11.5: Electron Spin

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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,

$\left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldm... ...mbox{\boldmath$\alpha$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^2,$ \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 $\left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\... ...,\mbox{\boldmath$\Sigma$}\cdot ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}).$ \ref{1193}

However,

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