Quantum Foundations · Chapter 12
Could measurement outcomes be fixed in advance, and are probabilities other than the Born rule possible?
Starting point: Chapters 8–11: projective measurements, observables, and compatibility.
Anchor example: The Peres–Mermin square of nine two-qubit observables.
Principle at work: Probabilities are assigned to projectors independently of the measurement in which they occur.
Foil: Noncontextual hidden-variable models.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We derived that the pure states of the theory give random outcomes with no ignorance behind them. We have also found that observables need not commute, and that a hidden-value model would have to supply one joint distribution for all of them.
Could the randomness still reflect ignorance of deeper variables that fix every outcome in advance? Our derivation concerned the theory’s own states, so it leaves this open. We therefore ask what such variables would have to be like. If each observable carried a value that does not depend on what else is measured with it, the values would have to obey the algebra of commuting observables, and a nine-observable square shows that they cannot. The same assumption, applied to probabilities rather than values, turns out to give the probability rule again, by a different route. Finally we ask what this rules out and what it leaves open.
The next chapter carries the question to pairs of distant systems.
Suppose every observable had a value before it was measured. What rules would those values have to obey? We first state the rules, including the requirement that a value not depend on its context. Then we test them on two qubits and find that no assignment satisfies them.
What would it mean for outcomes to be fixed in advance? A value assignment gives each observable \(A\) a definite value \(v(A)\), one of its eigenvalues, which any measurement of \(A\) would reveal. For it to be consistent with the observed statistics, it must respect the relations among compatible observables. If \(A\) and \(B\) commute, they can be measured together, and the product of their outcomes is the outcome of \(AB\) measured in the same run. So we require \(v(AB)=v(A)v(B)\) for commuting \(A\) and \(B\).
Could the value of \(A\) depend on what else is measured with it? Not if the value was there before the measurement. Then it is a property of the system, not of the experimental arrangement. We call an assignment noncontextual if \(v(A)\) does not depend on which other compatible observables are measured alongside \(A\). A contextual assignment allows \(v(A)\) to depend on the whole arrangement.
Can a noncontextual assignment exist? We test it on two qubits. Arrange nine observables, each with eigenvalues \(\pm1\):
\[\begin{array}{ccc} X\otimes I & I\otimes X & X\otimes X\\ I\otimes Z & Z\otimes I & Z\otimes Z\\ X\otimes Z & Z\otimes X & Y\otimes Y\end{array}\]
The three observables in each row commute, and so do the three in each column. Multiplying them, using \(XZ=-iY\) and \(ZX=iY\), gives \(+I\) for each row and for the first two columns, and \(-I\) for the third column: \((X\otimes X)(Z\otimes Z)(Y\otimes Y)=(XZY)\otimes(XZY)=(-iI)\otimes(-iI)=-I\).
What must the values satisfy? A noncontextual assignment gives each observable one value, used in both its row and its column, so it must satisfy six equations: the product of the values in each row is \(+1\), the product in the first two columns is \(+1\), and in the third column it is \(-1\). Multiplying the three row equations gives the product of all nine values as \(+1\). Multiplying the three column equations gives \(-1\). Each value appears once in a row and once in a column, so these are the same product. No assignment exists, whatever the state. Asher Peres and David Mermin found this argument in 1990.
Problem 12.1. Verify that the third row multiplies to \(+I\).
\((X\otimes Z)(Z\otimes X)(Y\otimes Y)=(XZY)\otimes(ZXY)\). Now \(XZY=-iYY=-iI\) and \(ZXY=iYY=iI\), so the product is \((-i)(i)I=I\).
The square shows that values cannot be noncontextual. What if we ask the same of probabilities instead of values? We first state that demand as a property of probability assignments. A theorem of Andrew Gleason then shows that it alone gives the probability rule, which we derived earlier from the shape of the state space. It excludes deterministic assignments as a by-product.
How much of the probability rule follows from the projective measurements alone? Suppose we know only which projective measurements exist on \(\mathbb C^N\): every orthogonal decomposition of the identity into projections. Ask for an assignment of probabilities \(f(P)\in[0,1]\) to projections such that, for every projective measurement \(\{P_k\}\), the probabilities sum to one. We require that \(f(P)\) depend only on \(P\), not on the measurement it belongs to. Such an \(f\) is called a frame function.
Is that requirement an assumption? In the operational framework this independence is automatic: an effect is defined by its probabilities, so a projection that occurs in two measurements is the same effect in both. Here it is a hypothesis, the noncontextuality of probabilities, and Gleason asks what it implies on its own.
What does the hypothesis give on its own?
Gleason’s theorem (1957). If \(N\ge3\), every frame function has the form \(f(P)=\tr(\rho P)\) for a unique density operator \(\rho\).
The proof is long and uses the continuity of frame functions on the sphere of unit vectors in three real dimensions, extended to all \(N\ge3\). The theorem fails for \(N=2\): a qubit’s projective measurements are just antipodal pairs of points on the Bloch sphere, and any function with \(g(\mathbf n)+g(-\mathbf n)=1\) is a frame function, including wildly discontinuous ones. Paul Busch showed in 2003 that the conclusion holds for every \(N\), including \(2\), if probabilities are assigned to all effects \(0\le E\le I\) and are additive over all general measurements.
Does the theorem say anything about predetermined outcomes? It excludes them. A noncontextual value assignment to projections would be a frame function taking only the values \(0\) and \(1\). But \(\tr(\rho P)\) varies continuously as \(P\) is rotated, and a continuous function on a connected set that takes both values must take everything in between. So for \(N\ge3\) no such assignment exists.
Problem 12.2. For a qubit, define \(g(\mathbf n)=1\) if \(n_z>0\), \(g(\mathbf n)=0\) if \(n_z<0\), and on the equator choose \(1\) on one half-circle and \(0\) on the other. Show that \(g\) is a frame function, and explain why it is not of Born form.
Every projective measurement on a qubit is a pair \(\{\mathbf n,-\mathbf n\}\), and the definition makes exactly one of them \(1\), so the values sum to one. A Born-form function \(\tr(\rho P_{\mathbf n})=\tfrac12(1+\mathbf r\cdot\mathbf n)\) is continuous in \(\mathbf n\), while \(g\) jumps. This is why Bell could build a deterministic model for a single qubit.
The square and Gleason’s theorem both show that quantum outcomes cannot be preassigned in a noncontextual way. Does contextuality rule out hidden variables altogether? We first state the general theorem behind the square, and its experimental tests. Then we see which hidden-variable theories survive it.
Does the square generalize beyond two qubits? It does.
Kochen–Specker theorem (1967). For \(N\ge3\), there is no assignment of values \(0\) or \(1\) to all projections on \(\mathbb C^N\) such that, in every projective measurement, exactly one projection receives the value \(1\).
Simon Kochen and Ernst Specker proved it with 117 vectors in three real dimensions; later proofs use fewer, for example 18 vectors in four dimensions found by Adán Cabello and collaborators. The Peres–Mermin square is a proof in terms of observables in dimension four. These proofs are state-independent: they show that no value assignment exists at all, not merely that one fails for some state. Experiments have tested the square directly, for example with trapped ions by Gerhard Kirchmair and collaborators in 2009, confirming the quantum predictions that make noncontextual assignments impossible.
Does the theorem exclude hidden variables? No. It excludes only those whose values are independent of the measurement context. A theory in which the outcome of measuring \(A\) depends on which compatible observables are measured with it, or on the details of the apparatus, escapes the argument. Bohmian mechanics is such a theory: particles have definite positions, but the outcomes of other measurements depend on the whole experimental arrangement.
So the situation is this. The states of quantum theory are intrinsically random, as continuous reversibility forced. Deeper variables could be added, but only at the price of contextuality. For distant pairs of systems they carry a second price, nonlocality, as we see next. Whether such variables exist is not decided by the principles of this course. What is decided is that a noncontextual, predetermined world would not be described by quantum theory.
Problem 12.3. Why does the Peres–Mermin argument fail for a single qubit?
On a single qubit, two observables commute only if one is a function of the other: distinct spin components never commute. There is no grid of observables with nontrivial compatible rows and columns, and indeed noncontextual models such as Bell’s exist for one qubit. Contextuality proofs need at least three dimensions.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Verify that the first column of the Peres–Mermin square multiplies to \(+I\).
The column is \(X\otimes I\), \(I\otimes Z\), \(X\otimes Z\).
Multiply factor by factor.
\((X\otimes I)(I\otimes Z)(X\otimes Z)=(X\cdot I\cdot X)\otimes(I\cdot Z\cdot Z)=I\otimes I\).
Why does Gleason’s theorem fail for a qubit?
What are the projective measurements of a qubit?
Do any two of them share a projection?
A qubit’s projective measurements are pairs of antipodal points on the Bloch sphere, and no two different measurements share a projection. A frame function only needs \(g(\mathbf n)+g(-\mathbf n)=1\), which allows non-Born functions. In three or more dimensions, measurements share projections and interlock, which forces the Born form.
Is Bohmian mechanics ruled out by the Kochen–Specker theorem? Why or why not?
The theorem excludes noncontextual assignments.
In Bohmian mechanics, outcomes depend on the apparatus.
No. Its outcomes depend on the whole experimental arrangement, so its value assignments are contextual, and the theorem excludes only noncontextual ones.
Show that \(XZ=-iY\) and \(ZX=iY\), and use them to verify that the third column of the Peres–Mermin square multiplies to \(-I\).
Multiply the matrices.
\(Y^2=I\).
\(XZ=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=-iY\) and \(ZX=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=iY\). Then \((X\otimes X)(Z\otimes Z)(Y\otimes Y)=(XZY)\otimes(XZY)=(-iY^2)\otimes(-iY^2)=(-i)^2I=-I\).
Let \(\rho=\mathrm{diag}(0.5,0.3,0.2)\) on \(\mathbb C^3\) and let \(\ket{v_1},\ket{v_2},\ket{v_3}\) be any orthonormal basis. Show that \(f(P)=\tr(\rho P)\) sums to one over the basis, and that the probability of a projection is the same in every basis containing it.
\(\sum_k\ketbra{v_k}{v_k}=I\).
\(f\) depends only on \(P\).
\(\sum_k\tr(\rho\ketbra{v_k}{v_k})=\tr(\rho I)=1\). The value \(\tr(\rho P)\) is computed from \(P\) alone, so it does not depend on the rest of the basis. Born-rule probabilities are frame functions, and Gleason’s theorem says that for \(N\ge3\) they are the only ones.
Try the assignment: all nine Peres–Mermin observables equal \(+1\) except \(Y\otimes Y=-1\). Which of the six constraints fail? Show that every assignment fails at least one.
Check each row and column product.
Compare the product of all nine computed by rows and by columns.
Rows 1 and 2 give \(+1\); row 3 gives \(-1\), violating its constraint. Columns 1 and 2 give \(+1\), and column 3 gives \(-1\) as required. So one constraint fails. In general the product of all nine values is the product of the row products and also of the column products; the constraints require these to be \(+1\) and \(-1\), so at least one constraint always fails.
For \(\rho=\ketbra00\) on \(\mathbb C^3\), let \(\ket{v(t)}=\cos t\ket0+\sin t\ket1\). Compute \(f(t)=\tr(\rho\ketbra{v(t)}{v(t)})\) and explain why no frame function taking only the values \(0\) and \(1\) can be of Born form.
\(f(t)=\cos^2t\).
A Born-form function is continuous.
\(f(t)=\cos^2t\), which passes continuously from \(1\) to \(0\), taking every value in between, such as \(\tfrac12\) at \(t=\pi/4\). A continuous function on the connected set of rays that takes only the values \(0\) and \(1\) must be constant, which is impossible for a frame function, since every basis contains one ray with value \(1\) and others with value \(0\). By Gleason’s theorem every frame function for \(N\ge3\) is of Born form, so no deterministic assignment exists.
A student says: “The Kochen–Specker theorem proves that hidden variables are impossible.” Correct the statement.
What assumption does the theorem need?
Recall what contextuality leaves open for hidden variables.
The theorem excludes noncontextual hidden variables: preassigned values independent of which compatible observables are measured together. Contextual hidden-variable theories, such as Bohmian mechanics, are consistent with it. The theorem constrains what deeper variables must be like; it does not exclude them.
Define \(\chi=\langle R_1\rangle+\langle R_2\rangle+\langle R_3\rangle+\langle C_1\rangle+\langle C_2\rangle-\langle C_3\rangle\), where \(R_i\), \(C_j\) are the products of the outcomes in row \(i\) and column \(j\) of the Peres–Mermin square. Show that noncontextual models give \(\chi\le4\) and quantum theory gives \(\chi=6\) in every state.
A noncontextual assignment satisfies at most five of the six constraints.
In quantum theory each product is fixed by an operator identity.
Each term is at most \(1\). A noncontextual assignment violates at least one constraint, so at least one term is \(-1\), giving at most \(5-1=4\), and mixtures of assignments cannot exceed this. In quantum theory the row and first two column products equal \(+I\) and the third equals \(-I\), so each term equals \(1\) and \(\chi=6\) for every state. Experiments with imperfect devices test this inequality, as Kirchmair and collaborators did in 2009.
Explain why Busch’s version of Gleason’s theorem, which assigns probabilities to all effects, works for a qubit.
Effects of a qubit form a four-dimensional convex set.
Additivity over general measurements links many effects.
General measurements combine effects in many more ways than projective ones: for example, \(\tfrac12P_{\mathbf n}\) appears in measurements together with other effects in all directions. Requiring additivity over all of them forces the probability assignment to be linear on effects, and a linear functional on \(2\times2\) Hermitian matrices that is positive and normalized is \(\tr(\rho E)\). The extra structure that Gleason gets from three dimensions comes, for a qubit, from general measurements.