Spin Expectation Values

The spin operators

The intrinsic spin of a spin-1/2 particle is described by three Hermitian operators $S_x$ , $S_y$ , $S_z$ , each built from the corresponding Pauli matrix $\sigma_i$ :

$S_i = \frac{\hbar}{2}\,\sigma_i.$

In the $z$ -basis $\{|{+}z\rangle, |{-}z\rangle\}$ , the explicit matrices are

\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad \sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ The eigenvalues of every $S_i$ are $\pm\hbar/2$: the only possible outcomes when you measure any component of spin are "spin up" and "spin down" along that axis. ## General spin state Any spin-1/2 state can be written as a two-component column vector $$|\psi\rangle = \alpha|{+}z\rangle + \beta|{-}z\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix}, \quad |\alpha|^2 + |\beta|^2 = 1.$$ Here $|\alpha|^2$ is the probability of finding $+\hbar/2$ along $z$, and $|\beta|^2$ is the probability of finding $-\hbar/2$ along $z$. ## Computing expectation values Given $|\psi\rangle$ as above, the expectation value $\langle A \rangle = \langle\psi|A|\psi\rangle$ of each spin component follows directly from matrix multiplication. **Along $z$:** $$\langle S_z \rangle = \frac{\hbar}{2}\bigl(|\alpha|^2 - |\beta|^2\bigr).$$ This is just the weighted average of the two eigenvalues, as expected from the Born rule. **Along $x$:** Applying $S_x$ and contracting with $\langle\psi|$ gives $$\langle S_x \rangle = \frac{\hbar}{2}(\alpha^*\beta + \alpha\beta^*) = \hbar\operatorname{Re}(\alpha^*\beta).$$ **Along $y$:** A similar calculation with $S_y$ (which has the factor $-i$ in its off-diagonal entries) yields $$\langle S_y \rangle = \frac{\hbar}{2}(-i\,\alpha^*\beta + i\,\alpha\beta^*) = \hbar\operatorname{Im}(\alpha^*\beta).$$ These three formulas are worth memorising. They say that once you know $\alpha$ and $\beta$, you can read off the mean spin along any axis. ## A worked example Consider the state aligned with the $+z$ axis, $|\psi\rangle = |{+}z\rangle$, so $\alpha = 1$ and $\beta = 0$: $$\langle S_z \rangle = \frac{\hbar}{2}(1 - 0) = \frac{\hbar}{2}, \quad \langle S_x \rangle = \hbar\operatorname{Re}(0) = 0, \quad \langle S_y \rangle = \hbar\operatorname{Im}(0) = 0.$$ The spin vector points entirely along $+z$ — exactly what you would expect for the eigenstate of $S_z$ with eigenvalue $+\hbar/2$. Now take $|\psi\rangle = \frac{1}{\sqrt{2}}|{+}z\rangle + \frac{1}{\sqrt{2}}|{-}z\rangle$ (the $+x$ eigenstate). Here $\alpha = \beta = 1/\sqrt{2}$: $$\langle S_z \rangle = \frac{\hbar}{2}\!\left(\frac{1}{2} - \frac{1}{2}\right) = 0, \quad \langle S_x \rangle = \hbar\operatorname{Re}\!\left(\frac{1}{2}\right) = \frac{\hbar}{2}, \quad \langle S_y \rangle = 0.$$ The mean spin now points entirely along $+x$, consistent with this state being an eigenstate of $S_x$. <Callout type="tip"> The Bloch vector $(n_x, n_y, n_z)$ of a pure spin state satisfies $\langle S_i \rangle = \frac{\hbar}{2} n_i$, so the expectation values directly give the direction the spin "points" on the Bloch sphere. For a mixed state $n_i$ would be strictly less than 1 in magnitude. </Callout> ## Try it This is a numerical exercise — your code should `return` a number, not a circuit. A spin-1/2 particle is prepared in the polar state $$|\psi\rangle = \cos\!\left(\frac{\pi}{6}\right)|{+}z\rangle + \sin\!\left(\frac{\pi}{6}\right)|{-}z\rangle,$$ where $\cos(\pi/6) = \sqrt{3}/2$ and $\sin(\pi/6) = 1/2$. Compute $\langle S_z \rangle / (\hbar/2)$, the dimensionless ratio of the $z$-expectation value to the spin quantum $\hbar/2$. Using the formula above, this equals $|\alpha|^2 - |\beta|^2$. <RunnableExample />