# A How the g factor comes from QFT?

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1. Jul 5, 2017

### leo.

I'm reading the book Quantum Field Theory and the Standard Model by Matthew Schwartz and currently I'm studying the chapter 17 titled "The anomalous magnetic moment" which is devoted to computing the corrections due to QFT to the g factor.

My main issue is in the begining of the chapter, where the author tries to establish a connection between the g factor and QED. First the author presents the non-relativistic limit of the Hamiltonian predicted by Dirac's equaton $$H = \dfrac{P^2}{2m}+V(R)+\dfrac{e}{2m}\mathbf{B}\cdot(\mathbf{L}+g\mathbf{S})$$. Then he says he wants to extract $g$ without having to take this non-relativistic limit.

First, given Dirac's equation $(i\gamma^\mu D_\mu-m)\psi=0$ one can get a Klein-Gordon type equation which is $(D_\mu^2+m^2+\frac{e}{2}F_{\mu\nu}\sigma^{\mu\nu})\psi=0$ where $D_\mu = \partial_\mu + ieA_\mu$ is the covariant derivative and $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$.

The last term is then what tells the difference between the scalar and spinor field. In the Weyl representation one can then compute $$\dfrac{e}{2}F_{\mu\nu}\sigma^{\mu\nu}=-e\begin{pmatrix}(\vec{B}+i\vec{E})\cdot\vec{\sigma} & 0 \\ 0 & (\vec{B}-i\vec{E})\cdot \vec{\sigma}\end{pmatrix}$$

My doubt then is that the author says that "going to momentum space, $(D_\mu^2+m^2+\frac{e}{2}F_{\mu\nu}\sigma^{\mu\nu})\psi=0$ implies" $$\dfrac{\left(H-eA_0\right)^2}{2m}\psi=\left(\dfrac{m}{2}+\dfrac{(\mathbf{p}-e\mathbf{A})^2}{2m}-2\dfrac{e}{2m}\mathbf{B}\cdot\mathbf{S}\pm i\dfrac{e}{m}\mathbf{E}\cdot\mathbf{S}\right)\psi$$

This is my first issue. How this is true? I mean, what is this Hamiltonian $H$ in the LHS and how this equation follows from the first KG type equation derived from the Dirac equation? This is not the classical limit, so I'm missing something here.

After that, the author says that "this can be compared directly to the classical limit to read off the strenght of the magnetic dipole intraction $ge\mathbf{B}\cdot\mathbf{S}$ and since $\mathbf{S}=\frac{\vec{\sigma}}{2}$ we find $g=2$"

Here's my second doubt, how the author is making this comparison? I mean it doesn't seem so direct as he says.

So how he derives that equation from the KG type equation, and how this allows him to read of the value of $g$?

2. Jul 11, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jul 12, 2017

### vanhees71

One should note that you can do the analogous calculation directly for the Pauli equation. The trick is to write
$$2m \hat{H}_1=[\vec{\sigma} \cdot (\hat{\vec{p}}-q \vec{A})]^2$$
with the Pauli matrices $\vec{\sigma}$. Multiplying this out gives
$$2m \hat{H}_1=(\vec{\sigma} \cdot \hat{\vec{p}})^2 + q^2 (\vec{\sigma} \cdot \vec{A})^2 + q [(\vec{\sigma} \cdot \hat{\vec{p}})(\vec{\sigma} \cdot \vec{A}) + (\vec{\sigma} \cdot\vec{A})(\vec{\sigma} \cdot \hat{\vec{p}})].$$
Since
$$\sigma_j \sigma_k+\sigma_k \sigma_j=2 \delta_{kj}$$
and since the $\hat{\vec{p}}$ commute and also the $\vec{A}$ commute with each other you get
$$2m \hat{H}_1 = \hat{\vec{p}}^2 + q^2 \vec{A}^2 q [(\vec{\sigma} \cdot \hat{\vec{p}})(\vec{\sigma} \cdot \vec{A}) + (\vec{\sigma} \cdot\vec{A})(\vec{\sigma} \cdot \hat{\vec{p}})].$$
For the last term we have
$$\sigma_j \sigma_k (\hat{p}_j A_k+A_j \hat{p}_k)=\frac{1}{2} \left [\{\sigma_j,\sigma_k \} + [\sigma_j,\sigma_k] \right] (\hat{p}_j A_k + A_j \hat{p}_k)= [\delta_{jk} + \mathrm{i} \epsilon_{jkl} \sigma_l] (\hat{p}_j A_k + A_j \hat{p}_k)=\hat{\vec{p}} \cdot \vec{A}+\vec{A} \cdot \vec{p} + \mathrm{i} \epsilon_{jkl} \sigma_l [-\mathrm{i} \partial_j A_k + A_k \hat{p}_j + A_j \hat{p}_k]= \hat{\vec{p}} \cdot \vec{A}+\vec{A} \cdot \vec{p} + \vec{\sigma} \cdot \vec{B}.$$
So we finally have
$$\hat{H}=H_1+q \phi=\frac{1}{2m} [(\hat{\vec{p}}-q \vec{A})^2 + 2 q \vec{S} \cdot \vec{B}]+q \phi.$$
We have used that the spin is $\vec{S}=\vec{\sigma}/2$. This shows that the gyrofactor should be 2.