A Stern–Gerlach magnet can be turned to any direction $\mathbf{n}$, and it then measures the spin component $\mathbf{n}\cdot\hat{\mathbf{S}}$. Its outcomes are again $\pm\hbar/2$, because no direction is special, and its eigenstates are up and down along $\mathbf{n}$. A spin prepared up along one axis and measured along another at angle $\alpha$ comes out up with probability $\cos^{2}(\alpha/2)$: certainly at 0, half and half at $90^\circ$, never at $180^\circ$. The average of the measured component is $\cos\alpha$ — exactly the projection a classical arrow of unit length would have.

Shine light polarised at some angle through a polariser: the fraction passing depends only on the angle between them. A spin through a turned Stern–Gerlach magnet obeys the same kind of rule, with half angles in place of whole ones.

A spin prepared up along $z$ and measured along a direction $\mathbf{n}$ at angle $\alpha$ from $z$, for $\alpha$ from 0 to $\pi$: up along with probability , down with . The two curves cross at , where the outcome is a coin toss.

For a unit vector $\mathbf{n}=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)$ the component of spin along $\mathbf{n}$ is , with eigenvalues .

• $(\mathbf{n}\cdot\vec{\sigma})^{2}=1$, so its eigenvalues are ; its eigenstate is .

Write the eigenstate of $\mathbf{n}\cdot\vec{\sigma}$ with eigenvalue $+1$ and take its overlap with $\ket{\uparrow}$: only the first amplitude survives, and its square is the probability.

The $+1$ eigenstate of $\mathbf{n}\cdot\vec{\sigma}$, as multiplying out the matrix confirms.

The bra conjugates the amplitudes, and only the $\ket{\uparrow}$ term overlaps.

The measurement postulate, and a double-angle identity.

Outcomes $\pm1$ with their probabilities: the projection of $\hat{\mathbf{z}}$ on $\mathbf{n}$.

Up along $z$, measured along a direction at angle $\alpha$, gives up with probability $\cos^{2}(\alpha/2)$.

The spin component along any direction has the same two eigenvalues, $\pm\hbar/2$: $(\mathbf{n}\cdot\vec{\sigma})^{2}=1$.

A spin up along $\mathbf{n}$ is never found down along $\mathbf{n}$; measured along the opposite direction it is never found up.

In the state up along $\mathbf{m}$, the average of $\mathbf{n}\cdot\vec{\sigma}$ is $\mathbf{n}\cdot\mathbf{m}$, the cosine of the angle between the two directions.

Great work! You now know how to:

• write the spin component along any direction as a matrix
• compute the probabilities $\cos^{2}(\alpha/2)$ and $\sin^{2}(\alpha/2)$
• find the average of a spin component in any state