Eigenstates Along Any Axis

We have seen that spin along $z$ has eigenstates $|0\rangle$ and $|1\rangle$, and spin along $x$ has eigenstates $|{\pm}\rangle$. What about an arbitrary direction $\hat{n} = (\sin\theta\cos\phi,\,\sin\theta\sin\phi,\,\cos\theta)$? The answer ties the abstract spin operator directly to a point on the Bloch sphere.

The spin operator along ĥ

The component of spin along $\hat{n}$ is the projection

Writing it out as a matrix,

Because $(\hat{n}\cdot\boldsymbol{\sigma})^2 = I$ for any unit vector (it follows from the anticommutator relations), this operator has the same eigenvalues $\pm 1$ as a single Pauli, so $S_{\hat{n}}$ has eigenvalues $\pm\tfrac{\hbar}{2}$ — two outcomes along every axis, as expected.

The eigenstates

Diagonalizing $\hat{n}\cdot\boldsymbol{\sigma}$ gives the spin-up and spin-down states along $\hat{n}$:

The half-angle $\theta/2$ is the signature of spin-½: rotating the measurement axis by $\theta$ in real space rotates the state by only $\theta/2$ in Hilbert space. This is why a spin-½ must be turned through $4\pi$, not $2\pi$, to return to its original state.

Quick checks

• $\theta = 0$: $|{+\hat{n}}\rangle = |0\rangle$ — spin up along $z$. ✓
• $\theta = \pi$: $|{+\hat{n}}\rangle = e^{i\phi}|1\rangle$ — spin down along $z$ (up along $-z$), up to an irrelevant phase. ✓
• $\theta = \tfrac{\pi}{2},\ \phi = 0$: $|{+\hat{n}}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |{+x}\rangle$. ✓

Try it

Prepare the spin-up eigenstate along the axis at polar angle $\theta = \pi/3$ with $\phi = 0$. After running, the Bloch arrow should tilt $60^\circ$ from the $+z$ axis toward $+x$.

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