Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Zeeman energy with an effective mass

  1. Dec 20, 2016 #1
    Hi all,

    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:

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

    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:

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

    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:

    E_{Zeeman}=\pm \frac{1}{2} g_{effective} \mu_B B,
    where the bohr magneton is then the nature constant defined with the free electron mass.
  2. jcsd
  3. Dec 20, 2016 #2
    My immediate impression of why this is bogus is that the spin is not a property described in any way by Schroedinger's equation, which is the one from which your typical treatment of electrons in condensed matter comes from. The spin is something we attach on top of it later on as an afterthought because we realise it is necessary to describe physical systems, in that view. So obviously it can not be influenced by the 'effective' mass, which is only effective in regards to how it acts on the kinetic term. You would certainly see the effect on the orbital magnetic momentum, but not on the intrinsic one.

    Spin is described as an emergent property only by Dirac's equation (there are also some non-relativistic equivalents to it but I can't find any good references as they're virtually useless beyond being physical curiosities). So if you were to redo the treatment of electrons in solid state through the Dirac equation you might have a better appreciation of what's going on. My guess is you'd just see how the 'effective' mass would only feature in the part concerning the dispersion relation of the free particle-like waves and somehow cancel out when spin is concerned. In order to see some weird 'effective' spin you would need to have a situation where multiple particles form one 'pseudo' particle with a different spin (for example electrons binding into Cooper pairs in superconductors).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Zeeman energy with an effective mass
  1. Zeeman Effect (Replies: 5)

  2. Effective mass (Replies: 1)

  3. Effective mass (Replies: 4)