Checkpoint: Apply the Postulates

This checkpoint pulls together the measurement machinery of Module 3: the Born rule, the projection postulate, expectation values, and a joint observable on an entangled state. Work each sub-problem from the definitions, then return the sum so the grader can check all four at once.

(A) Born rule in the computational basis

For $|\psi\rangle = \cos\theta\,|0\rangle + \sin\theta\,|1\rangle$ measured in the $Z$ basis, the projector for outcome $1$ is $|1\rangle\langle 1|$ and

With $\theta = \pi/6$, $p(1) = \sin^2(\pi/6) = (1/2)^2 = 1/4$.

(B) Expectation value of $Z$

The expectation value weights each eigenvalue by its probability:

For $\theta = \pi/6$, $\langle\hat Z\rangle = \cos(\pi/3) = 1/2$. (Consistency check: $\langle\hat Z\rangle = p(0) - p(1) = \tfrac34 - \tfrac14 = \tfrac12$.)

(C) Projection postulate and a conditional probability

Take the normalized two-qubit state

Measure qubit 0 and condition on the outcome $0$. The projector $|0\rangle\langle 0|\otimes\hat I$ keeps the $|00\rangle$ and $|01\rangle$ terms. After renormalizing, the conditional probability that qubit 1 reads $1$ is

(D) A joint observable on a Bell state

The operator $\hat Z\otimes\hat Z$ has eigenvalue $+1$ on $|00\rangle$ and $|11\rangle$, and $-1$ on $|01\rangle$ and $|10\rangle$. For the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$, the Born rule gives probability $\tfrac12$ on each of $|00\rangle$ and $|11\rangle$ and zero on the others, so

The two qubits always agree when both are measured in $Z$ — the correlation that defines this Bell state.

Try it

Compute the four quantities and return their sum: $A = p(1)$, $B = \langle\hat Z\rangle$, $C = p(q_1{=}1\mid q_0{=}0)$, and $D = \langle\hat Z\otimes\hat Z\rangle$. The values are $0.25 + 0.5 + 0.8 + 1 = 2.55$.

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