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Next: Motion in Central Field Up: Relativistic Electron Theory Previous: Free Electron Motion

Electron Spin

According to Equation (11.39), 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}.$ (11.94)


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

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

for $ i=1,3$ , where

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


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

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

It follows from Equation (11.95) that

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

Now, a straightforward generalization of Equation (5.93) 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}),$ (11.101)

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$}\cdot ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}).$ (11.102)


$\displaystyle ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A})$ $\displaystyle = e\,{\bf p}\times {\bf A} +e\,{\bf A}\times {\bf p} = -{\rm i}\,e\,\hbar\,\nabla\times {\bf A} -{ \rm i}\,e\,\hbar\,{\bf A}\times \nabla$    
  $\displaystyle = -{\rm i}\,e\,\hbar\,{\bf B},$ (11.103)

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}.$ (11.104)

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

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

$\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}.$ (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

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

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

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

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

The Heisenberg equation of motion for this quantity is

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

However, it is easily demonstrated that

$\displaystyle [L_x,r]$ $\displaystyle =0,$ (11.111)
$\displaystyle [L_x,p_x]$ $\displaystyle =0,$ (11.112)
$\displaystyle [L_x,p_y]$ $\displaystyle = {\rm i}\,\hbar\,p_z,$ (11.113)
$\displaystyle [L_x,p_z]$ $\displaystyle = -{\rm i}\,\hbar\,p_y.$ (11.114)

Hence, we obtain

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

which implies that

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

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


$\displaystyle [\Sigma_{\,1},\beta]$ $\displaystyle =0,$ (11.118)
$\displaystyle [\Sigma_{\,1},\gamma^{\,5}]$ $\displaystyle =0,$ (11.119)
$\displaystyle [\Sigma_{\,1},\Sigma_{\,1}]$ $\displaystyle =0,$ (11.120)
$\displaystyle [\Sigma_{\,1},\Sigma_{\,2}]$ $\displaystyle =2\,{\rm i}\,\Sigma_{\,3},$ (11.121)
$\displaystyle [\Sigma_{\,1},\Sigma_{\,3}]$ $\displaystyle =-2\,{\rm i}\,\Sigma_{\,2},$ (11.122)


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

which implies that

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

Hence, we deduce that

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

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

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

$\displaystyle g= 2.$ (11.127)

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

next up previous
Next: Motion in Central Field Up: Relativistic Electron Theory Previous: Free Electron Motion
Richard Fitzpatrick 2016-01-22