If magnetic fields don't do any work how does the zeeman effect work

Question

I am trying to understand exactly why we have terms such as $-\mu\cdot B$ in a Hamiltonian, as magnetic fields should not do any work and thus not change the energy of a system. I understand that classically if we have a magnetic dipole (a circular wire or a rotating ball of charge) They wiggle and oscillate while trying to align their moments with the magnetic field, but I always thought that if you were to follow all the motions of the charged particles and add their kinetic energy you'd find that the total energy has not changed.

Answer 1

I am trying to understand exactly why we have terms such as $-\mu\cdot B$ in a Hamiltonian, as magnetic fields should not do any work and thus not change the energy of a system.

The thing about magnetic force doing no work is often misunderstood and overrated.

The correct statement is that in classical physics, magnetic part of the Lorentz force on a charged particle does no work. This is trivial consequence of the fact magnetic part of the Lorentz force is perpendicular to the particle velocity.

So the statement is about work of force, just like defined in mechanics, not about "work of a field". The latter does not seem to be a clear and useful concept.

For this "theorem" to be relevant, the subject of the force has to be single charged particle with zero magnetic moment, and the "magnetic force" has to be the magnetic part of the Lorentz force. Otherwise "magnetic force" may be something different, possibly with a component along velocity of the subject body, and capable of doing work.

For example, the statement does not hold when we talk about work of macroscopic magnetic force on a current-carrying wire, or magnet; macroscopic magnetic forces of these kinds can do non-zero work on the body carrying current or magnetic moment.

Under some assumptions, one can derive that potential energy of a body with magnetic moment $\boldsymbol{\mu}$ in external magnetic field is $-\boldsymbol{\mu}\cdot \mathbf B$. This does not mean magnetic part of a Lorentz force can do work on a charged particle in that body, it means that changing magnetic moment orientation in space implies some work of electric part of the Lorentz forces, which is non-zero only when magnetic field is present. It's like with motional EMF; it is doing non-zero work on current only when the conductor moves in external magnetic field. This work is not work of magnetic Lorentz forces, but is of the other forces there (electric and non-electromagnetic), when magnetic field is present and allows this energy transformation and work to happen.

Answer 2

If magnetic fields don't do any work how does the zeeman effect work

The question is about the Zeeman effect, which is a quantum mechanical effect related to a system with a magnetic moment in a magnetic field. I will first discuss the classical description of a dipole in a magnetic field and then return to the quantum case.

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Classical Case:

In his Lectures on Physics Volume 2 (Section 15-1), Richard Feynman writes about the classical term $$ U_{mech} = -\vec \mu \cdot \vec B\;, $$ saying:

"... this energy is not the total energy of the current loop... $U_{mech}$... is not the real energy. It can, however, be used in computing forces, by the principle of virtual work..." (Emphasis in original.)

Classically, if you use the potential term $$ U_{mech} = -\vec \mu\cdot\vec B $$ to calculate the mechanical force required to move a current loop through an inhomogeneous magnetic field, you will find that a mechanical force is required, and you will find a mechanical work done in moving the loop in from infinity that is non-zero and equal to $U_{mech}$.

(If you look at Feynman's Vol 2 Section 15-2, you will see that the total classical energy of the classical current loop (including the current loop power source keeping the current constant) and the external field source turns out to be the negative of the mechanical energy, which is a bit surprising and can also lead to confusion.)

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Quantum Case:

Again, quoting Feynman, this time from Vol 3 Section 12-4:

"What, then is the Hamiltonian? We'll just tell you the answer, because we can't give you any 'proof' except to say that is the way the atom works. The Hamiltonian is $$ \hat H = A(\vec \sigma_e\cdot\vec \sigma_p) - \mu_e\vec \sigma_e\cdot\vec B - \mu_p\vec \sigma_p\cdot\vec B\;." $$

The term you are interested in is $$ \hat H_e = -\mu_e\vec\sigma_e\cdot \vec B\;, \tag{1} $$ and if we write $$ \vec \mu \equiv \mu_e \vec \sigma_e $$ then the expression for the Hamiltonian energy contribution of Eq. (1) looks a lot like the classical expression for $U_{mech}$.

The similarity between $\hat H_e$ and $U_{mech}$ should be considered more as a mnemonic device than a meaningful correspondence.