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 \(\bmu\) of the loop via

\[ \bmu = -\frac{e}{2\,m_e}\,{\bf L}, \label{e10.57} \]

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

\[ \bmu = -\frac{g\,e}{2\,m_e}\,{\bf S}, \label{e10.58} \]

where \(g\) is called the gyromagnetic ratio. Classically, we would expect \(g=1\). In fact,

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

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 \(\bmu\) in a uniform magnetic field \({\bf B}\) is

\[ H = - \bmu\cdot {\bf B}. \label{e10.60a} \]

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

\[ H = \Omega\,S_z, \label{e10.60} \]

where

\[\Omega = \frac{g\,e\,B}{2\,m_e}. \nonumber \]

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

Schrödinger’s equation can be written [see Equation \ref{etimed}]

\[ {\rm i}\,\hbar\,\frac{\partial\chi}{\partial t} = H\,\chi, \label{e10.62} \]

where the spin state of the electron is characterized by the spinor \(\chi\). Adopting the Pauli representation, we obtain

\[ \chi = \left(\begin{array}{c}c_+(t)\\c_-(t)\end{array}\right), \label{e10.63} \]

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 \ref{e10.46}, \ref{e10.53}, \ref{e10.60}, \ref{e10.62}, and \ref{e10.63} that

\[{\rm i}\,\hbar\left(\begin{array}{c}\dot{c}_+\\\dot{c}_-\end{array}\right) =\frac{\Omega\,\hbar}{2} \left(\begin{array}{cc}1,&0\\ 0,& -1\end{array}\right)\left(\begin{array}{c}c_+\\c_-\end{array}\right), \nonumber \]

where \(\dot{~}\equiv d/dt\). Hence,

\[ \dot{c}_\pm = \mp {\rm i}\,\frac{\Omega}{2}\,c_\pm. \label{e10.65} \]

Let

\[\begin{align} c_+(0) &= \cos(\alpha/2),\label{e10.66}\\[4pt] c_-(0) &= \sin(\alpha/2).\label{e10.67}\end{align} \]

The significance of the angle \(\alpha\) will become apparent presently. Solving Equation \ref{e10.65}, subject to the initial conditions \ref{e10.66} and \ref{e10.67}, we obtain

\[\begin{align} \label{e10.68} c_+(t) &= \cos(\alpha/2)\,\exp(-{\rm i}\,\Omega\,t/2),\\[4pt] c_-(t)&= \sin(\alpha/2)\,\exp(+{\rm i}\,\Omega\,t/2).\label{e10.69}\end{align} \]

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 \ref{e3.55}, the expectation value of a general spin operator \(A\) is simply

\[\langle A \rangle = \chi^\dagger\,A\,\chi. \nonumber \]

Hence, the expectation value of \(S_z\) is

\[\langle S_z\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\right) \left(\begin{array}{cc}1,&0\\ 0,& -1\end{array}\right)\left(\begin{array}{c}c_+\\ c_-\end{array}\right), \nonumber \]

which reduces to

\[ \langle S_z \rangle = \frac{\hbar}{2}\,\cos\alpha \label{e10.72} \]

with the help of Equations \ref{e10.68} and \ref{e10.69}. Likewise, the expectation value of \(S_x\) is

\[\langle S_x\rangle= \frac{\hbar}{2}\left(c_+^\ast, c_-^\ast\right) \left(\begin{array}{cc}0,&1\\ 1,& 0\end{array}\right)\left(\begin{array}{c}c_+\\ c_-\end{array}\right), \nonumber \]

which reduces to

\[ \langle S_x\rangle = \frac{\hbar}{2}\,\sin\alpha\,\cos(\Omega\,t). \label{e10.74} \]

Finally, the expectation value of \(S_y\) is

\[ \langle S_y\rangle = \frac{\hbar}{2}\,\sin\alpha\,\sin(\Omega\,t). \label{e10.75} \]

According to Equations \ref{e10.72}, \ref{e10.74}, and \ref{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

\[\Omega \simeq \frac{e\,B}{m_e}. \nonumber \]

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.