Quantum Foundations · Chapter 03
What does a measurement outcome do to a state, mathematically, and how many states can be told apart?
Starting point: Chapter 2: convex state spaces, fiducial measurements, and linearity.
Anchor example: Four state spaces with two perfectly distinguishable states: a segment, a square, a disc, and a ball.
Principle at work: Outcome probabilities are linear functions of the state, and the outcomes of a measurement exhaust all possibilities.
Foil: The square bit (gbit), whose state space is a square.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
States have become vectors. Every theory satisfying finiteness describes its states as a convex set in some \(\mathbb R^K\), and every outcome probability as a linear function on that set. So far we have looked only at the states.
But a theory is tested through its measurements, and the same linear functions that gave probabilities can be studied as objects of their own. Once outcomes are objects, we can ask what a single run of one system can reveal. Answering that needs a second number besides \(K\): the largest number of states one measurement can tell apart without error. The two numbers need not agree, and how they differ is the first sign that a theory is not classical. With both numbers in hand, we can lay out a small gallery of theories whose states and measurements we know completely.
The next chapter uses the gallery to ask how two systems combine.
Every outcome’s probability is a linear function of the state. Can we turn this around, and treat the functions themselves as the theory’s measurements? We first identify outcomes that no statistics can tell apart, which gives effects. Then we ask what a whole measurement is, in terms of its effects.
When are two outcomes, perhaps of different measurements, really the same? Suppose they have the same probability on every state. Then no statistics can tell them apart, and, as we did for preparations, we identify them. The resulting class is called an effect. Since outcome probabilities are linear, an effect is a linear functional \(e\) on the space spanned by the states, with \(0\le e(\omega)\le1\) for every normalized state \(\omega\).
Can effects be mixed, as states can? Suppose we run one measurement with probability \(\lambda\) and another otherwise, then report a chosen outcome of whichever was run. This realizes the effect \(\lambda e_1+(1-\lambda)e_2\). So effects form a convex set too.
Take the classical bit, with state vector \((p,1-p)\). The effect “heads” is the functional \((p,1-p)\mapsto p\), and “tails” is \((p,1-p)\mapsto1-p\). Now suppose a noisy detector reports heads correctly with probability \(0.9\), and falsely, on tails, with probability \(0.2\). It realizes the effect \(0.9p+0.2(1-p)\). So effects describe imperfect measurements as well as ideal ones.
How can we tell an ideal effect from a noisy one? The noisy effect above never reaches \(0\) or \(1\): it lies between \(0.2\) and \(0.9\) on every state. The effect “heads” reaches \(1\) on one state and \(0\) on another. We call an effect sharp if it takes the value \(1\) on some state and the value \(0\) on some other state. A sharp effect gives a certain answer on at least one state each way, so its randomness on other states cannot come from noise in the detector alone.
An effect is one outcome. What makes a list of effects into a whole measurement? One effect occurs in every theory: “some outcome occurs.” Its probability is the total probability of any complete measurement, which is \(1\) on normalized states and \(s\) on a state of success probability \(s\). We call it the unit effect \(u\). Every complete measurement has total probability \(u(\omega)\), so a measurement with outcomes \(1,\dots,m\) is a list of effects with
\[e_1+e_2+\dots+e_m=u .\]
Can we merge outcomes? Reporting “1 or 2” realizes the effect \(e_1+e_2\). This is called coarse graining. Every effect \(e\) also has a complement \(u-e\). Can every effect then be completed to a two-outcome measurement \(\{e,u-e\}\)? Only if the theory makes that measurement available. Whether it does is a genuine question, and one of our principles will settle it. Until then we work with the effects a theory provides, and we say when an argument needs more.
Problem 3.1. For the classical bit, is the functional \((p,1-p)\mapsto 1.2p-0.1\) an effect? What about \((p,1-p)\mapsto0.5\)?
The first takes the value \(-0.1\) at \(p=0\) and \(1.1\) at \(p=1\), outside \([0,1]\), so it is not an effect. The second is an effect: it is realized by ignoring the system and tossing a fair coin. It equals \(u/2\).
A state space says what preparations predict over many runs. But a message is read from one run. So we ask which states a single run can tell apart, and find that such states are always linearly independent. Their largest number is a second measure of a system, which we then compare with \(K\).
How many different messages can one run of one system carry? Suppose a sender prepares one of the states \(\omega_1,\dots,\omega_n\) to send symbol \(j\). The receiver can decode without error only if some measurement, with effects \(e_1,\dots,e_n\), satisfies \(e_i(\omega_j)=1\) when \(i=j\) and \(0\) otherwise. States with such a measurement are called perfectly distinguishable. Each effect \(e_i\) is then sharp.
How many states can be perfectly distinguishable? Suppose \(c_1\omega_1+\dots+c_n\omega_n=0\). Applying the linear functional \(e_i\) to both sides gives \(c_i=0\), for each \(i\). So perfectly distinguishable states are linearly independent. A \(K\)-dimensional space holds at most \(K\) linearly independent vectors, so at most \(K\) states can be perfectly distinguishable.
We now have a bound, \(K\). Is it reached, and does the number of messages deserve its own name? Give it one, and state its finiteness as the second part of our first principle.
Finiteness (second part). For each type of system, the largest number \(N\) of perfectly distinguishable states is finite.
We have just shown that \(N\le K\), so the second part follows from the first. We state it separately because \(N\) has its own meaning. It is the number of symbols one system can carry with no error, so \(\log_2N\) is its capacity in bits. A system with \(N=2\) is the simplest nontrivial system of a theory, called a bit of that theory. Two theories can both have bits and yet differ in \(K\), in the shape of their states, and in everything else.
What does a gap between \(K\) and \(N\) mean? \(K\) counts the numbers needed to describe a state, and \(N\) counts the symbols it can carry. In a classical theory \(K=N\): each state is a probability list over \(N\) distinguishable configurations, and needs no more numbers than that. Suppose instead \(K>N\). Then the states are not all mixtures of one distinguishable set. Their description contains information that cannot be read out in one run. We will find that the principles force \(K=N^2\).
Problem 3.2. In the disc theory, every pure state has an antipode from which it is perfectly distinguishable. Show that no three states of the disc are perfectly distinguishable, so \(N=2\).
Write a state as a vector \(\mathbf r\) with \(|\mathbf r|\le1\); an affine functional is \(e(\mathbf r)=a+\mathbf b\cdot\mathbf r\). Its largest and smallest values on the disc are \(a+|\mathbf b|\) and \(a-|\mathbf b|\), so it is an effect exactly when \(a-|\mathbf b|\ge0\) and \(a+|\mathbf b|\le1\). Suppose \(e(\mathbf r_1)=1\) and \(e(\mathbf r_2)=0\). The first forces \(a+|\mathbf b|=1\) with \(\mathbf r_1\) the unit vector along \(\mathbf b\); the second forces \(a-|\mathbf b|=0\) with \(\mathbf r_2\) the unit vector opposite to \(\mathbf b\). Hence any two perfectly distinguishable states are antipodal pure states. Three states would need to be pairwise antipodal, which is impossible, so \(N=2\).
So far the framework allows any closed and bounded convex set as a state space. To see what the principles will exclude, we need examples whose states and effects we know completely. We take a few with small \(K\) and ask of each the same question: which pure states can one measurement separate? Classical probability theory answers one way. The other shapes answer differently, and they become the foils against which the principles are tested.
Which pure states of a classical system can one measurement separate? Start with the shapes. The classical bit has states \((p,1-p)\), a segment with two pure ends, and \(K=N=2\). The classical trit has states \((p_1,p_2,p_3)\), a triangle with three pure corners, and \(K=N=3\). A classical system with \(N\) configurations has a simplex with \(N\) corners, the higher-dimensional version of a triangle. Every state is a mixture of corners in exactly one way.
What are its effects? Every effect is a combination of the indicator effects “configuration \(j\).” So an effect is a list of numbers \(f_j\) between \(0\) and \(1\), with probability \(\sum_jf_jp_j\). The indicators separate all the corners in one measurement. So every pure state is perfectly distinguishable from all the others at once. Classical pure states never overlap.
Can a theory have more pure states than distinguishable ones? The square bit can. Its normalized states are \((p_X,p_Z)\in[0,1]^2\), with \(K=3\) and four pure states at the corners. Which pairs can one measurement separate? Every pair: \((0,0)\) and \((1,1)\) are separated by the effect \(p_X\), and so are \((0,0)\) and \((1,0)\). Can three corners be separated at once? Any three corners include one, say \((0,0)\), whose two neighbours are the other two. An effect equal to \(1\) at it and \(0\) at both neighbours must be the affine function \(1-p_X-p_Z\), which is \(-1\) at the fourth corner. So no measurement separates three corners, and \(N=2\). The square bit has more pure states than distinguishable ones, but no two of its pure states overlap partially. It will reappear as the single system of a foil theory for pairs.
What happens when the boundary is smooth? Take the disc or the ball. Describe a state by a vector \(\mathbf r\) with \(|\mathbf r|\le1\), in two or three dimensions; pure states have \(|\mathbf r|=1\). For each unit vector \(\mathbf n\), consider the functional
\[e_{\mathbf n}(\mathbf r)=\tfrac12\,(1+\mathbf n\cdot\mathbf r).\]
It lies between \(0\) and \(1\) on every state, and \(e_{\mathbf n}+e_{-\mathbf n}=u\). It equals \(1\) at the pure state \(\mathbf n\) and \(0\) at \(-\mathbf n\), so opposite pure states are perfectly distinguishable. What does it give on a pure state at angle \(\theta\) from \(\mathbf n\)? It gives \(\tfrac12(1+\cos\theta)\), which lies strictly between \(0\) and \(1\) unless \(\theta\) is \(0\) or \(\pi\). So here, unlike the square, neighbouring pure states overlap. Since \(\tfrac12(1+\cos\theta)=\cos^2(\theta/2)\), this is the dependence on angle we met in the two-device experiment, appearing as pure geometry of a ball. The disc has \(K=3\) and the ball \(K=4\); both have \(N=2\).
Problem 3.3. In the square bit, try to distinguish the three corners \((0,0)\), \((1,0)\), \((0,1)\): find the affine function equal to \(1\) at \((0,0)\) and \(0\) at the other two, and check whether it is an effect.
The function is \(1-p_X-p_Z\). At the fourth corner \((1,1)\) it equals \(-1\), so it is not an effect. No measurement singles out \((0,0)\) against both other corners, consistent with \(N=2\).
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
In the disc theory, what probability does the effect \(e_{\mathbf n}\) give on a pure state at \(60^\circ\) from \(\mathbf n\)? At \(120^\circ\)?
\(e_{\mathbf n}(\mathbf r)=\tfrac12(1+\mathbf n\cdot\mathbf r)\).
\(\cos60^\circ=\tfrac12\).
At \(60^\circ\): \(\tfrac12(1+\tfrac12)=\tfrac34\). At \(120^\circ\): \(\tfrac12(1-\tfrac12)=\tfrac14\). These equal \(\cos^2(30^\circ)\) and \(\cos^2(60^\circ)\).
A detector reports “heads” correctly with probability \(0.9\) when the coin is heads and falsely with probability \(0.2\) when it is tails. Write its effect, its complement, and the probability of “heads” for a fair coin.
An effect on the classical bit is \(f_Hp+f_T(1-p)\).
The complement is \(u-e\).
\(e(p)=0.9p+0.2(1-p)\); the complement is \(0.1p+0.8(1-p)\). For \(p=\tfrac12\), \(e=0.55\). Effects describe imperfect detectors as naturally as ideal ones.
The three-dimensional ball has \(K=4\). What upper bound on \(N\) does \(N\le K\) give, and what is the actual value?
Distinguishable states are linearly independent.
In a ball, only antipodal states are perfectly distinguishable.
\(N\le4\), but the actual value is \(N=2\): two perfectly distinguishable states must be antipodal pure states, and three points cannot be pairwise antipodal. The bound \(N\le K\) is tight only for classical systems.
(a) Prove that perfectly distinguishable states are linearly independent. (b) Conclude \(N\le K\), and check that the classical trit achieves equality.
Apply the distinguishing effect \(e_i\) to a vanishing linear combination.
The trit’s corners are perfectly distinguishable.
(a) If \(\sum_jc_j\omega_j=0\), applying the linear functional \(e_i\) gives \(\sum_jc_je_i(\omega_j)=c_i=0\) for each \(i\). (b) At most \(K\) linearly independent vectors fit in a \(K\)-dimensional space, so \(N\le K\). The trit’s three corners are perfectly distinguishable and \(K=3\), so \(N=K=3\).
Characterize all effects of the disc theory, \(e(\mathbf r)=a+\mathbf b\cdot\mathbf r\), and find the range of probabilities of the effect with \(a=0.3\), \(|\mathbf b|=0.3\).
The extreme values on the disc are \(a\pm|\mathbf b|\).
Both must lie in \([0,1]\).
An affine functional is an effect exactly when \(0\le a-|\mathbf b|\) and \(a+|\mathbf b|\le1\), that is, \(|\mathbf b|\le\min(a,1-a)\). For \(a=|\mathbf b|=0.3\) the probabilities range over \([0,0.6]\): zero at the pure state opposite to \(\mathbf b\), and \(0.6\) at the pure state along \(\mathbf b\). Its complement ranges over \([0.4,1]\).
In the square bit, decide which are effects: (a) \((p_X+p_Z)/2\); (b) \(p_X+p_Z\); (c) \(p_Xp_Z\); (d) \(1-p_X\).
An effect is affine and takes values in \([0,1]\) at all four corners.
An affine function on a square is extreme at the corners.
(a) Affine with corner values \(0,\tfrac12,\tfrac12,1\): an effect. (b) Equals \(2\) at \((1,1)\): not an effect. (c) Not affine, so not an effect, although it lies in \([0,1]\); mixing \((0,0)\) and \((1,1)\) evenly would require \(\tfrac12\) but the formula gives \(\tfrac14\). (d) An effect, the complement of \(p_X\).
Two pure states of the ball have Bloch vectors \(\mathbf r_1\), \(\mathbf r_2\) at angle \(\theta\). With equal priors, a two-outcome measurement \(\{e_{\mathbf n},e_{-\mathbf n}\}\) guesses state 1 on outcome \(\mathbf n\). Find the best success probability over \(\mathbf n\) and evaluate it for \(\theta=90^\circ\).
Success is \(\tfrac12[e_{\mathbf n}(\mathbf r_1)+e_{-\mathbf n}(\mathbf r_2)]\).
\(|\mathbf r_1-\mathbf r_2|=2\sin(\theta/2)\).
Success is \(\tfrac12\big[\tfrac12(1+\mathbf n\cdot\mathbf r_1)+\tfrac12(1-\mathbf n\cdot\mathbf r_2)\big]=\tfrac12+\tfrac14\mathbf n\cdot(\mathbf r_1-\mathbf r_2)\), largest for \(\mathbf n\) along \(\mathbf r_1-\mathbf r_2\): \(\tfrac12+\tfrac12\sin(\theta/2)\). For \(\theta=90^\circ\), about \(0.854\). Only \(\theta=180^\circ\) gives certainty, which is perfect distinguishability.
A student claims: “If two states give different probabilities for some outcome, a measurement of that outcome tells them apart.” Correct the claim using the states \(p=0.4\) and \(p=0.6\) of a classical bit.
Distinguish single-shot from statistical discrimination.
What does one run reveal?
One run gives heads with probability \(0.4\) or \(0.6\), so guessing from a single outcome succeeds only \(60\%\) of the time. The states are distinguishable statistically, given many copies, but not perfectly distinguishable. Perfect distinguishability requires an effect equal to \(1\) on one state and \(0\) on the other, which is what \(N\) counts.
The extreme effects of a classical trit are the indicator effects of subsets of its configurations. How many are there? Show that every effect is a mixture of them.
A subset of three configurations: \(2^3\) choices.
An effect is a list \((f_1,f_2,f_3)\) in \([0,1]^3\).
There are \(2^3=8\), including the zero effect and the unit effect. Effects of the trit are points \((f_1,f_2,f_3)\) of the unit cube, whose corners are these indicators, and every point of a cube is a mixture of its corners. So every classical effect is a randomized choice among sharp tests.
Show that any two distinct pure states of the square bit are perfectly distinguishable, while two distinct pure states of the disc are perfectly distinguishable only if they are antipodal. What does this suggest about copying pure states in each theory?
Check adjacent and opposite corners separately.
Partial overlap of pure states is what makes copying impossible in quantum theory.
Adjacent corners differ in one coordinate, distinguished by \(p_X\) or \(p_Z\); opposite corners differ in both. So all six pairs are distinguishable. In the disc, only antipodal pairs are. In the square, a single measurement can identify which of any two pure states is present, so copying a pure state known to be one of two is possible; in the disc, pure states overlap partially, the situation in which quantum theory forbids cloning.