# Projective Measurement

> Measurement is where the deterministic, unitary world of quantum mechanics hands you a *random*
> classical number — and then violently rewrites the state to match. This lesson makes that precise.
> The **Born rule** assigns the probabilities, **Hermitian observables** package the question you ask,
> and **projection** describes the collapse. Get this exactly right and the rest of quantum information
> — uncertainty, POVMs, entanglement, every algorithm's read-out step — follows without mysticism.

## Learning Objectives

After this lesson you will be able to:

1. State the **Born rule** and compute outcome probabilities $p_k = \langle\psi|P_k|\psi\rangle$.
2. Decompose an **observable** $A=A^\dagger$ into its spectral projectors $A=\sum_k\lambda_k P_k$ and
   read off measurement outcomes and probabilities.
3. Apply the **projection postulate**: the post-measurement state $|\psi\rangle\to P_k|\psi\rangle/\sqrt{p_k}$.
4. Prove **repeatability** of projective measurement and explain measuring in **different bases**
   (computational $Z$ vs. $X$).
5. Show that **orthogonal states are perfectly distinguishable** by a single projective measurement.

---

## Intuition

A measurement asks the state a *question* with a fixed list of allowed answers. The question is an
observable; the allowed answers are its eigenvalues. Quantum mechanics adds two twists that classical
physics does not have:

1. **The answer is random** unless the state was already an eigenstate of the question. The amount of
   "overlap" the state has with each eigenstate sets the probability — that's the Born rule.
2. **Asking changes the state.** Once you get answer $\lambda_k$, the state snaps onto the subspace of
   states that *would have given* $\lambda_k$ with certainty. This is **collapse**, and it is described
   by a **projector**.

Geometrically: a normalized state $|\psi\rangle$ is a unit vector. Measuring in an orthonormal basis
$\{|k\rangle\}$ drops a perpendicular onto each axis; the squared length of the shadow on axis $k$ is
the probability of outcome $k$, and after the measurement the vector *is* that axis. Pythagoras
($\sum_k |\langle k|\psi\rangle|^2 = 1$) is exactly why the probabilities sum to one.

---

## Theory

### The Born rule

> **Born rule (basis form).** If a system is in normalized state $|\psi\rangle$ and we measure in the
> orthonormal basis $\{|k\rangle\}$, the probability of outcome $k$ is
> $$ p_k = |\langle k|\psi\rangle|^2 = \langle\psi|k\rangle\langle k|\psi\rangle. $$

These are genuine probabilities: each $p_k = |\langle k|\psi\rangle|^2 \ge 0$, and using completeness
$\sum_k |k\rangle\langle k| = I$ ([0.1.3](../../term-0-foundations/c01-linear-algebra/03-dirac-notation.md)),

$$
\sum_k p_k = \sum_k \langle\psi|k\rangle\langle k|\psi\rangle
= \langle\psi|\Big(\sum_k |k\rangle\langle k|\Big)|\psi\rangle
= \langle\psi|I|\psi\rangle = \langle\psi|\psi\rangle = 1.
$$

So **normalization is conservation of probability**. This is also why states are only defined up to a
**global phase**: $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ give $|\langle k|\psi\rangle|^2$
identically (the $|e^{i\gamma}|^2 = 1$ cancels), so no measurement can ever distinguish them.

### Observables and spectral decomposition

A measurable quantity is an **observable**: a Hermitian operator $A = A^\dagger$. By the spectral
theorem for Hermitian operators ([0.1.5](../../term-0-foundations/c01-linear-algebra/05-eigen-spectral-theorem.md),
[0.1.6](../../term-0-foundations/c01-linear-algebra/06-special-operators.md)) it has **real**
eigenvalues and an orthonormal eigenbasis. Grouping eigenvectors that share an eigenvalue into
**eigenspaces**, write the **spectral decomposition**

$$
\boxed{\;A = \sum_k \lambda_k\, P_k,\qquad P_k = \sum_{j:\,\lambda_j=\lambda_k} |v_j\rangle\langle v_j|\;}
$$

where the $\lambda_k$ are the *distinct* eigenvalues and $P_k$ is the **orthogonal projector** onto the
$\lambda_k$-eigenspace. These projectors form a **complete set of orthogonal projectors**:

$$
P_k = P_k^\dagger,\qquad P_j P_k = \delta_{jk} P_k,\qquad \sum_k P_k = I.
$$

The first two say each $P_k$ is a projector ($P_k^2=P_k$) and distinct eigenspaces are orthogonal; the
last is completeness (the eigenvectors span $\mathcal H$). The eigenvalues $\lambda_k$ are the
**possible measured values**; the projectors $P_k$ encode the question's answer-slots.

### Projective (von Neumann) measurement

> **Projective measurement postulate.** A projective measurement is specified by a complete set of
> orthogonal projectors $\{P_k\}$ ($P_jP_k=\delta_{jk}P_k$, $\sum_k P_k=I$), one per outcome. Measuring
> state $|\psi\rangle$ yields outcome $k$ with probability
> $$ p_k = \langle\psi|P_k|\psi\rangle = \|P_k|\psi\rangle\|^2, $$
> and the state immediately after, *given* outcome $k$, is
> $$ |\psi'\rangle = \frac{P_k|\psi\rangle}{\sqrt{p_k}}\qquad (p_k>0). $$

The basis-form Born rule is the special case where every $P_k = |k\rangle\langle k|$ is rank one: then
$p_k = \langle\psi|k\rangle\langle k|\psi\rangle = |\langle k|\psi\rangle|^2$, recovering the boxed rule
above. The projector form is more general because it allows **degenerate** outcomes (rank $> 1$): a
measurement need not resolve which vector *inside* an eigenspace you landed on.

**Why these are probabilities, again.** Positivity: $p_k = \langle\psi|P_k|\psi\rangle =
\langle\psi|P_k^\dagger P_k|\psi\rangle = \|P_k|\psi\rangle\|^2 \ge 0$, using $P_k=P_k^\dagger=P_k^2$.
Normalization: $\sum_k p_k = \langle\psi|(\sum_k P_k)|\psi\rangle = \langle\psi|\psi\rangle = 1$.

**Renormalization of the collapsed state.** The raw post-measurement vector $P_k|\psi\rangle$ has norm
$\|P_k|\psi\rangle\| = \sqrt{p_k}$, so dividing by $\sqrt{p_k}$ restores a unit vector — collapse keeps
us on the unit sphere. Note this is a **non-linear, non-unitary, irreversible** update: information
about the components in the other eigenspaces is gone.

> **Expectation preview.** Once you have $p_k$, the average measured value is
> $\langle A\rangle = \sum_k \lambda_k p_k = \langle\psi|A|\psi\rangle$. We make this — and its spread —
> the whole subject of [the next lesson](02-expectation-uncertainty.md).

### Repeatability

Projective measurement is **repeatable**: measure the same observable twice in immediate succession and
the second result equals the first, with certainty.

**Proof.** Suppose the first measurement returned $k$, collapsing to $|\psi'\rangle = P_k|\psi\rangle/\sqrt{p_k}$.
The probability that an immediate second measurement of the same $\{P_j\}$ returns outcome $j$ is

$$
p_{j\mid k} = \langle\psi'|P_j|\psi'\rangle
= \frac{\langle\psi|P_k\,P_j\,P_k|\psi\rangle}{p_k}
= \frac{\langle\psi|P_k\,(\delta_{jk}P_k)|\psi\rangle}{p_k}
= \delta_{jk}\,\frac{\langle\psi|P_k|\psi\rangle}{p_k} = \delta_{jk},
$$

using $P_jP_k = \delta_{jk}P_k$ and $P_k^2=P_k$. So $p_{j\mid k}=1$ if $j=k$ and $0$ otherwise. ∎

Repeatability is what makes the eigenvalue a *property* of the post-measurement state: after measuring,
the system is in an eigenstate of $A$ with definite value $\lambda_k$. (This idealization assumes no
evolution between the two measurements; dynamics — [Term 1.6](../c06-dynamics/01-time-evolution.md) —
will generally move the state off the eigenstate.)

### Measuring in different bases

The *same* state gives *different* statistics depending on which observable you measure — i.e. which
orthonormal basis you project onto.

- **Computational ($Z$) basis.** $Z = (+1)|0\rangle\langle0| + (-1)|1\rangle\langle1|$, so the
  projectors are $P_0 = |0\rangle\langle0|$, $P_1 = |1\rangle\langle1|$. Outcome "$0$" ($\lambda=+1$)
  with probability $|\langle 0|\psi\rangle|^2$, outcome "$1$" ($\lambda=-1$) with $|\langle1|\psi\rangle|^2$.
- **Hadamard ($X$) basis.** $X = (+1)|+\rangle\langle+| + (-1)|-\rangle\langle-|$ with
  $|\pm\rangle = \tfrac1{\sqrt2}(|0\rangle\pm|1\rangle)$. The projectors are $P_+ = |+\rangle\langle+|$,
  $P_- = |-\rangle\langle-|$, and outcome probabilities are $|\langle\pm|\psi\rangle|^2$.

Hardware almost always measures in the **computational basis only**. To measure in another basis you
**rotate that basis onto the computational basis first**, then measure $Z$. Since $H|+\rangle=|0\rangle$
and $H|-\rangle=|1\rangle$ (the Hadamard swaps the $X$- and $Z$-bases, $HXH=Z$;
[Appendix E](../../appendix/E-math-identities.md)), measuring $X$ on $|\psi\rangle$ is identical to
applying $H$ and measuring $Z$:

$$
|\langle\pm|\psi\rangle|^2 = |\langle 0/1|H|\psi\rangle|^2 .
$$

This "rotate, then read $Z$" pattern is how *all* basis changes (and expectation-value estimation) are
implemented on real devices — we use it from [Term 1.2.3](../c02-qubits-bloch/03-bases-and-expectation.md)
onward and in the Braket code below.

### Perfect distinguishability of orthogonal states

> **Claim.** Two orthogonal states $|\psi_0\rangle\perp|\psi_1\rangle$ can be distinguished with
> **certainty** by a single projective measurement. Non-orthogonal states cannot
> (proved in [Lesson 3](03-povm-generalized.md)).

**Proof.** Extend $\{|\psi_0\rangle,|\psi_1\rangle\}$ to an orthonormal basis and take the projective
measurement $\{P_0,P_1,P_{\text{rest}}\}$ with $P_i = |\psi_i\rangle\langle\psi_i|$ and
$P_{\text{rest}} = I - P_0 - P_1$. If the system is prepared in $|\psi_0\rangle$, then
$p_0 = \langle\psi_0|P_0|\psi_0\rangle = |\langle\psi_0|\psi_0\rangle|^2 = 1$, while
$p_1 = |\langle\psi_1|\psi_0\rangle|^2 = 0$. Symmetrically for $|\psi_1\rangle$. So the outcome reveals
the prepared state with zero error. ∎

This is the bedrock fact behind quantum communication and read-out: an orthonormal basis is a set of
*perfectly reliable codewords*. The moment states are non-orthogonal, perfect discrimination is
impossible — the seed of the uncertainty principle and of quantum cryptography.

---

## Worked Examples

### Example 1 — Measuring $|+\rangle$ in the $Z$ basis

Let $|\psi\rangle = |+\rangle = \tfrac1{\sqrt2}(|0\rangle + |1\rangle)$ and measure the observable $Z$,
i.e. project with $P_0 = |0\rangle\langle0|$, $P_1=|1\rangle\langle1|$.

**Probabilities.**

$$
p_0 = \langle+|P_0|+\rangle = |\langle 0|+\rangle|^2 = \Big|\tfrac1{\sqrt2}\Big|^2 = \tfrac12,
\qquad
p_1 = |\langle 1|+\rangle|^2 = \tfrac12 .
$$

A coin flip, as expected for an equatorial Bloch state. The associated eigenvalues are $\lambda_0=+1$
and $\lambda_1=-1$, so $\langle Z\rangle = (+1)\tfrac12 + (-1)\tfrac12 = 0$.

**Collapse.** If we get outcome "$0$",

$$
|\psi'\rangle = \frac{P_0|+\rangle}{\sqrt{p_0}}
= \frac{\tfrac1{\sqrt2}|0\rangle}{\sqrt{1/2}} = |0\rangle.
$$

The superposition is destroyed; a re-measurement of $Z$ now yields "$0$" with certainty (repeatability).

### Example 2 — Measuring $|+\rangle$ in the $X$ basis

Same state $|\psi\rangle = |+\rangle$, now measure the observable $X$ with projectors
$P_+ = |+\rangle\langle+|$, $P_- = |-\rangle\langle-|$.

**Probabilities.** Because $|+\rangle$ is itself an eigenstate of $X$ (with eigenvalue $+1$),

$$
p_+ = \langle+|P_+|+\rangle = |\langle+|+\rangle|^2 = 1,
\qquad
p_- = |\langle-|+\rangle|^2 = 0 .
$$

So in the $X$ basis the outcome is **deterministic**: always "$+$". The state did not need to collapse —
$P_+|+\rangle/\sqrt{1} = |+\rangle$ leaves it unchanged.

**The lesson.** A single state $|+\rangle$ is *maximally uncertain* in $Z$ ($50/50$) yet *perfectly
certain* in $X$. "Definiteness" is a relationship between a state and a chosen observable, not an
intrinsic property of the state. This complementarity is quantified by the uncertainty relation next
lesson, and the "rotate then read $Z$" recipe ($X$-measurement $=$ apply $H$, measure $Z$) is exactly
the Braket experiment below.

---

## Hands-on (Python)

A self-contained NumPy simulator for projective measurement, then a brief Braket sanity check. We do
all quantitative work in NumPy; Braket here is only an illustration of sampling on the (free) local
simulator.

```python
import numpy as np

# ---- States and observables (big-endian; |0> is index 0) -----------------
ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)
plus = (ket0 + ket1) / np.sqrt(2)          # |+>
minus = (ket0 - ket1) / np.sqrt(2)         # |->

Z = np.array([[1, 0], [0, -1]], dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)


def projectors_from_observable(A):
    """Spectral decomposition A = sum_k lambda_k P_k.

    Returns (distinct eigenvalues, projectors) with eigenvalues grouped so that
    degenerate eigenspaces yield a single (possibly rank>1) projector.
    """
    A = (A + A.conj().T) / 2                 # symmetrize for numerical safety
    eigvals, eigvecs = np.linalg.eigh(A)     # eigh: Hermitian, orthonormal columns
    distinct = []
    projectors = []
    for lam, vec in zip(eigvals, eigvecs.T):
        P = np.outer(vec, vec.conj())        # |v><v|
        # merge into an existing eigenspace if this eigenvalue already appeared
        for i, d in enumerate(distinct):
            if np.isclose(lam, d):
                projectors[i] = projectors[i] + P
                break
        else:
            distinct.append(lam.real)
            projectors.append(P)
    return np.array(distinct), projectors


def born_probabilities(psi, projectors):
    """p_k = <psi|P_k|psi> for each projector."""
    return np.array([np.real(psi.conj() @ (P @ psi)) for P in projectors])


def collapse(psi, P, p):
    """Post-measurement state P|psi>/sqrt(p)."""
    return (P @ psi) / np.sqrt(p)


def measure_once(psi, projectors, eigvals, rng):
    """Sample one projective-measurement outcome; return (eigenvalue, new_state)."""
    p = born_probabilities(psi, projectors)
    k = rng.choice(len(projectors), p=p)
    return eigvals[k], collapse(psi, projectors[k], p[k])


# ---- Measuring |+> in the Z basis vs the X basis -------------------------
valsZ, projZ = projectors_from_observable(Z)
valsX, projX = projectors_from_observable(X)

print("Z-basis probs on |+>:", np.round(born_probabilities(plus, projZ), 6))  # [0.5 0.5]
print("X-basis probs on |+>:", np.round(born_probabilities(plus, projX), 6))  # deterministic

# Sanity checks against the postulate
for P in projZ:
    assert np.allclose(P @ P, P) and np.allclose(P, P.conj().T)   # projector
assert np.allclose(sum(projZ), np.eye(2))                         # completeness
assert np.isclose(born_probabilities(plus, projZ).sum(), 1.0)     # normalization
```

```python
# ---- Empirical sampling: 50/50 for Z, and repeatability ------------------
rng = np.random.default_rng(0)

shots = 10_000
counts = {+1: 0, -1: 0}
for _ in range(shots):
    val, _ = measure_once(plus, projZ, valsZ, rng)
    counts[int(round(val.real))] += 1
print("Z-measurement of |+> :", counts)        # ~5000 / ~5000

# Repeatability: after collapse, remeasure -> same outcome, every time
val1, psi_after = measure_once(plus, projZ, valsZ, rng)
val2, _ = measure_once(psi_after, projZ, valsZ, rng)
assert val1 == val2                              # always equal
print("Collapsed eigenvalue", int(val1.real), "is repeatable:", val1 == val2)
```

A brief look on **Braket's local simulator** (free — no AWS charges, no account needed). Measuring
$|+\rangle$ in $Z$ is `H` on $|0\rangle$ then a default (computational-basis) measurement; we expect a
roughly even split. To measure $X$ instead, you would apply a second `H` before measuring (rotate the
$X$-basis onto $Z$).

```python
# Braket LocalSimulator: prepare |+> = H|0>, sample in the computational (Z) basis.
from braket.circuits import Circuit
from braket.devices import LocalSimulator

device = LocalSimulator()                  # free local state-vector simulator
circ = Circuit().h(0)                       # |0> -> |+>
counts = device.run(circ, shots=1000).result().measurement_counts
print(counts)                               # e.g. Counter({'0': ~500, '1': ~500})
# Big-endian: the single bit is qubit 0. Outcome '0' <-> Z eigenvalue +1.
```

> The NumPy `measure_once` is the *reference semantics*; Braket samples the same Born distribution on a
> high-performance simulator. We keep expectation/probability computations in NumPy (exact), and use
> Braket only for sampling — the deep Braket introduction is [Term 2.1](../../term-2-qc-core/c01-gates-circuits/04-braket-first-circuits.md).

---

## Exercises

**E1 (easy).** A qubit is in $|\psi\rangle = \tfrac35|0\rangle + \tfrac45|1\rangle$. Compute the
probabilities of outcomes "$0$" and "$1$" in a computational-basis measurement, and the post-measurement
state for each.

<details><summary>Solution</summary>

$p_0 = |\tfrac35|^2 = \tfrac9{25}$, $p_1 = |\tfrac45|^2 = \tfrac{16}{25}$; they sum to $1$. ✓
Collapse: outcome "$0$" gives $\dfrac{P_0|\psi\rangle}{\sqrt{p_0}} = \dfrac{\frac35|0\rangle}{3/5} = |0\rangle$;
outcome "$1$" gives $\dfrac{\frac45|1\rangle}{4/5} = |1\rangle$. (Computational-basis projectors are
rank one, so collapse always lands exactly on $|0\rangle$ or $|1\rangle$.)
</details>

**E2 (easy).** Measure the state $|0\rangle$ in the $X$ basis. What are the probabilities, and what is
the collapsed state for each outcome?

<details><summary>Solution</summary>

$\langle+|0\rangle = \langle-|0\rangle = \tfrac1{\sqrt2}$, so $p_+ = p_- = \tfrac12$. Outcome "$+$"
collapses to $|+\rangle$, outcome "$-$" to $|-\rangle$. ($|0\rangle$ is unbiased in the $X$ basis,
mirroring Example 2 with the roles of $X$ and $Z$ swapped.)
</details>

**E3 (medium).** Let $A = \begin{psmallmatrix}2&0\\0&5\end{psmallmatrix}$ and
$|\psi\rangle = \tfrac1{\sqrt5}(|0\rangle + 2|1\rangle)$. Find the spectral projectors of $A$, the
probability of each measured value, and the expectation $\langle A\rangle$ two ways (via $\sum_k\lambda_k p_k$
and via $\langle\psi|A|\psi\rangle$).

<details><summary>Solution</summary>

$A$ is already diagonal: eigenvalues $\lambda=2$ (eigenvector $|0\rangle$) and $\lambda=5$ ($|1\rangle$),
so $P_2 = |0\rangle\langle0|$, $P_5 = |1\rangle\langle1|$. Probabilities:
$p_2 = |\tfrac1{\sqrt5}|^2 = \tfrac15$, $p_5 = |\tfrac2{\sqrt5}|^2 = \tfrac45$.
Then $\langle A\rangle = 2\cdot\tfrac15 + 5\cdot\tfrac45 = \tfrac{2+20}{5} = \tfrac{22}{5}$. Directly,
$\langle\psi|A|\psi\rangle = \tfrac15(1\cdot2 + 4\cdot5) = \tfrac{22}{5}$. ✓ Agreement is the identity
$\langle A\rangle = \sum_k\lambda_k p_k$.
</details>

**E4 (medium).** Consider the degenerate observable on two qubits $A = Z\otimes I$ acting on
$|\psi\rangle = \tfrac1{\sqrt2}(|00\rangle + |11\rangle)$. Find the spectral projectors of $A$ (note the
eigenvalues are degenerate, rank-2), the outcome probabilities, and the collapsed states.

<details><summary>Solution</summary>

$Z\otimes I$ has eigenvalue $+1$ on the span of $\{|00\rangle,|01\rangle\}$ and $-1$ on
$\{|10\rangle,|11\rangle\}$, so $P_{+1} = |0\rangle\langle0|\otimes I$ and $P_{-1} = |1\rangle\langle1|\otimes I$
(each rank 2). On $|\psi\rangle$: $P_{+1}|\psi\rangle = \tfrac1{\sqrt2}|00\rangle$, giving
$p_{+1} = \tfrac12$ and collapse $|00\rangle$; likewise $p_{-1}=\tfrac12$ with collapse $|11\rangle$.
The degenerate measurement learns only qubit 0's value, but here that already pins the full state
because of the entanglement — a preview of [Term 1.4](../c04-entanglement/01-multi-qubit-states.md).
</details>

**E5 (hard).** Prove that the **average post-measurement state** (averaging over outcomes, weighted by
their probabilities, before you learn the outcome) is the "decohered" operator $\sum_k P_k\,\rho\,P_k$
where $\rho = |\psi\rangle\langle\psi|$. Then show its diagonal in the measurement basis matches the
Born probabilities.

<details><summary>Solution</summary>

With probability $p_k$ the state becomes $\rho_k = \dfrac{P_k|\psi\rangle\langle\psi|P_k}{p_k}$ (a pure
state). Averaging over outcomes: $\bar\rho = \sum_k p_k\,\rho_k = \sum_k P_k|\psi\rangle\langle\psi|P_k
= \sum_k P_k\,\rho\,P_k$ (the $p_k$ cancels). For rank-1 projectors $P_k = |k\rangle\langle k|$,
$\bar\rho = \sum_k |\langle k|\psi\rangle|^2\,|k\rangle\langle k| = \sum_k p_k|k\rangle\langle k|$, whose
$(k,k)$ entry is $p_k$ and whose off-diagonal entries vanish: measurement (unconditioned) *destroys
coherences*. This map $\rho\mapsto\sum_k P_k\rho P_k$ is the prototype of a quantum channel —
[Term 4.1](../../term-4-noise-hardware/c01-open-systems/02-kraus-operators-channels.md). ∎
</details>

**E6 (hard).** Prove the perfect-distinguishability claim **in reverse**: if a projective measurement
$\{P_0, P_1\}$ distinguishes states $|\psi_0\rangle, |\psi_1\rangle$ with zero error (outcome $0\Rightarrow$
prepared $\psi_0$, outcome $1\Rightarrow\psi_1$, each with certainty), then $\langle\psi_0|\psi_1\rangle = 0$.

<details><summary>Solution</summary>

Zero error means $\langle\psi_0|P_0|\psi_0\rangle = 1$ and $\langle\psi_1|P_1|\psi_1\rangle = 1$, with
$\langle\psi_0|P_1|\psi_0\rangle = \langle\psi_1|P_0|\psi_1\rangle = 0$ (since $P_0+P_1=I$ implies the
"wrong" probabilities are $1-1=0$). From $\langle\psi_0|P_0|\psi_0\rangle = \|P_0|\psi_0\rangle\|^2 = 1$
and $\|\,|\psi_0\rangle\| = 1$ with $\|P_0\|\le 1$, equality forces $P_0|\psi_0\rangle = |\psi_0\rangle$;
likewise $P_1|\psi_1\rangle = |\psi_1\rangle$ and $P_0|\psi_1\rangle = 0$. Therefore
$\langle\psi_0|\psi_1\rangle = \langle\psi_0|P_0|\psi_1\rangle = \langle\psi_0|(P_0|\psi_1\rangle) =
\langle\psi_0|0\rangle = 0$. ∎ So *only* orthogonal states are perfectly distinguishable — non-orthogonal
states need POVMs and still cannot reach zero error ([Lesson 3](03-povm-generalized.md)).
</details>

---

## Checkpoint

1. State the Born rule and prove that the outcome probabilities sum to $1$.
2. What three algebraic conditions define a complete set of orthogonal projectors $\{P_k\}$?
3. Write the post-measurement state after outcome $k$, and explain the role of the $\sqrt{p_k}$.
4. Prove that projective measurement is repeatable.
5. How do you measure the observable $X$ on hardware that can only measure $Z$?
6. Why are orthogonal states perfectly distinguishable but $|0\rangle$ and $|+\rangle$ are not?

<details><summary>Answers</summary>

1. $p_k = \langle\psi|P_k|\psi\rangle$. $\sum_k p_k = \langle\psi|\sum_k P_k|\psi\rangle =
   \langle\psi|I|\psi\rangle = 1$ by completeness and normalization.
2. $P_k=P_k^\dagger$ (Hermitian), $P_jP_k=\delta_{jk}P_k$ (orthogonal idempotents), and $\sum_k P_k=I$
   (complete).
3. $|\psi'\rangle = P_k|\psi\rangle/\sqrt{p_k}$. The raw vector $P_k|\psi\rangle$ has norm $\sqrt{p_k}$;
   dividing renormalizes it to a unit vector so it is a valid state.
4. After collapse to $P_k|\psi\rangle/\sqrt{p_k}$, a second measurement gives outcome $j$ with
   probability $\langle\psi|P_kP_jP_k|\psi\rangle/p_k = \delta_{jk}$ using $P_jP_k=\delta_{jk}P_k$.
5. Rotate the $X$-eigenbasis onto the computational basis (apply $H$, since $H|\pm\rangle = |0/1\rangle$),
   then measure $Z$; $|\langle\pm|\psi\rangle|^2 = |\langle 0/1|H\psi\rangle|^2$.
6. A projective measurement assigns $|\psi_0\rangle\mapsto$ outcome 0 with certainty only if
   $P_0|\psi_0\rangle=|\psi_0\rangle$; orthogonality lets $P_0,P_1$ be the corresponding rank-1
   projectors with no overlap. $|0\rangle,|+\rangle$ have $\langle0|+\rangle=\tfrac1{\sqrt2}\ne0$, so no
   projective measurement (indeed no measurement) separates them with zero error.

</details>

---

## Further Reading

- **[NC]** Nielsen & Chuang, §2.2.3–2.2.5 — the measurement postulate, projective measurements, and the
  global-phase / observable discussion.
- **[Sak]** Sakurai & Napolitano, §1.4 — measurements, observables, and the collapse postulate, with the
  Stern–Gerlach intuition.
- **[Pre]** Preskill, *Ph219*, Ch. 2 — the axioms of measurement and the orthogonal-projector formalism.

---

← Prev: [Bases & Expectation Values](../c02-qubits-bloch/03-bases-and-expectation.md) · Up: [Term 1](../README.md) · Next: [Expectation & Uncertainty](02-expectation-uncertainty.md) →