Electron Spin

According to Equation (11.39), the relativistic Hamiltonian of an electron in an electromagnetic field is

$$H =-e\,\phi + c\, \mbox{\boldmath$\alpha$} \cdot({\bf p}+e\,{\bf A})+ \beta\,m_e\,c^{\,2}.$$ (11.94)

Hence,

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldm... ...{\boldmath$\alpha$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^{\,2},$$ (11.95)

where use has been made of Equations (11.25) and (11.26). Now, we can write

$$\alpha_i = \gamma^{\,5}\,\Sigma_{\,i},$$ (11.96)

for $i=1,3$ , where

$$\gamma^{\,5} = \left(\begin{array}{cc} 0,& 1\\ [0.5ex]1, & 0\end{array}\right),$$ (11.97)

and

$$\Sigma_{\,i} = \left(\begin{array}{cc} \sigma_i,& 0\\ [0.5ex]0,& \sigma_i\end{array}\right).$$ (11.98)

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

$$[\gamma^{\,5}, \Sigma_{\,i}]=0.$$ (11.99)

It follows from Equation (11.95) that

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^{\,2}.$$ (11.100)

Now, a straightforward generalization of Equation (5.93) gives

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf a} ) \,( \mbox{\boldmath$\Sigma$} \cdot {\bf b}) = {\bf a} \cdot {\bf b} +{\rm i}\, \mbox{\boldmath$\Sigma$} \cdot ({\bf a} \times {\bf b}),$$ (11.101)

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

$$\left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\... ...,\mbox{\boldmath$\Sigma$}\cdot ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}).$$ (11.102)

However,

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

$$= -{\rm i}\,e\,\hbar\,{\bf B},$$ (11.103)

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

$$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = ({\bf p}+e\,{\bf A})^2 + m_e^{\,2}\,c^{\,2}+e\,\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf B}.$$ (11.104)

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

$$H = m_e\,c^{\,2}+ \delta H,$$ (11.105)

where $\delta H$ is small compared to $m_e\,c^{\,2}$ . Substituting into Equation (11.104), and neglecting $\delta H^{\,2}$ , and other terms involving $c^{\,-2}$ , we get

$$\delta H\simeq -e\,\phi + \frac{1}{2\,m_e}\,({\bf p} + e\,{\bf A})^2 + \frac{e\,\hbar}{2\,m_e}\, \mbox{\boldmath$\Sigma$} \cdot{\bf B}.$$ (11.106)

This Hamiltonian is the same as the classical Hamiltonian of a non-relativistic electron, except for the final term. (See Section 3.6.) This term may be interpreted as arising from the electron having an intrinsic magnetic moment

$$\mbox{\boldmath$\mu$} = - \frac{e\,\hbar}{2\,m_e}\, \mbox{\boldmath$\Sigma$} .$$ (11.107)

(See Section 5.6.)

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: that is, $\phi=\phi(r)$ and ${\bf A}={\bf0}$ . In this case, the Hamiltonian (11.94) becomes

$$H = - e\,\phi(r) + c\,\gamma^{\,5}\, \mbox{\boldmath$\Sigma$} \cdot{\bf p} + \beta\,m_e\,c^{\,2}.$$ (11.108)

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

$$L_x = y\,p_z-z\,p_y = {\rm i}\,\hbar\left(z\,\frac{\partial}{\partial y} - y\,\frac{\partial}{\partial z}\right).$$ (11.109)

The Heisenberg equation of motion for this quantity is

$${\rm i}\,\hbar\,\dot{L}_x = [L_x,H].$$ (11.110)

However, it is easily demonstrated that

$$[L_x,r] =0,$$ (11.111)

$$[L_x,p_x] =0,$$ (11.112)

$$[L_x,p_y] = {\rm i}\,\hbar\,p_z,$$ (11.113)

$$[L_x,p_z] = -{\rm i}\,\hbar\,p_y.$$ (11.114)

Hence, we obtain

$$[L_x,H] = {\rm i}\,\hbar\,c\,\gamma^{\,5}\,(\Sigma_{\,2}\,p_z-\Sigma_{\,3}\,p_y),$$ (11.115)

which implies that

$$\dot{L}_x = c\,\gamma^{\,5}\,(\Sigma_{\,2}\,p_z-\Sigma_{\,3}\,p_y).$$ (11.116)

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,

$${\rm i}\,\hbar\,\dot{\Sigma}_{\,1}= [\Sigma_{\,1},H].$$ (11.117)

However,

$$[\Sigma_{\,1},\beta] =0,$$ (11.118)

$$[\Sigma_{\,1},\gamma^{\,5}] =0,$$ (11.119)

$$[\Sigma_{\,1},\Sigma_{\,1}] =0,$$ (11.120)

$$[\Sigma_{\,1},\Sigma_{\,2}] =2\,{\rm i}\,\Sigma_{\,3},$$ (11.121)

$$[\Sigma_{\,1},\Sigma_{\,3}] =-2\,{\rm i}\,\Sigma_{\,2},$$ (11.122)

so

$$[\Sigma_{\,1},H] = 2\,{\rm i}\,c\,\gamma^{\,5}\,(\Sigma_{\,3}\,p_y-\Sigma_{\,2}\,p_z),$$ (11.123)

which implies that

$$\frac{\hbar}{2}\,\dot{\Sigma}_1 = -c\,\gamma^{\,5}\,(\Sigma_{\,2}\,p_z-\Sigma_{\,3}\,p_y).$$ (11.124)

Hence, we deduce that

$$\dot{L}_x +\frac{\hbar}{2}\,\dot{\Sigma}_{\,1} = 0.$$ (11.125)

Because 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 Equation (11.107), the relationship between the electron's spin angular momentum and its intrinsic (i.e., non-orbital) magnetic moment is

$$\mbox{\boldmath$\mu$} = - \frac{e\,g}{2\,m_e}\,{\bf S},$$ (11.126)

where the gyromagnetic ratio, $g$ , takes the value

$$g= 2.$$ (11.127)

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