11.5: Electron Spin

According to Equation $\ref{1132}$, the relativistic Hamiltonian of an electron in an electromagnetic field is

\( \mbox{\boldmath\)$\cdot({\bf p}+e\,{\bf A})+ \beta\,m_e\,c^2.$ $\ref{1185}$

Hence,

$\ref{1186}$

where use has been made of Equations $\ref{1118}$ and $\ref{1119}$. Now, we can write

$i=1,3$, where $\Sigma_i = \left(\begin{array}{cc} \sigma_i& 0\\ [0.5ex]0& \sigma_i\end{array}\right).$ $\ref{1189}$

Here, 0 and $1$ denote $\sigma_i$ are conventional $\gamma^5\,\gamma^5=1$ , and

\( \left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldmath\) $\ref{1191}$

Now, a straightforward generalization of Equation $\ref{508}$ gives

\( \mbox{\boldmath\)\( \mbox{\boldmath\)\( \mbox{\boldmath\)${\bf a}$ and $\Sigma$ . It follows that $\ref{1193}$

However,

$\ref{1194}$

where $\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = ({\bf p}+e\,{\bf A})^2 + m_e^{\,2}\,c^2+e\,\hbar\,$$\cdot{\bf B}.$

$\ref{1195}$

Consider the non-relativistic limit. In this case, we can write

$\delta H$ is small compared to $\delta H^{\,2}$ , and other terms involving $\delta H\simeq -e\,\phi + \frac{1}{2\,m_e}\,({\bf p} + e\,{\bf A})^2 + \frac{e\,\hbar}{2\,m_e}\,$$\cdot{\bf B}.$ $\ref{1197}$

This Hamiltonian is the same as the classical Hamiltonian of a non-relativistic electron, except for the final term. This term may be interpreted as arising from the electron having an intrinsic magnetic moment

$= - \frac{e\,\hbar}{2\,m_e}\,$$.$ $\ref{1198}$

In order to demonstrate that the electron's intrinsic magnetic moment is associated with an intrinsic angular momentum, consider the motion of an electron in a central electrostatic potential: i.e., ${\bf A}={\bf0}$ . In this case, the Hamiltonian $\ref{1185}$ becomes

\( \mbox{\boldmath\)$x$ component of the electron's orbital angular momentum, ${\rm i}\,\hbar\,\dot{L}_x = [L_x,H].$ $\ref{1201}$

However, it is easily demonstrated that

$= 0,$ $\ref{1202}$ $= 0,$ $\ref{1203}$ $= {\rm i}\,\hbar\,p_z,$ $\ref{1204}$ $= -{\rm i}\,\hbar\,p_y.$ $\ref{1205}$

Hence, we obtain

$\dot{L}_x = c\,\gamma^5\,(\Sigma_2\,p_z-\Sigma_3\,p_y).$ $\ref{1207}$

It can be seen that $x$ -component of the total angular momentum of the system must be a constant of the motion (because a central electrostatic potential exerts zero torque on the system). Hence, we deduce that the electron possesses additional angular momentum that is not connected with its motion through space. Now,

$[\Sigma_1,\gamma^5]$ $[\Sigma_1,\Sigma_1]$ $[\Sigma_1,\Sigma_2]$ $[\Sigma_1,\Sigma_3]$ $[\Sigma_1,H] = 2\,{\rm i}\,c\,\gamma^5\,(\Sigma_3\,p_y-\Sigma_2\,p_z),$ $\ref{1213}$

which implies that

$\dot{L}_x +\frac{\hbar}{2}\,\dot{\Sigma}_1 = 0.$ $\ref{1215}$

Since there is nothing special about the ${\bf L} + (\hbar/2)\,$$\Sigma$ is a constant of the motion. We can interpret this result by saying that the electron has a spin angular momentum $\Sigma$ , which must be added to its orbital angular momentum in order to obtain a constant of the motion. According to $\ref{1198}$, the relationship between the electron's spin angular momentum and its intrinsic (i.e., non-orbital) magnetic moment is

$= - \frac{e\,g}{2\,m_e}\,{\bf S},$ $\ref{1216}$

where the gyromagnetic ratio $g= 2.$

$\ref{1217}$

As explained in Section 5.5, this is twice the value one would naively predict by analogy with classical physics.

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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