Checkpoint: Spin Measurements

This checkpoint synthesises every tool developed in the module: the Born rule, the eigenbases of $S_x$ , $S_y$ , and $S_z$ , the collapse postulate, and sequential measurements. Work through each part carefully before computing the final answer.

Recap: the key eigenstates

The $S_z$ eigenstates $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ (equivalently $|0\rangle$ and $|1\rangle$ in qubit notation) form the standard basis. Every other spin direction has eigenstates that are superpositions of these two. For $S_x$ and $S_y$ :

|{+}\rangle_x = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle + |{\downarrow}\rangle\bigr), \qquad |{-}\rangle_x = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle - |{\downarrow}\rangle\bigr),

|{+}\rangle_y = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle + i|{\downarrow}\rangle\bigr), \qquad |{-}\rangle_y = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle - i|{\downarrow}\rangle\bigr).

These relations follow directly from diagonalising the Pauli matrices $\sigma_x$ and $\sigma_y$ . You can verify normalisation: $|1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1$ for both $|{\pm}\rangle_x$ , and $|1/\sqrt{2}|^2 + |i/\sqrt{2}|^2 = 1/2 + 1/2 = 1$ for both $|{\pm}\rangle_y$ .

The Born rule

For any state $|\psi\rangle$ and any eigenstate $|e\rangle$ of an observable, the probability of obtaining the corresponding eigenvalue is

$P = |\langle e|\psi\rangle|^2.$

The inner product $\langle e|\psi\rangle$ is the probability amplitude. Its squared modulus gives the probability. Because probabilities are real and non-negative, only the modulus of the amplitude matters for a single outcome — the phase is irrelevant.

The collapse postulate

After a measurement yields eigenvalue $\lambda$ , the state of the system collapses to the corresponding eigenstate. Any subsequent measurement treats the collapsed state as the new initial condition. Sequential measurements therefore require two independent applications of the Born rule.

Sequential measurements: the joint probability

Suppose a spin- $\frac{1}{2}$ particle is prepared in

$|\psi\rangle = \frac{1}{\sqrt{5}}|{\uparrow}\rangle_z + \frac{2}{\sqrt{5}}|{\downarrow}\rangle_z.$

Check normalization: $|1/\sqrt{5}|^2 + |2/\sqrt{5}|^2 = 1/5 + 4/5 = 1$ . Good.

Step 1. The particle enters a Stern-Gerlach device oriented along $x$ . The probability of obtaining $S_x = +\hbar/2$ is

$P_1 = |{}_x\langle{+}|\psi\rangle|^2.$

Compute the amplitude:

{}_x\langle{+}|\psi\rangle = \frac{1}{\sqrt{2}}\bigl(\langle{\uparrow}| + \langle{\downarrow}|\bigr) \left(\frac{1}{\sqrt{5}}|{\uparrow}\rangle + \frac{2}{\sqrt{5}}|{\downarrow}\rangle\right) = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}}\right) = \frac{3}{\sqrt{10}}.

Therefore

$P_1 = \left|\frac{3}{\sqrt{10}}\right|^2 = \frac{9}{10}.$

Step 2. Only the particles that gave $S_x = +\hbar/2$ continue. Their state has collapsed to $|{+}\rangle_x$ . These particles now enter a second device, oriented along $z$ . The probability of obtaining $S_z = +\hbar/2$ is

$P_2 = |{}_z\langle{\uparrow}|{+}\rangle_x|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}.$

Note that $P_2$ does not depend on the original state $|\psi\rangle$ at all — the collapse to $|{+}\rangle_x$ in step 1 has erased all information about the initial amplitudes.

Joint probability. The events are sequential and the conditional probabilities are computed from collapsed states, so the joint probability of both outcomes in order is

= P_1 \times P_2 = \frac{9}{10} \times \frac{1}{2} = \frac{9}{20} = 0.45.$$ <Callout type="tip"> The result $P_2 = 1/2$ regardless of the initial state illustrates a deep feature of quantum measurement: sending a particle through an x-filter completely randomises its z-spin. This is why sequential Stern-Gerlach measurements reveal the incompatibility of spin components — no hidden variable can restore the original z-information after the x-measurement. </Callout> ## Try it This is a numerical exercise — your code should `return` a number. Reproduce the two-step calculation above and return the joint probability $P(S_x = +\hbar/2,\,\text{then}\,S_z = +\hbar/2)$ for the state $|\psi\rangle = \frac{1}{\sqrt{5}}|{\uparrow}\rangle_z + \frac{2}{\sqrt{5}}|{\downarrow}\rangle_z$. <RunnableExample />