# Orbital motion problem

## Main Question or Discussion Point

I am analysing the motion of the electron and the nucleus of a hydrogenic atom. I know that the two particles, so to speak, rotate their common centre of mass and that the two-body system is identical with respect to the predictions (e.g. energy) borne out by the system to a one-body system of reduced mass mu. Correct me if I have been wrong on this.

Now my question is this:

In deriving the energy levels of the hydrogenic atom, we assume that the nucleus is infinitely heavy, so that the reduced mass becomes the mass of the electron. Why do we do so and what follows next?

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dextercioby
Homework Helper
We don't assume that if we want to keep the mathematical model as close to reality as possible. That's why, the full energy spectrum of the hydrogen atom (without relativistic and all other corrections) is composed by the reunion of 3 sets. One set is the positive real axis corresponding to the motion of the COM, one is the infinite discrete set of negative real spectral values (associated to the RHS of the <dummy> particle of mass = reduced mass) and the last one is a positive real axis of spectral values, again associated to the <dummy> particle of mass = reduced mass.

I hope one can make sense of what I wrote :uhh:

In deriving the energy levels of the hydrogenic atom, we assume that the nucleus is infinitely heavy, so that the reduced mass becomes the mass of the electron. Why do we do so and what follows next?

You don't even need to assume that it is infinite, just know that the mass of the proton is a lot larger than the mass of the electron. This basically means that the reduce mass becomes the mass of the electron and that you can treat the system as having a center of mass on the proton. I think this simplifies the problem because your electrostatic potential will begin at the center of mass which makes the hamiltonian a bit easier to write (or is it a lot easier?).

alxm
I'd prefer to look at it this way, since it covers the general case, and not just hydrogen:

If you start with the Schrödinger equation for the an atom or molecule, then you have 3-4 sets of terms: Nuclear kinetic energy, electronic kinetic energy, the nuclear-electronic potential, and if you have more than one electron, the electronic-electronic potential. You have the nuclear coordinates and the electronic coordinates as variables. For other systems than the hydrogenic atom, you can't solve this equation analytically, even if they only have one electron, since the electronic coordinates depend on the nuclear coordinates and vice-versa. Through the potential, the nuclear kinetic energy is coupled to the electronic kinetic energy, so it's a many-body problem.

The first approximation/simplification you can do here is the Born-Oppenheimer approximation. The nuclei are much heavier than the electrons and correspondingly move much slower. That means the potential 'felt' by the electrons varies very slowly. If a potential varies sufficiently slowly, there's no change of state for the electrons - That's the Adiabatic Theorem (adiabatic meaning 'no transfer of energy' in this case). The B-O approximation is an application of that. Another way of looking at this, is that the nuclei are so slow that they can be viewed as stationary in the electron's inertial frame (or, 'from their point of view'). Which means you can, to a good approximation, neglect the coupling of nuclear kinetic energy to electronic kinetic energy. Note that this doesn't mean you're assuming the nuclei are stationary relative your system's frame of motion, just that they're stationary relative the electrons. The nuclei can still have kinetic energy, just not transfer it.

So if you get rid of that coupling, you can separate the Schrödinger Equation into two PDEs, for the nuclear and electronic motion, respectively. In the electronic equation, the nuclear coordinates are no longer variable, but constants set to their average locations. So you can then solve the two equations separately, and the total energy is the sum of the electronic and nuclear energies.

Now, there's another approximation you can do here, which is to assume the nuclei have infinite mass i.e. that they're truly stationary, not just stationary-relative-the-electrons. That means you just neglect the nuclear kinetic-energy contribution to the total energy. This is usually called the 'clamped-nucleus Hamiltonian'. It's an additional approximation on top of the B-O approximation, and often confused with it, because both assume the nuclei are stationary (or 'infinitely heavy'), but in different inertial frames.

Okay, so that's how it works in general. But you have additional simplifications that apply in the case of the hydrogenic atom. First there's the situation for single atoms in-general: In the case where you have a single atom, it's a free particle. The nucleus only has translational degrees of freedom. The electronic energy is the same no matter how fast the nucleus is moving, and the translational kinetic energy of the nucleus is arbitrary and set to zero. So the clamped nucleus approximation isn't an approximation for a single atom, where the nucleus doesn't have any kinetic energy. It comes into play for molecules, where the nuclei are bound and have vibrational degrees of freedom. Due to zero-point vibrations, their nuclear kinetic energy is always greater than zero (but small).

The other thing that applies for the hydrogenic atom (but not in general), is that you only have one independent coordinate; the nuclear-electronic distance. This means that by introducing a reduced-mass coordinate system, you can eliminate the nuclear-electronic kinetic energy coupling and make the equation separable. For the hydrogenic atom then, you don't need the B-O approximation to make the thing separable, it's enough to used reduced-mass coordinates. But if you do use the B-O approximation, that's when you don't change the masses and assume it's separable anyway. And that is equivalent to assuming the nucleus is stationary relative the electron. And then the result is only approximate.

So you have to be careful about assuming the nuclei are stationary/infinitely massive, because it can mean two different approximations depending on the context. And you have to be a bit careful with the hydrogenic atom, since it happens to be separable without the B-O approximation, which isn't the case in general.