Homework Help: Hamiltonian For Two-Particle System

1. Mar 9, 2010

McCoy13

1. The problem statement, all variables and given/known data
Show that the time-independent Schrödinger equation becomes
$$-\frac{h^{2}}{2(m_{1}+m_{2})}\nabla^{2}_{R}\psi-\frac{h^{2}}{2\mu}\nabla^{2}_{r}\psi+V(r)\psi = E\psi$$

2. Relevant equations
$$-\frac{h^{2}}{2m_{1}}\nabla^{2}_{1}\psi-\frac{h^{2}}{2m_{2}}\nabla^{2}_{2}\psi+V(r)\psi = E\psi$$

3. The attempt at a solution
$$\nabla^{2}_{1} = (\frac{\mu}{m_{2}})\nabla^{2}_{R}+\nabla^{2}_{r}$$
$$\nabla^{2}_{2} = (\frac{\mu}{m_{1}})\nabla^{2}_{R}-\nabla^{2}_{r}$$

Substituting into the Hamiltonian, I got
$$-\frac{h^{2}}{2(m_{1}+m_{2})}\nabla^{2}_{R}\psi-(\frac{h^{2}}{2})(\frac{m_{2}-m{1}}{m_{1}m_{2}})\nabla^{2}_{r}\psi+V(r)\psi = E\psi$$

I think I must've calculated $$\nabla^{2}_{1} = (\frac{\mu}{m_{2}})\nabla^{2}_{R}+\nabla^{2}_{r}$$ and $$\nabla^{2}_{2} = (\frac{\mu}{m_{1}})\nabla^{2}_{R}-\nabla^{2}_{r}$$ incorrectly, but I can't for the life of me figure out why. I used chain rule to get $$\nabla_{R}$$ and $$\nabla_{r}$$ and then product rule to get the second derivatives. Since the derivatives of R and r with respect to r1 and r2 are simply constants, they don't contribute to the second derivative.

Thoughts?

2. Mar 9, 2010

nickjer

You need to show us the work you did to calculate those new laplacians, so that we can guide you where you went wrong. But you are right about those laplacians being wrong, I just can't tell you where you went wrong without seeing the work.

3. Mar 9, 2010

McCoy13

I started with:
$$\nabla_{1} = \frac{\mu}{m_{2}}\nabla_{R}+\nabla_{r}$$
$$\nabla_{2} = \frac{\mu}{m_{1}}\nabla_{R}-\nabla_{r}$$

Then, for the second derivative (where I may be going wrong is that $$\nabla(\nabla) \neq \nabla^{2}$$?) I used the product rule since $$\frac{\mu}{m_{2}}\nabla_{R}+\nabla_{r} = \frac{\partial R}{\partial r_{1}}\nabla_{R}+\frac{\partial r}{\partial r_{1}}\nabla_{r}$$ (that last partial should be partial of r with respect to r1, it's not showing up for some reason) and the same for $$\nabla_{2}$$. Since $$\frac{\partial R}{\partial r_{1}}$$ is a constant and the same for partial r with respect to r1 (and the same again for r2) then those terms disappear and you are left with the equations I had in my original post.

Last edited: Mar 9, 2010
4. Mar 9, 2010

nickjer

The first two lines are correct. Not sure what you are doing in the following paragraph.

Since you know $$\nabla_1$$, you just square it to get $$\nabla_1^2$$. Then it becomes:

$$\nabla_1^2 = \left(\frac{\mu}{m_{2}}\nabla_{R}+\nabla_{r}\right)^2$$

You then just expand it like you would for (a+b)^2. Do the same for the 2nd gradient.

5. Mar 9, 2010

McCoy13

This is what I suspected. Thanks.