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Given the Schrödinger equation

$i\hbar \frac{\partial}{\partial t}\Psi(\vec r_1,\vec r_2,t) =\left [ -\frac{\hbar^2}{2m_1}\nabla_{r_1}^2-\frac{\hbar^2}{2m_2}\nabla_{r_2}^2 +V(\vec r_1,\vec r_2)\right ]\Psi(\vec r_1,\vec r_2,t)$

where $r_{1,2}$ is the radial distance of the electron and proton respectively. A coordinate transformation $(\vec r_1,\vec r_2)\mapsto(\vec r,\vec R)$ with

$\vec r=\vec r_1-\vec r_2,\ \vec R=\frac{m_1\vec r_1 +m_2\vec r_2}{m_1+m_2}$

should yield

$i\hbar \frac{\partial}{\partial t}\Psi(\vec r,\vec R,t) =\left [ -\frac{\hbar^2}{2M}\nabla_{R}^2-\frac{\hbar^2}{2\mu}\nabla_{r}^2 +V(\vec r)\right ]\Psi(\vec r,\vec R,t)$

with $M=m_1+m_2, \ \mu =\frac{m_1m_2}{m_1+m_2}$

When I tried to transform the nabla operators I got a very different result. Therefore I am not sure how this transformation was done.

Here is my failed attempt:

$\frac{\partial }{\partial \vec r_2}=\frac{\partial \vec R}{\partial \vec r_2}\frac{\partial}{\partial \vec R},\ \frac{\vec R}{\partial \vec r_2}=\frac{m_2}{m_1+m_2}$

and

$\frac{\partial \vec r}{\partial \vec r_1}=1$

If I now insert I get the wrong result.

EpsilonDelta
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    The derivation should be in many books on atomic and molecular physics. If your question is related to the steps of your own calculation, then you should post that as part of your question so that someone can help you with potential algebraic mistakes. – CuriousOne Sep 17 '15 at 13:09
  • Only found it in Bransden only, but with no derivation. Other books avoid this transformation completely by assuming a resting proton... – EpsilonDelta Sep 17 '15 at 13:13
  • Related: http://physics.stackexchange.com/q/78664/2451 – Qmechanic Sep 17 '15 at 13:20
  • @EpsilonDelta $\frac{\partial}{\partial \vec{r}_2} \neq \frac{\partial \vec{R}}{\partial \vec{r}_2} \frac{\partial}{\partial \vec{R}}$. Can you tell why? – udrv Sep 17 '15 at 13:44
  • Because $\vec r_2$ is now also dependend on $\vec r$ as well. – EpsilonDelta Sep 17 '15 at 14:20
  • Ok, now I used $\frac{\delta}{\delta r_2}=\frac{\partial}{\partial R}\frac{\partial R}{\partial r_2}+\frac{\partial}{\partial r}\frac{\partial r}{\partial r_2}$ and got the right result, thank you. – EpsilonDelta Sep 17 '15 at 14:30
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    It is also worth commenting that the transformation yields the result only if the potential $V$ depends only on the relative distance $r=r_1-r_2$. – yuggib Sep 17 '15 at 14:31
  • I believe it would be valid to leave this question open (in a slightly edited form asking for the proper derivation), and let @EpsilonDelta add his derivation as answer (or would this be too close to a complete solution of a homework-like question?). After all the aim of this site is to create a useful question/answer repository. (If Wikipedia has the derivation I would no longer suggest this course of action). – Sebastian Riese Sep 17 '15 at 17:54

1 Answers1

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Your Transformation is wrong and it is easy to see that. Your transformed derivative of $\partial_{r_2}$ does not depend on the $r = r_1-r_2$ part of the transformation at all. In other words with your logic I could equally write $\partial_{r_2} = \frac{\partial r}{\partial r_2} \partial_{r} = - \partial_{r}$. The correct transformation is $$\partial_{r_2} = \frac{\partial r}{\partial r_2} \partial_{r} + \frac{\partial R}{\partial r_2} \partial_{R} = \frac{m_1}{m_1+m_2} \partial_R - \partial_r \\ \partial_{r_1} = \frac{m_1}{m_1+m_2} \partial_R + \partial_r $$ The second derivatives can be calculated by consecutive application of the derivative operators $$\partial_{r_1}^2 = \left(\frac{m_1}{m_1+m_2} \partial_R - \partial_r \right)\left(\frac{m_1}{m_1+m_2} \partial_R - \partial_r \right)\\ = \left( \frac{m_1}{m_1 + m_2} \right)^2 \partial_R^2 - \frac{2m_1}{m_1+m_2} \partial_R \partial_r + \partial_r^2\\ \partial_{r_2}^2 =\left( \frac{m_1}{m_1 + m_2} \right)^2 \partial_R^2 + \frac{2m_1}{m_1+m_2} \partial_R \partial_r + \partial_r^2 $$ The kinetic energy part of the hamiltonian now yields $$H_{\rm kin} = \frac{-\hbar^2}{2} \left( \frac{1}{m_1} \partial_{r_1}^2 + \frac{1}{m_2} \partial_{r_2}^2 \right)\\ = \frac{-\hbar^2}{2} \left( \frac{1}{m_1+m_2} \partial_R^2 + \left( \frac{1}{m_1}+\frac{1}{m_2} \right)\partial_r^2 \right) \\ = \frac{-\hbar^2}{2 M}\partial_R^2 - \frac{-\hbar^2}{2 \mu}\partial_r^2 $$

Jannick
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