Quantum Foundations · Chapter 02
What is a state, if it is defined only by what it predicts?
Starting point: Chapter 1: preparations, measurements, and relative frequencies.
Anchor example: A coin, a die, and a polarizer, each described only by outcome probabilities.
Principle at work: Mixing: any two preparations can be combined by a random choice.
Foil: A theory in which mixing is forbidden, so that states need not form a convex set.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We have a language for every testable theory: preparations, transformations, measurements, and the probabilities of their outcomes. A state is whatever a preparation implies about the statistics of future measurements. That definition says what a state does, but not what it looks like.
Can the framework alone tell us anything about the shape of a theory’s states? It can, because the framework lets us choose preparations at random, and ordinary probability then says exactly what such a choice predicts. This forces the states of every theory to form a convex set. A state still assigns a probability to infinitely many outcomes, so we next ask how many numbers are really needed to fix one. Our first principle answers that only finitely many are. Together, mixing and finiteness force every outcome probability to be a linear function of a finite list of numbers, so that states become vectors.
The next chapter turns from states to the outcomes that measure them.
The only freedom the framework gives us beyond running recipes is choosing among them by chance. We first work out what such a chance choice predicts, which shows that states form a convex set. Then we ask which states cannot be produced by chance at all.
Suppose we have two preparations and a number \(\lambda\) between \(0\) and \(1\). Can we build a third? We can: run the first with probability \(\lambda\) and the second otherwise, using an independent random device, and discard the record of which was run. The random device is itself a laboratory procedure, so the framework allows this. We take it as part of the framework that such mixing is always possible.
What does the mixture predict? Fix any outcome of any measurement, and let its probabilities be \(p_1\) and \(p_2\) for the two preparations. Among many runs of the mixture, a fraction \(\lambda\) used the first preparation and a fraction \(1-\lambda\) used the second. Within each group the outcome occurs with its own probability. So overall it occurs with probability
\[p_{\text{mix}}=\lambda p_1+(1-\lambda)p_2 .\]
This is the law of total probability, applied to a recipe, and it holds for every outcome of every measurement at once. So if we describe states by their outcome probabilities, the state of the mixture is the weighted average of the two states. A set that contains every weighted average of any two of its members is called convex. The states of every theory in the framework form a convex set.
Mixing produces new states from old ones. Are there states that no mixing can produce? Call a state mixed if it is a mixture of two different states with \(0<\lambda<1\), and pure otherwise. A pure state is a preparation that cannot be understood as ignorance about which of several other preparations was made. Geometrically, pure states are the extreme points of the convex set: the corners of a polygon, or every point on the boundary circle of a disc.
Take a coin first. Its state is the probability \(p\) of heads, a point of the interval \([0,1]\). The only pure states are \(p=0\) and \(p=1\). Every other state is a mixture of them in exactly one way: \(p\) of heads-for-certain and \(1-p\) of tails-for-certain. For a die, the states are probability lists \((p_1,\dots,p_6)\), and the pure states are the six certain outcomes. Again each state is a mixture of pure states in exactly one way.
Is that uniqueness forced by the framework? No. Picture a theory whose states fill a disc, with its pure states on the circle. The center is the even mixture of any two opposite points on the circle, and there are infinitely many such pairs. A state records only predictions, so all these recipes produce the same state. No measurement can reveal which mixture was used. We will find that quantum theory is of this kind, and that this costs us something in what can be learned and copied.
Problem 2.1. A preparation gives probabilities \((0.9,0.1)\) for a two-outcome measurement, another gives \((0.3,0.7)\). What does the mixture with \(\lambda=1/4\) on the first give? Can the state \((0.95,0.05)\) be obtained by mixing these two?
The mixture gives \(0.25(0.9)+0.75(0.3)=0.45\) for the first outcome and \(0.55\) for the second. No mixture of the two gives \(0.95\): weighted averages lie between \(0.3\) and \(0.9\). Mixing only moves inward, which is why pure states cannot be produced from mixtures.
A state assigns a probability to every outcome of every possible measurement. That is a list with infinitely many entries, and no laboratory can check infinitely many numbers. So we ask whether a finite list can stand in for the whole, which leads to our first principle. Then we count how long that list must be.
Could a finite laboratory ever confirm which state it has prepared, if a state needed infinitely many numbers? It could not. Every estimate of a probability takes many runs, and a laboratory can run only finitely many experiments. If no finite set of probabilities ever settled the state, two preparations could agree on every test we had made and still differ on the next one. Identifying a state would never be finished.
So suppose instead that some finite set of tests always suffices. Then a state is completely described by finitely many numbers. We raise this to a principle.
Finiteness (first part). For each type of system there is a finite set of measurement outcomes, called fiducial outcomes, whose probabilities determine the probability of every outcome of every measurement.
The argument above makes the principle reasonable but does not prove it. The framework allows theories that need ever more numbers. A particle that can sit anywhere on a line is described in quantum theory by a function, which no finite list fixes, and we return to such systems at the end of the course. Finiteness is a demand that we place on systems with finitely many distinguishable configurations, and that foil shows what it excludes.
Estimating the fiducial probabilities from repeated runs, and so identifying an unknown state, is called tomography. For a coin, one fiducial outcome suffices: the probability of heads fixes everything. For the disc theory, one two-outcome measurement does not suffice. Two states on the circle can give the same probability for that measurement and differ for another.
How many fiducial numbers does a system need? To count them cleanly, we allow preparations that sometimes fail. Suppose a source emits a system only with probability \(s\), and records failures as “no system.” Its outcome probabilities are then \(s\) times those of a successful run. Such subnormalized states form a cone: multiplying a state by a number between \(0\) and \(1\) gives another. The probabilities of all outcomes of a measurement on such a state add up to \(s\), the probability that the system is there.
Let \(K\) be the smallest number of fiducial probabilities that fix every subnormalized state. This counts normalization as one of the numbers, because \(s\) itself must be specified. For normalized states one number is fixed, and \(K-1\) remain free. A coin has \(K=2\): the probabilities of heads and of tails, which for a normalized coin sum to one. A die has \(K=6\). A classical system with \(N\) configurations has \(K=N\).
Could a cleverer choice of fiducial outcomes give a smaller count? We will see that it cannot: once states are vectors, \(K\) is the dimension of the space they span, so every minimal fiducial set has the same size. Later we compare \(K\) with \(N\), the number of states that one measurement can tell apart perfectly. The relation between these two numbers will separate classical theory from quantum theory.
Problem 2.2. A system admits two two-outcome measurements, X and Z. Experiments show that the probability of the first outcome of X and of Z can be any two numbers in \([0,1]\), independently, and that these two numbers fix every other prediction. What is \(K\)? What is the shape of the set of normalized states?
A subnormalized state needs the success probability and the two first-outcome probabilities, so \(K=3\). Normalized states correspond to all pairs \((p_X,p_Z)\in[0,1]^2\): the state space is a square. This is the square bit, a foil we meet again when we study measurements and pairs of systems.
We now have two facts about states: chance combinations average them, and a finite list fixes them. What do the two together say about the probability of an arbitrary outcome? They force it to be a linear function of the list, without any probability rule assumed. That turns each theory into a convex set of vectors, and we then compare the shapes such sets can take.
Write a state as its list of fiducial probabilities, a vector \(\omega\) in \(\mathbb R^K\). Any other outcome \(e\) has a probability that depends only on the state, so it is some function \(e(\omega)\). What kind of function can it be? The mixing rule applies to every outcome, fiducial or not, so
\[e\big(\lambda\omega_1+(1-\lambda)\omega_2\big)=\lambda\,e(\omega_1)+(1-\lambda)\,e(\omega_2).\]
A function on a convex set with this property is called affine. Subnormalized states add one more condition. Running a preparation with success probability \(s\) scales all probabilities, so \(e(s\omega)=s\,e(\omega)\).
Does an affine function with this scaling extend to a linear function on all of \(\mathbb R^K\)? First extend it to every nonnegative multiple \(t\omega\) by \(e(t\omega)=t\,e(\omega)\). Then \(e\) adds on the cone: for normalized \(\omega_1,\omega_2\) and \(a,d\geq0\), the vector \(a\omega_1+d\omega_2\) is \((a+d)\) times a mixture, so mixing and scaling give \(e(a\omega_1+d\omega_2)=a\,e(\omega_1)+d\,e(\omega_2)\). Now define \(e\) on any difference \(v=a\omega_1-b\omega_2\) by \(a\,e(\omega_1)-b\,e(\omega_2)\). Could two ways of writing \(v\) give different values? Suppose \(a\omega_1-b\omega_2=c\omega_3-d\omega_4\). Rearranging gives \(a\omega_1+d\omega_4=c\omega_3+b\omega_2\), an equality between vectors of the cone. Additivity on the cone gives \(a\,e(\omega_1)+d\,e(\omega_4)=c\,e(\omega_3)+b\,e(\omega_2)\), which rearranges to the same value both ways. So every outcome probability is given by a linear function on \(\mathbb R^K\), a linear functional.
What has this excluded? Suppose a theory claimed that an outcome had probability \(p^2\), where \(p\) is a fiducial probability. Mix a state with \(p=0\) and one with \(p=1\) equally. The recipe then gives the outcome probability \(\tfrac12(0)+\tfrac12(1)=\tfrac12\), but the rule gives \((1/2)^2=1/4\). No consistent theory in the framework has such a rule.
What does a theory look like now? Every theory satisfying finiteness has the same outline. The normalized states form a convex set \(\Omega\) in a \((K-1)\)-dimensional slice of \(\mathbb R^K\), and the subnormalized states form the cone of its multiples. Outcomes act as linear functionals that take values between \(0\) and \(1\) on every state. At this level, a theory is specified by the shape of \(\Omega\) and by which functionals are real outcomes.
This also settles the question left open about \(K\). If the states of a minimal fiducial set did not span \(\mathbb R^K\), some linear relation would hold among the fiducial probabilities of every state, and one of them could be dropped. So the states span \(\mathbb R^K\). Any other fiducial set gives probabilities that are linear in \(\omega\) and still distinguish all states. Could such a linear map send some nonzero vector \(v\) to zero? Take a state \(\omega\) in the interior of the cone. Then \(\omega+\varepsilon v\) is also in the cone for small \(\varepsilon\), and the map would not tell it from \(\omega\). So the map is one-to-one on all of \(\mathbb R^K\), and it needs at least \(K\) outputs. Every minimal fiducial set therefore has exactly \(K\) members, the dimension of the space spanned by the states.
Which shapes are possible? Several are already on the table. The classical bit has a segment, the classical trit a triangle, and a classical \(N\)-level system a simplex with \(N\) corners. The square bit has a square. A disc and a three-dimensional ball are also convex sets, with smooth boundaries. Each shape defines a different theory with different predictions. The framework permits all of them, and the principles will select one.
Is this linearity a hint of waves or amplitudes underneath? No. It is a property of probabilities as functions of the state, derived from mixing alone. Any theory in the framework has it, whatever lies underneath.
Problem 2.3. In the square bit, an outcome \(e\) has probability \(e(p_X,p_Z)\) on normalized states. Suppose \(e(0,0)=0\), \(e(1,0)=0.5\), and \(e(0,1)=0.5\). What is \(e(1,1)\), and what is \(e(0.5,0.5)\)?
An affine function on the plane has the form \(a+bp_X+cp_Z\). The data give \(a=0\), \(b=c=0.5\), so \(e(1,1)=1\) and \(e(0.5,0.5)=0.5\). The value at \((1,1)\) is forced by linearity once three non-collinear points are known.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Three preparations give an outcome with probabilities \(0.1\), \(0.6\), and \(0.9\). They are mixed with weights \(0.2\), \(0.3\), \(0.5\). What probability does the mixture give?
Apply the law of total probability.
Weight each probability by how often its preparation is used.
\(0.2(0.1)+0.3(0.6)+0.5(0.9)=0.02+0.18+0.45=0.65\). The rule depends only on the recipe, so it holds in every theory.
What is \(K\) for a die, and how many free parameters does a normalized state of a die have? What is \(K\) for a pair of classical bits?
For a classical system \(K=N\).
Normalization fixes one number.
A die has \(K=6\) and \(5\) free parameters. A pair of classical bits is a classical system with \(4\) configurations, so \(K=4\).
Which points of a triangle are pure states? Which points of a disc? Is the midpoint of an edge of the triangle pure?
Pure means not a mixture of two different states.
A point strictly between two states is a mixture.
In a triangle only the three corners are pure; the midpoint of an edge is the even mixture of its two corners. In a disc every point of the boundary circle is pure, so there are infinitely many pure states.
(a) Decompose the classical trit state \((0.2,0.3,0.5)\) into pure states and explain why the decomposition is unique. (b) For the disc, write its center as an even mixture of pure states in two different ways.
The pure states of a trit are \((1,0,0)\), \((0,1,0)\), \((0,0,1)\).
Any diameter of a disc has the center as its midpoint.
(a) \((0.2,0.3,0.5)=0.2(1,0,0)+0.3(0,1,0)+0.5(0,0,1)\). The weights must equal the coordinates, since each pure state contributes only to its own coordinate, so the decomposition is unique. (b) The center is the midpoint of the horizontal diameter and also of the vertical one: \(\tfrac12(1,0)+\tfrac12(-1,0)=\tfrac12(0,1)+\tfrac12(0,-1)\). The two recipes prepare the same state, and no measurement can reveal which was used.
Show that if a set contains every mixture of any two of its members, it contains every mixture \(\lambda_1\omega_1+\lambda_2\omega_2+\lambda_3\omega_3\) of three, with nonnegative weights summing to one.
Group the last two terms.
Assume \(\lambda_1<1\).
If \(\lambda_1=1\) the mixture is \(\omega_1\). Otherwise write it as \(\lambda_1\omega_1+(1-\lambda_1)\big(\tfrac{\lambda_2}{1-\lambda_1}\omega_2+\tfrac{\lambda_3}{1-\lambda_1}\omega_3\big)\). The bracket is a two-state mixture, hence in the set, and the whole is a two-state mixture of \(\omega_1\) and the bracket. Operationally: pick \(\omega_1\) with probability \(\lambda_1\); otherwise choose between the other two with the conditional weights.
A source emits a coin-like system with probability \(0.8\), and when it does, the probability of heads is \(0.25\). (a) Write the subnormalized state as the pair (probability of heads, probability of tails). (b) What is the unit effect’s value? (c) Mix this source equally with a perfect source of heads. What is the result?
Subnormalized probabilities are the success probability times the conditional ones.
Mixing averages the vectors.
(a) \((0.8\times0.25,\ 0.8\times0.75)=(0.2,0.6)\). (b) \(u=0.2+0.6=0.8\), the probability that a system is present. (c) \(\tfrac12(0.2,0.6)+\tfrac12(1,0)=(0.6,0.3)\), with total \(0.9\). Including subnormalized states lets us treat “no system” uniformly and makes \(K\) count normalization as one of the parameters.
For polarized light, a polarizer at one angle passes a fraction \(\cos^2\alpha\) of light polarized at angle \(\alpha\) to it. (a) Find two different linear polarizations that a single polarizer at \(0^\circ\) cannot tell apart statistically. (b) Experiments show that transmission at \(0^\circ\), at \(45^\circ\), and through a circular-polarization filter, together with intensity, fix all other predictions. What is \(K\)?
Try \(\pm\alpha\).
Count all fiducial numbers, including normalization.
(a) Polarizations at \(+30^\circ\) and \(-30^\circ\) both pass with probability \(\cos^230^\circ=0.75\), yet a polarizer at \(30^\circ\) passes one fully and the other with probability \(\cos^260^\circ=0.25\). One setting is not a fiducial set. (b) Three transmission probabilities plus normalization: \(K=4\), the value derived for the qubit.
A student says: “A mixed state means the system is really in one of the pure states of the mixture, and we just don’t know which.” Using the disc, explain why the state itself does not support this reading.
How many decompositions does the center of the disc have?
A state records only predictions.
The center of the disc is an even mixture of every pair of antipodal pure states. If it meant “really in one of the pure states of the mixture,” we would have to say which pair, but all pairs give identical predictions for every measurement, so the state contains no fact about it. The ignorance reading is consistent only for a particular preparation recipe, not for the state. Classical theory hides this issue because its decompositions are unique.
Let \(f\) be a function on \([0,1]\) with values in \([0,1]\) satisfying \(f\big(\tfrac{p+q}2\big)=\tfrac12f(p)+\tfrac12f(q)\) for all \(p,q\). Show that \(f(p)=f(0)+\big(f(1)-f(0)\big)p\) at every dyadic rational \(p=k/2^m\), and explain why boundedness then forces \(f\) to be affine everywhere.
Use induction on \(m\), halving intervals.
A midpoint-affine function that is bounded on an interval is continuous.
The formula holds at \(0\) and \(1\). If it holds at all multiples of \(2^{-m}\), each multiple of \(2^{-(m+1)}\) is a midpoint of two of them, where the midpoint rule gives the same linear value. So it holds at all dyadic rationals. A midpoint-affine function bounded on an interval is continuous (a classical result for Jensen’s equation), and dyadic rationals are dense, so \(f\) is affine. This is the one-dimensional core of the linearity of probabilities.
A theory has two two-outcome measurements, X and Z, and its normalized states are the points \((p_X,p_Z)\) with \((p_X-\tfrac12)^2+(p_Z-\tfrac12)^2\le\tfrac14\). (a) What is \(K\)? (b) If \(p_X=1\), what are the possible values of \(p_Z\)? (c) Compare with the square bit.
The state space is a disc inscribed in the unit square.
Substitute \(p_X=1\).
(a) \(K=3\). (b) \(p_X=1\) forces \((p_Z-\tfrac12)^2\le0\), so \(p_Z=\tfrac12\): certainty about X implies complete randomness for Z. (c) In the square bit both outcomes can be certain at once, at a corner. The disc builds a trade-off between X and Z into the geometry of the state space, a first glimpse of uncertainty relations.