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Measurement and Uncertainty

Measurement and uncertainty describe how the abstract quantum state connects to experimental outcomes, and why certain pairs of properties cannot be simultaneously sharp. This class is relevant whenever you need to predict what a detector will read, understand why a quantum computer’s output is probabilistic, or explain why a classical trajectory emerges from a quantum state. It bridges the formalism of Quantum states to the observable world.

The Born rule and projective measurement

Definition. If a system in state ∣ψ⟩|\psi\rangle is measured with a projective measurement {P^i}\{\hat{P}_i\} (orthogonal projectors summing to I^\hat{I}), the outcome ii occurs with probability pi=⟨ψ∣P^i∣ψ⟩p_i = \langle\psi|\hat{P}_i|\psi\rangle, and the post-measurement state is ∣ψi⟩=P^i∣ψ⟩/pi|\psi_i\rangle = \hat{P}_i|\psi\rangle / \sqrt{p_i}.

Details

From state to outcome

The quantum state ∣ψ⟩|\psi\rangle (see Quantum states) encodes all information, but a measurement extracts only one number (or a finite set). The Born rule is the bridge: it assigns probabilities to outcomes based on the overlap between the state and the measurement basis.

Step 1: Projective (von Neumann) measurement.

A projective measurement is specified by a set of orthogonal projectors P^i=∣i⟩⟨i∣\hat{P}_i = |i\rangle\langle i| that satisfy P^iP^j=δijP^i\hat{P}_i\hat{P}_j = \delta_{ij}\hat{P}_i and ∑iP^i=I^\sum_i \hat{P}_i = \hat{I}. The measurement asks: “which basis state is the system in?” The answer is probabilistic, and the state collapses to the measured outcome.

Step 2: Why collapse?

If the system is in ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle and we measure in the {∣0⟩,∣1⟩}\{|0\rangle, |1\rangle\} basis, we get 0 with probability ∣α∣2|\alpha|^2 and 1 with probability ∣β∣2|\beta|^2. After the measurement, the state is the eigenstate corresponding to the outcome. This is not a unitary evolution — it is an additional postulate. The physical mechanism behind collapse is decoherence (below).

Example: Stern–Gerlach experiment.

A spin-½ particle in state ∣ψ⟩=13∣↑z⟩+23∣↓z⟩|\psi\rangle = \frac{1}{\sqrt{3}}|\uparrow_z\rangle + \sqrt{\frac{2}{3}}|\downarrow_z\rangle enters a Stern–Gerlach apparatus measuring SzS_z. The beam splits into two: the upper beam (outcome +ℏ/2+\hbar/2) carries probability 1/31/3, the lower beam (−ℏ/2-\hbar/2) carries probability 2/32/3. If the upper beam is selected, the particle’s state is now ∣↑z⟩|\uparrow_z\rangle — the superposition is destroyed.

Example problem.

Problem

A qubit is in state ∣ψ⟩=12∣0⟩+32∣1⟩|\psi\rangle = \frac{1}{2}|0\rangle + \frac{\sqrt{3}}{2}|1\rangle. A projective measurement in the computational basis is performed. What are the outcomes and probabilities? If the outcome is 0, what is the state afterward?

Solution

Outcomes: 0 or 1.

P(0)=∣1/2∣2=1/4P(0) = |1/2|^2 = 1/4, P(1)=∣3/2∣2=3/4P(1) = |\sqrt{3}/2|^2 = 3/4.

If outcome is 0: post-measurement state is ∣0⟩|0\rangle.

Generalized measurement (POVM)

Definition. A positive-operator-valued measure (POVM) is a set {E^k}\{\hat{E}_k\} of positive semi-definite operators with ∑kE^k=I^\sum_k \hat{E}_k = \hat{I}. The probability of outcome kk is pk=Tr(ρ^ E^k)p_k = \text{Tr}(\hat{\rho}\,\hat{E}_k). POVMs generalize projective measurements to include inefficient detectors, indirect measurements, and measurements on subsystems of entangled states.

Details

Why go beyond projective measurements?

A projective measurement requires the system to be in a known basis and the detector to be perfect. In practice:

POVMs capture all of these. Every POVM can be realized by a projective measurement on a larger system (the system plus an ancilla), so POVMs are the most general measurement possible in quantum theory.

Example: measuring a qubit with an inefficient detector.

Suppose a detector clicks with probability η\eta when the qubit is in ∣1⟩|1\rangle and never clicks for ∣0⟩|0\rangle. The POVM elements are:

These are positive and sum to I^\hat{I}. For η<1\eta < 1, this is not a projective measurement — the “no-click” outcome does not uniquely determine the state.

Example problem.

Problem

A qubit in state ∣ψ⟩=12(∣0⟩+∣1⟩)|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) is measured with a POVM {E^1=12∣0⟩⟨0∣,  E^2=12∣1⟩⟨1∣,  E^3=12∣0⟩⟨0∣+12∣1⟩⟨1∣}\{\hat{E}_1 = \frac{1}{2}|0\rangle\langle 0|,\; \hat{E}_2 = \frac{1}{2}|1\rangle\langle 1|,\; \hat{E}_3 = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1|\}. Find the three outcome probabilities.

Solution

p1=⟨ψ∣E^1∣ψ⟩=12⋅12=14p_1 = \langle\psi|\hat{E}_1|\psi\rangle = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}

p2=⟨ψ∣E^2∣ψ⟩=12⋅12=14p_2 = \langle\psi|\hat{E}_2|\psi\rangle = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}

p3=⟨ψ∣E^3∣ψ⟩=12⋅12+12⋅12=12p_3 = \langle\psi|\hat{E}_3|\psi\rangle = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{2}

Check: p1+p2+p3=1p_1 + p_2 + p_3 = 1. ✓

Heisenberg uncertainty principle

Definition. For any two observables A^,B^\hat{A}, \hat{B} with standard deviations ΔA,ΔB\Delta A, \Delta B in state ∣ψ⟩|\psi\rangle: ΔA⋅ΔB≥12∣⟨[A^,B^]⟩∣\Delta A \cdot \Delta B \geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|. For position and momentum: Δx⋅Δp≥ℏ/2\Delta x \cdot \Delta p \geq \hbar/2.

Details

From commutators to uncertainty

The uncertainty principle is not a limitation of measurement technology — it is a statement about the state. A state that is sharply localized in position is necessarily delocalized in momentum, and vice versa. The mathematical origin is the non-commutativity of the operators.

Derivation sketch.

Define A^′=A^−⟨A⟩I^\hat{A}' = \hat{A} - \langle A\rangle\hat{I} and B^′=B^−⟨B⟩I^\hat{B}' = \hat{B} - \langle B\rangle\hat{I}. By the Cauchy–Schwarz inequality:

(ΔA)2(ΔB)2=⟨ψ∣A^′2∣ψ⟩⟨ψ∣B^′2∣ψ⟩≥∣⟨ψ∣A^′B^′∣ψ⟩∣2(\Delta A)^2(\Delta B)^2 = \langle\psi|\hat{A}'^2|\psi\rangle\langle\psi|\hat{B}'^2|\psi\rangle \geq |\langle\psi|\hat{A}'\hat{B}'|\psi\rangle|^2

Decompose A^′B^′=12{A^′,B^′}+12[A^′,B^]\hat{A}'\hat{B}' = \frac{1}{2}\{\hat{A}',\hat{B}'\} + \frac{1}{2}[\hat{A}',\hat{B}]. The anticommutator part is real, the commutator part is imaginary. Taking the modulus squared and dropping the (positive) anticommutator term gives:

(ΔA)2(ΔB)2≥14∣⟨[A^,B^]⟩∣2(\Delta A)^2(\Delta B)^2 \geq \frac{1}{4}|\langle[\hat{A},\hat{B}]\rangle|^2

For [x^,p^]=iℏ[\hat{x},\hat{p}] = i\hbar: Δx⋅Δp≥ℏ/2\Delta x \cdot \Delta p \geq \hbar/2.

Physical interpretation.

A wave packet localized to width Δx\Delta x in position space has a momentum-space spread Δp∼ℏ/Δx\Delta p \sim \hbar/\Delta x (Fourier duality). The more you squeeze the position wavefunction, the broader its Fourier transform. This is a property of waves, not of quantum mechanics per se — but in quantum mechanics, the wavefunction is the state, so the trade-off is fundamental.

Connection to the correspondence principle.

When the action SS of a system is much larger than ℏ\hbar (e.g., a baseball, a planet), the relative uncertainty Δx⋅Δp/S≪1\Delta x \cdot \Delta p / S \ll 1, and the uncertainty principle becomes negligible. The system follows an approximately classical trajectory. This is the bridge to Mechanics: classical mechanics is the large-action limit of quantum mechanics.

Example problem.

Problem

An electron is confined to a region of size Δx=10−10\Delta x = 10^{-10} m (atomic scale). Estimate the minimum momentum uncertainty and the corresponding kinetic energy. Compare to the rest energy mc2=511mc^2 = 511 keV.

Solution

Δp≥ℏ/(2Δx)=1.055×10−34/(2×10−10)=5.3×10−25\Delta p \geq \hbar/(2\Delta x) = 1.055\times10^{-34}/(2\times10^{-10}) = 5.3\times10^{-25} kg·m/s.

Kinetic energy: E≈(Δp)2/(2m)=(5.3×10−25)2/(2×9.11×10−31)≈1.5×10−19E \approx (\Delta p)^2/(2m) = (5.3\times10^{-25})^2/(2\times9.11\times10^{-31}) \approx 1.5\times10^{-19} J ≈1\approx 1 eV.

This is much less than 511 keV, so the non-relativistic treatment is valid. The electron’s kinetic energy in an atom is of order eV, consistent with atomic binding energies.

Decoherence

Definition. Decoherence is the loss of coherence (off-diagonal terms in the density matrix) of a system due to entanglement with its environment. It explains why macroscopic superpositions are not observed: the environment continuously “measures” the system, selecting a preferred basis (the pointer basis) and suppressing interference between branches.

Details

From unitary evolution to apparent collapse

The full system + environment evolves unitarily (see Quantum dynamics). If the system starts in a superposition α∣0⟩+β∣1⟩\alpha|0\rangle + \beta|1\rangle and the environment starts in ∣E0⟩|E_0\rangle, the joint evolution gives:

(α∣0⟩+β∣1⟩)∣E0⟩⟶α∣0⟩∣E0(0)⟩+β∣1⟩∣E0(1)⟩(\alpha|0\rangle + \beta|1\rangle)|E_0\rangle \longrightarrow \alpha|0\rangle|E_0^{(0)}\rangle + \beta|1\rangle|E_0^{(1)}\rangle

If the environment states become orthogonal (⟨E0(0)∣E0(1)⟩≈0\langle E_0^{(0)}|E_0^{(1)}\rangle \approx 0), the reduced density matrix of the system (tracing out the environment) becomes:

ρ^S=∣α∣2∣0⟩⟨0∣+∣β∣2∣1⟩⟨1∣\hat{\rho}_S = |\alpha|^2|0\rangle\langle 0| + |\beta|^2|1\rangle\langle 1|

The off-diagonal terms αβ∗∣0⟩⟨1∣\alpha\beta^*|0\rangle\langle 1| have vanished. The system behaves as if it is in a classical mixture, even though the global state is still a pure superposition.

Timescale.

For a macroscopic object (e.g., a dust grain of mass 10−1210^{-12} kg in air at room temperature), the decoherence time in the position basis is ∼10−31\sim 10^{-31} s — effectively instantaneous. For a superconducting qubit isolated from its environment, coherence times can reach milliseconds. The difference is the coupling strength to the environment.

Decoherence is not the same as collapse.

Decoherence explains why interference is unobservable in practice, but it does not select a single outcome. The global state still contains all branches. The measurement problem (why we see one outcome and not a superposition) requires an additional interpretive step (Copenhagen, many-worlds, objective collapse, etc.). Decoherence is a many-body phenomenon: it requires the system to be coupled to many environmental degrees of freedom (see Quantum many-body).

Example problem.

Problem

A qubit in state ∣ψ⟩=(∣0⟩+∣1⟩)/2|\psi\rangle = (|0\rangle + |1\rangle)/\sqrt{2} is coupled to an environment such that after time tt, the joint state is (∣0⟩∣e0(t)⟩+∣1⟩∣e1(t)⟩)/2( |0\rangle|e_0(t)\rangle + |1\rangle|e_1(t)\rangle )/\sqrt{2} where ⟨e0(t)∣e1(t)⟩=e−t/τ\langle e_0(t)|e_1(t)\rangle = e^{-t/\tau}. Find the off-diagonal element of the reduced density matrix as a function of tt.

Solution

The reduced density matrix is ρ^S=TrE[∣Ψ⟩⟨Ψ∣]\hat{\rho}_S = \text{Tr}_E[|\Psi\rangle\langle\Psi|]:

ρ^S=12(1⟨e1∣e0⟩⟨e0∣e1⟩1)=12(1e−t/τe−t/τ1)\hat{\rho}_S = \frac{1}{2}\begin{pmatrix} 1 & \langle e_1|e_0\rangle \\ \langle e_0|e_1\rangle & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & e^{-t/\tau} \\ e^{-t/\tau} & 1 \end{pmatrix}

The off-diagonal element decays as e−t/τe^{-t/\tau}. At t=τt = \tau, coherence is reduced to 1/e≈371/e \approx 37%; at t=10τt = 10\tau, it is ∼5×10−5\sim 5\times10^{-5} — effectively a classical mixture.

Quiz

Test your understanding

Problem. A particle is in a Gaussian wave packet ψ(x)=(2πσ2)−1/4exp⁡(−x2/4σ2)\psi(x) = (2\pi\sigma^2)^{-1/4}\exp(-x^2/4\sigma^2).

  1. Compute Δx\Delta x.
  2. Compute Δp\Delta p by Fourier transforming to momentum space.
  3. Verify the uncertainty relation.
Solution

  1. (Δx)2=⟨x2⟩−⟨x⟩2=2σ2−0=2σ2(\Delta x)^2 = \langle x^2\rangle - \langle x\rangle^2 = 2\sigma^2 - 0 = 2\sigma^2, so Δx=2 σ\Delta x = \sqrt{2}\,\sigma.

  2. The Fourier transform of a Gaussian is a Gaussian: ϕ(p)∝exp⁡(−p2σ2/ℏ2)\phi(p) \propto \exp(-p^2\sigma^2/\hbar^2). This gives (Δp)2=ℏ2/(4σ2)(\Delta p)^2 = \hbar^2/(4\sigma^2), so Δp=ℏ/(2σ)\Delta p = \hbar/(2\sigma).

  3. Δx⋅Δp=2 σ⋅ℏ/(2σ)=ℏ/2≈0.707 ℏ>ℏ/2\Delta x \cdot \Delta p = \sqrt{2}\,\sigma \cdot \hbar/(2\sigma) = \hbar/\sqrt{2} \approx 0.707\,\hbar > \hbar/2. ✓

The Gaussian is a minimum-uncertainty state only up to the numerical factor; the true minimum is ℏ/2\hbar/2, achieved by a slightly different (complex) Gaussian. The real Gaussian above gives ℏ/2\hbar/\sqrt{2}.

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