Does a wave function describe the motion of electrons or atoms?

Question

I took the course of quantum mechanics a while ago. I do not quite remember all the detail on how to derive the wave function for hydrogen but I still remember the general picture. I think the text always start the discussion with hydrogen because this atom contains 1 electron only. It is a simple 2-body model so we could solve the problem in the frame of center of mass. In other situation for atom has more than 1 electron but the outermost shell contains only 1 electron, we could reduce it as a hydrogen-like ion with 1 electron. Since the nucleus or the hydrogen-like ion is much heavier than the electron, we consider it a rest ion so the wave function we derive is mostly describing the motion of the electron? But in the book, they always mention the wave function of an atom. Is it the same thing to say the wave function for the atom or the electron (in the atom)?

As I still remember, if the wave function is used to describe the atom, the modulus square of the wave function is interpreted as the probability to find the atom in space; but if it is for electron, we should say it is interpreted as the probability of finding the electron in space. It is confusing.

Answer 1

The time-independent Schroedinger equation for the hydrogen atom is

$$ H \ \Psi(\vec r_N, \vec r_e) = E \ \Psi(\vec r_N, \vec r_e)$$

where $\vec r_N$ is the position of the nucleus, $\vec r_e$ the position of the electron and

$$H = \frac{p_N^2}{2 m_N} + \frac{p_e^2}{2 m_e} - \frac{e^2}{4 \pi \epsilon_0 \mid \vec r_N - \vec r_e\mid} \, .$$

We usually use center of mass coordinates:

$$\vec r = \vec r_e - \vec r_N\\\vec R= \frac{m_N\ \vec r_N + m_e \vec r_e}{m_N+m_e}$$

so that the Schroedinger equation becomes

$$H \ \Psi(\vec R,\vec r) = E \ \Psi (\vec R,\vec r)$$

where the hamiltonian is now

$$H = \frac{P^2}{2 M} + \frac{p^2}{2 \mu} + \frac{e^2}{4 \pi \epsilon_0 \mid \vec r\mid} $$

where $M=m_N+m_e$ is the total mass, $\vec P = \vec p_N + \vec p_e$ is the total momentum, $\mu \equiv (m_N m_e)/(m_N+m_e)$ is the reduced mass and $\vec p = (m_N \vec p_e - m_e \vec p_N)/(m_N+m_e)$ is the relative momentum.

The total wave function is the product of the nuclear and the electronic wave function ($\Phi$ and $\psi$ respectively):

$$\Psi(\vec R, \vec r) = \Phi(\vec R) \ \psi(\vec r)$$

where

$$\frac{P^2}{2M} \ \Phi (\vec R) = E_{CM} \ \Phi (\vec R)$$

and

$$\left(\frac{p^2}{2 \mu} + \frac{e^2}{4 \pi \epsilon_0 \mid \vec r\mid}\right)\ \psi (\vec r)= E \ \psi (\vec r) \, .$$

The orbitals we always talk about and that you have seen depicted (see following picture for example) are the $\psi(\vec r)$ and are usually expressed in polar coordinates: $\psi(r,\theta,\phi)$.

So what we usually refer to as "orbital" is not the wave function of the whole atom (which would be $\Psi$) nor the electronic wave function, but the wave function of a fictitous particle with mass $\mu$ , position $\vec r$ and momentum $\vec p$.

However, since $m_N \simeq 1800 \ m_e$, you can see that $\vec R \simeq \vec r_N$, $M \simeq m_N$ and $\mu \simeq m_e$, so that $\Phi$ is in a certain way close to the nuclear wave function and $\psi$ to the electronic wave function. Also notice that the Schroedinger equation for $\Phi$ is a free particle equation, which is trivial to solve and will held a plane wave solution.

Answer 2

Atoms.

We sometimes use somewhat loose language when we speak of the wave function of the electron. We typically either have chosen the center of mass frame, or pretended that the nuclei are fixed (as you point out). It's the atom that has energy levels.

We can get away with the loose language because if we really were able to fix the positions of the atoms.