Showing that relative motion is independent of center of mass motion? I am trying to understand the concept behind separating the center of mass motion and the relative motion in the Schrödinger equation for the Hydrogen atom. The idea is that the Hamiltonian given by (in atomic units)
$$
H = -\frac{1}{2m_1}\nabla^2_1-\frac{1}{2m_2}\nabla^2_2 + V(\boldsymbol{r_1}-\boldsymbol{r_2})
$$
may be separated into a center of mass motion and a relative motion. Defining
$$
\boldsymbol{R} = \frac{m_1}{M}\boldsymbol{r}_1 + \frac{m_2}{M}\boldsymbol{r}_2 \quad \text{and} \quad 
\boldsymbol{r} = \boldsymbol{r_1}-\boldsymbol{r_2}\thinspace,
$$
where $M=m_1+m_2$, this operator may be written as
$$
H = -\frac{1}{2M}\nabla_R^2-\frac{1}{2\mu}\nabla_r^2 + V(\boldsymbol{r})
$$
with $\mu^{-1} = m_1^{-1}+m_2^{-1}$. This is somewhat easy to show just by standard differentiation rules. Then one proceeds with separation of variables in $\boldsymbol{R}$ and $\boldsymbol{r}$.
What confuses me is this: To be able to separate the variables $\boldsymbol{R}$ and $\boldsymbol{r}$, one should first show that these variables are indeed independent. Correct? If so, how does one go about doing that? I thought about taking the derivative of $\boldsymbol{r}$ with respect to $\boldsymbol{R}$, but I cannot figure out of to do it. Any advice or guidance would be much appreciated.
 A: I think it would help here to first explore what we really mean by "independent" variables. To begin with, we have a a pair of coordinates $(\boldsymbol{r_1},\boldsymbol{r_2})$. We then define a new set of coordinates (very much like going to new "generalized" coordinates as we do in classical mechanics)
$$
(\boldsymbol{R}(\boldsymbol{r_1},\boldsymbol{r_2}),\boldsymbol{r}(\boldsymbol{r_1},\boldsymbol{r_2})).
$$
Note that the new coordinates are functions of the old coordinates. The coordinate transformation is a "good" one, meaning it will do everything we need it to including what you're calling independence, if it is possible to invert this functional relation and write the old coordinates as functions of the new ones:
$$
(\boldsymbol{r_1}(\boldsymbol{R},\boldsymbol{r}),\boldsymbol{r_2}(\boldsymbol{R},\boldsymbol{r})).
$$
From the relations you have defined, this can actually be done very explicitly since the equations relating the new and old variables are all linear.
So in this sense you have defined a good coordinate change and so these coordinates will do the trick. How does this relate to some notion of independence? The implicit function theorem essentially tells us that if we write $(\boldsymbol{R}(\boldsymbol{r_1},\boldsymbol{r_2}),\boldsymbol{r}(\boldsymbol{r_1},\boldsymbol{r_2}))$ as we have done, then the inverse will exist (under some mild assumptions on differentiability and such) if and only if the Jacobian of the transformation is invertible. The Jacobian, being the matrix of derivatives of the coordinate change, will be invertible if its determinant is non-zero, which is the usual way of stating invertibility, but it would be equivalent to say that the Jacobian will be invertible if all its rows (or columns) are linearly independent vectors.
So the invertibility of a coordinate change, and hence its suitability as one, is essentially equivalent to the linear independence of the vector of derivatives of the components. That is, in this particular example, if the vectors
$$
\left\langle \frac{\partial R_i}{\partial\boldsymbol{r_1}},\frac{\partial R_i}{\partial\boldsymbol{r_2}}\right\rangle,\qquad\left\langle \frac{\partial r_i}{\partial\boldsymbol{r_1}},\frac{\partial r_i}{\partial\boldsymbol{r_2}}\right\rangle
$$
for each value of $i$ indicating the components of the coordinates we are transforming to (so if you're in 3 dimensions, $i=1,2,3$).
A: Hint: $(\boldsymbol{r}_1,\boldsymbol{r}_2)$ and $(\boldsymbol{R},\boldsymbol{r})$ are both valid coordinate systems for the 6D configuration space. In particular each coordinate system consists of 6 independent variables, and there is a bijective transformation between the 2 coordinate systems.
