Quantum Foundations · Chapter 08
How do qubits determine the description of every finite system?
Starting point: Chapter 7: the qubit, its density matrices, and the probability rule.
Anchor example: A spin-1 particle with three perfectly distinguishable states.
Principle at work: Equivalence of subspaces, and the availability of all measurements consistent with the state space.
Foil: Theories with restricted measurements, and Hardy’s classical branch K = N.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We now know the qubit completely. Its states form the Bloch ball, they are written as complex \(2\times2\) density matrices, and the probability of an outcome is a trace.
But most systems are not qubits. An atom has many levels, and a pair of qubits has four distinguishable states. Does the qubit fix the description of these larger systems too? It does if every pair of alternatives inside a larger system behaves like a qubit, and that is the content of a further principle. With it, counting gives \(K=N^2\) for every system, and a reconstruction theorem turns this count into complex Hilbert space, for single systems and for pairs. One question then remains: which of the mathematically possible measurements are physically available? The last principle answers it, and the general probability rule follows.
The next chapter asks how these states change in time.
How can a qubit tell us anything about a system with three or more distinguishable states? Only if a larger system contains qubits. We first find the parts of a state space that hold fewer alternatives, called faces. Then we state the principle that makes every face behave like a smaller system, and look at what it says about the faces that hold two alternatives.
Where, inside a larger system, could a smaller one sit? Let \(\omega_1,\dots,\omega_N\) be perfectly distinguishable pure states with distinguishing effects \(e_1,\dots,e_N\). Choose a subset, say the first \(k\). Consider the states that give probability zero to all of \(e_{k+1},\dots,e_N\). They form a convex set. Suppose a mixture lies in this set. A mixture can give probability zero only if every component does, so each component lies in the set too. A convex subset closed under decomposition in this way is called a face. Physically, it is the set of states that pass a filter blocking the last \(N-k\) outcomes with certainty.
Must a face behave like a system with \(k\) alternatives? The framework does not say. We raise the answer “yes” to a principle.
Equivalence of subspaces. Every face generated by \(k\) perfectly distinguishable pure states, with the effects and reversible transformations that act within it, is equivalent to the state space of a system with \(N=k\). In particular, all systems with the same \(N\) are equivalent.
The principle says that a system confined to \(k\) of its alternatives behaves like any other system with \(k\) alternatives. It retains no trace of being part of something larger. Hardy called it the subspace axiom. It is the general form of the counting assumption we used to treat \(K\) as a function of \(N\) alone.
What would its failure look like? Imagine a theory in which the two-outcome faces of a three-level system were squares, while its genuine bits were balls. Then how a message behaved would depend on what else its carrier could have done. We give no worked consistent foil of this kind, so here we can only describe what the principle excludes, not exhibit a complete rival theory.
What does the principle say about the faces with two alternatives? Any two perfectly distinguishable pure states, together with all states that pass the corresponding filter, form a Bloch ball. Take a three-level system, with distinguishable states labelled \(1\), \(2\), \(3\). Each of the three pairs spans a qubit face. It is a ball whose poles are the two chosen states, and whose equator consists of pure states that give each of the two outcomes with probability one half.
Do these faces fill the state space? No. The faces overlap only in mixtures of their shared states. A pure state can give all three outcomes with nonzero probability, so it lies in none of the two-outcome faces. How, then, do three qubit faces fit into one state space?
Problem 8.1. In a system with \(N=4\), how many two-outcome faces are generated by a fixed set of four distinguishable pure states? How many three-outcome faces?
Choosing \(2\) of \(4\) gives \(6\) qubit faces, and choosing \(3\) of \(4\) gives \(4\) faces equivalent to a system with \(N=3\).
To see how the faces fit together, we start by counting. The count shows that pairs of alternatives carry all the information beyond the outcome probabilities. A reconstruction theorem then identifies the state space with the density matrices of complex Hilbert space, and the same theorem tells us how to describe two systems together.
How many parameters does a system with \(N\) distinguishable states need? We already have the answer in principle. The counting assumption, now a consequence of equivalence of subspaces, gives \(K(N)=N^r\). The qubit has \(K(2)=4\), so \(r=2\) and
\[K(N)=N^2 .\]
Where are those \(N^2\) parameters? Hardy noticed that the faces account for all of them. A state needs \(N\) probabilities for the distinguishing measurement. Each pair of distinguishable states spans a qubit face. A qubit has \(K=4\), of which two parameters are the probabilities of its two outcomes, already counted, and two are new. There are \(N(N-1)/2\) pairs, so the faces supply
\[N+2\cdot\frac{N(N-1)}{2}=N^2\]
parameters. Could there be further parameters belonging to three or more alternatives together? The power law says no: pairs already supply all \(N^2\). Both counts rely on local tomography, the first through the qubit and the second through the composition rule.
The count says how many parameters there are. Does it say what the states are? For that we need a theorem.
Theorem (Masanes and Müller, 2011). Suppose a theory satisfies finiteness, local tomography, continuous reversibility, equivalence of subspaces, and all measurements allowed. Then the states of a system with \(N\) perfectly distinguishable states are the \(N\times N\) complex Hermitian matrices with unit trace and nonnegative eigenvalues, the density operators on \(\mathbb C^N\). Its reversible transformations are \(\rho\mapsto U\rho U^\dagger\) with \(U\) unitary.
The proof extends the qubit to larger \(N\) by induction, using the faces and the classification of groups that act transitively; we cite it. Is the result consistent with what we know? Hermitian \(N\times N\) matrices have \(N\) real diagonal entries and \(N(N-1)/2\) complex off-diagonal entries, so \(N^2\) real parameters. The \(2\times2\) block for any pair of basis states, with its two diagonal entries, is exactly a Bloch ball, as equivalence of subspaces requires.
What are the pure states? They are the projections \(\ketbra\psi\psi\) onto unit vectors \(\ket\psi\in\mathbb C^N\), and vectors that differ by a phase factor give the same state. A pure state is thus a ray, a line through the origin, in a complex vector space with an inner product, called a Hilbert space. Write \(\ket\psi=\sum_jc_j\ket j\) in the basis of distinguishable states. This displays superposition. The \(|c_j|^2\) are the outcome probabilities, while the relative phases of the \(c_j\) carry the information in the qubit faces. The transition probability between pure states is \(\tr(\ketbra\phi\phi\,\ketbra\psi\psi)=|\braket\phi\psi|^2\). So the usual Hilbert-space description of states now follows from the principles, via the Masanes–Müller theorem.
How do we describe two systems together? A pair with \(N_A\) and \(N_B\) distinguishable states has \(N_{AB}=N_AN_B\) by the composite capacity assumption. So the theorem says its states are density operators on \(\mathbb C^{N_AN_B}\), and local tomography agrees: \(K_{AB}=(N_AN_B)^2=K_AK_B\). But how do the parts sit inside the pair? Label the distinguishable states of A as \(\ket i\) and of B as \(\ket j\). Their products are \(N_AN_B\) perfectly distinguishable pure states of the pair, so they form a basis, which we write \(\ket{ij}\).
What is the effect “A gives \(i\), whatever B gives”? Call its matrix \(E\). It gives \(1\) on every \(\ket{ij}\) and \(0\) on every \(\ket{i'j}\) with \(i'\neq i\). An effect with \(E\le I\) that gives \(1\) on a unit vector must leave that vector unchanged, and one with \(E\ge0\) that gives \(0\) must send it to zero. So \(E=\sum_j\ketbra{ij}{ij}\), which is the Kronecker product \(\ketbra ii\otimes I\) in this basis. The same theorem fixes the rest.
Tensor-product rule (consequence of the Masanes–Müller theorem). With a suitable choice of phases for the basis vectors \(\ket{ij}\), the pure product states of the pair are the Kronecker products \(\ket\psi\otimes\ket\phi\) of the local vectors, and the joint effect “\(E\) on A and \(F\) on B” is \(E\otimes F\). The pair is described on \(\mathbb C^{N_A}\otimes\mathbb C^{N_B}=\mathbb C^{N_AN_B}\), with \(\ket{ij}=\ket i\otimes\ket j\).
This is the same tensor product of local spaces that local tomography produced for the arrays of fiducial probabilities, now written for amplitudes. Not every pure state of the pair is a product. For example, \((\ket{00}+\ket{11})/\sqrt2\) is not: a product \(\ket\psi\otimes\ket\phi\) with nonzero amplitudes on \(\ket{00}\) and \(\ket{11}\) would also have a nonzero amplitude \(\psi_0\phi_1\) on \(\ket{01}\).
Problem 8.2. For a three-level system, compare the pure state \(\ket\psi=(1,1,1)^T/\sqrt3\) with the maximally mixed state \(I/3\). Which measurement distinguishes them, and with what probabilities?
Both give probability \(1/3\) for each basis outcome. The test \(\{\ketbra\psi\psi,\ I-\ketbra\psi\psi\}\) gives the first outcome with probability \(1\) on \(\ket\psi\) and \(\tr(\ketbra\psi\psi/3)=1/3\) on \(I/3\). The off-diagonal entries of \(\ketbra\psi\psi\), all equal to \(1/3\), carry the difference.
States are now fixed, but the theorem used a principle about measurements that we have so far only named. Which linear functionals are physical effects? The last principle says: all that the states permit. That turns effects into matrices and gives the probability rule for every finite system. Then we see that measurements need not be the projective tests we have used so far.
Could a theory have a rich state space but few measurements? Then states that differ in principle could never be told apart. We exclude this.
All measurements allowed. Every linear functional that takes values between \(0\) and \(1\) on all states is an effect of some measurement, and every list of effects summing to the unit effect is a measurement.
The principle says that the only limits on measurement are those that the states themselves impose. Its foils are theories with the same states and fewer measurements.
What do these effects look like as matrices? The expression \(\tr(AB)\) is an inner product on the real space of Hermitian matrices. So every linear functional on that space can be written as \(\rho\mapsto\tr(\rho E)\) for a unique Hermitian \(E\). Every density operator is a mixture of pure states. So the condition \(0\le\tr(\rho E)\le1\) for all density operators holds exactly when \(0\le\braket\psi{E\psi}\le1\) for all unit vectors, that is, when \(E\) has eigenvalues between \(0\) and \(1\). So the effects are the operators with \(0\le E\le I\), the unit effect is \(I\), and the probability of an outcome is
\[p=\tr(\rho E).\]
This is the Born rule in its general form, now derived for every finite system.
So far our measurements were tests that separate distinguishable states. What else does the principle make available? A measurement is any list of effects \(E_1,\dots,E_m\) with \(\sum_kE_k=I\), often called a positive-operator-valued measure, or POVM. When each \(E_k\) is a projection and they are mutually orthogonal, the measurement is projective. Projective measurements onto a basis distinguish its states perfectly. General measurements can have more outcomes than \(N\), and their effects need not be projections.
Can a qubit measurement have three outcomes? The trine measurement does. Take three unit vectors \(\mathbf n_0,\mathbf n_1,\mathbf n_2\) in a plane of the Bloch ball at \(120^\circ\) to each other, so that \(\mathbf n_0+\mathbf n_1+\mathbf n_2=0\), and set \(E_k=\tfrac13(I+\mathbf n_k\cdot\vec\sigma)\). Each \(E_k\) has eigenvalues \(2/3\) and \(0\), and the three sum to \(I\). The state \(\mathbf n_0\) gives outcome \(0\) with probability \(\tfrac13(1+1)=\tfrac23\) and each other outcome with probability \(\tfrac13(1-\tfrac12)=\tfrac16\). The trine is a valid measurement because the principle makes every such list available. It is not a projective measurement on the qubit alone. When we study systems coupled to their surroundings, we will see how to realize it as a projective measurement on a larger system.
Problem 8.3. Check that the trine effects sum to \(I\), and find the probability of each outcome for the maximally mixed state.
\(\sum_kE_k=I+\tfrac13(\mathbf n_0+\mathbf n_1+\mathbf n_2)\cdot\vec\sigma=I\). For \(\rho=I/2\), \(\tr(\rho E_k)=\tfrac12\tr E_k=\tfrac13\) for each outcome.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
What is \(K\) for systems with \(N=3\) and \(N=4\), and how many parameters does a normalized state have?
\(K=N^2\).
Subtract one for normalization.
\(K=9\) and \(16\); normalized states have \(8\) and \(15\) parameters.
Is \(E=\mathrm{diag}(0.5,0.5,1)\) an effect on a three-level system? Is it a projection?
Check the eigenvalues.
A projection has eigenvalues \(0\) or \(1\).
Its eigenvalues lie in \([0,1]\), so it is an effect. It is not a projection, because of the eigenvalue \(0.5\): it describes a test that passes level \(3\) for sure and the others with probability one half.
Compute the transition probability between \((1,1,0)^T/\sqrt2\) and \((1,0,1)^T/\sqrt2\).
\(|\braket\phi\psi|^2\).
The inner product is \(1/2\).
\(\braket\phi\psi=\tfrac12\), so the probability is \(\tfrac14\).
For a spin-1 system with basis \(\ket{+1},\ket0,\ket{-1}\), describe the face of states that give probability zero for outcome \(-1\). Show that it is a Bloch ball.
Zero probability for a basis outcome forces a zero diagonal entry.
A positive matrix with a zero diagonal entry has zeros in that row and column.
Probability zero for \(-1\) means \(\rho_{-1,-1}=0\), and positivity then forces the whole row and column of \(-1\) to vanish. The remaining \(2\times2\) block on \(\ket{+1},\ket0\) is an arbitrary qubit density matrix, so the face is a Bloch ball, as the equivalence of subspaces requires.
Use Hardy’s face count to compute \(K\) for \(N=3\) and \(N=4\), and prove \(K(N)=N^2\) in general.
\(N\) probabilities plus \(2\) new parameters per pair.
There are \(N(N-1)/2\) pairs.
\(K(3)=3+2\cdot3=9\) and \(K(4)=4+2\cdot6=16\). In general \(K(N)=N+2\cdot\tfrac{N(N-1)}2=N^2\). The \(2\) per pair correspond to the real and imaginary parts of an off-diagonal density-matrix entry.
The three trine states of a qubit are the pure states \(\mathbf n_k\) at \(120^\circ\) in a plane. With equal priors, a party measures the trine POVM \(E_k=\tfrac13(I+\mathbf n_k\cdot\vec\sigma)\) and guesses \(k\). Find the success probability.
\(\tr(\rho_kE_k)=\tfrac13(1+\mathbf n_k\cdot\mathbf n_k)\).
Average over \(k\).
Each state gives the matching outcome with probability \(\tfrac13(1+1)=\tfrac23\), so the success probability is \(\tfrac23\). This is the optimal value for the trine with equal priors; a two-outcome projective measurement cannot even produce three guesses.
For a three-level system, let \(\ket\psi=(1,e^{i\phi},0)^T/\sqrt2\). Find the probability of passing a test for \((1,1,0)^T/\sqrt2\), and interpret it using the qubit face.
Compute the inner product.
The state lies in the face of the first two levels.
The inner product is \((1+e^{i\phi})/2\), so the probability is \(\tfrac12(1+\cos\phi)\). Both states lie in the qubit face of levels 1 and 2, and the result is the Bloch-sphere formula for two equatorial states at angle \(\phi\). Relative phases in a qubit face produce interference.
A student says: “A general measurement is just a projective measurement with added noise.” Is the trine a counterexample?
Can the trine be obtained by randomly choosing projective measurements and relabeling outcomes?
Each trine effect has rank one.
Yes. The trine’s effects are rank-one and it is an extremal measurement: it cannot be written as a random choice among projective measurements followed by classical processing of their outcomes. It is genuinely more general, though, by Naimark’s theorem, it is a projective measurement on a larger system.
A modified qubit theory has the Bloch ball as its state space but allows only the measurement \(\{\tfrac12(I+Z),\tfrac12(I-Z)\}\) and its noisy versions. (a) Which states can it distinguish? (b) What would an observer conclude about the state space from measurements alone?
Effects depend only on \(z\).
States with the same \(z\) give identical statistics.
(a) Only states with different \(z\), and perfectly only \(z=\pm1\). (b) States with the same \(z\) can never be told apart, so operationally they are the same state; the effective state space collapses to the segment \(-1\le z\le1\), a classical bit. The principle that all measurements are allowed ensures that states and effects match, so that every difference in the state space is observable.
Suppose the two-outcome faces of every system were discs instead of balls. Redo Hardy’s face count and identify the theory.
A disc has \(K=3\), so one new parameter per pair.
Compare with the parameter count of real symmetric matrices.
\(K(N)=N+1\cdot\tfrac{N(N-1)}2=\tfrac{N(N+1)}2\), the count of real symmetric \(N\times N\) matrices: real quantum theory. Its faces are discs, its parameter count is not multiplicative, and it violates local tomography, as the parameter count and the construction of the real pair found.