Uncertainty relation for a spin one-half particle

Let $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$, both real. In the basis $(\ket+,\ket-)$, the state is the column vector $(c,s)^{\mathsf T}$. We calculate

$$\sigma_x\begin{pmatrix}c\\s\end{pmatrix}=\begin{pmatrix}s\\c\end{pmatrix},\qquad \sigma_y\begin{pmatrix}c\\s\end{pmatrix}=\begin{pmatrix}-is\\ic\end{pmatrix},\qquad \sigma_z\begin{pmatrix}c\\s\end{pmatrix}=\begin{pmatrix}c\\-s\end{pmatrix}.$$

Taking the inner product with $(c,s)$ gives

$$\langle\sigma_x\rangle=2cs=\sin\theta,\qquad \langle\sigma_y\rangle=-ics+ics=0,\qquad \langle\sigma_z\rangle=c^2-s^2=\cos\theta .$$

Thus $\langle S_x\rangle=\frac\hbar2\sin\theta$, $\langle S_y\rangle=0$, and $\langle S_z\rangle=\frac\hbar2\cos\theta$. The vector $\langle\mathbf S\rangle$ has magnitude $\hbar/2$, lies in the $(x,z)$ plane, and makes an angle $\theta$ with the $z$-axis: it is the Bloch vector of the state, multiplied by $\hbar/2$.

Each Pauli matrix satisfies $\sigma_j^2=\mathbf{1}$, so $S_j^2=\frac{\hbar^2}{4}\mathbf{1}$, and $\langle S_j^2\rangle=\hbar^2/4$ in any normalised state. Consequently,

$$\sigma_{S_x}^2=\frac{\hbar^2}{4}-\frac{\hbar^2}{4}\sin^2\theta=\frac{\hbar^2}{4}\cos^2\theta, \qquad \sigma_{S_y}^2=\frac{\hbar^2}{4}-0=\frac{\hbar^2}{4},$$

so $\sigma_{S_x}=\frac\hbar2|\cos\theta|$ and $\sigma_{S_y}=\frac\hbar2$.

The product of the standard deviations is $\sigma_{S_x}\sigma_{S_y}=\frac{\hbar^2}{4}|\cos\theta|$. The right-hand side of the inequality is

$$\frac12\bigl|\langle[S_x,S_y]\rangle\bigr|=\frac12\bigl|i\hbar\langle S_z\rangle\bigr|=\frac\hbar2\cdot\frac\hbar2|\cos\theta|=\frac{\hbar^2}{4}|\cos\theta| .$$

The inequality is satisfied, and is in fact an equality for every value of $\theta$: all the states in this family are minimum-uncertainty states for the pair $(S_x,S_y)$.

For $\theta=\pi/2$, the state is $(\ket++\ket-)/\sqrt2$, namely the eigenstate of $S_x$ with eigenvalue $+\hbar/2$. We then have $\sigma_{S_x}=0$: a measurement of $S_x$ yields $+\hbar/2$ with certainty. The right-hand side is also zero, since $\langle S_z\rangle=0$, and the inequality reduces to $0\geq0$. This does not mean that $S_x$ and $S_y$ are compatible: their commutator $i\hbar S_z$ is not the zero operator, and they have no common eigenvector. Rather, the Robertson bound depends on the state, through $\langle[\hat A,\hat B]\rangle$, and vanishes here. Indeed, $\sigma_{S_y}=\hbar/2$ is maximal: in an eigenstate of $S_x$, the component $S_y$ is completely undetermined. The position—momentum relation is a special case in which the commutator $i\hbar\mathbf{1}$ is proportional to the identity, so the bound $\hbar/2$ does not depend on the state.