Uncertainty for Spin

Spin-1/2 is the simplest quantum system with non-commuting observables, making it the ideal laboratory for the Robertson uncertainty relation. Every result here follows from three $2\times 2$ matrices and straightforward algebra.

The Pauli operators

The spin- $\tfrac{1}{2}$ observables along the three coordinate axes are proportional to the Pauli matrices:

\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad \sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ Their commutators are $$[\sigma_x,\sigma_y] = 2i\sigma_z, \quad [\sigma_y,\sigma_z] = 2i\sigma_x, \quad [\sigma_z,\sigma_x] = 2i\sigma_y,$$ and each Pauli matrix squares to the identity: $\sigma_i^2 = I$. ## The Robertson uncertainty relation For any two observables $\hat{A}$ and $\hat{B}$ and any state $|\psi\rangle$, the **Robertson inequality** (1929) states $$\Delta A \cdot \Delta B \;\geq\; \frac{1}{2}\bigl|\langle[\hat{A},\hat{B}]\rangle\bigr|,$$ where the standard deviation of $\hat{A}$ is defined as $\Delta A = \sqrt{\langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2}$. Applied to $\sigma_x$ and $\sigma_y$ with the commutator $[\sigma_x,\sigma_y] = 2i\sigma_z$, the bound becomes $$\Delta\sigma_x \cdot \Delta\sigma_y \;\geq\; \bigl|\langle\sigma_z\rangle\bigr|.$$ ## Applying the bound to $|0\rangle$ Take the computational basis state $|0\rangle$ — spin-up along $z$. This is an eigenstate of $\sigma_z$ with eigenvalue $+1$, so $\langle\sigma_z\rangle = 1$. The Robertson bound therefore demands $$\Delta\sigma_x \cdot \Delta\sigma_y \;\geq\; 1.$$ Now compute each factor explicitly. Because $\sigma_x^2 = I$, we have $\langle\sigma_x^2\rangle = 1$ for **any** state. The expectation value of $\sigma_x$ in $|0\rangle$ is $$\langle 0|\sigma_x|0\rangle = \langle 0|\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \langle 0|\begin{pmatrix}0\\1\end{pmatrix}\rangle = \langle 0|1\rangle = 0.$$ Therefore $$\Delta\sigma_x = \sqrt{1 - 0^2} = 1.$$ By the same calculation with $\sigma_y$, $\langle\sigma_y\rangle = 0$ for $|0\rangle$, giving $\Delta\sigma_y = 1$ as well. The product $$\Delta\sigma_x \cdot \Delta\sigma_y = 1 \cdot 1 = 1$$ exactly matches the lower bound $|\langle\sigma_z\rangle| = 1$. The state $|0\rangle$ is a **minimum-uncertainty state** for this pair of observables. <Callout type="tip"> Eigenstates of one Pauli operator always saturate the Robertson bound for the other two. This is because for an eigenstate of $\sigma_z$ both transverse spreads take their maximal value $\Delta\sigma_x = \Delta\sigma_y = 1$, while $|\langle\sigma_z\rangle| = 1$ is also maximal — so both sides of the inequality equal $1$. </Callout> ## What changes in a superposition? Now consider the equal superposition $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$, which is an eigenstate of $\sigma_x$ with eigenvalue $+1$. For this state $\langle\sigma_z\rangle = 0$, so the Robertson bound is trivially $0 \leq \Delta\sigma_x\cdot\Delta\sigma_y$. Yet $|+\rangle$ is an eigenstate of $\sigma_x$, so $\Delta\sigma_x = 0$, and the product really is $0$. The state $|+\rangle$ has perfectly sharp spin along $x$, but completely uncertain spin along $z$ ($\Delta\sigma_z = 1$). The broader lesson: which pair of observables is simultaneously uncertain depends entirely on the state. The Robertson relation tells you the minimum price you must pay, but the state determines which directions bear the cost. ## Try it This is a numerical exercise — your code should `return` a number. For the spin-up state $|0\rangle$, compute the standard deviation $\Delta\sigma_x$. <RunnableExample />