Observables & the Measurement Postulate

Postulate 2 connects the abstract state vector to the laboratory. It tells us what is measurable (the eigenvalues of Hermitian operators), with what probability (the Born rule), and what happens to the state afterward (collapse). This is the postulate that makes quantum mechanics probabilistic and irreversible-on-measurement — and it is built entirely from the spectral theorem you proved in Term 0. Course 1.3 expands every piece; here we state the axiom and its core consequences.

Learning Objectives

After this lesson you will be able to:

1. State Postulate 2 using the spectral decomposition of a Hermitian observable.
2. Compute outcome probabilities via the Born rule and the expectation value $\langle A\rangle$.
3. Determine the post-measurement state (collapse) and explain repeatability.
4. Connect "observable = Hermitian operator" to the spectral theorem and real eigenvalues.
5. Simulate projective measurements in NumPy.

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Intuition

Classically, you read off a property and the system is unchanged. Quantum measurement is different in three ways. First, the possible results of measuring a quantity are constrained — they are exactly the eigenvalues of the operator representing that quantity. Second, the result is random, with probabilities set by how much the state "overlaps" each eigenspace (the Born rule). Third, the act of measuring changes the state: it collapses onto the eigenspace of the result you got. Measure again immediately and you get the same answer — measurement is repeatable, but the first one already disturbed the system.

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Theory

Postulate 2 (Observables and measurement)

Postulate 2. An observable is a Hermitian operator $A = A^\dagger$ on the state space, with spectral decomposition (from 0.1.5)

$$A = \sum_k \lambda_k P_k, \qquad P_k P_l = \delta_{kl}P_k,\quad \sum_k P_k = I,$$

where $\lambda_k$ are the (real, distinct) eigenvalues and $P_k$ the orthogonal projectors onto their eigenspaces. Measuring $A$ on state $\lvert\psi\rangle$ yields outcome $\lambda_k$ with probability

$$\boxed{\,p(\lambda_k) = \langle\psi| P_k |\psi\rangle = \lVert P_k\lvert\psi\rangle\rVert^2\,} \quad(\textbf{Born rule}),$$

and immediately after a result $\lambda_k$, the state collapses to the normalized projection

$$\lvert\psi'\rangle = \frac{P_k\lvert\psi\rangle}{\sqrt{p(\lambda_k)}} .$$

Why Hermitian? Because measured values are real numbers, and Hermitian operators are exactly those with real eigenvalues and an orthonormal eigenbasis (0.1.6). The orthonormal eigenbasis guarantees the outcomes are perfectly distinguishable and the projectors resolve the identity.

The Born rule is a probability distribution

The numbers $p(\lambda_k)$ are a valid distribution: each $p(\lambda_k) = \langle\psi|P_k|\psi\rangle = \langle\psi|P_k^\dagger P_k|\psi\rangle = \lVert P_k\lvert\psi\rangle\rVert^2 \ge 0$ (projectors are Hermitian idempotents), and they sum to one,

$$\sum_k p(\lambda_k) = \sum_k \langle\psi|P_k|\psi\rangle = \langle\psi|\Big(\sum_k P_k\Big)|\psi\rangle = \langle\psi|I|\psi\rangle = \langle\psi|\psi\rangle = 1,$$

using completeness $\sum_k P_k = I$ and normalization. This is the physical meaning of the Parseval identity and of normalization from 1.1.1. For a non-degenerate observable with eigenvectors $\lvert v_k\rangle$, $P_k = \lvert v_k\rangle\langle v_k|$ and the Born rule takes its familiar form $p(\lambda_k) = |\langle v_k|\psi\rangle|^2$.

Expectation value

The expectation value (mean outcome over many measurements on identically prepared states) is

$$\langle A\rangle_\psi := \sum_k \lambda_k\, p(\lambda_k) = \sum_k \lambda_k \langle\psi|P_k|\psi\rangle = \langle\psi|\Big(\sum_k\lambda_k P_k\Big)|\psi\rangle = \langle\psi|A|\psi\rangle .$$

So the operator does double duty: its eigenvalues are the outcomes, and the sandwich $\langle\psi|A|\psi\rangle$ is their average. (This is real because $A$ is Hermitian.) The variance and the uncertainty relation are built from this in 1.3.2.

Collapse and repeatability

After outcome $\lambda_k$, the state is $P_k\lvert\psi\rangle/\sqrt{p(\lambda_k)}$, which lies entirely in the $\lambda_k$-eigenspace. Measure $A$ again: now $P_k\lvert\psi'\rangle = \lvert\psi'\rangle$ so $p(\lambda_k) = 1$ — the same result with certainty. Measurement is repeatable, but the first measurement has irreversibly disturbed any prior superposition. This irreversibility is unique to measurement; ordinary evolution (Postulate 3) is reversible.

Compatibility (preview). Two observables can be measured simultaneously with definite values iff they commute ($[A,B]=0$), i.e. share an eigenbasis (0.1.5). Incompatible observables (like $X$ and $Z$) obey an uncertainty relation — 1.3.2.

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Worked Examples

Example 1 — Measuring $Z$ on $\lvert+\rangle$

The observable $Z = (+1)\lvert0\rangle\langle0| + (-1)\lvert1\rangle\langle1|$ has projectors $P_{+1} = \lvert0\rangle\langle0|$,$P_{-1} = \lvert1\rangle\langle1|$. On$\lvert+\rangle = \tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle)$:

$$p(+1) = \langle+|P_{+1}|+\rangle = |\langle0|+\rangle|^2 = \tfrac12, \qquad p(-1) = |\langle1|+\rangle|^2 = \tfrac12 .$$

Expectation $\langle Z\rangle = (+1)\tfrac12 + (-1)\tfrac12 = 0$. If the result is $+1$, the state collapses to $P_{+1}\lvert+\rangle/\sqrt{1/2} = \lvert0\rangle$.

Example 2 — Same state, different observable ($X$ on $\lvert+\rangle$)

Measure $X = (+1)\lvert+\rangle\langle+| + (-1)\lvert-\rangle\langle-|$ on $\lvert+\rangle$. Since $\lvert+\rangle$ is the $+1$ eigenvector, $p(+1) = |\langle+|+\rangle|^2 = 1$ and $p(-1) = 0$: a certain outcome, $\langle X\rangle = 1$, and no collapse. The same state is "random" for $Z$ but "definite" for $X$ — measurement outcomes are observable-relative, the seed of complementarity.

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Hands-on (Python)

import numpy as np

ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)
plus = (ket0 + ket1) / np.sqrt(2)

def projective_measure(psi, projectors, eigenvalues, rng=None):
    """Sample one outcome via the Born rule and return (eigenvalue, collapsed state)."""
    rng = rng or np.random.default_rng()
    probs = [np.real(psi.conj() @ (P @ psi)) for P in projectors]   # <ψ|P_k|ψ>
    k = rng.choice(len(projectors), p=probs)
    collapsed = (projectors[k] @ psi) / np.sqrt(probs[k])
    return eigenvalues[k], collapsed

# Z-measurement on |+>:
P0 = np.outer(ket0, ket0.conj())     # projector for eigenvalue +1
P1 = np.outer(ket1, ket1.conj())     # projector for eigenvalue -1
val, post = projective_measure(plus, [P0, P1], [+1, -1],
                               rng=np.random.default_rng(0))
print("outcome:", val, " collapsed to:", np.round(post, 3))   # +1 -> |0>  (rng-dependent)

# Expectation <Z> = <ψ|Z|ψ> directly:
Z = np.array([[1, 0], [0, -1]], dtype=complex)
print("〈Z〉 =", np.real(plus.conj() @ Z @ plus))               # 0.0

# Empirical Born statistics over many shots match the predicted 50/50:
from collections import Counter
rng = np.random.default_rng(1)
counts = Counter(projective_measure(plus, [P0, P1], [+1, -1], rng)[0] for _ in range(2000))
print(counts)        # ~ Counter({1: ~1000, -1: ~1000})

On Braket. A real device or simulator only ever returns samples (shots), exactly like projective_measure above — never the amplitudes. You estimate probabilities and expectation values from finite shots, with the $1/\sqrt{M}$ error you derived in 0.2.2. You'll measure $\lvert+\rangle$ on the local simulator in 1.2.3.

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Exercises

E1 (easy). For $\lvert\psi\rangle = \tfrac1{\sqrt5}(\lvert0\rangle + 2\lvert1\rangle)$, give the $Z$-measurement probabilities and $\langle Z\rangle$.

Solution

$p(+1) = |1/\sqrt5|^2 = 1/5$, $p(-1) = |2/\sqrt5|^2 = 4/5$. $\langle Z\rangle = (1)(1/5)+(-1)(4/5) = -3/5$.

E2 (easy). Show the Born probabilities sum to $1$ for any normalized state and any observable, in one line.

Solution

$\sum_k\langle\psi|P_k|\psi\rangle = \langle\psi|\sum_k P_k|\psi\rangle = \langle\psi|I|\psi\rangle = 1$ by completeness and normalization. ∎

E3 (medium). Prove $\langle A\rangle = \langle\psi|A|\psi\rangle$ is real for any Hermitian $A$ (without invoking the eigenvalue formula).

Solution

$\overline{\langle\psi|A|\psi\rangle} = \langle\psi|A^\dagger|\psi\rangle = \langle\psi|A|\psi\rangle$ using $A=A^\dagger$ (and $\overline{\langle\psi|M|\psi\rangle} = \langle\psi|M^\dagger|\psi\rangle$). A number equal to its own conjugate is real. ∎

E4 (medium). A qubit is measured with the observable $A = 3\lvert+\rangle\langle+| - 1\lvert-\rangle\langle-|$ in state $\lvert0\rangle$. Find the outcome probabilities, $\langle A\rangle$, and the post-measurement states.

Solution

$\lvert0\rangle = \tfrac1{\sqrt2}(\lvert+\rangle+\lvert-\rangle)$, so $p(3) = |\langle+|0\rangle|^2 = 1/2$ and $p(-1) = |\langle-|0\rangle|^2 = 1/2$. $\langle A\rangle = 3(\tfrac12) + (-1)(\tfrac12) = 1$. Post-measurement: outcome $3 \Rightarrow \lvert+\rangle$; outcome $-1 \Rightarrow \lvert-\rangle$.

E5 (hard). Show that if $\lvert\psi\rangle$ is an eigenvector of observable $A$, the measurement of $A$ is deterministic and leaves the state unchanged. Conversely, argue that zero variance implies the state is an eigenvector.

Solution

If $A\lvert\psi\rangle = \lambda_j\lvert\psi\rangle$ then $P_k\lvert\psi\rangle = \delta_{kj}\lvert\psi\rangle$, so $p(\lambda_j) = 1$ and all others $0$; the collapse $P_j\lvert\psi\rangle/\sqrt1 = \lvert\psi\rangle$ is unchanged. Conversely the variance $(\Delta A)^2 = \langle A^2\rangle - \langle A\rangle^2 = \sum_k(\lambda_k-\langle A\rangle)^2 p(\lambda_k)$; being a sum of nonnegative terms, it is $0$ iff all probability sits on a single eigenvalue, i.e. $\lvert\psi\rangle$ lies in one eigenspace — an eigenvector. ∎ (Formalized in 1.3.2.)

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Checkpoint

1. State Postulate 2, including the role of the spectral decomposition.
2. Write the Born rule and prove the probabilities sum to $1$.
3. Give two equivalent formulas for the expectation value $\langle A\rangle$.
4. What is the post-measurement state, and what does repeatability mean?
5. Why must observables be Hermitian?

Answers

1. Observables are Hermitian $A=\sum_k\lambda_k P_k$; measuring gives outcome $\lambda_k$ w.p. $\langle\psi|P_k|\psi\rangle$ and collapses the state to $P_k\lvert\psi\rangle/\sqrt{p_k}$.
2. $p(\lambda_k)=\langle\psi|P_k|\psi\rangle$; $\sum_k p_k = \langle\psi|\sum_k P_k|\psi\rangle = \langle\psi|\psi\rangle = 1$.
3. $\langle A\rangle = \sum_k\lambda_k p(\lambda_k) = \langle\psi|A|\psi\rangle$.
4. $P_k\lvert\psi\rangle/\sqrt{p_k}$, lying in the $\lambda_k$-eigenspace; an immediate re-measurement returns $\lambda_k$ with certainty.
5. Measured values are real and outcomes must be perfectly distinguishable; Hermitian operators have real eigenvalues and an orthonormal eigenbasis.

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Further Reading

• [NC] Nielsen & Chuang, §2.2.3 (Postulate 3, measurement) and §2.2.5 (projective measurements).
• [Sak] Sakurai & Napolitano, §1.4 — measurements, observables, expectation values.
• [Pre] Preskill, Ph219, Ch. 2 — the measurement axiom.

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