# Quantum Mechanics - Dynamics?

1. Jan 4, 2009

### joker_900

1. The problem statement, all variables and given/known data

Consider a system of two particles of mass m that each move in one dimension along a given rod. Let |x> be the state of the first particle when it’s at x and |y> be the state of the second particle when it’s at y.

A complete set of states of the pair of particles is {|xy>} = {|x>|y>}. Write down the Hamiltonian of this system given that the particles attract one another with a force that’s equal to C times their separation.

Suppose the particles experience an additional potential

V (x, y) = 0.5C(x + y)^2.

Show that the dynamics of the two particles is now identical with the dynamics of a single particle that moves in two dimensions in a particular potential psi(x, y), and give the form of psi.

2. Relevant equations
None?

3. The attempt at a solution

Really I have no idea, I don't want someone to do it for me but just to give me an idea of where to start. Just in case I've written my ramblings below but feel free to ignore.

Basically I don't really know what "dynamics" means in this context (which may be laughable, I'm not sure). The only thing I can think of is finding the expected position of the centre of mass? In which case I would need the general state of the system. Would this be:

int[phi(x)|x>]int[phi(y)|y>] where phi(x) is the amplitude of the first particle being at x, and then would I find this by solving the TDSE?

I'm really stumped

Thanks!

2. Jan 4, 2009

### tpg

I don't think you necessarily need even to consider the centre of mass - I think you can just show that the TDSE turns out the same in both cases. I'm not entirely sure though, I'll try to have a look at this problem later. I think section 2.4 of "The physics of Quantum Mechanics", Binney & Skinner should help, if you happen to have that book.

3. Jan 4, 2009

### joker_900

I had a look at that chapter, thanks. It seems I can show that the Hamiltonian of both situations are the same fairly trivially. But then this doesn't use the states at all. Hmm

4. Jan 5, 2009

### tpg

I don't think you need to use the state labels actually - I think it is actually as trivial as it looks!