Quantum Foundations · Chapter 05
Classical probability satisfies every principle so far. What can it not do?
Starting point: Chapters 2–4: states, effects, distinguishability, and composition.
Anchor example: A classical bit that must change from 0 to 1 within one second.
Principle at work: The question: can every pure state be reached from every other by a continuous reversible change?
Foil: Classical probability theory itself, whose reversible transformations are permutations.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We have the framework, finiteness, and local tomography, and from them the rule \(K_{AB}=K_AK_B\) for pairs. Classical probability theory satisfies all of it. So does boxworld, and so, we will find, does quantum theory.
If classical theory passes every test so far, why is the world not classical? We must find something classical theory cannot do. Its states have a special shape, the simplex, and every mixed state has exactly one reading as ignorance. That shape leaves room for very few reversible transformations, and so classical systems cannot change continuously without losing information. This is the crack through which quantum theory enters. Before following it, we count parameters: composition and finiteness already restrict how \(K\) can depend on \(N\) to a short list, with classical theory first and a candidate for quantum theory second.
The next chapter accepts continuous reversible change as a principle and follows its consequences.
To find what classical theory cannot do, we first need its state space exactly. It is a simplex, whose pure states are fully deterministic. We then ask what singles out the simplex among all convex sets, and find that it is the only shape in which every mixture has one reading.
What does the state space of a classical system with \(N\) configurations look like? Its states are lists \((p_1,\dots,p_N)\) of nonnegative numbers summing to one. Geometrically this is a simplex: a segment for \(N=2\), a triangle for \(N=3\), a tetrahedron for \(N=4\). Its corners are the \(N\) pure states. They are perfectly distinguishable, so \(K=N\).
In what sense are classical pure states deterministic? Consider the effects that cannot be obtained by mixing other effects, the extreme effects. For a classical system they are the indicator effects of sets of configurations: “the configuration is in set \(S\).” Every pure state gives each of them probability \(0\) or \(1\). So probabilities strictly between \(0\) and \(1\) arise in only two ways. Either the state is mixed, which expresses ignorance of the configuration, or the effect is noisy, which expresses an imperfect detector. A classical world can therefore always be read as a world of definite configurations that we know imperfectly.
Is this ignorance reading available in every theory? It needs each mixed state to be a mixture of pure states in only one way. Every classical state is, with weights \(p_j\). Does any other shape share this?
No. A standard theorem of finite-dimensional convex geometry says that a closed, bounded convex set in which every point has a unique decomposition into extreme points is a simplex. The plane case shows why. A polygon that is not a triangle has at least four corners. Take four of them, in order around the boundary. The two diagonals of the quadrilateral they form cross at an interior point. That point is a mixture of the two ends of one diagonal, and also of the two ends of the other. So its decomposition is not unique. The disc, the square, and the ball all fail in this way.
What does the failure cost? If a state can be prepared as a mixture of pure states in two different ways, the state itself holds no fact about which pure state the system “really” has. In classical theory this question never arises. In every other theory of the gallery it does.
Problem 5.1. Write the center of the square bit, \((p_X,p_Z)=(\tfrac12,\tfrac12)\), as an even mixture of pure states in two different ways.
It is the midpoint of the diagonal from \((0,0)\) to \((1,1)\), and also of the diagonal from \((1,0)\) to \((0,1)\). Both even mixtures give the same state, so the square is not a simplex.
A transformation is reversible if another transformation undoes it, returning every state to itself. Why care about reversible transformations? Consider a system that interacts with nothing else. If its evolution lost information, that information would have to go somewhere, and nothing is there to receive it. So we expect the evolution of isolated systems to be reversible, and we will raise this to a principle. We first ask which reversible transformations a classical system has, and then whether any of them can act continuously in time.
What can a reversible transformation do to the states? A transformation maps states to states and respects mixing, so it acts linearly on the state space. Suppose \(T\) is reversible. Both \(T\) and its inverse map the state space into itself, so \(T\) maps it onto itself. Can \(T\) send a pure state to a mixed one? If \(T\omega\) were a mixture, applying \(T^{-1}\) would exhibit \(\omega\) as a mixture. So a reversible transformation permutes the pure states.
For a simplex, a linear map is fixed by what it does to the corners. So the reversible transformations of a classical system with \(N\) configurations are exactly the \(N!\) permutations of its configurations. A bit has two: the identity and the swap. A trit has six.
Can a classical bit change continuously and reversibly in time? Suppose it does. Then the transformation \(T_t\) from time \(0\) to time \(t\) depends continuously on \(t\), and \(T_0\) is the identity. Each \(T_t\) is one of the two permutations. A continuous path in a set of two isolated points cannot move from one to the other. So \(T_t\) is the identity for all \(t\), and the bit can never change. The same argument applies for every \(N\): continuous reversible dynamics of a finite classical system is trivial.
But surely a bit can change from 0 to 1 continuously? It can, through the mixtures \((1-t,t)\). A map that takes the pure state \((1,0)\) to the mixture \((1-t,t)\), however, is not reversible for \(0<t<1\). It loses information, since it must also squeeze the whole segment into a shorter one. A finite classical world must choose between continuity and reversibility.
How does classical mechanics escape this? Only because its systems are not finite. A particle has a continuum of configurations, its possible positions and velocities, and the equations of motion move them continuously and reversibly. Finiteness excludes this: a system with finitely many distinguishable states has no continuum of perfectly distinguishable configurations to flow between. So finiteness and continuous reversibility together are incompatible with classical theory. Next we accept both and follow the consequences.
Problem 5.2. Show directly that the map on the classical bit \(p\mapsto(1-t)p+t(1-p)\) is not reversible for \(0<t<1\).
The map sends \(p=0\) to \(t\) and \(p=1\) to \(1-t\), so its image is the interval between \(t\) and \(1-t\), shorter than \([0,1]\). Its linear inverse would have to stretch that interval back to \([0,1]\), sending some valid states, such as \(p=0\), to numbers outside \([0,1]\). So the inverse is not a transformation.
Before we follow continuous reversibility, how much do the principles we have already restrict \(K\) as a function of \(N\)? The composition rules turn \(K\) into a function that maps products to products. Such functions are powers, so the possible theories form a short list. This argument is due to Lucien Hardy, and it places classical theory first and a candidate for quantum theory second.
Can we treat \(K\) as a function of \(N\) alone? Not without an assumption, because two systems with the same \(N\) might need different numbers of parameters. We make the assumption explicit. It anticipates the principle of equivalence of subspaces, which we state when we study larger systems.
Counting assumption (after Hardy). \(K\) depends only on \(N\). This holds also for the states of a larger system that never give certain of its distinguishable outcomes: if \(N\) outcomes remain possible, those states form a system with \(K(N)\) parameters.
Does \(K\) then grow with \(N\)? Take a system with \(N+1\) perfectly distinguishable states, and the measurement that distinguishes them. The states that never give the last outcome form a system with \(N\) possible outcomes, so they span a space of dimension \(K(N)\). The state that gives the last outcome for certain is not in that span, because the last effect is \(0\) on the whole span and \(1\) on it. So \(K(N+1)\ge K(N)+1\), and \(K\) increases strictly with \(N\).
What does composition add? The composite capacity assumption \(N_{AB}=N_AN_B\), together with local tomography, gives
\[K(N_AN_B)=K(N_A)\,K(N_B).\]
Which functions behave this way? A strictly increasing function on the positive integers that turns products into products must be a power, \(K(N)=N^r\) for some \(r>0\). This is a result of number theory, proved by Paul Erdős in 1946. Since \(K\) is an integer for every \(N\), \(r\) is a positive integer, as Hardy argued in 2001. So the possible theories are indexed by \(r=1,2,3,\dots\).
Which members of the list survive? For \(r=1\), \(K=N\): each state needs only as many numbers as there are distinguishable states. This is classical probability theory, which, as we just saw, has no continuous reversible change. For \(r=2\), \(K=N^2\). A bit then has \(K=4\), so its normalized states form a three-dimensional convex set. Quantum theory has exactly this count, as we will derive.
How do we choose \(r=2\) over larger values? Hardy closed the argument with a simplicity axiom: \(K\) takes the smallest value consistent with the other principles. Continuity excludes \(r=1\), so simplicity selects \(r=2\). We will not need that axiom. When we reach the simplest non-classical system, we derive \(K=4\) for the bit from local tomography and continuous reversible interaction, and so obtain \(r=2\) without appeal to simplicity.
Does the counting exclude any known theory? It excludes quantum theory with real numbers in place of complex ones. We construct that theory later; for now we quote its count. Its bit has \(K(2)=3\), and a pair of bits has \(K(4)=10\), not \(3\times3=9\). So real quantum theory cannot satisfy local tomography: its pairs carry information not visible in local statistics.
Problem 5.3. Under \(K=N^2\), how many real parameters fix a normalized state of a system with three distinguishable states? Of two such systems together?
A single system has \(K=9\), so \(8\) parameters for a normalized state. The pair has \(N=9\) and \(K=81\), so \(80\) parameters, consistent with \(K_{AB}=9\times9\).
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
How many reversible transformations does a classical system with four configurations have? Which of them can be reached continuously from the identity?
Reversible maps permute the corners of the simplex.
A continuous path in a finite set is constant.
\(4!=24\) permutations. Only the identity can be reached continuously, since a continuous path through a finite set of isolated points cannot leave its starting point.
Under \(K=N^r\) with \(r=3\), what would \(K\) be for a bit, and what would be the dimension of its set of normalized states?
Substitute \(N=2\).
Normalization removes one parameter.
\(K=8\), so normalized states would form a \(7\)-dimensional convex set. The theorem on reversible interaction of two bits excludes this, leaving \(r=2\).
In a classical system with four configurations, which sets of pure states are perfectly distinguishable?
Classical pure states have no overlap.
Use indicator effects.
Every set: the indicator effects of the configurations distinguish all four corners at once, and hence any subset. Classical pure states never overlap.
(a) Prove that a reversible transformation maps pure states to pure states. (b) Use this to show that a linear map of the classical trit sending the corner \((1,0,0)\) to the midpoint \((\tfrac12,\tfrac12,0)\) cannot be reversible.
Apply \(T^{-1}\) to a decomposition of \(T\omega\).
The midpoint is a mixture.
(a) If \(T\omega=\lambda\sigma_1+(1-\lambda)\sigma_2\) with distinct states, then \(\omega=\lambda T^{-1}\sigma_1+(1-\lambda)T^{-1}\sigma_2\) with distinct states, contradicting purity. (b) The image of a pure corner would be a mixture, which (a) forbids. Such a map loses information and its linear inverse would send some states outside the simplex.
Quantum theory with real amplitudes has \(K=N(N+1)/2\). Compute \(K\) for \(N=2,3,4,6\) and test the composition rule \(K(N_AN_B)=K(N_A)K(N_B)\) for \(2\times2\) and \(2\times3\).
\(K(2)=3\), \(K(3)=6\).
Compare \(K(4)\) with \(K(2)^2\) and \(K(6)\) with \(K(2)K(3)\).
\(K=3,6,10,21\). Then \(K(4)=10\ne9=K(2)^2\) and \(K(6)=21\ne18=K(2)K(3)\). In each case the composite has more parameters than local tomography allows: real quantum theory satisfies only \(K_{AB}\ge K_AK_B\).
Classical Hamiltonian flow moves points of phase space continuously and reversibly. (a) Why does this not contradict the absence of continuous reversible change for finite classical systems? (b) Divide phase space into \(N\) cells and suppose the flow over one step maps cells to cells. What kind of map is the step, and what happens to continuity?
How many perfectly distinguishable configurations does a point particle have?
A bijection of finitely many cells is a permutation.
(a) A classical particle has a continuum of perfectly distinguishable configurations, so \(N\) is infinite and finiteness fails; the flow connects them continuously. (b) The step is a permutation of cells, and the evolution in time can only jump from one permutation to another. Continuity is recovered only in the limit of infinitely many cells. With finitely many distinguishable states, classical theory must choose between continuity and reversibility.
Decompose the trit state \((0.5,0.3,0.2)\) into pure states, and prove that no other decomposition exists, using the fact that its coordinates are probabilities of the indicator effects.
In a mixture of pure states, the indicator effect of configuration \(j\) picks out the weight of corner \(j\).
Pure states of the trit are its corners.
\(0.5(1,0,0)+0.3(0,1,0)+0.2(0,0,1)\). In any decomposition into pure states, which are corners, the indicator effect of configuration \(j\) gives the total weight on corner \(j\), and it must equal the \(j\)th coordinate. So the weights are forced. This barycentric uniqueness is what characterizes a simplex.
A student argues: “A classical bit can change continuously from 0 to 1 by a stochastic process, so continuity is no problem for classical theory.” What is missing?
Is the map that takes \((1,0)\) to \((1-t,t)\) reversible?
What happens to distinguishability?
The process is continuous but irreversible. For \(0<t<1\) it sends the pure state to a mixture and squeezes the state space, losing information that no transformation can restore. Continuous reversibility asks for continuity and reversibility together, which a finite classical system cannot provide.
Let \(K\) be strictly increasing on the positive integers with \(K(mn)=K(m)K(n)\) and \(K(2)=4\). Using \(2^4<3^3<2^5\), \(2^7<3^5<2^8\), and \(2^{11}<3^7<2^{12}\), bound \(K(3)\) and conclude that it is \(9\) if it is an integer.
\(K(2^k)=4^k\) and \(K(3^k)=K(3)^k\).
Monotonicity turns \(2^a<3^b<2^c\) into \(4^a<K(3)^b<4^c\).
From \(16<27<32\): \(256<K(3)^3<1024\), so \(6.35<K(3)<10.08\). From \(128<243<256\): \(4^7<K(3)^5<4^8\), so \(6.96<K(3)<9.19\). From \(2048<2187<4096\): \(4^{11}<K(3)^7<4^{12}\), so \(8.83<K(3)<10.77\). Together \(8.83<K(3)<9.19\), and the only integer is \(9=3^2\). Continuing with better rational approximations of \(\log_23\) pins \(K(3)=3^2\) exactly, the idea behind Erdős’s theorem.
Hardy’s continuity axiom requires a continuous reversible path between any two pure states. Show that the case \(r=1\) violates it, and that the Bloch ball (\(r=2\), \(N=2\)) satisfies it.
For \(r=1\) the state space is a simplex.
Rotations of the ball form a connected group.
For \(r=1\), \(K=N\) and the state space is a simplex, whose reversible maps are permutations, so no continuous path joins distinct pure states. For the ball, any two points of the sphere are related by a rotation, and rotations by \(t\) times the angle, \(0\le t\le1\), form a continuous path of reversible maps from the identity. So \(r=2\) passes where \(r=1\) fails.