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Next: Exercises Up: Spin Angular Momentum Previous: Pauli Representation

Spin Precession

According to classical physics, a small current loop possesses a magnetic moment of magnitude $\mu=I A$, where $I$ is the current circulating around the loop, and $A$ the area of the loop. The direction of the magnetic moment is conventionally taken to be normal to the plane of the loop, in the sense given by a standard right-hand circulation rule. Consider a small current loop consisting of an electron in uniform circular motion. It is easily demonstrated that the electron's orbital angular momentum ${\bf L}$ is related to the magnetic moment $\mbox{\boldmath$\mu$}$ of the loop via
\mbox{\boldmath$\mu$}= -\frac{e}{2 m_e} {\bf L},
\end{displaymath} (758)

where $e$ is the magnitude of the electron charge, and $m_e$ the electron mass.

The above expression suggests that there may be a similar relationship between magnetic moment and spin angular momentum. We can write

\mbox{\boldmath$\mu$}= -\frac{g e}{2 m_e} {\bf S},
\end{displaymath} (759)

where $g$ is called the gyromagnetic ratio. Classically, we would expect $g=1$. In fact,
g = 2\left(1+\frac{\alpha}{2\pi}+\cdots\right) = 2.0023192,
\end{displaymath} (760)

where $\alpha= e^2/(2 \epsilon_0 h c) \simeq 1/137$ is the so-called fine-structure constant. The fact that the gyromagnetic ratio is (almost) twice that expected from classical physics is only explicable using relativistic quantum mechanics. Furthermore, the small corrections to the relativistic result $g=2$ come from quantum field theory.

The energy of a classical magnetic moment $\mbox{\boldmath$\mu$}$ in a uniform magnetic field ${\bf B}$ is

H = - \mbox{\boldmath$\mu$}\cdot {\bf B}.
\end{displaymath} (761)

Assuming that the above expression also holds good in quantum mechanics, the Hamiltonian of an electron in a $z$-directed magnetic field of magnitude $B$ takes the form
H = \Omega S_z,
\end{displaymath} (762)

\Omega = \frac{g e B}{2 m_e}.
\end{displaymath} (763)

Here, for the sake of simplicity, we are neglecting the electron's translational degrees of freedom.

Schrödinger's equation can be written [see Eq. (199)]

{\rm i} \hbar \frac{\partial\chi}{\partial t} = H \chi,
\end{displaymath} (764)

where the spin state of the electron is characterized by the spinor $\chi$. Adopting the Pauli representation, we obtain
\chi = \left(\begin{array}{c}c_+(t)\ c_-(t)\end{array}\right),
\end{displaymath} (765)

where $\vert c_+\vert^2+\vert c_-\vert^2=1$. Here, $\vert c_+\vert^2$ is the probability of observing the spin-up state, and $\vert c_-\vert^2$ the probability of observing the spin-down state. It follows from Eqs. (748), (755), (762), (764), and (765) that
{\rm i} \hbar\left(\begin{array}{c}\dot{c}_+\ \dot{c}_-\en...
...rray}\right)\left(\begin{array}{c}c_+\ c_-\end{array}\right),
\end{displaymath} (766)

where $\dot{ }\equiv d/dt$. Hence,
\dot{c}_\pm = \mp {\rm i} \frac{\Omega}{2} c_\pm.
\end{displaymath} (767)

$\displaystyle c_+(0)$ $\textstyle =$ $\displaystyle \cos(\alpha/2),$ (768)
$\displaystyle c_-(0)$ $\textstyle =$ $\displaystyle \sin(\alpha/2).$ (769)

The significance of the angle $\alpha $ will become apparent presently. Solving Eq. (767), subject to the initial conditions (768) and (769), we obtain
$\displaystyle c_+(t)$ $\textstyle =$ $\displaystyle \cos(\alpha/2) \exp(-{\rm i} \Omega t/2),$ (770)
$\displaystyle c_-(t)$ $\textstyle =$ $\displaystyle \sin(\alpha/2) \exp(+{\rm i} \Omega t/2).$ (771)

We can most easily visualize the effect of the time dependence in the above expressions for $c_\pm$ by calculating the expectation values of the three Cartesian components of the electron's spin angular momentum. By analogy with Eq. (192), the expectation value of a general spin operator $A$ is simply

\langle A \rangle = \chi^\dag A \chi.
\end{displaymath} (772)

Hence, the expectation value of $S_z$ is
\langle S_z\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\...
...rray}\right)\left(\begin{array}{c}c_+\ c_-\end{array}\right),
\end{displaymath} (773)

which reduces to
\langle S_z \rangle = \frac{\hbar}{2} \cos\alpha
\end{displaymath} (774)

with the help of Eqs. (770) and (771). Likewise, the expectation value of $S_x$ is
\langle S_x\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\...
...rray}\right)\left(\begin{array}{c}c_+\ c_-\end{array}\right),
\end{displaymath} (775)

which reduces to
\langle S_x\rangle = \frac{\hbar}{2} \sin\alpha \cos(\Omega t).
\end{displaymath} (776)

Finally, the expectation value of $S_y$ is
\langle S_y\rangle = \frac{\hbar}{2} \sin\alpha \sin(\Omega t).
\end{displaymath} (777)

According to Eqs. (774), (776), and (777), the expectation value of the spin angular momentum vector subtends a constant angle $\alpha $ with the $z$-axis, and precesses about this axis at the frequency
\Omega \simeq \frac{e B}{m_e}.
\end{displaymath} (778)

This behaviour is actually equivalent to that predicted by classical physics. Note, however, that a measurement of $S_x$, $S_y$, or $S_z$ will always yield either $+\hbar/2$ or $-\hbar/2$. It is the relative probabilities of obtaining these two results which varies as the expectation value of a given component of the spin varies.

next up previous
Next: Exercises Up: Spin Angular Momentum Previous: Pauli Representation
Richard Fitzpatrick 2010-07-20