What Does 'Mutual Potential Energy' Refer to in a Hydrogen Atom?

[Erik 05]

Homework Statement

Hence calculate the expectation value and uncertainty of the mutual potential energy of the electron and proton for a hydrogen atom in this state.

This is the Coulomb model, and that is the question. I just need to know if 'mutual potential energy' means the effective potential energy, or just the Coulomb potential energy.

Homework Equations

The Attempt at a Solution

I think probably it's just Coulomb...

 

[kuruman]

The word "Hence" suggests that there is more statement of the problem preceding what you posted which might be relevant to answering your question. Specifically what "this state" refers to. From the looks of it, however, I would agree that it's just the Coulomb potential energy.

 

[Erik 05]

The part before is "Calculate the expectation values of 1/r and 1/r^2 in the state described by ψ2,1,−1."

 

[ehild]

The part before is "Calculate the expectation values of 1/r and 1/r^2 in the state described by ψ2,1,−1."

And what is the mutual potential energy function of a proton and an electron?

 

[Erik 05]

Coulomb, probably. What is throwing me is that the effective potential energy also includes a term for the angular momentum, but I would have thought the angular momentum energy was a part of kinetic energy.

 

[kuruman]

Coulomb, probably. What is throwing me is that the effective potential energy also includes a term for the angular momentum, but I would have thought the angular momentum energy was a part of kinetic energy.

Classically, the rotational part of kinetic energy is $K_{rot}=\frac{1}{2}mr^2 \dot{\phi}^2$. Because angular momentum is conserved in a central potential, $L=mr^2 \dot{\phi}=const.~$This allows rewriting $K_{rot.}=\frac{L^2}{2mr^2}$. It looks like a potential, but it isn't.

Quantum mechanically, the potential term in the Hamiltonian is $V(r)=-\frac{e^2}{4\pi \epsilon_0 r}$. You put that in the time independent Schrodinger equation, turn the crank to separate variables, and out comes a term with $\frac{L(L+1)}{2mr^2}$ in the radial part of the equation. Still not a potential; you have already taken the potential into account as $V(r)$.

 

[Erik 05]

Ok, that's clear, thanks.