Quantum Foundations · Chapter 10
Why are physical quantities represented by self-adjoint operators?
Starting point: Chapters 8 and 9: projective measurements and Hamiltonians as generators.
Anchor example: Spin components along different axes.
Principle at work: An observable is a measurement whose outcomes are labelled by real numbers.
Foil: Hidden-value models in which every quantity has a value before it is measured.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
The formalism is now complete for finite systems. States are density operators, outcomes are effects, reversible change is unitary, and its generator, the Hamiltonian, is a Hermitian matrix that we have called energy.
But physics speaks of quantities, such as energy, spin components, or positions, and we have not yet said what it means to measure a quantity. So we ask how a measurement with numerical outcomes is packaged, and find that a single Hermitian matrix does it. Hermitian matrices need not commute, so we then ask when two quantities can be measured together. The answer is commutation, and it shows that quantum theory cannot simply be a theory of quantities with hidden definite values. Incompatibility also has a size, which we measure by a bound on the spreads of two quantities in one state.
The next chapter asks what happens to the state when a measurement is made.
We have effects and measurements, but no notion of a measured quantity. We first attach numbers to the outcomes of a projective measurement, and find that one Hermitian matrix records both the measurement and its numbers. Then we check that every Hermitian matrix arises in this way.
Suppose a measurement’s outcomes are labelled by real numbers, such as an energy reading. What is the average reading? Take a projective measurement, with projections \(P_1,\dots,P_m\) that are mutually orthogonal and sum to \(I\), and give its outcomes distinct labels \(a_1,\dots,a_m\). The expected value of the label in state \(\rho\) is
\[\langle A\rangle=\sum_ka_k\tr(\rho P_k)=\tr(\rho A),\qquad A=\sum_ka_kP_k .\]
We call such a labelled projective measurement an observable. The single Hermitian matrix \(A\) packages the measurement and its labels. Is it a new physical ingredient? No: everything it predicts was already available from the effects \(P_k\). It earns its place because sums, products, and commutators of such matrices express physical relations compactly, and because Hermitian matrices also generate reversible transformations, as the Hamiltonian does.
Why use projective measurements here? Each projection \(P_k\) is a sharp effect: it gives \(1\) on vectors in its range and \(0\) on vectors in the other ranges. So a state in one range gives its label with certainty, and the readings carry no detector noise. Projective measurements are also repeatable, as we will see when we study how measurement changes the state. Measurements with numerical labels whose effects are not projections also exist. They are called unsharp observables, and they describe imperfect or deliberately gentle measurements.
Does every Hermitian matrix arise as an observable? Yes. By the spectral theorem, a Hermitian matrix \(A\) can be written uniquely as \(A=\sum_ka_kP_k\), where the \(a_k\) are its distinct eigenvalues and \(P_k\) are the projections onto the corresponding eigenspaces. The eigenvalues are the possible outcomes, and the numbers \(\tr(\rho P_k)\) are their probabilities. (Hermitian matrices are also called self-adjoint.) A function of an observable is defined by relabelling: \(f(A)=\sum_kf(a_k)P_k\). The spread of the readings is measured by the variance \((\Delta A)^2=\langle A^2\rangle-\langle A\rangle^2\).
What are the observables of a qubit? The spin component along a unit vector \(\mathbf n\), in units of \(\hbar/2\), is \(\mathbf n\cdot\vec\sigma\), with eigenvalues \(\pm1\). In the state with Bloch vector \(\mathbf r\) its expectation is \(\tr(\rho\,\mathbf n\cdot\vec\sigma)=\mathbf n\cdot\mathbf r\). So the Bloch vector is the list of expected spin components along \(x\), \(y\), and \(z\), a list of fiducial numbers for the qubit.
Problem 10.1. In the state \(\mathbf r=(0.6,0,0.8)\), find the probabilities of \(\pm1\) for the spin component along \(x\) and its variance.
\(\langle X\rangle=0.6\), so the probabilities are \((1\pm0.6)/2\): \(0.8\) and \(0.2\). Since \(X^2=I\), \((\Delta X)^2=1-0.36=0.64\).
In classical theory any collection of quantities can be measured together, because all are functions of the configuration. Is the same true of observables? We first say what measuring two observables together would mean, and see that it can fail. Then we find the algebraic condition that decides it, which is commutation.
What would it mean to measure two observables at once? There would be a single measurement from whose outcomes both can be read off. It has outcomes \((j,k)\), and merging over \(k\) gives the first observable while merging over \(j\) gives the second. Two observables with such a measurement are called compatible, or jointly measurable. For two qubits, \(Z\otimes I\) and \(I\otimes Z\) are compatible: measuring each qubit in its basis gives both. For a single qubit, \(X\) and \(Z\) are not.
Compatibility is a property of the observables, not of the state. Can incompatible observables ever have definite values together? For the qubit’s \(X\) and \(Z\), never: a state with a definite value of \(Z\) gives each value of \(X\) with probability \(1/2\). With three or more distinguishable states, two incompatible observables can share some states in which both are definite. What they cannot share is a complete basis of such states, so there are states in which they cannot both be definite. Compatible observables share a complete basis. This is the operational content of the statement that some quantities “cannot be known simultaneously.”
How can we tell from the matrices whether two observables are compatible? For observables given by projective measurements, compatibility is equivalent to commutation: \(AB=BA\).
Suppose first that \(A\) and \(B\) commute. Then they have a common basis of eigenvectors, so their spectral projections \(P_j\) and \(Q_k\) commute, and each product \(P_jQ_k\) is a projection. The measurement onto these products reads both observables.
Now suppose a joint measurement exists, with effects \(M_{jk}\) such that \(\sum_kM_{jk}=P_j\) and \(\sum_jM_{jk}=Q_k\). Must the projections commute? Each \(M_{jk}\) satisfies \(0\le M_{jk}\le P_j\). An effect below a projection lives inside the projection’s range: \((I-P_j)M_{jk}(I-P_j)\) is both \(\ge0\) and \(\le0\), so it vanishes. Then \(M_{jk}^{1/2}(I-P_j)=0\), so \(M_{jk}(I-P_j)=0\), and \(M_{jk}=P_jM_{jk}P_j\). Hence \(M_{j'k}P_j=0\) for \(j'\neq j\), and
\[Q_kP_j=\sum_{j'}M_{j'k}P_j=M_{jk}.\]
The same argument with the roles exchanged gives \(P_jQ_k=M_{jk}\). So \(P_jQ_k=Q_kP_j\), the projections commute, and the joint effects are their products. Commuting projections give commuting observables.
What does this exclude? Consider a hidden-value model, in which every observable has a definite value in every individual run and measurement merely reveals it. Such a model assigns, in each run, values to all observables at once. So it predicts a joint probability distribution for all of them, whose marginals are the observed statistics. Can quantum statistics be the marginals of one such joint distribution? For noncommuting observables there is no joint measurement to supply it, but a hidden-value model does not need one. Whether such a distribution can exist is a sharper question, and we answer it when we study contextuality.
Problem 10.2. Show that \(X\) and \(Z\) have no common eigenvector, and compute \(XZ-ZX\).
The eigenvectors of \(Z\) are \(\ket0\) and \(\ket1\); \(X\ket0=\ket1\), so neither is an eigenvector of \(X\). \(XZ-ZX=\begin{pmatrix}0&-2\\2&0\end{pmatrix}=-2iY\).
Incompatibility is a yes-or-no property. Does it also have a size? The best-known measure is a bound on the spreads of two observables in the same state. Then we ask what that bound does and does not say about measurement itself.
How small can the spreads of two observables be in one state? The commutator sets the limit.
Theorem (Robertson, 1929). For observables \(A\), \(B\) and any state \(\rho\),
\[\Delta A\,\Delta B\ \ge\ \tfrac12\,\big|\langle[A,B]\rangle\big|.\]
Proof. For matrices \(M\), \(M'\), set \(\langle M,M'\rangle_\rho=\tr(\rho M^\dagger M')\). This is linear in \(M'\), conjugate-linear in \(M\), and \(\langle M,M\rangle_\rho\ge0\), so the Cauchy–Schwarz inequality holds for it. Let \(A_0=A-\langle A\rangle\) and \(B_0=B-\langle B\rangle\). Then \((\Delta A)^2=\langle A_0,A_0\rangle_\rho\) and \((\Delta B)^2=\langle B_0,B_0\rangle_\rho\), so
\[\Delta A\,\Delta B\ge\big|\langle A_0,B_0\rangle_\rho\big|\ge\big|\mathrm{Im}\,\langle A_0,B_0\rangle_\rho\big|.\]
Since \(\langle B_0,A_0\rangle_\rho\) is the complex conjugate of \(\langle A_0,B_0\rangle_\rho\), their difference is \(2i\,\mathrm{Im}\,\langle A_0,B_0\rangle_\rho\). It also equals \(\tr(\rho[A_0,B_0])=\langle[A,B]\rangle\). So the imaginary part is \(\tfrac1{2i}\langle[A,B]\rangle\). \(\square\)
What does it give for spin? Here \([X,Y]=2iZ\), so \(\Delta X\,\Delta Y\ge|\langle Z\rangle|\). In \(\ket0\), \(\Delta X=\Delta Y=1\) and \(\langle Z\rangle=1\), so the bound is reached. In \(\ket+\), \(\Delta X=0\) and \(\langle Z\rangle=0\), so the bound is zero and gives no constraint.
Can we get a bound that does not depend on the state? We need a measure of spread other than the variance. The Shannon entropy of a probability distribution \(p\) is \(H(p)=-\sum_kp_k\log_2p_k\), measured in bits. It is \(0\) for a certain outcome and \(1\) for two equally likely outcomes, since \(-2\cdot\tfrac12\log_2\tfrac12=1\). Hans Maassen and Jos Uffink proved in 1988 that, for \(X\) and \(Z\) on a qubit, the entropies of the two outcome distributions sum to at least one bit in every state.
Does the Robertson relation say that measuring \(A\) disturbs \(B\)? No. It concerns preparations. It compares the spread of \(A\) in one set of runs with the spread of \(B\) in another set of runs on identically prepared systems. It says that no preparation makes both sharp. It does not, by itself, say that measuring \(A\) disturbs \(B\), or that a joint approximate measurement must be inaccurate.
Those are separate questions about measurement uncertainty. One asks how accurately a single device can approximate two incompatible observables. The other asks how much measuring one disturbs the other. They have their own relations, developed by Masanao Ozawa and by Paul Busch, Pekka Lahti, and Reinhard Werner, among others. Their qualitative core, that any measurement that gains information about a state must in general disturb it, needs a rule for how the state changes after a measurement.
Problem 10.3. Check the entropic relation for the states \(\ket0\) and \(\mathbf r=(1/\sqrt2,0,1/\sqrt2)\).
For \(\ket0\), the \(Z\) distribution is \((1,0)\), entropy \(0\), and the \(X\) distribution is \((\tfrac12,\tfrac12)\), entropy \(1\): the sum is exactly \(1\). For the second state both distributions are \(\big(\tfrac{1+1/\sqrt2}2,\tfrac{1-1/\sqrt2}2\big)\approx(0.854,0.146)\), each with entropy about \(0.60\), so the sum is about \(1.20\), above one bit.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
For the observable \(A=2\ketbra00-\ketbra11\), list its outcomes and compute \(\langle A\rangle\) in \(\ket+\).
Outcomes are eigenvalues.
Each outcome occurs with probability \(\tfrac12\) in \(\ket+\).
Outcomes \(2\) and \(-1\). \(\langle A\rangle=\tfrac12(2)+\tfrac12(-1)=\tfrac12\).
Do \(X\otimes X\) and \(Z\otimes Z\) commute?
\(XZ=-ZX\).
Two sign changes cancel.
Yes: \((X\otimes X)(Z\otimes Z)=XZ\otimes XZ=(-ZX)\otimes(-ZX)=(Z\otimes Z)(X\otimes X)\). They are jointly measurable, by a measurement in the Bell basis.
What is the expected spin component along \(x\) in the state \(\mathbf r=(0,0,1)\)? Along \(\mathbf n=(1/\sqrt2,0,1/\sqrt2)\)?
\(\langle\mathbf n\cdot\vec\sigma\rangle=\mathbf n\cdot\mathbf r\).
Compute dot products.
\(0\) along \(x\), and \(1/\sqrt2\approx0.71\) along \(\mathbf n\).
Apply the Robertson relation to \(X\) and \(Z\), using \([X,Z]=-2iY\). Evaluate both sides for \(\mathbf r=(1/\sqrt2,0,1/\sqrt2)\) and for \(\mathbf r=(0,1,0)\).
\(\Delta X\,\Delta Z\ge|\langle Y\rangle|\).
For a \(\pm1\) observable, \((\Delta A)^2=1-\langle A\rangle^2\).
The bound is \(\Delta X\,\Delta Z\ge|\langle Y\rangle|\). For the first state, \(\Delta X=\Delta Z=\sqrt{1/2}\), product \(\tfrac12\ge0\). For the second, \(\Delta X=\Delta Z=1\) and \(\langle Y\rangle=1\): equality. The Robertson bound depends on the state and can be zero even when both spreads are positive.
Construct the joint measurement of \(Z\otimes I\) and \(I\otimes Z\), and show that its coarse-grainings give each observable.
Use the products of spectral projections.
Sum over one index.
The effects are \(P_{ab}=\ketbra aa\otimes\ketbra bb\) for \(a,b\in\{0,1\}\). Summing over \(b\) gives \(\ketbra aa\otimes I\), the spectral projections of \(Z\otimes I\); summing over \(a\) gives those of \(I\otimes Z\). Commuting observables are measured together by measuring in a common eigenbasis.
A three-level system has \(H=\mathrm{diag}(0,1,3)\) in some units and is in \(\ket\psi=(1,1,1)^T/\sqrt3\). Find \(\langle H\rangle\), \(\langle H^2\rangle\), and the variance.
Each level has probability \(1/3\).
\(H^2=\mathrm{diag}(0,1,9)\).
\(\langle H\rangle=\tfrac13(0+1+3)=\tfrac43\), \(\langle H^2\rangle=\tfrac13(0+1+9)=\tfrac{10}3\), so the variance is \(\tfrac{10}3-\tfrac{16}9=\tfrac{14}9\approx1.56\).
Check the Maassen–Uffink relation for \(X\) and \(Z\) in the state \(\mathbf r=(0,1,0)\).
Find the outcome distributions of \(X\) and \(Z\).
A fair two-outcome distribution has entropy one bit.
Both distributions are fair, with entropy \(1\) bit each, so the sum is \(2\ge1\). This state has \(Y\) definite and both \(X\) and \(Z\) maximally uncertain; the entropic bound is far from tight here, while it is tight for eigenstates of \(X\) or \(Z\).
A student says: “The uncertainty relation means that measuring \(X\) disturbs \(Z\).” What does the Robertson relation actually say?
Are \(X\) and \(Z\) measured on the same systems in the relation?
Distinguish preparation from measurement uncertainty.
The Robertson relation compares the spread of \(X\) in one set of runs with the spread of \(Z\) in another, on identically prepared systems. It limits preparations: no state makes both sharp. Disturbance by measurement is a different question, with different relations, and the analysis of state update derives its qualitative form.
A noisy measurement of \(Z\) has effects \(\tfrac12(I\pm\eta Z)\) with \(0\le\eta\le1\). (a) What is the expectation of its outcome \(\pm1\)? (b) Paul Busch showed that noisy measurements of \(X\) and \(Z\) with sharpnesses \(\eta_X\), \(\eta_Z\) are jointly measurable exactly when \(\eta_X^2+\eta_Z^2\le1\). Check that \(\eta_X=\eta_Z=1/\sqrt2\) satisfies this, and explain what it shows about incompatibility.
\(\tr\big(\rho\,\tfrac12(I+\eta Z)\big)-\tr\big(\rho\,\tfrac12(I-\eta Z)\big)\).
\(\tfrac12+\tfrac12=1\).
(a) \(\eta\langle Z\rangle\): the noisy measurement shrinks the signal by \(\eta\). (b) \(\tfrac12+\tfrac12=1\), on the boundary. Incompatible sharp observables become compatible when made sufficiently unsharp: incompatibility is a matter of degree, and joint measurement is possible at a precise cost in sharpness.
Suppose a hidden-value model assigned values \(v(X),v(Z)\in\{\pm1\}\) and also required \(v\big((X+Z)/\sqrt2\big)=\big(v(X)+v(Z)\big)/\sqrt2\). Show this is impossible, and explain why the requirement was unreasonable for noncommuting observables.
\((X+Z)/\sqrt2\) has eigenvalues \(\pm1\).
Can \((v(X)+v(Z))/\sqrt2\) be \(\pm1\)?
\((v(X)+v(Z))/\sqrt2\) is \(\pm\sqrt2\) or \(0\), never \(\pm1\), so no assignment works. But \(X\), \(Z\), and \((X+Z)/\sqrt2\) are measured by three different experiments, so nothing requires their individual values to add, only their expectations. This was Bell’s criticism of von Neumann’s early no-go argument. The Peres–Mermin argument requires additivity only among commuting observables.