- #1

McCoy13

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## Homework Statement

Show that the time-independent Schrödinger equation becomes

[tex]-\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[/tex]

## Homework Equations

[tex]-\frac{h^{2}}{2m_{1}}\nabla^{2}_{1}\psi-\frac{h^{2}}{2m_{2}}\nabla^{2}_{2}\psi+V(r)\psi = E\psi[/tex]

## The Attempt at a Solution

[tex]\nabla^{2}_{1} = (\frac{\mu}{m_{2}})\nabla^{2}_{R}+\nabla^{2}_{r}[/tex]

[tex]\nabla^{2}_{2} = (\frac{\mu}{m_{1}})\nabla^{2}_{R}-\nabla^{2}_{r}[/tex]

Substituting into the Hamiltonian, I got

[tex]-\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[/tex]

I think I must've calculated [tex]\nabla^{2}_{1} = (\frac{\mu}{m_{2}})\nabla^{2}_{R}+\nabla^{2}_{r}[/tex] and [tex]\nabla^{2}_{2} = (\frac{\mu}{m_{1}})\nabla^{2}_{R}-\nabla^{2}_{r}[/tex] incorrectly, but I can't for the life of me figure out why. I used chain rule to get [tex]\nabla_{R}[/tex] and [tex]\nabla_{r}[/tex] 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?