Electron Spin

According to Equation (1132), the relativistic Hamiltonian of an electron in an electromagnetic field is

$$H =-e\,\phi + c\, \mbox{\boldmath$\alpha$} \cdot({\bf p}+e\,{\bf A})+ \beta\,m_e\,c^2.$$ (1185)

Hence,

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldm... ...mbox{\boldmath$\alpha$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^2,$$ (1186)

where use has been made of Equations (1118) and (1119). Now, we can write

$$\alpha_i = \gamma^5\,\Sigma_i,$$ (1187)

for $i=1,3$ , where

$$\gamma^5 = \left(\begin{array}{cc} 0& 1\\ [0.5ex]1 & 0\end{array}\right),$$ (1188)

and

$$\Sigma_i = \left(\begin{array}{cc} \sigma_i& 0\\ [0.5ex]0& \sigma_i\end{array}\right).$$ (1189)

Here, 0 and $1$ denote $2\times 2$ null and identity matrices, respectively, whereas the $\sigma_i$ are conventional $2\times 2$ Pauli matrices. Note that $\gamma^5\,\gamma^5=1$ , and

$$[\gamma^5, \Sigma_i]=0.$$ (1190)

It follows from (1186) that

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^2.$$ (1191)

Now, a straightforward generalization of Equation (508) gives

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf a} ) \,( \mbox{\boldmath$\Sigma$} \cdot {\bf b}) = {\bf a} \cdot {\bf b} +{\rm i}\, \mbox{\boldmath$\Sigma$} \cdot ({\bf a} \times {\bf b}),$$ (1192)

where ${\bf a}$ and ${\bf b}$ are any two three-dimensional vectors that commute with $\Sigma$ . It follows that

$$\left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\... ...,\mbox{\boldmath$\Sigma$}\cdot ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}).$$ (1193)

However,

$$({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}) = e\,{\bf p}\time... ...\bf A} -{ \rm i}\,e\,\hbar\,{\bf A}\times \nabla = -{\rm i}\,e\,\hbar\,{\bf B},$$ (1194)

where ${\bf B} = \nabla\times {\bf A}$ is the magnetic field strength. Hence, we obtain

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = ({\bf p}+e\,{\bf A})^2 + m_e^{\,2}\,c^2+e\,\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf B}.$$ (1195)

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

$$H = m_e\,c^2+ \delta H,$$ (1196)

where $\delta H$ is small compared to $m_e\,c^2$ . Substituting into (1195), and neglecting $\delta H^{\,2}$ , and other terms involving $c^{\,-2}$ , we get

$$\delta H\simeq -e\,\phi + \frac{1}{2\,m_e}\,({\bf p} + e\,{\bf A})^2 + \frac{e\,\hbar}{2\,m_e}\, \mbox{\boldmath$\Sigma$} \cdot{\bf B}.$$ (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

$$\mbox{\boldmath$\mu$} = - \frac{e\,\hbar}{2\,m_e}\, \mbox{\boldmath$\Sigma$} .$$ (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., $\phi=\phi(r)$ and ${\bf A}={\bf0}$ . In this case, the Hamiltonian (1185) becomes

$$H = - e\,\phi(r) + c\,\gamma^5\, \mbox{\boldmath$\Sigma$} \cdot{\bf p} + \beta\,m_e\,c^2.$$ (1199)

Consider the $x$ component of the electron's orbital angular momentum,

$$L_x = y\,p_z-z\,p_y = {\rm i}\,\hbar\left(z\,\frac{\partial}{\partial y} - y\,\frac{\partial}{\partial z}\right).$$ (1200)

The Heisenberg equation of motion for this quantity is

$${\rm i}\,\hbar\,\dot{L}_x = [L_x,H].$$ (1201)

However, it is easily demonstrated that

$$[L_x,r] = 0,$$ (1202)

$$[L_x,p_x] = 0,$$ (1203)

$$[L_x,p_y] = {\rm i}\,\hbar\,p_z,$$ (1204)

$$[L_x,p_z] = -{\rm i}\,\hbar\,p_y.$$ (1205)

Hence, we obtain

$$[L_x,H] = {\rm i}\,\hbar\,c\,\gamma^5\,(\Sigma_2\,p_z-\Sigma_3\,p_y),$$ (1206)

which implies that

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

It can be seen that $L_x$ is not a constant of the motion. However, the $x$ -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,

$${\rm i}\,\hbar\,\dot{\Sigma}_1= [\Sigma_1,H].$$ (1208)

However,

$$[\Sigma_1,\gamma^5] = 0,$$ (1209)

$$[\Sigma_1,\Sigma_1] = 0,$$ (1210)

$$[\Sigma_1,\Sigma_2] =2\,{\rm i}\,\Sigma_3,$$ (1211)

$$[\Sigma_1,\Sigma_3] =-2\,{\rm i}\,\Sigma_2,$$ (1212)

so

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

which implies that

$$\frac{\hbar}{2}\,\dot{\Sigma}_1 = -c\,\gamma^5\,(\Sigma_2\,p_z-\Sigma_3\,p_y).$$ (1214)

Hence, we deduce that

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

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

$$\mbox{\boldmath$\mu$} = - \frac{e\,g}{2\,m_e}\,{\bf S},$$ (1216)

where the gyromagnetic ratio $g$ takes the value

$$g= 2.$$ (1217)

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