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# 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

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

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

$\label{e10.58} \bmu = -\frac{g\,e}{2\,m_e}\,{\bf S},$ 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 $$\bmu$$ in a uniform magnetic field $${\bf B}$$ is

$\label{e10.60a} H = - \bmu\cdot {\bf B}.$ 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,

$\label{e10.65} \dot{c}_\pm = \mp {\rm i}\,\frac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/9:_Spin_Angular_Momentum/9.5:_Spin_Precession), /content/body/p[14]/span/span, line 1, column 1  {2}\,c_\pm.$ 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 $\langle A \rangle = \chi^\dag\,A\,\chi.$ 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

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