Spin ½

We saw that with angular momentum we got $l$ eigenstates that were either integers or ½-integers. Integer $l$ leads to orbital angular momentum, while $l=½$ leads to spin.

Instead of $\LL^2$ and $\LL_z$ as the simultaneous operators, we will write $S^2$ and $S_z$ , with simultaneous eigenstates $\ket{s=\half, m=\pm \half}$ .

\begin{align*} S^2\ket{s,m} &= \hbar^2 s(s+1) \ket{s,m} = \frac34 \hbar^2 \ket{s,m} \\ S_z\ket{s,m} &= m\hbar \ket{s,m} = \pm \frac12 \ket{s,m}. \end{align*}

We also write these eigenstates as vectors,

\begin{align*} \ket{s=\half,m=+\half} &=: \ket{+}_z = \vectwo{1}{0} \\ \ket{s=\half,m=-\half} &=: \ket{-}_z = \vectwo{0}{1}. \end{align*}

A spin state is a superposition of these two eigenvectors

\begin{align*} \ket{\psi} = c_1 \ket{+}_z + c_2 \ket{-}_z = \vectwo{c_1}{c_2}. \end{align*}

Using this vector notation, we can represent $S_z$ as a matrix. We require $S_z \vectwo{1}{0} = \frac\hbar2 \vectwo10$ and $S_z \vectwo01 = -\frac\hbar2 \vectwo01$ , which gives us

\begin{align*} S_z &= \frac\hbar2 \mat{1 & 0 \\ 0 & -1} =: \frac\hbar2 \sigma_z \end{align*}

where $\sigma_z$ is the $z$ Pauli matrix. The other pauli matrices $\sigma_x$ and $\sigma_y$ can be derived from the commutation relations between the operators (given here).

\begin{align*} \sigma_x &= \mat{0 & 1 \\ 1 & 0}, \quad \sigma_y = \mat{0 & -i \\ i & 0}, \quad \sigma_z = \mat{1 & 0 \\ 0 & -1} \\ S_i &= \sigma_i \text{ for } i = x,y,z. \end{align*}

This also gives us the spin states in the perpendicular directions

\begin{align*} \ket\pm_x &= \frac{1}{\sqrt{2}} \big(\ket+_z \pm \ket-_z\big) \\ \ket\pm_y &= \frac{1}{\sqrt{2}} \big(\ket+_z \pm i\ket-_z\big). \end{align*}

What about spin in some arbitrary direction? We can write

\begin{align*} S_n &= \vec n \cdot \vec S = n_x S_x + n_y S_y + n_z S_z = \frac\hbar2 \vec n \cdot \vec \sigma \\ \vec\sigma &= \mat{\sigma_x \\ \sigma_y \\ \sigma_z} \text{ is a vector of matrices (weird!)} \\ S_n &= \frac\hbar2 \mat{ n_z & n_x - in_y \\ n_x + in_y & -n_z } = \frac\hbar2 \mat{ \cos\theta & \sin\theta e^{-i\phi} \\ \sin\theta e^{i\phi} & -\cos\theta }. \end{align*}

We require that the eigenstates of $S_n$ be

\begin{align*} S_n \mat{c_1 \\ c_2} &\overset!= \frac\hbar2 \mat{c_1 \\ c_2} \\ \ket{+}_n &= \mat{\cos \frac\pi2 \\ e^{i\phi} \sin\frac{\theta}{2}} \\ \ket{-}_n &= \mat{\sin\frac\theta2 \\ -e^{i\phi} \cos\frac\theta2}. \end{align*}

Spin in a magnetic field

Consider a $Q$ charged particle with mass $m$ in a circular loop of radius $R$ . The current around the loop can be written as

\begin{align*} I = \frac{\text{charge}}{\text{time}} = \frac{Q}{2\pi R/v} = \frac{Q}{m} \frac{1}{2\pi R^2} mRv = \frac{Q}{2m} \frac{L}{A}. \end{align*}

The magnetic moment is given by

\begin{align*} \vec\mu = \frac{I\vec A}{c} = \frac{Q}{2mc} \vec L. \end{align*}

For an electron, $Q=-e$ . We define the Bohr magneton $\mu_B = \frac{e\hbar}{2mc}$ . We can then write the electron spin magnetic dipole moment as

\begin{align*} \vec \mu_{e} &= -g_e \mu_B \frac{\vec S}{\hbar}. \end{align*}

Now consider the energy of an electron in a magnetic field. We get a torque of a $\vec \mu$ and $\vec B$ are not aligned, and the energy is minimized when they are aligned. We can write the Hamiltonian as

\begin{align*} \HH = -\vec \mu \cdot \vec B. \end{align*}

For an electron,

\begin{align*} \HH = g_e \frac{\mu_b}{\hbar} \vec S \cdot \vec B = \underbrace{g_e \frac{\mu_B}{\hbar} B_z}_{\omega_L} S_z \text{ when } \vec B = B_z \vec z. \end{align*}

In this case, since the Hamiltonian is proportional to the $z$ spin, energy eigentsates are proportional to $z$ spin eigenstates.

\begin{align*} \HH \ket{\pm}_z = \HH \ket{\half, \pm \half}_z = \pm \omega_L \frac{\hbar}{2} \ket{\pm}_z. \end{align*}

We see that $\ket\pm_z$ are stationary states. Consider instead the time evolution of a perpendicular spin $\ket+_x$ .

\begin{align*} \ket{\psi(t=0)} &= \ket+_x = \frac{1}{\sqrt2} \big( \ket+_z + \ket-_z \big) \\ \ket{\psi(t)} &= \frac{1}{\sqrt2} \left( \ket+_z e^{i\omega_L t/2} + \ket-_z e^{-i\omega_L t/2} \right) \\ &= \frac{e^{-i\omega_L t/2}}{\sqrt2} \left( \ket+_z + \ket-_z e^{i\omega_L t} \right) \\ \ket{\psi(\omega_L t=\pi/2)} &= \frac{e^{-i\pi/4}}{\sqrt 2}\left(\ket+_z + i\ket-_z\right) = e^{-i\pi/4} \ket+_y \\ \ket{\psi(\omega_L t=\pi)} &= \frac{-i}{\sqrt{2}} \left(\ket+_z - \ket-_z\right) = -i\ket-_x \\ \ket{\psi(\omega_L t = 3\pi/2)} &= \frac{e^{-i3\pi/4}}{\sqrt2} \left(\ket+_z - i\ket-_z\right) = e^{-i3\pi/4}\ket-_z \\ \ket{\psi(\omega_L t=2\pi)} &= \frac{-1}{\sqrt2} \left(\ket+_z + \ket-_z\right) = -\ket+_x. \end{align*}

We see that a spin misaligned with the external magnetic field processes around the $\vec B$ axis. Intuitively, we know that a misaligned dipole experiences a torque which aligns it with the external field. But being aligned with the external field is a lower energy state, and if there’s no way for the dipole to dissipate energy (as is the case here for an electron), the dipole can’t align itself.

In fact, consider the energy of an electron with spin perpendicular to the magnetic field $\ket+_x = \frac{1}{\sqrt2} (\ket+_z + \ket-_z)$ . The possible energy measurements are $\pm \frac\hbar2$ with equal probability, so the expected energy $\braket E = 0$ . This is the case for any direction perpendicular to $\vec B$ , so energy is conserved during the dipole’s procession.