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

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.

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.

• I don't think this gets to the OP's question. I don't think the Born-Oppenheimer approximation relates to center-of-mass calculation. In the typical hydrogen atom calculation, a center-of-mass calculation, the wave function is 8not factored into a nuclear and electronic part. It's factored into a center-of-mass part and a relative motion part. The B-O approximation is applied on top of the c.o.m. calculation in the case of molecules or solids. – garyp Jul 16 '16 at 11:30
• The c.o.m. factorization is not an approximation. Except, of course, we are ignoring relativity and coupling to the radiation field. Sorry for the second comment. I'm having editing difficulties. :) – garyp Jul 16 '16 at 11:38
• @garyp You are right, B-O is more about molecules. I'll edit my answer. – valerio Jul 16 '16 at 11:42
• @garyp Edited. I got confused because the nucleus-electron decoupling in the c.o.m. solution is similar to the B-O decoupling, but they are in fact different. B-O is not needed in the hydrogen atom case. – valerio Jul 16 '16 at 12:18

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.

• Thanks for your reply. I am more or less understand the statement. However, it is quite misleading in the text why they assume the nucleus is fixed in space. Even in the extreme low temperature the atom should move, so does nucleus. But I understand that since the nucleus is heavier than electron, so should I say even the atom is moving, the relative motion between the nucleus to the electron is steady? – user1285419 Jul 16 '16 at 18:25
• Your mention of temperature implies that the atom in question has other atoms nearby, and interacts with them. That's a different problem. An ensemble of atoms will have collisions, and as atoms approach each other in collisions their wavefunctions and energy levels change. The continual change in speed of the atom means that it sees the radiation field Doppler shifted in complicated ways. The analysis in this thread is about isolated atoms. For more complicated systems, approximations such as the B-O approx must be used. The relative motion is steady only for isolated atoms. – garyp Jul 17 '16 at 21:27