Quantum Foundations · Chapter 04
How must a theory describe two systems together if it describes each alone?
Starting point: Chapters 2 and 3: states, effects, the parameters K and N.
Anchor example: Two separated laboratories that each measure one of two bits.
Principle at work: No-signaling and local tomography: joint states are fixed by local measurements and their correlations.
Foil: Boxworld, whose PR boxes reach the algebraic maximum of the CHSH expression.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We can now describe one system of any theory: a convex set of states, a set of effects, and two numbers, \(K\) for description and \(N\) for distinguishable messages. A small gallery of such theories is in hand.
But no system stays alone for long. Systems interact, and a description of two systems together must agree with the description of each one alone. What does the framework already demand of such a description? Causality forces each part to have a state of its own, whatever happens to the other. That raises a counting question: how many numbers does the pair need? Our second principle answers it, and gives the rule for combining the spaces of two systems. Even then, the principles so far leave open how strongly two systems can be correlated, and a simple game shows that they allow far more than we observe.
The next chapter tests classical theory against what we have so far, and finds where it fails.
Suppose two laboratories share a pair of systems. Before counting how many numbers the pair needs, we must know whether each part has a state of its own at all. Causality settles this: neither laboratory’s choices can change the other’s statistics. Then we look at the simplest joint states, and at the correlations that mixing them produces.
Let two systems A and B be prepared together. Laboratory A applies measurement \(x\) and obtains outcome \(a\). Laboratory B applies measurement \(y\) and obtains \(b\). The joint state predicts probabilities \(p(a,b\mid x,y)\). Laboratory A sees only its own outcome, whose probability is the sum over the distant one:
\[p(a\mid x,y)=\sum_b p(a,b\mid x,y).\]
Can this depend on \(y\)? If it could, B could send messages to A by its choice of measurement, instantly and at any distance. The causality assumption of the framework excludes this, in two steps. The two measurements act on separate systems, so by causality we may perform them in either order without changing any statistics. Let A measure first. Then B’s choice of \(y\) comes later, and causality says that later choices cannot change the probabilities of an earlier outcome. So \(p(a\mid x,y)\) does not depend on \(y\). The same argument with the order reversed applies to B. This consequence is called no-signaling.
What does no-signaling give us? The predictions for A alone are the same whatever is done to B. So they are the predictions of some state of A. This state is called the marginal or reduced state of A. It is what an observer with access to A alone would ascribe, and it exists in every theory that satisfies the framework, not only in quantum theory.
Each part now has its own state. When does the pair contain nothing more than the two parts? The simplest joint preparation prepares A and B independently, in states \(\omega_A\) and \(\omega_B\). Then the probability of recording effect \(e\) on A and effect \(f\) on B factorizes:
\[p(e,f)=e(\omega_A)\,f(\omega_B).\]
A joint state of this form is called a product state. It is fixed by the two local states.
Can we get correlation from product states? Yes, by mixing them. Toss a coin, and prepare both systems in state 0 on heads and both in state 1 on tails. The outcomes are then perfectly correlated, but the shared coin explains the correlation. Mixtures of product states are called separable, and every correlation in classical probability theory is of this kind. Does a theory permit joint states that are not mixtures of products? And what correlations would those states produce? The framework leaves both questions open.
Problem 4.1. A joint state of two classical bits gives \(p(0,0)=0.4\), \(p(0,1)=0.1\), \(p(1,0)=0.2\), \(p(1,1)=0.3\). Find the marginal states and decide whether the joint state is a product.
A’s marginal is \((0.5,0.5)\) and B’s is \((0.6,0.4)\). The product would give \(p(0,0)=0.3\), not \(0.4\), so the state is correlated. Like every classical joint state it is a mixture of products of pure states: \(0.4\) of \((0,0)\), \(0.1\) of \((0,1)\), and so on.
How many numbers does a joint state need? The marginals do not suffice, because correlated and uncorrelated states can have the same marginals. The correlations between local outcomes contain more. Whether they contain everything is the content of our second principle. If it holds, counting fixes the number \(K\) of the pair, and the rule for combining two spaces follows.
Suppose two observers share their records completely: every local outcome, and every correlation between them. Can they always reconstruct the joint state? The framework does not say. We raise the answer “yes” to a principle.
Local tomography. The state of a composite system is determined by the joint probabilities of outcomes of measurements performed separately on its components.
The principle says that a whole is no more than its parts and their correlations. What would its failure look like? There would be joint states that agree on every local measurement and every correlation between local measurements, yet differ in the prediction of some measurement on the pair as a whole. Some information in the pair would then be accessible only by bringing the systems together. Local tomography rules out this kind of holism. It is a claim about the world, testable in principle. A consistent theory that violates it exists: quantum theory with real numbers in place of complex ones, which we meet when we reach the simplest non-classical system.
How many numbers does local tomography leave for the pair? Choose fiducial outcomes \(f_1,\dots,f_{K_A}\) for A and \(g_1,\dots,g_{K_B}\) for B. By local tomography the joint probabilities \(p(f_i,g_j)\) fix the joint state. There are \(K_AK_B\) of them, so \(K_{AB}\le K_AK_B\).
Could fewer numbers suffice? Pick \(K_A\) linearly independent states of A and \(K_B\) of B. Put the fiducial probabilities of the A states into a \(K_A\times K_A\) matrix \(M_A\), with entry \((k,i)\) equal to \(f_i(\omega_k)\), and similarly \(M_B\). Each matrix is invertible, because its rows are independent. Now take the \(K_AK_B\) product states. On a product state the probability \(p(f_i,g_j)\) is \(f_i(\omega_k)\,g_j(\omega_l)\). So the matrix of their fiducial probabilities has entries \((M_A)_{ki}(M_B)_{lj}\). This is the Kronecker product \(M_A\otimes M_B\): the block matrix whose block \((k,i)\) is \((M_A)_{ki}\,M_B\). Its determinant is \(\det(M_A)^{K_B}\det(M_B)^{K_A}\), which is nonzero. So the product states are linearly independent, \(K_{AB}\ge K_AK_B\), and
\[K_{AB}=K_A\,K_B .\]
A small case makes this concrete. For a classical bit, take the fiducial outcomes “some outcome” and “heads,” and the states heads-for-certain and tails-for-certain. Then
\[M=\begin{pmatrix}1&1\\1&0\end{pmatrix},\qquad M\otimes M=\begin{pmatrix}1&1&1&1\\1&0&1&0\\1&1&0&0\\1&0&0&0\end{pmatrix},\]
with \(\det M=-1\) and \(\det(M\otimes M)=(-1)^2(-1)^2=1\). The four product states of two bits are independent, as they must be.
What kind of space do joint states live in? A joint state is fixed by its array \(p(f_i,g_j)\), and every such array is a combination of arrays of product states. The space of \(K_A\times K_B\) arrays is called the tensor product \(\mathbb R^{K_A}\otimes\mathbb R^{K_B}\) of the two local spaces, and its dimension is the product. We write \(\omega_A\otimes\omega_B\) for the array of a product state and \(e\otimes f\) for the joint effect “\(e\) on A and \(f\) on B,” so that \((e\otimes f)(\omega_A\otimes\omega_B)=e(\omega_A)f(\omega_B)\). The tensor product is not postulated here. It is the space of arrays that local tomography makes sufficient. Without local tomography only \(K_{AB}\ge K_AK_B\) survives.
Does the same kind of rule hold for distinguishable states? Products of perfectly distinguishable states are perfectly distinguishable by local measurements, so \(N_{AB}\ge N_AN_B\). Can a pair carry more messages than its parts? Nothing so far decides this, so we make an explicit assumption.
Composite capacity (assumption, after Hardy). A pair of systems has no further perfectly distinguishable states: \(N_{AB}=N_AN_B\).
It holds for classical and quantum theory, and says that a pair carries exactly the messages of its parts. Together with \(K_{AB}=K_AK_B\), it severely restricts how \(K\) can depend on \(N\).
Problem 4.2. Verify \(K_{AB}=K_AK_B\) and \(N_{AB}=N_AN_B\) for two classical trits, and compute \(K\) for a pair of square bits under local tomography.
A pair of trits is a classical system with \(9\) configurations, so \(N_{AB}=9=3\times3\) and \(K_{AB}=9=3\times3\). A pair of square bits has \(K_{AB}=3\times3=9\), with joint states fixed by the probabilities of pairs of local fiducial outcomes.
Local tomography and no-signaling say what a joint state is, but not which joint states exist. How strongly can two systems be correlated? We measure it with a simple game. Shared plans give one limit, and a hypothetical pair of boxes gives another, far above it. The theory built from those boxes is the foil that shows our principles are not yet enough.
How can we measure correlation strength in a way every theory understands? With a game. Each laboratory receives a random bit, \(x\) for A and \(y\) for B, and must output a bit, \(a\) and \(b\), without communicating. They win if \(a\oplus b=xy\), where \(\oplus\) is addition modulo 2. So the outputs should agree unless both inputs are \(1\), when they should differ. The players may share any joint state and choose measurements depending on their inputs. The game is named after Clauser, Horne, Shimony, and Holt.
How well can a shared plan do? Suppose the outputs are fixed in advance by a plan, possibly chosen at random. Write the outputs as signs \(A_x=(-1)^{a}\) and \(B_y=(-1)^{b}\). A deterministic plan assigns four signs \(A_0,A_1,B_0,B_1\), and
\[A_0B_0+A_0B_1+A_1B_0-A_1B_1=A_0(B_0+B_1)+A_1(B_0-B_1)=\pm2,\]
since one of \(B_0\pm B_1\) is zero and the other is \(\pm2\). Averaging over random plans, the expected value \(S\) of this combination satisfies \(|S|\le2\). In terms of the game, the winning probability is \(\tfrac12+S/8\), so shared plans win at most \(3/4\) of the time. Every separable state acts as a random plan of this kind, so classical theory obeys the same limit.
Can any pair beat \(3/4\) without signaling? Sandu Popescu and Daniel Rohrlich asked in 1994 whether causality alone limits the game. Consider the joint behavior
\[p(a,b\mid x,y)=\begin{cases}\tfrac12 & a\oplus b=xy,\\ 0 & \text{otherwise.}\end{cases}\]
Each local outcome is uniformly random whatever the other laboratory does, so it does not signal. Yet the game is won with certainty: \(S=4\), the largest value the expression can take. Quantum pairs, we will find, reach only \(S=2\sqrt2\approx2.83\), a bound proved by Boris Tsirelson.
Is there a whole theory with such boxes? There is. Boxworld is a theory whose single systems are square bits and whose pairs contain every joint state allowed by no-signaling and local tomography. The PR box is one of its joint states. Boxworld satisfies the framework, finiteness, and local tomography, so these principles do not yet select quantum theory.
What does boxworld lack? Reversible dynamics. David Gross, Markus Müller, Roger Colbeck, and Oscar Dahlsten proved in 2010 that its only reversible transformations are local operations and permutations of systems. So two square bits can never be brought into correlation by a reversible process. A principle that demands rich reversible change, continuous reversibility, is exactly what boxworld violates.
Problem 4.3. Show that the plan “always output 0” wins the CHSH game with probability \(3/4\), and that no plan with fixed outputs does better.
With \(a=b=0\), \(a\oplus b=0\), which equals \(xy\) for three of the four equally likely input pairs. For any deterministic plan \(S=\pm2\), so the winning probability \(\tfrac12+S/8\) is at most \(\tfrac12+\tfrac14=\tfrac34\).
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 both local outcomes of the PR box are uniformly random for every pair of inputs.
Sum \(p(a,b\mid x,y)\) over \(b\).
For each \(a\) exactly one \(b\) satisfies \(a\oplus b=xy\).
For fixed \(a\), \(x\), \(y\), exactly one value of \(b\) satisfies \(a\oplus b=xy\), with probability \(\tfrac12\). So \(p(a\mid x,y)=\tfrac12\) for both \(a\), independent of \(y\). The same holds for B. The PR box does not signal.
Under local tomography, what is \(K\) for a pair of balls (\(K=4\) each), a pair of square bits, and a pair of classical bits?
\(K_{AB}=K_AK_B\).
The square bit has \(K=3\).
\(16\), \(9\), and \(4\). For classical bits this matches the four configurations of the pair.
If a strategy achieves CHSH value \(S\), its winning probability is \(\tfrac12+S/8\). Evaluate it for \(S=2\), \(S=2\sqrt2\), and \(S=4\).
Substitute.
\(2\sqrt2/8=\sqrt2/4\).
\(0.75\), \(\tfrac12+\tfrac{\sqrt2}4\approx0.854\), and \(1\): the classical limit, the quantum limit, and the PR box.
A proposed joint behavior has \(a=y\) with certainty and \(b\) uniformly random. Show that it signals, and explain why the framework excludes it for separated laboratories.
Compute \(p(a\mid x,y)\).
Could B send a bit to A?
\(p(a=1\mid x,y)=y\), which depends on B’s choice. B could send a bit to A instantly by choosing \(y\). The framework forbids later choices from influencing earlier records, and for laboratories whose operations have no fixed order this forbids signaling in either direction. No-signaling ensures that A has a well-defined marginal state.
The joint state of two classical bits \(p(0,0)=p(1,1)=\tfrac12\), \(p(0,1)=p(1,0)=0\) has perfectly correlated outcomes. Write it as a mixture of products, and compute its marginals.
Each nonzero entry is a product of pure states.
Marginals sum over the other bit.
It is \(\tfrac12(1,0)\otimes(1,0)+\tfrac12(0,1)\otimes(0,1)\): a shared fair coin decides whether both are \(0\) or both \(1\). The marginals are \((\tfrac12,\tfrac12)\) each. The correlation has a classical explanation, and every classical joint state has one of this kind.
Show that if \(\omega_1,\dots,\omega_m\) are linearly independent states of A and \(\sigma_1,\dots,\sigma_n\) of B, the \(mn\) product states are linearly independent. Use fiducial probabilities and the Kronecker product.
Arrange the fiducial probabilities of the \(\omega_i\) in a matrix \(M\), and similarly \(N\) for B.
The fiducial probabilities of products form \(M\otimes N\).
With \(m=K_A\) and \(n=K_B\), \(M\) and \(N\) are invertible. The joint fiducial probabilities of \(\omega_i\otimes\sigma_j\) are \(M_{ia}N_{jb}\), the entries of \(M\otimes N\), whose determinant is \(\det(M)^{n}\det(N)^{m}\ne0\). So the products are linearly independent, giving \(K_{AB}\ge K_AK_B\) in every theory.
Mix the PR box with weight \(v\) and the uniformly random box (all outcomes probability \(\tfrac14\)) with weight \(1-v\). (a) Find \(S\). (b) For which \(v\) can the result be reproduced by local plans? (c) For which \(v\) do quantum pairs suffice?
The random box has all correlations zero.
Compare with \(2\) and \(2\sqrt2\).
(a) The PR box has correlations \(+1,+1,+1,-1\) giving \(S=4\), and the random box gives \(0\), so \(S=4v\). (b) Local plans reach \(S\le2\), so the CHSH test is passed for \(v\le\tfrac12\); in fact such noisy boxes are local exactly in that range. (c) Quantum theory reaches \(S=2\sqrt2\) at \(v=1/\sqrt2\approx0.71\), and noisy PR boxes with \(v\le1/\sqrt2\) are quantum-achievable. Between \(0.71\) and \(1\) lies correlation allowed by no-signaling but not by quantum theory.
A student says: “The outcomes of two distant measurements are correlated, so one measurement must influence the other.” Give a counterexample from classical theory, and say what no-signaling does and does not exclude.
Recall the shared coin.
Distinguish correlation from control.
A shared coin that prepares both systems in \(0\) or both in \(1\) produces perfect correlation with no influence. No-signaling excludes that one laboratory’s choice changes the other’s local statistics; it does not exclude correlation. Whether all no-signaling correlations have common-cause explanations is a separate question, answered negatively by Bell’s theorem.
Imagine a theory in which \(K_{AB}=K_AK_B+1\) for a pair. Describe an experiment that would reveal the extra parameter and explain why two observers who pooled all their local records could never detect it.
Local statistics fix only \(K_AK_B\) numbers.
Some effect on the pair must depend on the extra number.
Two joint states can agree on all \(K_AK_B\) local correlation probabilities and differ in the extra parameter. Only a global measurement on the pair, whose effect is not a combination of products of local effects, can reveal the difference. Pooled local records are functions of the local correlations, so they cannot. Real quantum theory, constructed with the qubit, behaves exactly like this, with \(K=10\) instead of \(9\) for two bits.
Show that \(N_{AB}\ge N_AN_B\): products of perfectly distinguishable local states are perfectly distinguishable, and the product effects \(e_i\otimes f_j\) form a measurement.
Evaluate \(e_i\otimes f_j\) on \(\omega_k\otimes\sigma_l\).
Sum the product effects.
\((e_i\otimes f_j)(\omega_k\otimes\sigma_l)=e_i(\omega_k)f_j(\sigma_l)\), which is \(1\) if \(i=k\) and \(j=l\), else \(0\). The sum \(\sum_{ij}e_i\otimes f_j=u_A\otimes u_B\) is the unit effect of the pair, so these effects form a measurement, realized by measuring each system separately. Hence \(N_AN_B\) states of the pair are perfectly distinguishable.