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11.5: Electron Spin

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

 

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

 

 

Hence,

 

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

 

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

 

 

for $ i=1,3$ , where

 

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

 

 

and

 

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

 

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

 

 

It follows from (1186) that

 

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

 

 

Now, a straightforward generalization of Equation (508) gives

 

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

 

$\displaystyle \left[\mbox{\boldmath$\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,

 

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

 

 

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

 

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

 

 

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

 

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

 

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

 

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

 

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

 

 

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

 

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

 

 

The Heisenberg equation of motion for this quantity is

 

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

 

 

However, it is easily demonstrated that

 

$\displaystyle [L_x,r]$ $\displaystyle = 0,$ (1202)
$\displaystyle [L_x,p_x]$ $\displaystyle = 0,$ (1203)
$\displaystyle [L_x,p_y]$ $\displaystyle = {\rm i}\,\hbar\,p_z,$ (1204)
$\displaystyle [L_x,p_z]$ $\displaystyle = -{\rm i}\,\hbar\,p_y.$ (1205)

 

 

Hence, we obtain

 

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

 

 

which implies that

 

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

 

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

 

 

However,

 

$\displaystyle [\Sigma_1,\gamma^5]$ $\displaystyle = 0,$ (1209)
$\displaystyle [\Sigma_1,\Sigma_1]$ $\displaystyle = 0,$ (1210)
$\displaystyle [\Sigma_1,\Sigma_2]$ $\displaystyle =2\,{\rm i}\,\Sigma_3,$ (1211)
$\displaystyle [\Sigma_1,\Sigma_3]$ $\displaystyle =-2\,{\rm i}\,\Sigma_2,$ (1212)

 

 

so

 

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

 

 

which implies that

 

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

 

 

Hence, we deduce that

 

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

 

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

 

 

where the gyromagnetic ratio $ g$ takes the value

 

$\displaystyle g= 2.$ (1217)

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

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