Quantum Foundations · Chapter 06
What follows if every pure state can be turned into any other continuously and reversibly?
Starting point: Chapter 5: the classical simplex and why its reversible maps are discrete.
Anchor example: A system that changes smoothly from one of two perfectly distinguishable states to the other.
Principle at work: Continuous reversibility: the reversible transformations form a connected group acting transitively on pure states.
Foil: Classical theory, which has distinguishable pure states but no continuous reversible path between them.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Classical theory has survived every test except one. A finite classical system has only finitely many reversible transformations, so it cannot change continuously without losing information. The parameter count left classical theory, \(K=N\), as the first of a short list of possibilities.
What if we demand the continuous reversible change that classical theory lacks? We raise it to a principle and ask what it costs. The first consequence is that some outcomes on pure states must be random, with no ignorance behind them, so the state space cannot be a simplex. The second is that the symmetry of reversible change makes the pure states of a bit lie on a sphere. A theorem then shows that the bit is a ball, of a dimension that one system alone cannot fix.
The next chapter fixes that dimension by asking how two bits combine.
Before we can demand continuous reversible change, we need a precise way to speak about all the reversible transformations of a system at once. They form a group, and the principle is a statement about that group. Then we ask why the demand is reasonable, and which theories in our gallery it excludes.
How should we describe all the reversible transformations of a system together? Compose two of them and we get another. The identity is one of them, and each has an inverse. A collection with these three properties is called a group. We call this one \(G\). Each element acts on the state space as a linear map carrying it onto itself, and so, as we found for classical systems, it permutes the pure states.
What else do we know about \(G\)? Its maps preserve a bounded set that spans the state space, so their matrix entries are bounded. We also assume, as is standard, that a limit of reversible transformations is again a reversible transformation. A group of matrices that is bounded and contains its limits in this way is called compact. Compactness will let us average over \(G\).
Now we can state the demand.
Continuous reversibility. For any two pure states there is a reversible transformation taking the first to the second that can be connected to the identity by a continuous path of reversible transformations.
Let \(G_0\) be the part of \(G\) that is connected to the identity by continuous paths. The principle says that \(G_0\) acts transitively on the pure states: any pure state can be carried to any other within it. So the principle combines two ideas. Transitivity says no pure state is special, since all are equivalent up to a reversible change. Connectedness says the change can happen gradually.
Why should reversible change be continuous? Physical evolution unfolds in time. Suppose a system evolves reversibly from time \(0\) to time \(t\). Stopping at intermediate times gives transformations that trace a path from the identity to the final one. If the dynamics depends continuously on time, this path is continuous. So every transformation that arises as continuous time evolution lies in \(G_0\).
And why transitivity? It is a statement of homogeneity. For a fundamental system, any pure preparation can be converted into any other by suitably arranged dynamics. It does not say that every system actually evolves this way, only that the theory allows it.
Which theories fail? Classical theory with finitely many configurations violates the principle, as we saw. So does boxworld, whose reversible transformations are discrete. The square bit fails too. The linear maps of a square onto itself are its eight rotations and reflections, and none except the identity lies on a continuous path.
Problem 6.1. In the disc theory, rotations about the center map the disc onto itself. Show that the disc satisfies continuous reversibility.
A rotation by angle \(\alpha\) is linear, maps the disc onto itself, and is undone by the rotation by \(-\alpha\). Rotation by \(t\alpha\), for \(t\) from \(0\) to \(1\), is a continuous path from the identity. Any pure state on the circle can be rotated to any other, so the rotations act transitively on pure states.
What does the principle cost? Its first consequence needs nothing beyond two distinguishable states. A continuous path between them must pass through pure states with uncertain outcomes. Then we ask whether that uncertainty could be mere ignorance, and find that within the theory it cannot.
Take two perfectly distinguishable pure states and a continuous reversible path from one to the other. What happens to the probability of the outcome that tells them apart along the way?
Theorem. In any theory with \(N\ge2\) that satisfies continuous reversibility, there is a pure state and an effect whose probability on it is exactly \(1/2\).
Proof. Let \(\omega_0\) and \(\omega_1\) be perfectly distinguishable pure states, with an effect \(e\) such that \(e(\omega_0)=0\) and \(e(\omega_1)=1\). By the principle there is a continuous path \(T_t\), \(0\le t\le1\), of reversible transformations with \(T_0\) the identity and \(T_1\omega_0=\omega_1\). Each \(\psi_t=T_t\omega_0\) is pure, because reversible maps send pure states to pure states. The function \(f(t)=e(\psi_t)\) is continuous, being a linear functional applied to a continuously varying state, with \(f(0)=0\) and \(f(1)=1\). By the intermediate value theorem there is a \(t\) with \(f(t)=1/2\). \(\square\)
What else follows? The same argument produces pure states with any probability between \(0\) and \(1\). Each value of \(f\) needs a different pure state, so the theory has infinitely many pure states. Yet it has only finitely many distinguishable ones. So \(K>N\), and the state space is not a simplex.
Could the probability \(1/2\) still be ignorance, as it always is in classical theory? In classical theory a probability of \(1/2\) on a pure state can arise only from a noisy measurement. Here neither escape is open. The effect \(e\) is sharp: it is \(1\) on \(\omega_1\) and \(0\) on \(\omega_0\), so its randomness does not come from detector noise. And \(\psi_t\) is pure, so it is not a mixture of states with definite outcomes. Within the theory, the randomness of the outcome cannot be read as ignorance of a finer state, because there is no finer state.
Does this rule out hidden states altogether? No. The conclusion concerns the theory’s own states. A different theory might add hidden states beneath these, as Bell’s hidden-direction model does. Such a theory has a different state space and must answer to its own principles, and we will see what it must give up when we study contextuality. What we have derived is that randomness without ignorance follows from finiteness and continuous reversibility. The framework did not assume it, since it treats deterministic and random theories alike.
Problem 6.2. In the disc, take \(\omega_0=-\mathbf n\), \(\omega_1=\mathbf n\), the effect \(e_{\mathbf n}(\mathbf r)=\tfrac12(1+\mathbf n\cdot\mathbf r)\), and the path of rotations by angle \(\pi t\). Compute \(f(t)\) and find where it equals \(1/2\).
The rotated state makes angle \(\pi-\pi t\) with \(\mathbf n\), so \(f(t)=\tfrac12(1+\cos(\pi-\pi t))=\tfrac12(1-\cos\pi t)\). It equals \(1/2\) at \(t=1/2\), where the state is perpendicular to \(\mathbf n\).
The second consequence concerns geometry. A group that can carry any pure state to any other makes all pure states look alike. Can we turn that symmetry into a statement about shape? We first find a state that every reversible change leaves fixed, and a notion of distance that every reversible change preserves. The pure states then lie on a sphere, and a theorem completes the picture for a bit.
Is there a state that no reversible change can move? Try averaging. Suppose first that \(G\) is finite, with \(|G|\) elements. Take any pure state \(\omega\) and form
\[\mu=\frac1{|G|}\sum_{g\in G}g\,\omega .\]
This is a mixture, so it is a state. Apply any \(h\) in \(G\). As \(g\) runs over the group, so does \(hg\), so the sum is unchanged and \(h\mu=\mu\). For a compact group with infinitely many elements, the sum becomes an integral against a uniform weighting of the group, its Haar measure. Alfréd Haar and John von Neumann showed in the 1930s that such a weighting exists for every compact group. The same argument then gives a state \(\mu\) that every reversible transformation leaves unchanged. Because \(G\) acts transitively, \(\mu\) is the same whichever pure state we start from. We call it the maximally mixed state.
Can we also find a distance that every reversible change preserves? Average again. Take any inner product \(\langle x,y\rangle\) on the state space and define
\[\langle x,y\rangle_G=\frac1{|G|}\sum_{g\in G}\langle gx,gy\rangle .\]
Then \(\langle hx,hy\rangle_G=\frac1{|G|}\sum_g\langle ghx,ghy\rangle=\langle x,y\rangle_G\), because \(gh\) also runs over the group. So every \(h\) preserves \(\langle\cdot,\cdot\rangle_G\), and the Haar measure gives the same result for compact groups.
Now measure each state by its distance from \(\mu\) in this inner product. A reversible \(T\) fixes \(\mu\) and preserves the inner product, so \(|T\omega-\mu|=|T(\omega-\mu)|=|\omega-\mu|\). Since any pure state can be carried to any other, all pure states lie at the same distance from \(\mu\). So the pure states lie on a sphere.
Does lying on a sphere rule out polygons? Not by itself: the corners of a square also lie on a circle about its centre. Polygons and polyhedra were already excluded by continuity, because their reversible transformations are discrete. What the sphere adds is a constraint on every surviving theory: all its pure states are equally far from \(\mu\). It does not yet say that every point of the sphere is a pure state, or what the sphere’s dimension is.
For a bit, can we say which points of the sphere are pure? Not with the tools of this chapter. The answer needs a deep theorem, and it uses two principles that we state formally only when we study larger systems: equivalence of subspaces, which says that systems with the same \(N\) behave alike, and all measurements allowed, which says that every effect consistent with the states can be measured.
Theorem (Masanes and Müller, 2011). Suppose a theory satisfies
Then the state space of a system with \(N=2\) is a Euclidean ball of some dimension \(d\). Every point of its boundary sphere is a pure state, and every pair of opposite points is perfectly distinguishable.
The proof uses the classification of compact groups that act transitively on spheres, and lies beyond this course. Borivoje Dakić and Časlav Brukner gave a related argument. We take the result as established, with its assumptions stated. Like Masanes and Müller, we first treat a single bit and then use pairs of bits to fix \(d\).
Which balls are candidates? Each \(d\) defines a candidate theory of a bit. For \(d=2\) the bit is the disc, which is the bit of quantum theory with real numbers. For \(d=3\) it is the ball of our gallery, which will turn out to be the bit of ordinary quantum theory. For \(d=5\) it is the bit of quantum theory built on quaternions, a four-dimensional extension of the complex numbers. Every \(d\ge2\) gives a consistent single system that satisfies everything assumed so far. The case \(d=1\), a segment, is the classical bit, already excluded. A single system cannot tell us which ball we inhabit, but pairs of systems can.
Problem 6.3. For the disc and for the three-dimensional ball, what is \(K\), and where is \(\mu\)?
A ball of dimension \(d\) has normalized states with \(d\) free parameters plus normalization, so \(K=d+1\): \(K=3\) for the disc and \(K=4\) for the ball. In both, \(\mu\) is the center, the only point fixed by every rotation.
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 the square bit have, and does it satisfy continuous reversibility?
Reversible maps are the linear symmetries of the square.
A finite group has no nontrivial continuous paths.
Eight: four rotations and four reflections. It violates continuous reversibility, since only the identity can be reached continuously and no pure state can be moved continuously to another.
Where is the maximally mixed state of the disc, and why is it the average of any pure state over all rotations?
Averaging over rotations gives a rotation-invariant state.
Which point of the disc is fixed by every rotation?
It is the center. Averaging a boundary point over all rotations spreads it evenly around the circle, and the average is the unique rotation-invariant point, the center, whichever pure state we start from.
What is \(K\) for the bit of quaternionic quantum theory, whose state space is a ball of dimension \(5\)?
A \(d\)-ball has \(K=d+1\).
Count normalization.
\(K=6\). Counting alone excludes it: \(6\) is not a power of \(2\), while \(K=N^r\) with \(N=2\) requires one.
In the Bloch ball, rotate the north pole \((0,0,1)\) about the \(x\)-axis by angle \(\pi t\). (a) Give the state at time \(t\). (b) Find the probability of the effect \(e_{\mathbf z}\) as a function of \(t\). (c) At what \(t\) does it equal \(0.9\)?
Rotation about \(x\) mixes \(y\) and \(z\).
\(e_{\mathbf z}(\mathbf r)=\tfrac12(1+z)\).
(a) \(\mathbf r(t)=(0,-\sin\pi t,\cos\pi t)\), for one sense of rotation. (b) \(\tfrac12(1+\cos\pi t)=\cos^2(\pi t/2)\). (c) \(\cos\pi t=0.8\) gives \(t=\arccos(0.8)/\pi\approx0.205\). Every probability in \([0,1]\) is reached by some pure state on the path.
Find all reversible transformations of the disc and identify those connected to the identity.
A reversible map is linear and maps the disc onto itself about its center.
Linear maps preserving the unit circle are orthogonal.
They are the linear maps preserving the unit disc: rotations and reflections, the orthogonal group \(O(2)\). The rotations form the connected component \(G_0=SO(2)\); reflections have determinant \(-1\) and cannot be reached continuously from the identity. Rotations alone act transitively on the circle, so the disc satisfies continuous reversibility.
The symmetry group of a triangle (all six permutations of its corners) acts transitively on its pure states, and all corners are at the same distance from the centroid. Does this triangle satisfy continuous reversibility? What does the example show about the argument that pure states lie on a sphere?
Is the permutation group connected?
Equal distance from \(\mu\) is necessary, not sufficient.
No. The group is finite, so only the identity is reachable continuously. The triangle passes the necessary test of the sphere argument, pure states equidistant from \(\mu\), but this test does not imply continuous reversibility. Continuity is what forces a continuum of pure states, and with it a round state space.
Using the intrinsic-randomness theorem, show that a theory satisfying continuous reversibility with \(N\ge2\) has infinitely many pure states, and conclude that \(K>N\).
Different probabilities need different states.
A simplex has exactly \(N\) pure states.
Along the path \(\psi_t\), the probability \(e(\psi_t)\) takes every value in \([0,1]\), and different values belong to different pure states, so there are infinitely many. If \(K=N\), the \(N\) perfectly distinguishable states would span the state space and every state would be a mixture of them, making them the only pure states. So \(K>N\), and the state space is not a simplex.
A student says: “If a pure state gives a random outcome, the theory must be incomplete: there must be more precise states that fix the outcome.” Respond using the argument for randomness without ignorance.
Within the theory, pure states cannot be decomposed.
Adding states changes the theory.
Within the theory, a pure state is not a mixture of other states, so its randomness is not ignorance of a finer state of the theory. One can postulate a different theory with extra hidden states, but it must answer to its own principles; contextuality shows that such states must be contextual, and Bell’s theorem that, for composites, they must be nonlocal. The claim “must be incomplete” assumes determinism, which the course does not assume and does not derive.
Bell’s model reproduces single-qubit statistics with predetermined outcomes. Explain why it does not contradict the intrinsic-randomness theorem.
Which state space does Bell’s model use?
Does the model satisfy continuous reversibility for its own states?
The theorem concerns the pure states of the theory satisfying the principle. Bell’s model has a larger state space, with hidden variables, and its hidden-variable states are deterministic. Those states do not form a theory satisfying continuous reversibility, since no continuous reversible path joins two deterministic states with different outcomes. The quantum states appear in the model only as mixtures, so there is no contradiction.
Explain why transitivity alone, without continuity, does not imply intrinsic randomness. Use the triangle as the example.
Where did the proof of intrinsic randomness use continuity?
Corners of a triangle are deterministic.
The proof used the intermediate value theorem along a continuous path. In the triangle, transitive permutations carry each corner to each other, but in discrete jumps, and every pure state assigns \(0\) or \(1\) to the indicator effects. Transitivity says pure states are equivalent; continuity is what forces pure states in between, and with them randomness.