9.5: 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 $\mu$ of the loop via

$$\label{e10.57} \mu = -\frac{e}{2\,m_e}\,{\bf L},$$

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

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

$$\label{e10.58} \mu = -\frac{g\,e}{2\,m_e}\,{\bf S},$$ where $g$ is called the gyromagnetic ratio. Classically, we would expect $g=1$. In fact,

$$\label{e10.59} g = 2\left(1+\frac{\alpha}{2\pi}+\cdots\right) = 2.0023192,$$

here $\alpha= e^{\,2}/(2\,\epsilon_0\,h\,c) \simeq 1/137$ 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 $\mu$ in a uniform magnetic field ${\bf B}$ is

$$\label{e10.60a} H = - \mu\cdot {\bf B}.$$ Assuming that the previous expression also holds good in quantum mechanics, the Hamiltonian of an electron in a $z$-directed magnetic field of magnitude $B$ takes the form

$$\label{e10.60} H = {\mit\Omega}\,S_z,$$ where $${\mit\Omega} = \frac{g\,e\,B}{2\,m_e}.$$ Here, for the sake of simplicity, we are neglecting the electron’s translational degrees of freedom.

Schrödinger’s equation can be written [see Equation ([etimed])]

$$\label{e10.62} {\rm i}\,\hbar\,\frac{\partial\chi}{\partial t} = H\,\chi,$$ where the spin state of the electron is characterized by the spinor $\chi$. Adopting the Pauli representation, we obtain

$$\label{e10.63} \chi = \left(\begin{array}{c}c_+(t)\\c_-(t)\end{array}\right),$$ where $|c_+|^{\,2}+|c_-|^{\,2}=1$. Here, $|c_+|^{\,2}$ is the probability of observing the spin-up state, and $|c_-|^{\,2}$ the probability of observing the spin-down state. It follows from Equations ([e10.46]), ([e10.53]), ([e10.60]), ([e10.62]), and ([e10.63]) that $${\rm i}\,\hbar\left(\begin{array}{c}\dot{c}_+\\\dot{c}_-\end{array}\right) =\frac{{\mit\Omega}\,\hbar}{2} \left(\begin{array}{cc}1,&0\\ 0,& -1\end{array}\right)\left(\begin{array}{c}c_+\\c_-\end{array}\right),$$ where $\dot{~}\equiv d/dt$. Hence,

$$\begin{equation}\dot{c}_{\pm}=\mp \mathrm{i} \frac{\Omega}{2} c_{\pm}\end{equation}$$ Let

$$\begin{aligned} c_+(0) &= \cos(\alpha/2),\label{e10.66}\\[0.5ex] c_-(0) &= \sin(\alpha/2).\label{e10.67}\end{aligned}$$ The significance of the angle $\alpha$ will become apparent presently. Solving Equation ([e10.65]), subject to the initial conditions ([e10.66]) and ([e10.67]), we obtain

$$\begin{aligned} \label{e10.68} c_+(t) &= \cos(\alpha/2)\,\exp(-{\rm i}\,{\mit\Omega}\,t/2),\\[0.5ex] c_-(t)&= \sin(\alpha/2)\,\exp(+{\rm i}\,{\mit\Omega}\,t/2).\label{e10.69}\end{aligned}$$

We can most easily visualize the effect of the time dependence in the previous expressions for $c_\pm$ by calculating the expectation values of the three Cartesian components of the electron’s spin angular momentum. By analogy with Equation ([e3.55]), the expectation value of a general spin operator $A$ is simply

$$\langle A \rangle = \chi^\dagger\,A\,\chi.$$ Hence, the expectation value of $S_z$ is $$\langle S_z\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\right) \left(\begin{array}{cc}1,&0\\ 0,& -1\end{array}\right)\left(\begin{array}{c}c_+\\ c_-\end{array}\right),$$ which reduces to

$$\label{e10.72} \langle S_z \rangle = \frac{\hbar}{2}\,\cos\alpha$$ with the help of Equations ([e10.68]) and ([e10.69]). Likewise, the expectation value of $S_x$ is $$\langle S_x\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\right) \left(\begin{array}{cc}0,&1\\ 1,& 0\end{array}\right)\left(\begin{array}{c}c_+\\ c_-\end{array}\right),$$ which reduces to

$$\label{e10.74} \langle S_x\rangle = \frac{\hbar}{2}\,\sin\alpha\,\cos({\mit\Omega}\,t).$$ Finally, the expectation value of $S_y$ is

$$\label{e10.75} \langle S_y\rangle = \frac{\hbar}{2}\,\sin\alpha\,\sin({\mit\Omega}\,t).$$ According to Equations ([e10.72]), ([e10.74]), and ([e10.75]), 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 $${\mit\Omega} \simeq \frac{e\,B}{m_e}.$$ This behavior 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 that varies as the expectation value of a given component of the spin varies.

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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