Center of mass in hydrogen atom I have few questions regarding quantum treatment of the hydrogen atom problem.

*

*Why does one changes coordinate from position vector of electron and nucleus to COM coordinates and relative position between electron and the nucleus? I know there are 6 position vectors and it will be difficult to solve but are there are any other reason of choosing them? We say that Schrödinger equation in the center of mass is fixed and doesn't provide that much information.


*I saw a video where the professor says that during relative motion of two body, COM doesn't change. I don't know why is that so.


*As we assume nucleus to be stationary and electron moves, don't we think its COM will change as electron moves? having different COM location, so what is the actual scenario? Doesn't that mean that as electron moves, the nucleus should also move a bit?


*I don't get the concept clear in my head that after change in coordinates, we keep COM location to be at origin. Why? And we study the final motion of COM of system and the electron, as electron is already in COM.
 A: The time-independent Schrödinger equation for the hydrogen atom
(i.e. nucleus and electron) is
$$\left(
  -\frac{\hbar^2}{2m_n}\nabla_n^2
  -\frac{\hbar^2}{2m_e}\nabla_e^2 -
  -\frac{e^2}{4\pi\epsilon_0|\mathbf{r}_e-\mathbf{r}_n|}
  \right)\Psi(\mathbf{r}_n,\mathbf{r}_e)
  = E \Psi(\mathbf{r}_n,\mathbf{r}_e) \tag{1}$$
where $\Psi(\mathbf{r}_n,\mathbf{r}_e)$ is the wavefunction of nucleus
and electron.


*

*Why does one changes coordinate from position vector of electron and
nucleus to COM coordinates and relative position between electron
and the nucleus.
Like I know there are 6 position vectors and it will be difficult
to solve but are there are any other reason of choosing them.
Like after some maths and physics we say that schrodinger equation
of center of mass is fixed and doesn't give that much of information.


Equation (1) is hard to solve because of the potential energy term depending
on both positions $\mathbf{r}_e$ and $\mathbf{r}_n$.
But at least, the potential energy depends only on the position difference
$\mathbf{r}_e-\mathbf{r}_n$.
That is why we decide to not attack this problem in the original coordinates
$\mathbf{r}_n$ and $\mathbf{r}_e$.
Instead we define the new coordinates:
$$\begin{align}
\mathbf{r} &= \mathbf{r}_e - \mathbf{r}_n 
  \quad &\text{position difference between electron and nucleus}\\
\mathbf{R} &= \frac{m_n\mathbf{r}_n + m_e\mathbf{r}_e}{m_n+m_e}
  \quad &\text{center-of-mass position}
\end{align} \tag{2}$$
That means we don't look for a wavefunction $\Psi(\mathbf{r}_n,\mathbf{r}_e)$,
but for a wavefunction $\psi(\mathbf{R},\mathbf{r})$.
And keep in mind, these positions $\mathbf{R}$ and $\mathbf{r}$ are just
abstract definitions. There is no particle at any of these two positions.
With these new coordinates we can transform (skipping the mathematical details)
the Schrödinger equation (1) to
$$\left(
  -\frac{\hbar^2}{2(m_n+m_e)}\nabla_R^2
  -\frac{\hbar^2(m_n+m_e)}{2m_nm_e}\nabla_r^2
  -\frac{e^2}{4\pi\epsilon_0|\mathbf{r}|}
  \right)\psi(\mathbf{R},\mathbf{r})
  = E \psi(\mathbf{R},\mathbf{r}) \tag{3}$$
This equation can be written a little bit simpler by defining
$M=m_n+m_e$ (the total mass) and $\mu=\frac{m_nm_e}{m_n+m_e}$
(the so-called reduced mass):
$$\left(
  -\frac{\hbar^2}{2M}\nabla_R^2
  -\frac{\hbar^2}{2\mu}\nabla_r^2
  -\frac{e^2}{4\pi\epsilon_0|\mathbf{r}|}
  \right)\Psi(\mathbf{R},\mathbf{r})
  = E \Psi(\mathbf{R},\mathbf{r}) \tag{4}$$
Now for solving this equation (4) we can use separation of variables,
i.e. we do the approach
$\psi(\mathbf{R},\mathbf{r}) = \psi_R(\mathbf{R}) \psi_r(\mathbf{r})$.
We get two separate equations:
$$-\frac{\hbar^2}{2M}\nabla_R^2 \psi_R(\mathbf{R})
 = E_R \psi_R(\mathbf{R}) \tag{5a}$$
and
$$\left(
  -\frac{\hbar^2}{2\mu}\nabla_r^2
  -\frac{e^2}{4\pi\epsilon_0|\mathbf{r}|}
  \right)\psi_r(\mathbf{r})
  = E_r \psi_r(\mathbf{r}) \tag{5b}$$
Unlike equation (1), these separate equations (5a, 5b) are easily solvable.
And that is the whole point why we used the coordinates defined in (2)
in the first place.
The solutions of (5a) are simple plane waves for the COM motion
(with any wave-vector $\mathbf{k}$, i.e. with constant total momentum
$\mathbf{P}=\hbar\mathbf{k}$):
$$\psi_R(\mathbf{R})=e^{i\mathbf{k}\mathbf{R}} \tag{6a}$$
And the solutions of (5b) are the hydrogen orbitals
(centered at $\mathbf{r}=\mathbf{0}$) for the relative motion:
$$\psi_r(\mathbf{r})=\psi_{nlm_l}(r,\theta,\phi) \tag{6b}$$
Also keep in mind, because $\mathbf{r}$ and $\mathbf{R}$ were only abstract
"positions", the wavefunctions $\psi_r(\mathbf{r})$ and $\psi_R(\mathbf{R})$
are abstract as well. Especially, $\psi_r(\mathbf{r})$ describes the movement
of the position difference between electron and nucleus, but not directly
the position of the electron.



*I saw one of the video where the professor says that during
relative motion of two body, COM doesn't changes.
I don't know why is that so.


That is not quite correct. Actually the COM does move,
but it moves only with constant speed in a straight line,
because there is no potential energy in equation (5a),
and hence no force acting on the COM.



*As we assume nucleus to be stationary and electron moves,
don't we think its COM will change as electron moves,
having different COM location, so what is the actual scenario.
Does that don't mean that as electron moves, the nucleus should also move a bit.


You are right, the nucleus moves a little bit relative to the COM.
You can see this when you solve the transformation (2)
for $\mathbf{r}_e$ and $\mathbf{r}_n$:
$$\begin{align}
\mathbf{r}_e = \mathbf{R} + \frac{m_n}{m_n+m_e}\mathbf{r} \\
\mathbf{r}_n = \mathbf{R} - \frac{m_e}{m_n+m_e}\mathbf{r}
\end{align} \tag{7}$$
From the second equation of (7) you see, the nucleus is moving a little bit
relative to the COM position $\mathbf{R}$, because $\frac{m_e}{m_n+m_e}\ll 1$.



*I don't get the concept clear in my head that after change in coordinates,
we keep COM location to be at origin. Why? And we study the final motion
of COM of system and the electron, as electron is already in COM thing.


You are confusing COM position and position difference.
The COM position is at $\mathbf{R}$.
The hydrogen orbital $\psi_r(\mathbf{r})$ is in terms of the position difference $\mathbf{r}$.
From equations (2) and (7) you can visualize it like this.

A: 1: Because the problem is separable in a free particle Hamiltonian for the COM, and then the hydrogen atom for the reduced mass coordinate.
2: For every force there is an equal and opposite force, so $f = ma = -F = -MA$ preserves the center of mass
3:We don't assume the nucleus to be stationary. It's in little tiny $Y_{lm}(\theta, \phi)$, too.
4: The COM is at the origin because $\vec r = 0$ means $\vec r_p = \vec r_e$ which must be the COM since all mass is at one point.
