omega x^2}{2}$.
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 $$H=H_1+H_2$$

I am assuming that it is asking for a way to derive this relation but I'm not sure where to start from?

2. Oct 21, 2007

christianjb

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?

3. Oct 21, 2007

indigojoker

this problem isn't unnecessarily a quantum mechanical problem, it's just in a QM book

4. Oct 21, 2007

christianjb

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
5. Oct 21, 2007

indigojoker

$$H=\frac{p_1^2}{2m}\frac{p_2^2}{2m}+\frac{m \omega^2 x_1^2}{2}\frac{m \omega^2 x_2^2}{2}$$

6. Oct 21, 2007

christianjb

Yes- except you seem to have missed out the + signs. It's p1^2/2m1 + p2^2/2m2 and
mw^2x1^2/2+mw^2x2^2/2

7. Oct 21, 2007

indigojoker

then isnt it trivial to show $$H=H_1+H_2$$

8. Oct 21, 2007

christianjb

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.