How to Derive the Hamiltonian for a Two-Particle Harmonic Oscillator?

[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?

 

[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?

 

[indigojoker]

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

 

[christianjb]

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

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)

 

[indigojoker]

Do you mean I should start with the Hamiltonian:
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}

 

[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

 

[indigojoker]

then isn't it trivial to show H=H_1+H_2

 

[christianjb]

then isn't it trivial to show H=H_1+H_2

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.