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This might seem like a stupid question, and maybe it is, but I've been stuck on this for quite a while: So I have the hamiltonian for 2 particles: $$-\frac{\hbar^2}{2m_1}\vec{\nabla^2_{r_1}} - \frac{\hbar^2}{2m_2}\vec{\nabla^2_{r_2}} + V(\vec{r_1}-\vec{r_2})$$. Now invoking the relative coördinate $$\vec{r} = \vec{r_1} - \vec{r_2}$$ and the center of mass coördinate: $$\vec{R} = \frac{m_1\vec{r_1} + m_2\vec{r_2}}{m_1 + m_2}$$ It is said that "with some easy algebra" it is found that: $$-\frac{\hbar^2}{2m_1}\vec{\nabla^2_{r_1}} - \frac{\hbar^2}{2m_2}\vec{\nabla^2_{r_2}} = -\frac{\hbar^2}{2M}\vec{\nabla^2_{R}} - \frac{\hbar^2}{2\mu}\vec{\nabla^2_{r}}$$ with $M = m_1 + m_2$ and $\mu = \frac{m1m2}{m1 + m2}$. Now my question is: what "easy algebra"? I've tried the one dimensional version with $\vec{R} \rightarrow X$ and $\vec{r} \rightarrow x$. And then it is as simple as 'using the chain rule' as to prove: $$\frac{\partial}{\partial x} = \frac{\partial x}{\partial x_1}\frac{\partial}{\partial x} + \frac{\partial X}{\partial x_1}\frac{\partial}{\partial X}$$ But now i'm stuck with another question.. How is this true? Like how is this the chain rule? I know when for example Z depends on y and y on x this is true: $$\frac{dZ}{dx} = \frac{dZ}{dy}\frac{dy}{dx}$$ and this is regarded as the 'chain rule' but how is the previous statement formed? Thanks in advance.

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    $\begingroup$ It seems like you're looking for the Multivariable Chain Rule. It can take many forms, such as the one with partial derivatives you describe above. $\endgroup$ Dec 12, 2020 at 16:26

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Oh I think @electronpusher has found it, it is indeed as with the Multivariable Chain Rule, as the wave function can be considered function of X and x: $\Psi(X,x)$ and thus using the multivariable chain rule on this function you get the above mentioned expression

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