Exam Prep — Measurement: Probability + Post-Measurement State

Method

Measurement procedure

Given: a quantum state $\ket{\psi}$ and a measurement basis $\{\ket{a}, \ket{b}\}$.

Step 1 — Rewrite the state in the measurement basis.
Express $\ket{\psi} = c_a\ket{a} + c_b\ket{b}$. If the measurement basis is not the computational basis, you need to do a basis change first.

Step 2 — Probability = |coefficient|².
$P(\text{outcome } a) = |c_a|^2$, $P(\text{outcome } b) = |c_b|^2$.

Step 3 — Post-measurement state = the basis vector you measured.
If outcome $a$ occurs, the state collapses to $\ket{a}$. That's it — the basis vector itself.

Partial measurement (one qubit of a multi-qubit state)

When measuring only one qubit of a multi-qubit state:

Group terms by the measured qubit's value in the measurement basis.
Probability = sum of |amplitude|² in that group.
Post-measurement state = the group, renormalized so the remaining amplitudes squared sum to 1.

The key skill is Step 1 — rewriting the state in a non-standard basis. Everything else is just reading off coefficients and squaring.

Worked Examples

Example 1 — Standard basis measurement (warm-up)

Measure $\ket{\psi} = \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}\ket{1}$ in the computational basis $\{\ket{0}, \ket{1}\}$.

Step 1 — Rewrite in measurement basis Already in the computational basis. \(c_0 = \frac{\sqrt{3}}{2}\), \(c_1 = \frac{1}{2}\). Step 2 — Probabilities $$P(0) = \left|\frac{\sqrt{3}}{2}\right|^2 = \frac{3}{4} \qquad P(1) = \left|\frac{1}{2}\right|^2 = \frac{1}{4}$$ Step 3 — Post-measurement states $$\text{If outcome } 0: \quad \ket{\psi'} = \ket{0}$$ $$\text{If outcome } 1: \quad \ket{\psi'} = \ket{1}$$

Sanity check: $\frac{3}{4} + \frac{1}{4} = 1$. Probabilities always sum to 1.

Example 2 — Non-standard basis ($\{+, -\}$)

Measure $\ket{\psi} = \ket{0}$ in the $\{\ket{+}, \ket{-}\}$ basis.

Step 1 — Rewrite in the measurement basis We need to express \(\ket{0}\) in terms of \(\ket{+}\) and \(\ket{-}\). Using the inverse relations: $$\ket{0} = \frac{1}{\sqrt{2}}\ket{+} + \frac{1}{\sqrt{2}}\ket{-}$$ So \(c_+ = \frac{1}{\sqrt{2}}\), \(c_- = \frac{1}{\sqrt{2}}\). Step 2 — Probabilities $$P(+) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} \qquad P(-) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$ Step 3 — Post-measurement states $$\text{If outcome } +: \quad \ket{\psi'} = \ket{+} = \frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$$ $$\text{If outcome } -: \quad \ket{\psi'} = \ket{-} = \frac{1}{\sqrt{2}}(\ket{0}-\ket{1})$$

Key insight: $\ket{0}$ is maximally uncertain in the $\{+, -\}$ basis. This is the complementarity principle in action.

Example 3 — Partial measurement (2-qubit, standard basis)

Measure the first qubit (in the computational basis) of:

$$\ket{\psi} = \frac{1}{\sqrt{3}}\ket{00} + \frac{1}{\sqrt{3}}\ket{01} + \frac{1}{\sqrt{3}}\ket{10}$$

Step 1 — Group by first qubit's value First qubit = \(\ket{0}\):   \(\frac{1}{\sqrt{3}}\ket{00} + \frac{1}{\sqrt{3}}\ket{01} = \frac{1}{\sqrt{3}}\ket{0}\bigl(\ket{0}+\ket{1}\bigr)\) First qubit = \(\ket{1}\):   \(\frac{1}{\sqrt{3}}\ket{10} = \frac{1}{\sqrt{3}}\ket{1}\ket{0}\) Step 2 — Probabilities $$P(0) = \left|\frac{1}{\sqrt{3}}\right|^2 + \left|\frac{1}{\sqrt{3}}\right|^2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$ $$P(1) = \left|\frac{1}{\sqrt{3}}\right|^2 = \frac{1}{3}$$ Step 3 — Post-measurement states (renormalize!) If outcome 0: The unnormalized remaining state is \(\frac{1}{\sqrt{3}}\ket{0}(\ket{0}+\ket{1})\). Renormalize by dividing by \(\sqrt{P(0)} = \sqrt{2/3}\): $$\ket{\psi'} = \ket{0} \otimes \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) = \ket{0}\ket{+}$$ If outcome 1: Only one term, already normalized: $$\ket{\psi'} = \ket{1}\ket{0}$$

The renormalization step is where most exam mistakes happen. Always check that your post-measurement state is normalized.

Example 4 — Non-standard partial measurement ($\{+i, -i\}$ basis, exam-level)

Measure the first qubit in the $\{\ket{+i}, \ket{-i}\}$ basis on the state:

$$\ket{\psi} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}$$

where $\ket{+i} = \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1})$ and $\ket{-i} = \frac{1}{\sqrt{2}}(\ket{0}-i\ket{1})$.

Step 1 — Express computational basis in terms of measurement basis Invert the definitions. From the basis change reference below: $$\ket{0} = \frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i}$$ $$\ket{1} = \frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}$$ Step 2 — Substitute into the state $$\ket{\psi} = \frac{1}{\sqrt{2}}\ket{0}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}\ket{1}$$ Replace the first qubit only (we're only measuring the first qubit): $$= \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i}\right)\ket{0} + \frac{1}{\sqrt{2}}\left(\frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}\right)\ket{1}$$ Expand and group by measurement basis state: $$= \ket{+i}\left(\frac{1}{2}\ket{0} - \frac{i}{2}\ket{1}\right) + \ket{-i}\left(\frac{1}{2}\ket{0} + \frac{i}{2}\ket{1}\right)$$ Step 3 — Probabilities $$P(+i) = \left|\frac{1}{2}\right|^2 + \left|\frac{-i}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ $$P(-i) = \left|\frac{1}{2}\right|^2 + \left|\frac{i}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ Step 4 — Post-measurement states (renormalize) If outcome \(+i\): Renormalize \(\frac{1}{2}\ket{0} - \frac{i}{2}\ket{1}\) by dividing by \(\sqrt{1/2}\): $$\ket{\psi'} = \ket{+i} \otimes \frac{1}{\sqrt{2}}(\ket{0} - i\ket{1}) = \ket{+i}\ket{-i}$$ If outcome \(-i\): Renormalize \(\frac{1}{2}\ket{0} + \frac{i}{2}\ket{1}\) by dividing by \(\sqrt{1/2}\): $$\ket{\psi'} = \ket{-i} \otimes \frac{1}{\sqrt{2}}(\ket{0} + i\ket{1}) = \ket{-i}\ket{+i}$$

Notice the anti-correlation: measuring $\ket{+i}$ on the first qubit forces the second into $\ket{-i}$, and vice versa. This is entanglement in action — the Bell state $\ket{\Phi^+}$ is correlated in every basis.

Watch Out

Common pitfalls

Forgetting to renormalize. After a partial measurement, the remaining state must be renormalized. Divide by $\sqrt{P(\text{outcome})}$. This is the #1 source of lost marks.
Confusing measurement basis with computational basis. If the problem says "measure in the $\{\ket{+}, \ket{-}\}$ basis," you must rewrite the state in that basis first. You cannot just read off $\ket{0}$ and $\ket{1}$ coefficients.
Complex modulus squared. Remember: $|a + bi|^2 = a^2 + b^2$, not $(a+bi)^2$. Equivalently, $|z|^2 = z \cdot z^*$. In particular: $|i|^2 = 1$, $|-i|^2 = 1$, $\left|\frac{1+i}{\sqrt{2}}\right|^2 = 1$.
Post-measurement state is the basis vector, not the coefficient. If you measure in $\{\ket{+i}, \ket{-i}\}$ and get outcome $+i$, the measured qubit collapses to $\ket{+i}$ — not to the coefficient that was in front of it.
Partial vs. full measurement. Measuring one qubit of a 2-qubit state does not collapse the other qubit to a basis state. The other qubit's state is whatever remains after grouping and renormalizing.

Basis Change Reference

Computational ↔ Hadamard basis ($\{+, -\}$)

Definitions:

$$\ket{+} = \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) \qquad \ket{-} = \frac{1}{\sqrt{2}}(\ket{0}-\ket{1})$$

Inverse (for rewriting in $\{+,-\}$ basis):

$$\ket{0} = \frac{1}{\sqrt{2}}\ket{+} + \frac{1}{\sqrt{2}}\ket{-} \qquad \ket{1} = \frac{1}{\sqrt{2}}\ket{+} - \frac{1}{\sqrt{2}}\ket{-}$$

Computational ↔ Circular basis ($\{+i, -i\}$)

Definitions:

$$\ket{+i} = \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) \qquad \ket{-i} = \frac{1}{\sqrt{2}}(\ket{0}-i\ket{1})$$

Inverse (for rewriting in $\{+i,-i\}$ basis):

$$\ket{0} = \frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i}$$

$$\ket{1} = \frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}$$

Derivation of $\ket{1}$: Solve the system. From the definitions, $\ket{+i}+\ket{-i} = \sqrt{2}\ket{0}$ and $\ket{+i}-\ket{-i} = i\sqrt{2}\ket{1}$, so $\ket{1} = \frac{1}{i\sqrt{2}}(\ket{+i}-\ket{-i}) = \frac{-i}{\sqrt{2}}(\ket{+i}-\ket{-i})$.

Practice

Problem 1

Measure $\ket{\psi} = \frac{1}{\sqrt{2}}\ket{0} + \frac{i}{\sqrt{2}}\ket{1}$ in the computational basis $\{\ket{0}, \ket{1}\}$. What are the probabilities, and what is the post-measurement state for each outcome?

Solution

Already in the computational basis. $c_0 = \frac{1}{\sqrt{2}}$, $c_1 = \frac{i}{\sqrt{2}}$.

$$P(0) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} \qquad P(1) = \left|\frac{i}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

Note: $|i|^2 = 1$, so the imaginary coefficient doesn't change the probability.

Post-measurement: outcome $0 \to \ket{0}$, outcome $1 \to \ket{1}$.

Problem 2

Measure $\ket{\psi} = \ket{1}$ in the $\{\ket{+}, \ket{-}\}$ basis. What are the probabilities and post-measurement states?

Solution

Rewrite $\ket{1}$ in the $\{+, -\}$ basis:

$$\ket{1} = \frac{1}{\sqrt{2}}\ket{+} - \frac{1}{\sqrt{2}}\ket{-}$$

$$P(+) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} \qquad P(-) = \left|-\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

Post-measurement: outcome $+ \to \ket{+}$, outcome $- \to \ket{-}$.

The minus sign in front of $\ket{-}$ only affects the coefficient's sign, not the probability (we take the modulus squared).

Problem 3

Measure $\ket{\psi} = \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}$ in the $\{\ket{+i}, \ket{-i}\}$ basis. What are the probabilities and post-measurement states?

Solution

The state is $\ket{+}$. Rewrite $\ket{0}$ and $\ket{1}$ in the $\{+i, -i\}$ basis:

$$\ket{0} = \frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i} \qquad \ket{1} = \frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}$$

Substitute:

$$\ket{\psi} = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i}\right) + \frac{1}{\sqrt{2}}\left(\frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}\right)$$

$$= \frac{1-i}{2}\ket{+i} + \frac{1+i}{2}\ket{-i}$$

$$P(+i) = \left|\frac{1-i}{2}\right|^2 = \frac{1+1}{4} = \frac{1}{2} \qquad P(-i) = \left|\frac{1+i}{2}\right|^2 = \frac{1+1}{4} = \frac{1}{2}$$

Post-measurement: outcome $+i \to \ket{+i}$, outcome $-i \to \ket{-i}$.

$\ket{+}$ is equally uncertain in the circular basis, just as $\ket{0}$ is equally uncertain in the Hadamard basis. Three mutually unbiased bases.

Problem 4

Measure the first qubit (in the computational basis) of:

$$\ket{\psi} = \frac{1}{2}\ket{00} + \frac{i}{2}\ket{01} + \frac{1}{2}\ket{10} + \frac{i}{2}\ket{11}$$

Find probabilities and post-measurement states.

Solution

Group by first qubit:

First qubit = $\ket{0}$: $\frac{1}{2}\ket{0}\ket{0} + \frac{i}{2}\ket{0}\ket{1} = \ket{0}\left(\frac{1}{2}\ket{0} + \frac{i}{2}\ket{1}\right)$

First qubit = $\ket{1}$: $\frac{1}{2}\ket{1}\ket{0} + \frac{i}{2}\ket{1}\ket{1} = \ket{1}\left(\frac{1}{2}\ket{0} + \frac{i}{2}\ket{1}\right)$

$$P(0) = \left|\frac{1}{2}\right|^2 + \left|\frac{i}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

$$P(1) = \left|\frac{1}{2}\right|^2 + \left|\frac{i}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Renormalize each group by dividing by $\sqrt{1/2} = \frac{1}{\sqrt{2}}$:

$$\text{If outcome } 0: \quad \ket{\psi'} = \ket{0} \otimes \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = \ket{0}\ket{+i}$$

$$\text{If outcome } 1: \quad \ket{\psi'} = \ket{1} \otimes \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = \ket{1}\ket{+i}$$

The second qubit ends up in $\ket{+i}$ regardless of the measurement outcome. This means the state was actually a product state: $\frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = \ket{+}\ket{+i}$. Measuring the first qubit tells us nothing about the second.

Problem 5

Measure the first qubit in the $\{\ket{+i}, \ket{-i}\}$ basis on the state:

$$\ket{\psi} = \frac{1}{\sqrt{2}}\ket{01} + \frac{1}{\sqrt{2}}\ket{10}$$

Find probabilities and post-measurement states. (This is $\ket{\Psi^+}$, a Bell state.)

Solution

Substitute the first qubit using the $\{+i, -i\}$ inverse relations:

$$\ket{0} = \frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i} \qquad \ket{1} = \frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}$$

$$\ket{\psi} = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\ket{+i} + \frac{1}{\sqrt{2}}\ket{-i}\right)\ket{1} + \frac{1}{\sqrt{2}}\left(\frac{-i}{\sqrt{2}}\ket{+i} + \frac{i}{\sqrt{2}}\ket{-i}\right)\ket{0}$$

Group by measurement basis state:

$$= \ket{+i}\left(\frac{-i}{2}\ket{0} + \frac{1}{2}\ket{1}\right) + \ket{-i}\left(\frac{i}{2}\ket{0} + \frac{1}{2}\ket{1}\right)$$

$$P(+i) = \left|\frac{-i}{2}\right|^2 + \left|\frac{1}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

$$P(-i) = \left|\frac{i}{2}\right|^2 + \left|\frac{1}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Renormalize (divide by $\sqrt{1/2} = \frac{1}{\sqrt{2}}$):

If outcome $+i$:

$$\ket{\psi'} = \ket{+i} \otimes \frac{1}{\sqrt{2}}(-i\ket{0}+\ket{1}) = \ket{+i} \otimes \frac{1}{\sqrt{2}}(\ket{1}-i\ket{0})$$

Reorder: $\frac{1}{\sqrt{2}}(-i\ket{0}+\ket{1})$. Factor out $-i$: $-i \cdot \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = -i\ket{+i}$. Up to global phase:

$$\ket{\psi'} = \ket{+i}\ket{+i}$$

If outcome $-i$:

$$\ket{\psi'} = \ket{-i} \otimes \frac{1}{\sqrt{2}}(i\ket{0}+\ket{1})$$

Factor out $i$: $i \cdot \frac{1}{\sqrt{2}}(\ket{0}-i\ket{1}) = i\ket{-i}$. Up to global phase:

$$\ket{\psi'} = \ket{-i}\ket{-i}$$

The Bell state $\ket{\Psi^+}$ shows perfect correlation in the circular basis: both qubits always end up in the same state. Compare with $\ket{\Phi^+}$ (Example 4 above), which showed anti-correlation in this basis.

Example 1 — Standard basis measurement (warm-up)

Example 2 — Non-standard basis (\(\{+, -\}\))

Example 3 — Partial measurement (2-qubit, standard basis)

Example 4 — Non-standard partial measurement (\(\{+i, -i\}\) basis, exam-level)