Is my interpretation of the g-factor correct?

[Frigorifico9]

TL;DR

Can the electron g factor be understood as electrons having roughly twice as much momentum as the Bohr Mageton?

I'm trying to make sure I understand the g-factor of the electron, so if my question is flawed please don't just point out my flaws, but help me correct my understanding

If I understand correctly the magnetic moment of an object depends on it charge, its mass and its momentum

$$
\mu = \frac{q}{2m} L
$$

Now, given that charge and angular momentum are quantized, you'd think the smallest magnetic moment would correspond to the smallest charge and the smallest momentum it is possible to have, this is Bohr's magneton

$$
\mu = \frac{e}{2m_e} \hbar
$$

But then it turns out electrons in the hydrogen atom have twice that much magnetic moment, and (this is the part where I think I may be wrong) this is due to the fact that electrons have $\hbar$ momentum around the nucleus AND $\hbar$ intrinsic momentum (spin)

$$
\mu = \frac{e}{2m_e} (\hbar+\hbar)=\frac{e}{m_e} \hbar
$$

So it makes sense it would be twice as much... But of course it then turns out they have slightly more than that because of self interactions in QED and this is still an active area of research

Did I say everything right? Could you correct me on my misunderstandings?

 

[vanhees71]

An electron has a gyro-factor of about $2$. That means that its intrinsic magnetic moment is
$$\hat{\vec{\mu}}=g \frac{-e}{2m} \hat{\vec{s}}.$$
since the electron has spin 1/2 this means with $g \simeq 2$ that its magnetic moment is about 1 Bohr magneton.

The gyro-factor of 2 follows from using the "minimal-coupling principle", i.e., to get a gauge invariant theory you substitute $\partial_{\mu}$ by the gauge-covariant derivative $D_{\mu}=\partial_{\mu} +\mathrm{i} q A_{\mu}$. For the electron, $q=-e$. In leading order perturbation theory ("tree level") this leads to a gyro-factor of 2 when applied to the Dirac equation.

Then there are radiative corrections, i.e., higher-order corrections in QED, involving Feynman diagrams with loops. The predicted "anomalous magnetic moment" of the electron is among the most accurate values ever calculated.

 

[Vanadium 50]

Can the electron g factor be understood as electrons having roughly twice as much momentum as the Bohr Mageton?

Thr Bohr Magnetron is not a unit of momentum, or even angular momentum. It's like comparing gallons to volts.

 

[Meir Achuz]

TL;DR Summary: Can the electron g factor be understood as electrons having roughly twice as much momentum as the Bohr Mageton?

due to the fact that electrons have $\hbar$ momentum around the nucleus

No. The fact that g =2 has nothing to do with orbital angular momentum. The 2 comes out of the Dirac equation, which surprised everyone at the time. (1929?)

 

[Frigorifico9]

Thr Bohr Magnetron is not a unit of momentum, or even angular momentum. It's like comparing gallons to volts.

I agree, that's why I said that it DEPENDS on angular momentum not that it was angular momentum