Quantum Foundations · Chapter 14
What changes when two systems cannot be told apart by any measurement?
Starting point: Chapters 9 and 13: symmetry and composite systems.
Anchor example: Two photons meeting at a beam splitter.
Principle at work: Operational indistinguishability: no measurement can depend on which system carries which label.
Foil: Labelled particles, and the parastatistics that the principle alone does not exclude.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Every pair we have studied had parts that could be told apart: one system in laboratory A, one in laboratory B. The tensor-product rule describes such a pair, with each factor belonging to one identifiable part.
But two electrons, or two photons, cannot be told apart by any measurement. Which of them is “the first”? We take that question operationally: if no measurement can read the labels, the description must not let them do physical work. This forces the states of a pair to split into a symmetric and an antisymmetric sector, between which no coherence can be detected. Assuming that each kind of particle uses one sector, we find exclusion for one kind and bunching for the other, both of which experiments confirm. We then ask how much of the usual treatment this argument actually reaches, and find two things it does not settle.
The next chapter asks what happens to a system that is part of a larger whole we cannot see.
What does it mean, in the language of effects, that no measurement can tell two systems apart? We first state it as a symmetry condition on every effect. The condition splits the pair’s states into two sectors and makes coherence between them undetectable.
How can we say “no measurement can read the labels” in mathematics? Let \(\mathcal H\) be the Hilbert space of one system, and describe two systems of the same type on \(\mathcal H\otimes\mathcal H\), labelling the factors 1 and 2 for bookkeeping. Let \(S\) be the swap operator, \(S\ket a\ket b=\ket b\ket a\). The systems are indistinguishable if no physical operation can depend on the labels: every effect \(E\) of the pair satisfies \(SES=E\), and so does every Hamiltonian.
Why is this reasonable? It is an operational principle in the spirit of the course. The labels are not properties of the systems; they belong to our description, and the principle says the description must not let them do physical work. Its foil is a theory of labeled particles, which is consistent and is the correct description whenever the particles differ in some property, such as mass or a long-lived internal state.
What does the principle do to the states of the pair? The swap satisfies \(S^2=I\), so its eigenvalues are \(\pm1\) and the space splits into a symmetric part (\(S=+1\)) and an antisymmetric part (\(S=-1\)), with projections \((I\pm S)/2\). Every allowed effect commutes with \(S\), so it maps each part into itself. Consequently no measurement can detect a coherent superposition of a symmetric and an antisymmetric state: such a superposition gives exactly the same statistics for all allowed effects as the corresponding mixture. The two sectors are separated by a superselection rule.
Does this force every state of a given kind of particle into one sector? No. The superselection rule only says that coherence between the sectors cannot be detected. That only one sector occurs is a further assumption, which we state explicitly.
Sector assumption. For a given kind of particle, all states that occur lie in one sector.
For two particles the sectors are just these two, and the particles are called bosons if their states are symmetric and fermions if antisymmetric. For two qubit-like systems the symmetric part has dimension \(3\) and the antisymmetric part dimension \(1\).
Problem 14.1. Show that \(\tfrac1{\sqrt2}(\ket s+\ket a)\) and \(\tfrac1{\sqrt2}(\ket s-\ket a)\), with \(\ket s\) symmetric and \(\ket a\) antisymmetric, give the same probability for every effect commuting with \(S\).
The probabilities differ by \(\pm\mathrm{Re}\,\bra sE\ket a\) terms. Since \(E\) commutes with \(S\), \(E\ket a\) is antisymmetric, so \(\bra sE\ket a=0\). Both states give \(\tfrac12(\bra sE\ket s+\bra aE\ket a)\).
Can two identical particles occupy the same state? The sector decides. We first write the joint states of each sector and find exclusion for one and sharing for the other. Then a beam splitter turns this difference into a counting experiment.
How do we put two identical particles into two given single-particle states \(\ket a\) and \(\ket b\), with \(\braket ab=0\)? The product \(\ket a\ket b\) is not allowed, since it lies in neither sector. Projecting it onto each sector and normalizing gives
\[\ket{a,b}_\pm=\tfrac1{\sqrt2}\big(\ket a\ket b\pm\ket b\ket a\big),\]
with \(+\) for bosons and \(-\) for fermions. For fermions with \(a=b\), the state vanishes: two fermions cannot occupy the same single-particle state. This is the Pauli exclusion principle, which underlies the structure of atoms and the stability of matter. For bosons, the state \(\ket a\ket a\) is allowed, and bosons tend to share states.
Are such pairs entangled? The reduced state of one particle in \(\ket{a,b}_\pm\) is \(\tfrac12(\ketbra aa+\ketbra bb)\), a mixture, even though nothing has interacted. Whether this formal entanglement is usable as a resource is a subtle question, because the labels it refers to cannot be accessed. Operationally meaningful entanglement between identical particles is defined between modes rather than between labels. A mode is a single-particle state that a detector can address, such as one of the two paths of a beam splitter; we then ask how many particles occupy each mode.
How can we see the difference between the sectors in a laboratory? Label the two paths of a beam splitter \(\ket1\) and \(\ket2\). A balanced beam splitter acts on each particle’s path as \(\ket1\mapsto(\ket1+\ket2)/\sqrt2\) and \(\ket2\mapsto(\ket1-\ket2)/\sqrt2\). Send one particle in each input. For distinguishable particles the input is \(\ket1\ket2\), and the output has amplitude for each of the four path combinations of magnitude \(1/2\): coincidences, one particle in each output, occur with probability \(1/2\).
For bosons the input is \(\ket{1,2}_+\). The two amplitudes for one particle in each output, one where both are transmitted and one where both are reflected, cancel exactly, and the coincidence probability is \(0\): the photons bunch. For fermions they add, and the particles always leave through different ports. These three answers, \(1/2\), \(0\), and \(1\), are direct tests of indistinguishability and of the sector. Hong, Ou, and Mandel observed the bosonic dip for photons in 1987.
Problem 14.2. Compute the bosonic output of the beam splitter for the input \(\ket{1,2}_+\) and confirm that both particles exit together.
\(\ket1\ket2\mapsto\tfrac12(\ket1+\ket2)(\ket1-\ket2)\) and \(\ket2\ket1\mapsto\tfrac12(\ket1-\ket2)(\ket1+\ket2)\). Adding and dividing by \(\sqrt2\), the \(\ket1\ket2\) and \(\ket2\ket1\) terms cancel, leaving \(\tfrac1{\sqrt2}(\ket1\ket1-\ket2\ket2)\): both particles in port 1 or both in port 2.
How much of the usual treatment has indistinguishability derived? It explains why coherence between symmetry sectors can never be detected. The rest needed further input. With three or more particles there are more sectors than two, and the principles do not say which particles use which sector.
Do three or more particles also have only two sectors? No. For three or more particles, permutations form a larger group, and the space splits into more sectors: the totally symmetric, the totally antisymmetric, and sectors of mixed symmetry in which permutations act by matrices rather than by signs. Indistinguishability alone permits all of them. Particles described by mixed sectors obey parastatistics.
The usual statement that only the totally symmetric and antisymmetric sectors occur is called the symmetrization postulate. Albert Messiah and Oscar Greenberg examined its status in 1964 and concluded that it is an additional assumption. In relativistic quantum field theory in three spatial dimensions, the analysis of Sergio Doplicher, Rudolf Haag, and John Roberts showed that parastatistics can be rewritten as ordinary Bose or Fermi statistics together with a hidden internal degree of freedom. Within the finite-dimensional principles of this course, the postulate is not derived.
Which particles are bosons and which are fermions? The principles do not settle this either. Experience says particles of integer spin are bosons and those of half-integer spin are fermions. This connection is a theorem of relativistic quantum field theory, due to Markus Fierz and Wolfgang Pauli around 1940, and it relies on Lorentz invariance, locality, and positivity of energy, which go beyond the setting of this course.
In two spatial dimensions, exchanging particles is described by paths that can wind around each other, and the possible statistics include anyons, whose exchange multiplies the state by an arbitrary phase or matrix. This shows that statistics depend on the geometry of space, not on the finite-dimensional state spaces alone, and is one more reason to treat spin and statistics as an input from outside the reconstruction.
Problem 14.3. For two particles, why are there no parastatistics?
The permutation group of two objects has only the identity and the swap, and \(S^2=I\) forces every sector to be an eigenspace of \(S\) with eigenvalue \(\pm1\). Mixed-symmetry sectors need at least three particles, where permutations need not commute.
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 are the dimensions of the symmetric and antisymmetric subspaces for two particles with three-dimensional single-particle spaces?
Symmetric: \(n(n+1)/2\); antisymmetric: \(n(n-1)/2\).
\(n=3\).
\(6\) and \(3\), adding up to \(9\).
What is \(\ket{a,a}_-\), and what principle does this express?
\(\ket a\ket a-\ket a\ket a\).
Two fermions in the same state.
It is zero: two fermions cannot occupy the same single-particle state. This is the Pauli exclusion principle.
What is the coincidence probability at a balanced beam splitter for distinguishable particles, identical bosons, and identical fermions?
Coincidence means one particle in each output.
Two amplitudes cancel or add.
\(\tfrac12\), \(0\), and \(1\).
Two photons have internal states \(\ket{\chi_1}\), \(\ket{\chi_2}\) with \(|\braket{\chi_1}{\chi_2}|^2=s\). Their coincidence probability at a balanced beam splitter is \((1-s)/2\). Check the limits \(s=0\) and \(s=1\), and explain the use of the formula in experiments.
\(s=0\) means distinguishable.
\(s=1\) means identical.
For \(s=0\) the probability is \(\tfrac12\), the distinguishable value; for \(s=1\) it is \(0\), perfect bunching. Measuring the depth of the dip estimates the overlap of the photons’ internal states, a standard test of how indistinguishable a photon source is.
For two qubits, show that \(E\otimes I+I\otimes E\) commutes with the swap \(S\), while \(E\otimes I\) generally does not. What does this mean for measurements on identical particles?
\(S(A\otimes B)S=B\otimes A\).
Take \(E=Z\).
\(S(E\otimes I+I\otimes E)S=I\otimes E+E\otimes I\), unchanged. But \(S(E\otimes I)S=I\otimes E\ne E\otimes I\) for \(E=Z\). The allowed observables measure properties of the pair without saying which particle has them, such as the total \(Z\), not the \(Z\) of “particle 1.”
Two fermions each have a two-dimensional single-particle space. Find all allowed states and the reduced state of one particle.
The antisymmetric subspace has dimension \(1\).
Compute the reduced state of \((\ket{01}-\ket{10})/\sqrt2\).
The only state is \((\ket{01}-\ket{10})/\sqrt2\), and the reduced state is \(I/2\). The labels carry no information that a measurement could access; the formal mixedness reflects the symmetry, and operational entanglement must be defined between modes.
For three qubits, the \(8\)-dimensional space decomposes under permutations. Given that the symmetric subspace has dimension \(4\) and the antisymmetric subspace dimension \(0\), how many dimensions belong to mixed-symmetry sectors?
Subtract.
Antisymmetric states need three different single-particle states.
\(8-4-0=4\) dimensions of mixed symmetry, forming two copies of a two-dimensional representation of the permutation group. Indistinguishability alone allows states there; the symmetrization postulate excludes them.
A student says: “Bosons bunch at a beam splitter because they attract each other.” Explain why not.
Is there any interaction in the calculation?
Bunching comes from interference.
The beam-splitter calculation involves no force: each photon passes the beam splitter independently. Bunching comes from the symmetry of the joint state, which makes the transmit–transmit and reflect–reflect amplitudes cancel. It is an interference effect of indistinguishability.
Explain why a coherent superposition of a bosonic and a fermionic state of the same pair cannot be prepared or detected, and relate this to the mass superselection rule found from the symmetries of space.
Allowed observables commute with \(S\).
A superselection rule forbids observable coherence between sectors.
All allowed observables commute with \(S\), so none connect the symmetric and antisymmetric sectors, and a superposition across sectors has the same statistics as the mixture. The sectors are superselected. Bargmann’s analysis of the Galilean group gives a similar rule for different total masses: symmetry decides which superpositions are physically meaningful.
In three dimensions, exchanging two particles twice is a loop that can be contracted to staying put; in two dimensions it cannot. Explain why this allows only the signs \(\pm1\) in three dimensions but arbitrary phases (anyons) in two.
If a double exchange is trivial, the exchange phase squares to one.
In two dimensions double exchange winds one particle around the other.
In three dimensions a double exchange is deformable to no motion, so the exchange phase \(e^{i\theta}\) must satisfy \(e^{2i\theta}=1\), giving \(\pm1\): bosons and fermions. In two dimensions the double exchange winds one particle around the other and cannot be undone, so \(e^{i\theta}\) is unconstrained. Statistics depend on the topology of space, not only on the finite-dimensional state spaces of the course.