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I am reading Schwartz's book on Quantum field theory. In the chapter anomalous magnetic moment (chapter 17) he mentions that Dirac equation naturally implies that the electron spin $g$-factor is 2. Can anyone please explain why does the Dirac equation implies that $g_s=2$?

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  • $\begingroup$ Do you want an explanation in words, or the actual math? $\endgroup$ – G. Smith Sep 16 at 17:41
  • $\begingroup$ Actual math would be better. But any explanation on why this is 'natural' is also very welcome. $\endgroup$ – abhijit975 Sep 16 at 18:08
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You can derive the electron's magnetic moment by starting with the Dirac equation for an electron in an electromagnetic field and looking at its non-relativistic limit. You find that there is an interaction energy of the form $-\vec{\mu}\cdot\vec{B}$ -- the interaction of a magnetic dipole with the magnetic field -- where $\vec{\mu}=q\vec{S}/m$. This means that $g_s=2$. Here are the details, using natural units where $\hbar$ and $c$ are $1$:

The Dirac equation for an electron with charge $q=-|e|$ in an electromagnetic field described by the four-potential $A^\mu=(\phi,\vec{A})$, $A_\mu=(\phi,-\vec{A})$ is

$$[\gamma^\mu(i\partial_\mu-qA_\mu)-m]\Psi=0\tag{1}$$

where $\Psi$ is the four-component Dirac spinor for the electron.

The standard representation of the $4\times 4$ Dirac gamma matrices is

$$\gamma^0=\begin{pmatrix}I_2 & 0\\0 & -I_2\end{pmatrix}\tag{2a}$$

$$\gamma^i=\begin{pmatrix}0 & \sigma_i\\-\sigma_i & 0\end{pmatrix}\tag{2b}$$

where $I_2$ is the $2\times 2$ identity matrix and the $\sigma_i$ are the $2\times 2$ Pauli matrices satifying

$$\sigma_j\sigma_k=\delta_{jk}I_2+i\epsilon_{jkl}\sigma_l\tag{3}.$$

To study the non-relativistic limit of the Dirac equation, write

$$\Psi=\begin{pmatrix}\Psi_+\\ \Psi_-\end{pmatrix}e^{-iEt}\tag{4}$$

where $\Psi_+$ and $\Psi_-$ have two components each.

One then finds two coupled equations for $\Psi_+$ and $\Psi_-$:

$$(E-m-q\phi)\Psi_++\vec{\sigma}\cdot(i\vec\nabla+q\vec{A})\Psi_-=0\tag{5a};$$

$$(E+m-q\phi)\Psi_-+\vec{\sigma}\cdot(i\vec\nabla+q\vec{A})\Psi_+=0\tag{5b}.$$

In the non-relativistic limit, one can make the approximation

$$E+m-q\phi\approx2m\tag{6}$$

and solve (5b) to get

$$\Psi_-\approx-\frac{1}{2m}\vec\sigma\cdot(i\vec\nabla+q\vec{A})\Psi_+\tag{7}.$$

Substituting this into (5a) gives

$$\left\{\frac{1}{2m}[\vec\sigma\cdot(i\vec\nabla+q\vec{A})]^2+q\phi\right\}\Psi_+\approx (E-m)\Psi_+\tag{8}.$$

Using (3) and $\vec{B}=\vec\nabla\times\vec{A}$, this becomes

$$\left[\frac{1}{2m}(-i\vec\nabla-q\vec{A})^2+q\phi-\frac{q}{2m}\vec\sigma\cdot\vec{B}\right]\Psi_+\approx (E-m)\Psi_+\tag{9}.$$

This looks like a non-relativistic time-independent Schrodinger equation with an extra interaction energy,

$$-\frac{q}{2m}\vec\sigma\cdot\vec{B}=-\frac{q}{m}\vec{S}\cdot\vec{B}\tag{10}$$

where

$$\vec{S}=\frac{1}{2}\vec\sigma\tag{11}$$

is the non-relativistic spin angular momentum operator.

Since the interaction energy of a magnetic dipole moment $\vec\mu$ in a magnetic field $\vec{B}$ is $-\vec{\mu}\cdot\vec{B}$, this means that the electron has a magnetic moment

$$\vec\mu=\frac{q}{m}\vec{S}\tag{12}$$

and thus $g_s=2$ since the definition of $g_s$ is

$$\vec\mu=g_s\frac{q}{2m}\vec{S}\tag{13}.$$

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  • $\begingroup$ I have a question about Eq. (6). In the non-relativistic limit, $E\approx m$ since the momentum is small. But why is the third term negligible? Is it because we can use gauge fixing to fix $A_0 = 0$ or because the electrostatic potential energy is small compared to the rest mass of the electron? $\endgroup$ – abhijit975 Sep 17 at 1:30
  • $\begingroup$ Because the electrostatic potential energy is small compared to the rest mass of the electron. $\endgroup$ – G. Smith Sep 17 at 2:11
  • $\begingroup$ Note: Since I don’t have a copy of Schwarz, I don’t know whether I’ve used the same sign conventions as in that book. How this derivation looks in various references depends both on the metric signature and on whether $e$ is a positive or negative number. For example, see physics.stackexchange.com/questions/54907/… $\endgroup$ – G. Smith Sep 17 at 4:23
  • $\begingroup$ That's fine. I needed a guideline on how to take the non-relativistic limit to obtain something equivalent to eq. (9) and your answer covers that perfectly. Thanks for the help! $\endgroup$ – abhijit975 Sep 17 at 18:34

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