- #1

Bjarke Nicolaisen

- 5

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Say we have a model in a solid state system where we have an effective mass, \begin{equation}m^*,\end{equation}some fraction of the free electron mass. Now we apply an external magnetic field. Then the Zeeman energy splitting should be:

\begin{equation}

E_{Zeeman}=\pm 1/2 g \mu_B B

\end{equation}

My question is: Does the effective mass also enter in the expression for the Zeeman energy? The bohr magneton is proportional to the inverse of the mass, as we know. If that is the case, then in a system where we have an effective mass, we could describe the dynamics as the g-factor changing into an effective g-factor I guess. So if the mass ratio is equal to, say, 10, then we could have an effective g-factor equal to 20, without spin-orbit coupling.

I tried to answer the question myself using an approach I saw in Blundell,

*Magnetism in Condensed Matter*exercise (1.7), where you start with the kinetic energy:

\begin{equation}

\frac{p^2}{2m} =

\frac{(\sigma \cdot p)^2}{2m} \rightarrow

\frac{(\sigma \cdot (p+eA))^2}{2m},

\end{equation}

and when you write out the above, using that p and A do not commute, you get additionally to the kinetic energy a zeeman term with the bohr magneton multiplied. Of course, if you started this argument with an effective mass:

\begin{equation}

\frac{p^2}{2m^*},

\end{equation}

this would propagate all the way through the derivation, leading to an effective g-factor (or, if you will, an effective bohr magneton). However, I am told that this derivation is bogus and makes no sense. However I do not really understand why. So if someone could make sense of this it would be nice.

Say that the effective mass does affect the g-factor; do you know if this feature is included in what people normally speak of when they talk about the effective g-factor size, or if this would be a separate feature? I would guess if the effective mass does affect the zeeman energy, people would normally include this feature into the effective g-factor, such that you would always have an expression like:

\begin{equation}

E_{Zeeman}=\pm \frac{1}{2} g_{effective} \mu_B B,

\end{equation}

where the bohr magneton is then the nature constant defined with the free electron mass.