Using chain rule on a two-particle quantum system. The 2 (two) particles have masses $m_1$,$m_2$ and their position and momentum variables are $\mathbf{r}_1$,$\mathbf{r}_2$,$\mathbf{p}_1$,$\mathbf{p}_2$ respectively with potential $V(\mathbf{r})$ with $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$.
The Hamiltonian is $$\hat{H}=\frac{1}{2m_1}\hat{\mathbf{p}}_1^2+\frac{1}{2m_2}\hat{\mathbf{p}}_2^2 + V(\hat{\mathbf{r}}_1-\hat{\mathbf{r}}_2)$$
The wave function $\Psi$ for this system obeys the SE
$$i\hbar\frac{\partial\Psi}{\partial t} = \hat{\mathbf{H}}\Psi$$
The problem I have is the actual maths behind the following question:
introducing COM and relative displacement coordinates, $$\mathbf{R}=\frac{m_1}{M}\mathbf{r_1}+\frac{m_2}{M}\mathbf{r_2}, \quad (M=m_1+m_2)$$
Use the extended chain rule to show $-i\hbar \frac{\partial}{\partial\mathbf{R}}\Phi=-i\hbar\frac{\partial}{\partial\mathbf{\mathbf{r_1}}}\Psi -i\hbar\frac{\partial}{\partial\mathbf{\mathbf{r_2}}}\Psi$ and from this conclude $\hat{\mathbf{P}}=\hat{\mathbf{p_1}}+\hat{\mathbf{p_2}}$
 A: (first off big up LLU ;) ), good luck with the resit. What we did was
reaarange $\underline{r_1}$, $\underline{r_2}$ for $\underline{R}$ and $\underline{r}$. So you should get
\begin{equation}
\begin{split}
\underline{r_1} &= \frac{\underline{R}M + m_2 \underline{r_2}}{M}\\
\underline{r_2} &= \frac{\underline{R}M + m_2 \underline{r_2}}{M} - \underline{r}\\
\end{split}
\end{equation}
then for the equations all you have to do is start with left hand side and work with right in the original equation $\psi(\underline{r_1},\underline{r_2};t)$ = $\phi(\underline{R},\underline{r};t)$
so you multiply by $-i\hbar \frac{\partial}{\partial \underline{R}}$
therefore we do that for the wave function $\psi$ but remember that $\underline{R}$ and $\underline{r}$ are parametrised by $\underline{r_1}$ and $\underline{r_2}$ so you need to implement a chain rule
\begin{equation}
\begin{split}
-i\hbar \frac{\partial}{\partial \underline{R}} \phi = -i\hbar \frac{\partial}{\partial \underline{R}} \psi \\
\end{split}
\end{equation}
but expand with chain rule so
\begin{equation}
\frac{\partial}{\partial \underline{R}} \phi = \frac{\partial \psi}{\partial \underline{r_1}} \frac{\partial \underline{r_1}}{\partial \underline{R}} +  
\frac{\partial \psi}{\partial \underline{r_2}} \frac{\partial \underline{r_2}}{\partial \underline{R}}
\end{equation}
If you work through with your definitions for $\underline{r_1}$ and $\underline{r_2}$ you'll get the $\frac{\partial \underline{r_1}}{\partial \underline{R}} = \frac{\partial \underline{r_2}}{\partial \underline{R}} = 1$
and thus you've proved your desired result. Just apply the same process again for the $\frac{\partial}{\partial \underline{r}}$ part and you get the results you're after. Hope this helps
