State Update After Measurement

The second half of Postulate 3 is the projection postulate: after a measurement that returns outcome $a_k$ , the state collapses to the normalized projection of the old state onto the eigenspace of $a_k$ . This lesson makes that update concrete and shows how it generates conditional probabilities for later measurements.

The update rule

If a measurement of $\hat{A} = \sum_k a_k\hat{P}_k$ on $|\psi\rangle$ returns $a_k$ , the new state is

|\psi'\rangle = \frac{\hat{P}_k|\psi\rangle}{\sqrt{\langle\psi|\hat{P}_k|\psi\rangle}} = \frac{\hat{P}_k|\psi\rangle}{\sqrt{p(a_k)}}.

The numerator $\hat{P}_k|\psi\rangle$ keeps only the part of $|\psi\rangle$ lying in the eigenspace; the denominator $\sqrt{p(a_k)}$ restores unit norm so $|\psi'\rangle$ is a legitimate state. Notice that $|\psi'\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_k$ , which is why re-measuring gives $a_k$ again with certainty.

Partial measurement on a composite system

A frequent situation is measuring one qubit of a multi-qubit register. Measuring qubit 0 in the $Z$ basis uses the two projectors

\hat{P}_0^{(0)} = |0\rangle\langle 0| \otimes \hat{I}, \qquad \hat{P}_0^{(1)} = |1\rangle\langle 1| \otimes \hat{I},

which act only on qubit 0 and leave qubit 1 untouched. The projector keeps every term of the state whose qubit-0 label matches the outcome, then we renormalize.

Worked example

Consider the normalized two-qubit state

|\psi\rangle = \frac{1}{\sqrt{10}}\big(|00\rangle + 2|01\rangle + 2|10\rangle + |11\rangle\big),

with $1^2 + 2^2 + 2^2 + 1^2 = 10$ confirming normalization. Suppose we measure qubit 0 in the $Z$ basis and obtain $0$ . Apply $\hat{P}_0^{(0)} = |0\rangle\langle 0|\otimes\hat{I}$ : only the terms with qubit 0 equal to $|0\rangle$ survive,

\hat{P}_0^{(0)}|\psi\rangle = \frac{1}{\sqrt{10}}\big(|00\rangle + 2|01\rangle\big).

The probability of this outcome is the squared norm of what survives:

p(q_0 = 0) = \frac{1^2 + 2^2}{10} = \frac{5}{10} = \frac12.

Renormalizing by $\sqrt{p} = \sqrt{1/2}$ gives the collapsed state

|\psi'\rangle = \frac{1}{\sqrt5}\big(|00\rangle + 2|01\rangle\big).

Now ask: in this collapsed state, what is the probability that qubit 1 reads $1$ ? The amplitude of $|01\rangle$ (qubit 1 = 1) is $2/\sqrt5$ , so

p(q_1 = 1 \mid q_0 = 0) = \left|\frac{2}{\sqrt5}\right|^2 = \frac{4}{5} = 0.8.

This is a genuine conditional probability: the first measurement reshaped the state, and the second probability is computed from the updated state, not the original.

Try it

Starting from $|\psi\rangle = \tfrac{1}{\sqrt{10}}(|00\rangle + 2|01\rangle + 2|10\rangle + |11\rangle)$ , measure qubit 0 and condition on the outcome $0$ . Project, renormalize, and return the conditional probability $p(q_1 = 1 \mid q_0 = 0)$ .

Run your code to see the quantum state.