QuantumCore

Lesson 11 · 60 min · core

Postulate 3: Projective Measurement & the Born Rule

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Measurement is the only way information leaves a quantum register. This lesson states the projective-measurement postulate through the spectral decomposition A=∑mmPmA=\sum_m mP_m of an observable, derives the Born rule P(m)=⟨ψ∣Pm∣ψ⟩=∥Pmψ∥2P(m)=\langle\psi\vert P_m\vert\psi\rangle=\lVert P_m\psi\rVert^2, the collapse rule, normalisation, the expectation value ⟨A⟩=⟨ψ∣A∣ψ⟩\langle A\rangle=\langle\psi\vert A\vert\psi\rangle and variance, repeatability, measurement in a rotated basis as "rotate, then measure ZZ", the Bloch-vector expectations ⟨X⟩,⟨Y⟩,⟨Z⟩\langle X\rangle,\langle Y\rangle,\langle Z\rangle, and proves the Robertson uncertainty relation from the Cauchy–Schwarz inequality.

Prerequisites: Postulate 1: State Space & Global Phase, Hermitian & Unitary Operators

Objectives and key results

  • Compute outcome probabilities P(m)=⟨ψ∣Pm∣ψ⟩P(m)=\langle\psi\vert P_m\vert\psi\rangle and post-measurement states Pm∣ψ⟩/P(m)P_m\lvert\psi\rangle/\sqrt{P(m)} for any observable given by its spectral decomposition.

  • Compute expectation values ⟨A⟩=⟨ψ∣A∣ψ⟩\langle A\rangle=\langle\psi\vert A\vert\psi\rangle and variances (ΔA)2=⟨A2⟩−⟨A⟩2(\Delta A)^2=\langle A^2\rangle-\langle A\rangle^2 of observables.

  • Measure in a rotated orthonormal basis {U∣i⟩}\lbrace U\lvert i\rangle\rbrace by applying U†U^\dagger and then measuring in the computational basis.

  • Derive ⟨X⟩=sin⁡θcos⁡ϕ\langle X\rangle=\sin\theta\cos\phi, ⟨Y⟩=sin⁡θsin⁡ϕ\langle Y\rangle=\sin\theta\sin\phi, ⟨Z⟩=cos⁡θ\langle Z\rangle=\cos\theta and prove ΔA ΔB≥12∣⟨[A,B]⟩∣\Delta A\,\Delta B\ge\tfrac12\lvert\langle[A,B]\rangle\rvert.

theorem · Born rule
P(m)=⟨ψ∣Pm∣ψ⟩=∥Pm∣ψ⟩∥2P(m)=\langle\psi\vert P_m\vert\psi\rangle=\lVert P_m\lvert\psi\rangle\rVert^2
formula · Post-measurement state (collapse)
∣ψ⟩ → m  Pm∣ψ⟩P(m)\lvert\psi\rangle\ \xrightarrow{\ m\ }\ \frac{P_m\lvert\psi\rangle}{\sqrt{P(m)}}
identity · Expectation value and variance
⟨A⟩=∑mm P(m)=⟨ψ∣A∣ψ⟩,(ΔA)2=⟨A2⟩−⟨A⟩2\langle A\rangle=\sum_m m\,P(m)=\langle\psi\vert A\vert\psi\rangle,\qquad (\Delta A)^2=\langle A^2\rangle-\langle A\rangle^2
formula · Bloch-vector expectations
(⟨X⟩,⟨Y⟩,⟨Z⟩)=(sin⁡θcos⁡ϕ, sin⁡θsin⁡ϕ, cos⁡θ)\big(\langle X\rangle,\langle Y\rangle,\langle Z\rangle\big)=(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)
theorem · Robertson uncertainty relation
ΔA ΔB ≥ 12∣⟨ψ∣[A,B]∣ψ⟩∣\Delta A\,\Delta B\ \ge\ \tfrac12\big\lvert\langle\psi\vert[A,B]\vert\psi\rangle\big\rvert

Observables, projectors and the Born rule#

Postulates 1 and 2 tell us what a closed quantum system is (a unit vector ∣ψ⟩\lvert\psi\rangle in a Hilbert space H\mathcal H) and how it changes (by a unitary UU). Neither says anything about what an experimenter sees. The third postulate closes the loop: it assigns probabilities to the outcomes of a measurement and says what the state becomes afterwards. Everything a quantum computer ever outputs — a bit string, a histogram, an estimated energy — is produced by this postulate, so every later topic (Grover, phase estimation, variational algorithms, kernels) ends with the formulas derived here.

Physical quantities are represented by observables: Hermitian operators A=A†A=A^\dagger. By the spectral theorem (Module 1.2) every Hermitian operator on a finite-dimensional space has real eigenvalues and can be written as a sum over its distinct eigenvalues mm of the orthogonal projectors PmP_m onto the corresponding eigenspaces. The eigenvalues are the possible measurement results; the projectors decide how likely each result is.

Definition

Projective measurement of an observable

A projective measurement is described by an observable with spectral decomposition A=∑mm PmA=\sum_m m\,P_m, where the mm are the distinct real eigenvalues and the PmP_m are orthogonal projectors satisfying Pm†=PmP_m^\dagger=P_m, PmPm′=δmm′PmP_mP_{m'}=\delta_{mm'}P_m and ∑mPm=I\sum_mP_m=I. Postulate 3: measuring AA on the state ∣ψ⟩\lvert\psi\rangle yields the outcome mm with probability P(m)=⟨ψ∣Pm∣ψ⟩P(m)=\langle\psi\vert P_m\vert\psi\rangle, and immediately afterwards the system is in the state Pm∣ψ⟩/P(m)P_m\lvert\psi\rangle/\sqrt{P(m)}. A computational-basis measurement of one qubit is the case A=Z=(+1)∣0⟩⟨0∣+(−1)∣1⟩⟨1∣A=Z=(+1)\lvert0\rangle\langle0\rvert+(-1)\lvert1\rangle\langle1\rvert; one conventionally reports the outcome +1+1 as bit 00 and −1-1 as bit 11.

Derivation

The Born rule in three equivalent forms

  1. Step 1:
    P(m)=⟨ψ∣Pm∣ψ⟩P(m)=\langle\psi\vert P_m\vert\psi\rangle
    Statement of the postulate.
  2. Step 2:
    =⟨ψ∣Pm†Pm∣ψ⟩=\langle\psi\vert P_m^\dagger P_m\vert\psi\rangle
    A projector is Hermitian and idempotent, so Pm=Pm2=Pm†PmP_m=P_m^2=P_m^\dagger P_m.
  3. Step 3:
    =(Pm∣ψ⟩)†(Pm∣ψ⟩)=∥Pm∣ψ⟩∥2 ≥0=\big(P_m\lvert\psi\rangle\big)^\dagger\big(P_m\lvert\psi\rangle\big)=\big\lVert P_m\lvert\psi\rangle\big\rVert^2\ \ge0
    The bra of Pm∣ψ⟩P_m\lvert\psi\rangle is ⟨ψ∣Pm†\langle\psi\vert P_m^\dagger; a squared norm is non-negative, so P(m)P(m) is a legitimate probability.
  4. Step 4:
    Pm=∑k=1gm∣m,k⟩⟨m,k∣ ⇒ P(m)=∑k=1gm∣⟨m,k∣ψ⟩∣2P_m=\sum_{k=1}^{g_m}\lvert m,k\rangle\langle m,k\rvert\ \Rightarrow\ P(m)=\sum_{k=1}^{g_m}\big\lvert\langle m,k\vert\psi\rangle\big\rvert^2
    Expand the projector in an orthonormal basis {∣m,k⟩}k=1gm\lbrace\lvert m,k\rangle\rbrace_{k=1}^{g_m} of the eigenspace (degeneracy gmg_m), and use ⟨ψ∣m,k⟩⟨m,k∣ψ⟩=∣⟨m,k∣ψ⟩∣2\langle\psi\vert m,k\rangle\langle m,k\vert\psi\rangle=\lvert\langle m,k\vert\psi\rangle\rvert^2.
  5. Step 5:
    P(i)=∣⟨i∣ψ⟩∣2=∣ψi∣2P(i)=\big\lvert\langle i\vert\psi\rangle\big\rvert^2=\lvert\psi_i\rvert^2
    Non-degenerate special case, e.g. a computational-basis measurement with Pi=∣i⟩⟨i∣P_i=\lvert i\rangle\langle i\rvert: the probability is the squared modulus of the amplitude.

Result

P(m)=⟨ψ∣Pm∣ψ⟩=∥Pmψ∥2=∑k∣⟨m,k∣ψ⟩∣2\boxed{P(m)=\langle\psi\vert P_m\vert\psi\rangle=\lVert P_m\psi\rVert^2=\sum_k\lvert\langle m,k\vert\psi\rangle\rvert^2}
Proposition

Born probabilities are normalised

For every unit vector ∣ψ⟩\lvert\psi\rangle and every projective measurement {Pm}\lbrace P_m\rbrace, ∑mP(m)=1\sum_mP(m)=1. The collapsed state Pm∣ψ⟩/P(m)P_m\lvert\psi\rangle/\sqrt{P(m)} is again a unit vector whenever P(m)>0P(m)>0.

Proof (4 steps)
  1. Step 1:
    ∑mP(m)=∑m⟨ψ∣Pm∣ψ⟩\sum_mP(m)=\sum_m\langle\psi\vert P_m\vert\psi\rangle
    Born rule for each outcome.
  2. Step 2:
    =⟨ψ∣∑mPm∣ψ⟩=\Big\langle\psi\Big\vert\sum_mP_m\Big\vert\psi\Big\rangle
    Linearity of the inner product in its second argument.
  3. Step 3:
    =⟨ψ∣I∣ψ⟩=⟨ψ∣ψ⟩=1=\langle\psi\vert I\vert\psi\rangle=\langle\psi\vert\psi\rangle=1
    Completeness ∑mPm=I\sum_mP_m=I (the eigenspaces of a Hermitian operator span H\mathcal H), and ∣ψ⟩\lvert\psi\rangle is normalised.
  4. Step 4:
    ∥Pm∣ψ⟩P(m)∥2=∥Pm∣ψ⟩∥2P(m)=P(m)P(m)=1\Big\lVert\frac{P_m\lvert\psi\rangle}{\sqrt{P(m)}}\Big\rVert^2=\frac{\lVert P_m\lvert\psi\rangle\rVert^2}{P(m)}=\frac{P(m)}{P(m)}=1
    Norm form of the Born rule: the division by P(m)\sqrt{P(m)} is exactly the renormalisation needed after projecting.

∎

Worked example. RY(θ)=(cos⁡θ2−sin⁡θ2sin⁡θ2cos⁡θ2)R_Y(\theta)=\begin{pmatrix}\cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2\end{pmatrix} applied to ∣0⟩\lvert0\rangle with θ=π/3\theta=\pi/3 gives ∣ψ⟩=cos⁡π6∣0⟩+sin⁡π6∣1⟩=32∣0⟩+12∣1⟩\lvert\psi\rangle=\cos\tfrac\pi6\lvert0\rangle+\sin\tfrac\pi6\lvert1\rangle=\tfrac{\sqrt3}2\lvert0\rangle+\tfrac12\lvert1\rangle. A ZZ-measurement returns bit 00 with probability P(0)=∣32∣2=34P(0)=\lvert\tfrac{\sqrt3}2\rvert^2=\tfrac34 and bit 11 with P(1)=∣12∣2=14P(1)=\lvert\tfrac12\rvert^2=\tfrac14; the sum is 11 as the proposition guarantees. If the outcome is 11, the post-measurement state is ∣1⟩⟨1∣ψ⟩/1/4=12∣1⟩/12=∣1⟩\lvert1\rangle\langle1\rvert\psi\rangle/\sqrt{1/4}=\tfrac12\lvert1\rangle/\tfrac12=\lvert1\rangle: every trace of the original amplitude 32\tfrac{\sqrt3}2 is gone. This is why measurement is irreversible and why algorithms delay it until the end.

Circuit · 1 qubit

Quantum circuit on 1 qubitRY(π/3) on q0q0RY(π/3)

RY(π/3)R_Y(\pi/3) on ∣0⟩\lvert0\rangle: the statevector is (32,12)≈(0.8660,0.5)(\tfrac{\sqrt3}2,\tfrac12)\approx(0.8660,0.5) and the Born-rule histogram shows P(0)=34P(0)=\tfrac34, P(1)=14P(1)=\tfrac14.

∣ψ⟩=32∣0⟩+12∣1⟩\lvert\psi\rangle = \tfrac{\sqrt{3}}{2}\lvert 0\rangle + \tfrac{1}{2}\lvert 1\rangle
  • |0⟩0.75
  • |1⟩0.25

Quick check

The state 32∣0⟩+12∣1⟩\tfrac{\sqrt3}2\lvert0\rangle+\tfrac12\lvert1\rangle is measured in the computational basis and the outcome is 00. What is the state immediately afterwards?

Select one answer

Expectation values, variance and repeatability#

A single shot returns one eigenvalue at random. Repeating the preparation and measurement NN times gives a sample whose mean converges to the expectation value of the observable. Variational algorithms (Stage 4) optimise exactly such expectation values, so we need a formula that does not require knowing the eigendecomposition of AA.

Derivation

E[A]=⟨ψ∣A∣ψ⟩\mathbb E[A]=\langle\psi\vert A\vert\psi\rangle and Var A=⟨A2⟩−⟨A⟩2\mathrm{Var}\,A=\langle A^2\rangle-\langle A\rangle^2

  1. Step 1:
    E[A]=∑mm P(m)=∑mm ⟨ψ∣Pm∣ψ⟩\mathbb E[A]=\sum_m m\,P(m)=\sum_m m\,\langle\psi\vert P_m\vert\psi\rangle
    Definition of the mean of a discrete random variable taking value mm with probability P(m)P(m), then the Born rule.
  2. Step 2:
    =⟨ψ∣∑mmPm∣ψ⟩=⟨ψ∣A∣ψ⟩≡⟨A⟩=\Big\langle\psi\Big\vert\sum_m mP_m\Big\vert\psi\Big\rangle=\langle\psi\vert A\vert\psi\rangle\equiv\langle A\rangle
    Linearity, then the spectral decomposition A=∑mmPmA=\sum_m mP_m.
  3. Step 3:
    A2=(∑mmPm)(∑m′m′Pm′)=∑m,m′mm′ PmPm′=∑mm2PmA^2=\Big(\sum_m mP_m\Big)\Big(\sum_{m'}m'P_{m'}\Big)=\sum_{m,m'}mm'\,P_mP_{m'}=\sum_m m^2P_m
    Expand the product and use orthogonality PmPm′=δmm′PmP_mP_{m'}=\delta_{mm'}P_m.
  4. Step 4:
    E[A2]=∑mm2P(m)=⟨ψ∣A2∣ψ⟩\mathbb E[A^2]=\sum_m m^2P(m)=\langle\psi\vert A^2\vert\psi\rangle
    Same argument as the first two lines applied to the decomposition of A2A^2.
  5. Step 5:
    Var A=∑m(m−⟨A⟩)2P(m)=∑mm2P(m)−2⟨A⟩∑mmP(m)+⟨A⟩2∑mP(m)\mathrm{Var}\,A=\sum_m\big(m-\langle A\rangle\big)^2P(m)=\sum_m m^2P(m)-2\langle A\rangle\sum_m mP(m)+\langle A\rangle^2\sum_mP(m)
    Definition of variance, expanding the square (m−μ)2=m2−2μm+μ2(m-\mu)^2=m^2-2\mu m+\mu^2 with μ=⟨A⟩\mu=\langle A\rangle.
  6. Step 6:
    =⟨A2⟩−2⟨A⟩2+⟨A⟩2=⟨A2⟩−⟨A⟩2=\langle A^2\rangle-2\langle A\rangle^2+\langle A\rangle^2=\langle A^2\rangle-\langle A\rangle^2
    Substitute the three sums: ⟨A2⟩\langle A^2\rangle, ⟨A⟩\langle A\rangle and 11 (normalisation).
  7. Step 7:
    (ΔA)2=∥(A−⟨A⟩I)∣ψ⟩∥2(\Delta A)^2=\big\lVert(A-\langle A\rangle I)\lvert\psi\rangle\big\rVert^2
    Equivalently, ⟨ψ∣(A−μ)2∣ψ⟩\langle\psi\vert(A-\mu)^2\vert\psi\rangle with A−μIA-\mu I Hermitian: the standard deviation is the length of the "deviation vector". It vanishes iff A∣ψ⟩=μ∣ψ⟩A\lvert\psi\rangle=\mu\lvert\psi\rangle, i.e. iff ∣ψ⟩\lvert\psi\rangle is an eigenstate.

Result

⟨A⟩=⟨ψ∣A∣ψ⟩,(ΔA)2=⟨A2⟩−⟨A⟩2\boxed{\langle A\rangle=\langle\psi\vert A\vert\psi\rangle,\qquad(\Delta A)^2=\langle A^2\rangle-\langle A\rangle^2}

Example. For ∣ψ⟩=32∣0⟩+12∣1⟩\lvert\psi\rangle=\tfrac{\sqrt3}2\lvert0\rangle+\tfrac12\lvert1\rangle and A=ZA=Z: ⟨Z⟩=(+1)34+(−1)14=12\langle Z\rangle=(+1)\tfrac34+(-1)\tfrac14=\tfrac12. Since Z2=IZ^2=I, ⟨Z2⟩=1\langle Z^2\rangle=1 and (ΔZ)2=1−14=34(\Delta Z)^2=1-\tfrac14=\tfrac34. With NN shots the sample mean has standard error ΔZ/N\Delta Z/\sqrt N; for N=1000N=1000 that is 0.75/1000≈0.027\sqrt{0.75/1000}\approx0.027. For a single outcome frequency the standard error is p(1−p)/N\sqrt{p(1-p)/N}, which the sampler below displays.

Proposition

Repeatability of projective measurements

If a projective measurement of {Pm}\lbrace P_m\rbrace gives outcome mm, then an immediate second measurement of the same observable gives mm again with probability 11.

Proof (4 steps)
  1. Step 1:
    ∣ψm⟩=Pm∣ψ⟩P(m)\lvert\psi_m\rangle=\frac{P_m\lvert\psi\rangle}{\sqrt{P(m)}}
    State after the first measurement (collapse rule), assuming P(m)>0P(m)>0.
  2. Step 2:
    P(m′∣m)=⟨ψm∣Pm′∣ψm⟩=⟨ψ∣PmPm′Pm∣ψ⟩P(m)P(m'\mid m)=\langle\psi_m\vert P_{m'}\vert\psi_m\rangle=\frac{\langle\psi\vert P_mP_{m'}P_m\vert\psi\rangle}{P(m)}
    Born rule on the new state; the bra of Pm∣ψ⟩P_m\lvert\psi\rangle is ⟨ψ∣Pm\langle\psi\vert P_m.
  3. Step 3:
    PmPm′Pm=δmm′Pm3=δmm′PmP_mP_{m'}P_m=\delta_{mm'}P_m^3=\delta_{mm'}P_m
    Orthogonality Pm′Pm=δmm′PmP_{m'}P_m=\delta_{mm'}P_m and idempotence Pm2=PmP_m^2=P_m.
  4. Step 4:
    P(m′∣m)=δmm′⟨ψ∣Pm∣ψ⟩P(m)=δmm′P(m)P(m)=δmm′P(m'\mid m)=\delta_{mm'}\frac{\langle\psi\vert P_m\vert\psi\rangle}{P(m)}=\delta_{mm'}\frac{P(m)}{P(m)}=\delta_{mm'}
    The second outcome equals the first with certainty; the post-measurement state is an eigenstate of AA.

∎

Quick check

For ∣ψ⟩=32∣0⟩+12∣1⟩\lvert\psi\rangle=\tfrac{\sqrt3}2\lvert0\rangle+\tfrac12\lvert1\rangle, what is the variance (ΔZ)2(\Delta Z)^2?

tolerance ±0.001

Measuring in other bases and the Bloch expectations#

Hardware can only measure ZZ on each qubit. How then does one "measure XX" or estimate ⟨Y⟩\langle Y\rangle? The answer is a change of basis: rotate the state so that the desired basis is mapped onto the computational basis, then measure ZZ. The following theorem makes this precise for any orthonormal basis.

Theorem

Measuring in the basis {U∣i⟩}\lbrace U\lvert i\rangle\rbrace equals applying U†U^\dagger then measuring computationally

Let UU be unitary and ∣ui⟩=U∣i⟩\lvert u_i\rangle=U\lvert i\rangle. A projective measurement in the orthonormal basis {∣ui⟩}\lbrace\lvert u_i\rangle\rbrace on ∣ψ⟩\lvert\psi\rangle has the same outcome probabilities as a computational-basis measurement on U†∣ψ⟩U^\dagger\lvert\psi\rangle, and the post-measurement states are related by UU: ∣ui⟩=U∣i⟩\lvert u_i\rangle=U\lvert i\rangle.

Proof (5 steps)
  1. Step 1:
    ⟨ui∣uj⟩=⟨i∣U†U∣j⟩=⟨i∣j⟩=δij\langle u_i\vert u_j\rangle=\langle i\vert U^\dagger U\vert j\rangle=\langle i\vert j\rangle=\delta_{ij}
    {∣ui⟩}\lbrace\lvert u_i\rangle\rbrace is orthonormal because U†U=IU^\dagger U=I; it is a basis since there are dim⁡H\dim\mathcal H of them.
  2. Step 2:
    Qi=∣ui⟩⟨ui∣=U∣i⟩⟨i∣U†=UPiU†Q_i=\lvert u_i\rangle\langle u_i\rvert=U\lvert i\rangle\langle i\rvert U^\dagger=UP_iU^\dagger
    Projectors of the rotated measurement, with Pi=∣i⟩⟨i∣P_i=\lvert i\rangle\langle i\rvert; the bra of U∣i⟩U\lvert i\rangle is ⟨i∣U†\langle i\vert U^\dagger. Completeness: ∑iQi=U(∑iPi)U†=UU†=I\sum_iQ_i=U\big(\sum_iP_i\big)U^\dagger=UU^\dagger=I.
  3. Step 3:
    Prot(i)=⟨ψ∣Qi∣ψ⟩=⟨ψ∣UPiU†∣ψ⟩=(⟨ψ∣U)Pi(U†∣ψ⟩)P_{\text{rot}}(i)=\langle\psi\vert Q_i\vert\psi\rangle=\langle\psi\vert UP_iU^\dagger\vert\psi\rangle=\big(\langle\psi\vert U\big)P_i\big(U^\dagger\lvert\psi\rangle\big)
    Born rule for the rotated measurement; regroup the product.
  4. Step 4:
    =⟨ψ′∣Pi∣ψ′⟩=∣⟨i∣ψ′⟩∣2,∣ψ′⟩=U†∣ψ⟩=\langle\psi'\vert P_i\vert\psi'\rangle=\lvert\langle i\vert\psi'\rangle\rvert^2,\qquad\lvert\psi'\rangle=U^\dagger\lvert\psi\rangle
    ⟨ψ∣U=(U†∣ψ⟩)†\langle\psi\vert U=(U^\dagger\lvert\psi\rangle)^\dagger: this is the Born rule for a computational measurement of ∣ψ′⟩\lvert\psi'\rangle.
  5. Step 5:
    Qi∣ψ⟩P(i)=U PiU†∣ψ⟩P(i)=U Pi∣ψ′⟩P(i)=U∣i⟩ eiarg⁡⟨i∣ψ′⟩\frac{Q_i\lvert\psi\rangle}{\sqrt{P(i)}}=U\,\frac{P_iU^\dagger\lvert\psi\rangle}{\sqrt{P(i)}}=U\,\frac{P_i\lvert\psi'\rangle}{\sqrt{P(i)}}=U\lvert i\rangle\,e^{i\arg\langle i\vert\psi'\rangle}
    Post-measurement states: the rotated collapse is the computational collapse of ∣ψ′⟩\lvert\psi'\rangle mapped back by UU (up to an unobservable global phase).

∎

Recipes. The XX basis is ∣±⟩=H∣0⟩,H∣1⟩\lvert\pm\rangle=H\lvert0\rangle,H\lvert1\rangle, so U=HU=H and U†=HU^\dagger=H: apply HH, then measure. The YY basis is ∣±i⟩=12(∣0⟩±i∣1⟩)=SH∣0⟩,SH∣1⟩\lvert\pm i\rangle=\tfrac1{\sqrt2}(\lvert0\rangle\pm i\lvert1\rangle)=SH\lvert0\rangle,SH\lvert1\rangle, so U=SHU=SH and U†=HS†U^\dagger=HS^\dagger: apply S†S^\dagger, then HH, then measure. In matrix form, the operator applied before the ZZ-measurement is (SH)†=HS†(SH)^\dagger=HS^\dagger; the matrix below is its adjoint U=SHU=SH with columns ∣+i⟩,∣−i⟩\lvert+i\rangle,\lvert-i\rangle.

H=12(111−1)=(∣+⟩  ∣−⟩),H†=HH=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\big(\lvert+\rangle\ \ \lvert-\rangle\big),\qquad H^\dagger=H
Hadamard (X-basis change)2×2

Unitary verified: U†U=IU^\dagger U = I (max deviation 2.2e-16)

SH=(100i)12(111−1)=12(11i−i)SH=\begin{pmatrix}1&0\\0&i\end{pmatrix}\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\frac1{\sqrt2}\begin{pmatrix}1&1\\i&-i\end{pmatrix}
SH (Y-basis change, columns |+i⟩ and |−i⟩)2×2

Unitary verified: U†U=IU^\dagger U = I (max deviation 2.2e-16)

Derivation

Pauli expectations of ∣ψ⟩=cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\lvert\psi\rangle=\cos\tfrac\theta2\lvert0\rangle+e^{i\phi}\sin\tfrac\theta2\lvert1\rangle

  1. Step 1:
    c≡cos⁡θ2,s≡sin⁡θ2,∣ψ⟩=(ceiϕs),⟨ψ∣=(ce−iϕs)c\equiv\cos\tfrac\theta2,\quad s\equiv\sin\tfrac\theta2,\qquad\lvert\psi\rangle=\begin{pmatrix}c\\e^{i\phi}s\end{pmatrix},\quad\langle\psi\rvert=\begin{pmatrix}c&e^{-i\phi}s\end{pmatrix}
    Abbreviations; c,sc,s are real, so the bra only conjugates the phase eiϕ↦e−iϕe^{i\phi}\mapsto e^{-i\phi}.
  2. Step 2:
    X∣ψ⟩=(0110)(ceiϕs)=(eiϕsc)X\lvert\psi\rangle=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}c\\e^{i\phi}s\end{pmatrix}=\begin{pmatrix}e^{i\phi}s\\c\end{pmatrix}
    XX swaps the two amplitudes.
  3. Step 3:
    ⟨X⟩=c eiϕs+e−iϕs c=cs(eiϕ+e−iϕ)=2cscos⁡ϕ\langle X\rangle=c\,e^{i\phi}s+e^{-i\phi}s\,c=cs\big(e^{i\phi}+e^{-i\phi}\big)=2cs\cos\phi
    Row times column; Euler's formula eiϕ+e−iϕ=2cos⁡ϕe^{i\phi}+e^{-i\phi}=2\cos\phi.
  4. Step 4:
    2cs=2sin⁡θ2cos⁡θ2=sin⁡θ ⇒ ⟨X⟩=sin⁡θcos⁡ϕ2cs=2\sin\tfrac\theta2\cos\tfrac\theta2=\sin\theta\ \Rightarrow\ \langle X\rangle=\sin\theta\cos\phi
    Double-angle identity sin⁡2α=2sin⁡αcos⁡α\sin2\alpha=2\sin\alpha\cos\alpha with α=θ/2\alpha=\theta/2.
  5. Step 5:
    Y∣ψ⟩=(0−ii0)(ceiϕs)=(−i eiϕsi c)Y\lvert\psi\rangle=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}c\\e^{i\phi}s\end{pmatrix}=\begin{pmatrix}-i\,e^{i\phi}s\\i\,c\end{pmatrix}
    Matrix–vector product with YY.
  6. Step 6:
    ⟨Y⟩=c (−i eiϕs)+e−iϕs (i c)=−i cs(eiϕ−e−iϕ)\langle Y\rangle=c\,(-i\,e^{i\phi}s)+e^{-i\phi}s\,(i\,c)=-i\,cs\big(e^{i\phi}-e^{-i\phi}\big)
    Row times column, factoring out −i cs-i\,cs.
  7. Step 7:
    =−i cs (2isin⁡ϕ)=2cssin⁡ϕ=sin⁡θsin⁡ϕ=-i\,cs\,(2i\sin\phi)=2cs\sin\phi=\sin\theta\sin\phi
    Euler: eiϕ−e−iϕ=2isin⁡ϕe^{i\phi}-e^{-i\phi}=2i\sin\phi; then −i⋅2i=2-i\cdot2i=2 and the double-angle identity again.
  8. Step 8:
    Z∣ψ⟩=(c−eiϕs),⟨Z⟩=c2−e−iϕeiϕs2=c2−s2Z\lvert\psi\rangle=\begin{pmatrix}c\\-e^{i\phi}s\end{pmatrix},\qquad\langle Z\rangle=c^2-e^{-i\phi}e^{i\phi}s^2=c^2-s^2
    ZZ flips the sign of the ∣1⟩\lvert1\rangle amplitude; e−iϕeiϕ=1e^{-i\phi}e^{i\phi}=1.
  9. Step 9:
    c2−s2=cos⁡2θ2−sin⁡2θ2=cos⁡θc^2-s^2=\cos^2\tfrac\theta2-\sin^2\tfrac\theta2=\cos\theta
    Double-angle identity cos⁡2α=cos⁡2α−sin⁡2α\cos2\alpha=\cos^2\alpha-\sin^2\alpha.
  10. Step 10:
    ⟨X⟩2+⟨Y⟩2+⟨Z⟩2=sin⁡2θ(cos⁡2ϕ+sin⁡2ϕ)+cos⁡2θ=1\langle X\rangle^2+\langle Y\rangle^2+\langle Z\rangle^2=\sin^2\theta(\cos^2\phi+\sin^2\phi)+\cos^2\theta=1
    The vector of expectations is the unit Bloch vector r⃗=(sin⁡θcos⁡ϕ,sin⁡θsin⁡ϕ,cos⁡θ)\vec r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta): a pure qubit state lies on the sphere.

Result

⟨X⟩=sin⁡θcos⁡ϕ,⟨Y⟩=sin⁡θsin⁡ϕ,⟨Z⟩=cos⁡θ\boxed{\langle X\rangle=\sin\theta\cos\phi,\qquad\langle Y\rangle=\sin\theta\sin\phi,\qquad\langle Z\rangle=\cos\theta}

Consequences. Measuring XX, YY and ZZ on many copies (with the basis-change recipes above) estimates the three coordinates of the Bloch vector — this is single-qubit state tomography. For the RY(π/3)R_Y(\pi/3) state (θ=π/3\theta=\pi/3, ϕ=0\phi=0): ⟨X⟩=sin⁡π3=32\langle X\rangle=\sin\tfrac\pi3=\tfrac{\sqrt3}2, ⟨Y⟩=0\langle Y\rangle=0, ⟨Z⟩=cos⁡π3=12\langle Z\rangle=\cos\tfrac\pi3=\tfrac12. The outcome probabilities of an XX-measurement follow from ⟨X⟩=P(+)−P(−)\langle X\rangle=P(+)-P(-) and P(+)+P(−)=1P(+)+P(-)=1: P(±)=12(1±⟨X⟩)P(\pm)=\tfrac12(1\pm\langle X\rangle), i.e. P(+)=12+34≈0.933P(+)=\tfrac12+\tfrac{\sqrt3}4\approx0.933.

Quick check

For θ=π/3\theta=\pi/3 and ϕ=0\phi=0, compute ⟨X⟩\langle X\rangle for ∣ψ⟩=cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\lvert\psi\rangle=\cos\tfrac\theta2\lvert0\rangle+e^{i\phi}\sin\tfrac\theta2\lvert1\rangle (three decimals).

tolerance ±0.001

The Robertson uncertainty relation#

Two observables that do not commute cannot, in general, both have sharp values: an eigenstate of ZZ gives completely random XX outcomes. The Robertson relation quantifies this trade-off using only the commutator [A,B]=AB−BA[A,B]=AB-BA and the Cauchy–Schwarz inequality ∣⟨f∣g⟩∣2≤⟨f∣f⟩⟨g∣g⟩\lvert\langle f\vert g\rangle\rvert^2\le\langle f\vert f\rangle\langle g\vert g\rangle from Module 1.1.

Theorem

Robertson uncertainty relation

For Hermitian A,BA,B and any unit vector ∣ψ⟩\lvert\psi\rangle, ΔA ΔB≥12∣⟨ψ∣[A,B]∣ψ⟩∣\Delta A\,\Delta B\ge\tfrac12\lvert\langle\psi\vert[A,B]\vert\psi\rangle\rvert, where ΔA=⟨A2⟩−⟨A⟩2\Delta A=\sqrt{\langle A^2\rangle-\langle A\rangle^2}.

Proof (7 steps)
  1. Step 1:
    A~=A−⟨A⟩I,B~=B−⟨B⟩I,∣f⟩=A~∣ψ⟩,∣g⟩=B~∣ψ⟩\tilde A=A-\langle A\rangle I,\quad\tilde B=B-\langle B\rangle I,\qquad\lvert f\rangle=\tilde A\lvert\psi\rangle,\quad\lvert g\rangle=\tilde B\lvert\psi\rangle
    Shifted operators are still Hermitian because ⟨A⟩,⟨B⟩\langle A\rangle,\langle B\rangle are real.
  2. Step 2:
    ⟨f∣f⟩=⟨ψ∣A~2∣ψ⟩=(ΔA)2,⟨g∣g⟩=(ΔB)2\langle f\vert f\rangle=\langle\psi\vert\tilde A^2\vert\psi\rangle=(\Delta A)^2,\qquad\langle g\vert g\rangle=(\Delta B)^2
    A~†A~=A~2\tilde A^\dagger\tilde A=\tilde A^2 and the variance identity ∥(A−⟨A⟩)ψ∥2=(ΔA)2\lVert(A-\langle A\rangle)\psi\rVert^2=(\Delta A)^2 derived above.
  3. Step 3:
    (ΔA)2(ΔB)2=⟨f∣f⟩⟨g∣g⟩ ≥ ∣⟨f∣g⟩∣2(\Delta A)^2(\Delta B)^2=\langle f\vert f\rangle\langle g\vert g\rangle\ \ge\ \lvert\langle f\vert g\rangle\rvert^2
    Cauchy–Schwarz inequality.
  4. Step 4:
    ∣z∣2=(Re⁡z)2+(Im⁡z)2 ≥ (Im⁡z)2=∣z−zˉ2i∣2,z=⟨f∣g⟩\lvert z\rvert^2=(\operatorname{Re}z)^2+(\operatorname{Im}z)^2\ \ge\ (\operatorname{Im}z)^2=\Big\lvert\frac{z-\bar z}{2i}\Big\rvert^2,\qquad z=\langle f\vert g\rangle
    Drop the non-negative real part; Im⁡z=(z−zˉ)/(2i)\operatorname{Im}z=(z-\bar z)/(2i).
  5. Step 5:
    z=⟨ψ∣A~B~∣ψ⟩,zˉ=⟨g∣f⟩=⟨ψ∣B~A~∣ψ⟩z=\langle\psi\vert\tilde A\tilde B\vert\psi\rangle,\qquad\bar z=\langle g\vert f\rangle=\langle\psi\vert\tilde B\tilde A\vert\psi\rangle
    ⟨f∣=⟨ψ∣A~†=⟨ψ∣A~\langle f\vert=\langle\psi\vert\tilde A^\dagger=\langle\psi\vert\tilde A, and conjugate symmetry ⟨f∣g⟩‾=⟨g∣f⟩\overline{\langle f\vert g\rangle}=\langle g\vert f\rangle.
  6. Step 6:
    z−zˉ=⟨ψ∣[A~,B~]∣ψ⟩=⟨ψ∣[A,B]∣ψ⟩z-\bar z=\langle\psi\vert[\tilde A,\tilde B]\vert\psi\rangle=\langle\psi\vert[A,B]\vert\psi\rangle
    Subtract; multiples of II commute with everything, so [A~,B~]=[A,B][\tilde A,\tilde B]=[A,B].
  7. Step 7:
    (ΔA)2(ΔB)2 ≥ ∣⟨[A,B]⟩∣24 ⇒ ΔA ΔB ≥ 12∣⟨[A,B]⟩∣(\Delta A)^2(\Delta B)^2\ \ge\ \frac{\lvert\langle[A,B]\rangle\rvert^2}{4}\ \Rightarrow\ \Delta A\,\Delta B\ \ge\ \tfrac12\lvert\langle[A,B]\rangle\rvert
    Chain the inequalities (∣2i∣2=4\lvert2i\rvert^2=4) and take non-negative square roots.

∎

Derivation

Example: XX and ZZ on a qubit

  1. Step 1:
    XZ=(0−110),ZX=(01−10),[X,Z]=(0−220)=−2iYXZ=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\quad ZX=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad[X,Z]=\begin{pmatrix}0&-2\\2&0\end{pmatrix}=-2iY
    Multiply the Pauli matrices; −2iY=−2i(0−ii0)=(0−220)-2iY=-2i\begin{pmatrix}0&-i\\i&0\end{pmatrix}=\begin{pmatrix}0&-2\\2&0\end{pmatrix}.
  2. Step 2:
    ΔX ΔZ ≥ 12∣−2i⟨Y⟩∣=∣⟨Y⟩∣=∣sin⁡θsin⁡ϕ∣\Delta X\,\Delta Z\ \ge\ \tfrac12\lvert-2i\langle Y\rangle\rvert=\lvert\langle Y\rangle\rvert=\lvert\sin\theta\sin\phi\rvert
    Robertson with the Bloch expectation of YY.
  3. Step 3:
    (ΔX)2=1−sin⁡2θcos⁡2ϕ,(ΔZ)2=1−cos⁡2θ(\Delta X)^2=1-\sin^2\theta\cos^2\phi,\qquad(\Delta Z)^2=1-\cos^2\theta
    X2=Z2=IX^2=Z^2=I, so (Δσ)2=1−⟨σ⟩2(\Delta\sigma)^2=1-\langle\sigma\rangle^2.
  4. Step 4:
    θ=π2, ϕ=π2:ΔX=ΔZ=1, ∣⟨Y⟩∣=1\theta=\tfrac\pi2,\ \phi=\tfrac\pi2:\quad\Delta X=\Delta Z=1,\ \lvert\langle Y\rangle\rvert=1
    For ∣+i⟩\lvert+i\rangle the bound 1≥11\ge1 is saturated.
  5. Step 5:
    θ=0:ΔZ=0, ΔX=1, ∣⟨Y⟩∣=0\theta=0:\quad\Delta Z=0,\ \Delta X=1,\ \lvert\langle Y\rangle\rvert=0
    For ∣0⟩\lvert0\rangle the right-hand side vanishes and the bound is trivially satisfied although XX is maximally uncertain: Robertson is a lower bound that can be loose.

Result

ΔX ΔZ ≥ ∣⟨Y⟩∣\boxed{\Delta X\,\Delta Z\ \ge\ \lvert\langle Y\rangle\rvert}

Check it numerically. The cell below prepares RY(π/3)∣0⟩R_Y(\pi/3)\lvert0\rangle, computes the Born probabilities, the three Pauli expectations and the two sides of ΔX ΔZ≥∣⟨Y⟩∣\Delta X\,\Delta Z\ge\lvert\langle Y\rangle\rvert.

Pythonrunnable
import numpy as np

theta = np.pi / 3
psi = np.array([np.cos(theta / 2), np.sin(theta / 2)], dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
Y = np.array([[0, -1j], [1j, 0]])
Z = np.diag([1, -1]).astype(complex)
print(np.round(np.abs(psi) ** 2, 4))
ev = [float(np.real(psi.conj() @ P @ psi)) for P in (X, Y, Z)]
print(np.round(ev, 4))
dX = np.sqrt(1 - ev[0] ** 2)
dZ = np.sqrt(1 - ev[2] ** 2)
print(round(float(dX * dZ), 4), round(abs(ev[1]), 4))

Expected output

[0.75 0.25]
[0.866 0.    0.5  ]
0.433 0.0

Quick check

For which pair of observables does the Robertson bound give no constraint for every state?

Select one answer

References

  1. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §2.2.3 (quantum measurement), §2.2.5 (projective measurements, Box 2.4 on the Heisenberg uncertainty principle).
  2. H. P. Robertson, "The Uncertainty Principle", Physical Review 34, 163–164 (1929).doi:10.1103/PhysRev.34.163
  3. M. Born, "Zur Quantenmechanik der Stoßvorgänge", Zeitschrift für Physik 37, 863–867 (1926).doi:10.1007/BF01397477
  4. J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press (2020), §1.4 (measurements, observables and the uncertainty relations).

Hands-on labs

  • Hit three expectation values

    1 step · auto-graded

    Use ⟨X⟩=sin⁡θcos⁡ϕ\langle X\rangle=\sin\theta\cos\phi, ⟨Y⟩=sin⁡θsin⁡ϕ\langle Y\rangle=\sin\theta\sin\phi, ⟨Z⟩=cos⁡θ\langle Z\rangle=\cos\theta to prepare a qubit with ⟨Z⟩=12\langle Z\rangle=\tfrac12, ⟨X⟩=32\langle X\rangle=\tfrac{\sqrt3}{2}, ⟨Y⟩=0\langle Y\rangle=0.

    Open lab