# Why $g$ is given by this coefficient of a term in the matrix element?

In the non-relativistic limit of the Dirac equation, we are led to the hamiltonian

$$H = \dfrac{P^2}{2m}+V(R)+\dfrac{e}{2m}\mathbf{B}\cdot \mathbf{L}+\dfrac{ge}{2m}\mathbf{B}\cdot \mathbf{S}$$

and we have this $g$ factor there.

Now I'm reading about the anomalous magnetic moment in QFT and the author starts discussing that charged spinors satisfy the KG equation $(\displaystyle{\not} D^2+m^2)\psi=0$ where we have

$$\displaystyle{\not} D^2=D_\mu^2+\dfrac{e}{2}F_{\mu\nu}\sigma^{\mu\nu}$$

He then computes this in the Weyl representation and says that the KG equation implies

$$\dfrac{(H-eA_0)^2}{2m}\psi=\left(\dfrac{m}{2}+\dfrac{(\mathbf{p}-e\mathbf{A})^2}{2m}-2\dfrac{e}{2m}\mathbf{B}\cdot\mathbf{S}\pm i\dfrac{e}{m}\mathbf{E}\cdot\mathbf{S}\right)\psi$$

which upon comparison with the other Hamiltonian leads the author to identify $g =2$. He then says that to look for corrections one needs to study loops that have the same effect as an additional $F_{\mu\nu}\sigma^{\mu\nu}$ term. What would mean "a loop that has the same effect as this additional term"?

Anyway, the author considers a process $e^{-}(q_1)\gamma(p)\to e^{-}(q_2)$ with on-shell spinors and off-shell photon.

He uses the Gordon identity to conclude that

$$\mathcal{M}_0^\mu=-e\left(\dfrac{q_1^\mu+q_2^\mu}{2m}\right)\overline{u}(q_2)u(q_1)-\dfrac{e}{2m}ip_\nu\overline{u}(q_2)\sigma^{\mu\nu}u(q_1)$$

He then says:

The first term is an interaction just like the scalar QED interaction: the photon couples to the momentum of the field, as in the $D_\mu^2$ term in the Klein-Gordon equation. The $q_1^\mu$ and $q_2^\mu$ in this first term are just the momentum factors that appear in the scalar QED Feynman rule. The second term is spin dependent and gives the magnetic moment. So we can identify $g$ as $\frac{4m}{e}$ times the coefficient of $ip_\nu \overline{u}\sigma^{\mu\nu}u.$

Now wait a minute, why $g$ is $\frac{4m}{e}$ times the coefficient of $i p_\nu\overline{u}\sigma^{\mu\nu} u$? This obviously gives $2$ in this case.

But what $g$ appearing in the non-relativistic hamiltonian I presented has to do with a factor in front of a term in a matrix element?

My main doubt is how can a term appearing in a matrix element be connected to a term appearing in a non-relativistic Hamiltonian. I believe this boils down to understanding what the author meant with "a loop that has the same effect as an additional $F_{\mu\nu}\sigma^{\mu\nu}$ term".

• The point is that lets say you perform a vertex correction calculation to some order and finds a $F_2 \epsilon_\mu(q)q_\nu \sigma^{\mu \nu}$ term. This term can then be seen to arise from an effective action. If you then apply an external magnetic field you can use the effective action term as an interaction Hamiltonian term for the magnetic moment calculation. – Lunaron Jul 6 '17 at 10:12

## 1 Answer

For a constant magnetic field in the z-direction the spin interaction term in the non-relativistic Hamiltonian is given by $$H_I=-g \frac{e}{2m} B S_z$$

Let us now look at the spin part of the matrix element $$-\frac{e}{2m}i p_\nu \bar{u}(q_2)\sigma^{\mu \nu}u(q_1)$$

If we re-insert the polarization vector $\epsilon_\mu$, this term arise from the action $$-\int d^4 x \frac{e}{4m}\bar{\Psi} F_{\mu \nu} \sigma^{\mu \nu}\Psi +\ldots$$

The above term will then contribute a factor of $$-\frac{e}{4m}F_{\mu \nu} \sigma^{\mu \nu}$$ to the eom.

In order to make the connection with the usual dirac equation result we take a constant magnetic field in the z-direction, or in terms of the vector potential $$\vec{A}=\frac{B}{2}(-y,x,0)$$. Recalling that $S_z=\frac{1}{2}\sigma ^{12}$, we find $$\frac{e}{4m}F_{\mu \nu} \sigma^{\mu \nu}=-\frac{e}{2m}(2)B S_z$$, and thus g=2.

The above explanation is however a bit unsatisfactory since we did not consider how we really measure the magnetic moment in field theory.

To this end recall the definition for the magnetic moment of the electron in an external magnetic field B: $$\mu B =-<e|H_I|e>$$. In QM for an electron with spin in the z-direction this directly gives $$\mu=\frac{eg}{4m}$$

In QFT the situation is a bit different, and the electron state is given as a wave packet with some distribution.

To be explicit consider an electron state at rest with the spin in the z-direction $$|e>=\int \frac{d^3p}{2E_p}f(\vec{p}) b^\dagger _{+}(\vec{p})|0>$$,

where $f$ is peaked around $\vec{p}=0$ and normalized such that $$<e|e>=1$$. Using the interaction Hamiltonian $$-\int d^4 x \frac{e}{4m}\bar{\Psi} F_{\mu \nu} \sigma^{\mu \nu}\Psi$$ for a spherically symmetric distribution(l=0) one finds after some algebra $g=2$.

Note that the spin-orbit coupling can be incorporated by choosing f to not be spherically symmetric. Higher order corrections can be incorporated by looking at their contributions to the $F_{\mu \nu} \sigma^{\mu \nu}$ term.