Question 3 of 7: Larmor Precession of a Spin-1/2 Particle

Question 3: Larmor Precession of a Spin-1/2 Particle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $H=-\vec\mu\cdot\vec B=-\mu\vec\sigma\cdot\vec B$ for a spin-1/2 particle in a uniform field; the field defines the axis named in the question, $\vec B=B\hat x$, and at $t=0$ the spin is fully polarized along the perpendicular axis $\hat z$ (the assignment consistent with $\langle s_x\rangle\equiv0$ and $\langle s_z\rangle(0)=\hbar/2$ — see the check note).

Find. $\langle s_x(t)\rangle$, $\langle s_y(t)\rangle$, $\langle s_z(t)\rangle$.

Approach. Diagonalize $H=-\mu B\sigma_x$ to get the time-dependent spinor $\chi(t)$, then sandwich each Pauli matrix between $\chi^\dagger$ and $\chi$.

1. Hamiltonian and its eigenstates. With $\vec B=B\hat x$: $H=-\mu B\sigma_x$. The eigenstates of $\sigma_x$ are $\chi_{+x}=\frac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}$ (energy $E_+=-\mu B$) and $\chi_{-x}=\frac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}$ (energy $E_-=+\mu B$).
2. Expand the initial state and evolve. The $\sigma_z$ eigenstate $\chi(0)=\begin{pmatrix}1\\0\end{pmatrix}$ decomposes as $\chi(0)=\frac{1}{\sqrt2}\left(\chi_{+x}+\chi_{-x}\right)$. Attaching the phase $e^{-iE_\pm t/\hbar}$ to each eigenstate, with $\omega\equiv\mu B/\hbar$: $$ \chi(t) = \frac{1}{\sqrt2}\left[e^{i\omega t}\chi_{+x}+e^{-i\omega t}\chi_{-x}\right] = \begin{pmatrix}\cos\omega t\\ i\sin\omega t\end{pmatrix}. $$
3. Expectation value $\langle s_x\rangle$. $$ \langle s_x\rangle = \frac{\hbar}{2}\chi^\dagger\sigma_x\chi = \frac{\hbar}{2}\big(\cos\omega t,\,-i\sin\omega t\big)\begin{pmatrix}i\sin\omega t\\ \cos\omega t\end{pmatrix} = \frac{\hbar}{2}\left[i\sin\omega t\cos\omega t - i\sin\omega t\cos\omega t\right] $$ $$ \boxed{\langle s_x\rangle = 0\ \text{for all }t} $$ — the spin component along the field axis is conserved, as it must be since $[H,\sigma_x]=0$.
4. Expectation value $\langle s_z\rangle$. $$ \langle s_z\rangle = \frac{\hbar}{2}\big(\cos\omega t,\,-i\sin\omega t\big)\begin{pmatrix}\cos\omega t\\ -i\sin\omega t\end{pmatrix} = \frac{\hbar}{2}\left[\cos^2\omega t - \sin^2\omega t\right] $$ $$ \boxed{\langle s_z\rangle = \frac{\hbar}{2}\cos(2\omega t)} $$ using the double-angle identity $\cos^2\theta-\sin^2\theta=\cos2\theta$.
5. Expectation value $\langle s_y\rangle$. $$ \langle s_y\rangle = \frac{\hbar}{2}\big(\cos\omega t,\,-i\sin\omega t\big)\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}\cos\omega t\\ i\sin\omega t\end{pmatrix} = \frac{\hbar}{2}\cdot2\sin\omega t\cos\omega t $$ $$ \boxed{\langle s_y\rangle = \frac{\hbar}{2}\sin(2\omega t)} $$ using $2\sin\theta\cos\theta=\sin2\theta$, reproducing all three stated results.