Quantum states and indistinguishable particles

In a quantum system, a physical state is described by a state vector. A pair of distinct state vectors are physically equivalent if they differ only by an overall phase factor, ignoring other interactions. A pair of indistinguishable particles such as this have only one state. This means that if the positions of the particles are exchanged (i.e., they undergo a permutation), this does not identify a new physical state, but rather one matching the original physical state. In fact, one cannot tell which particle is in which position.

While the physical state does not change under the exchange of the particles' positions, it is possible for the state vector to change sign as a result of an exchange. Since this sign change is just an overall phase, this does not affect the physical state.

The essential ingredient in proving the spin-statistics relation is relativity, that the physical laws do not change under Lorentz transformations. The field operators transform under Lorentz transformations according to the spin of the particle that they create, by definition.

Additionally, the assumption (known as microcausality) that spacelike-separated fields either commute or anticommute can be made only for relativistic theories with a time direction. Otherwise, the notion of being spacelike is meaningless. However, the proof involves looking at a Euclidean version of spacetime, in which the time direction is treated as a spatial one, as will be now explained.

Lorentz transformations include 3-dimensional rotations and boosts. A boost transfers to a frame of reference with a different velocity and is mathematically like a rotation into time. By analytic continuation of the correlation functions of a quantum field theory, the time coordinate may become imaginary, and then boosts become rotations. The new "spacetime" has only spatial directions and is termed Euclidean.

Exchange symmetry or permutation symmetry

Bosons are particles whose wavefunction is symmetric under such an exchange or permutation, so if we swap the particles, the wavefunction does not change. Fermions are particles whose wavefunction is antisymmetric, so under such a swap the wavefunction gets a minus sign, meaning that the amplitude for two identical fermions to occupy the same state must be zero. This is the Pauli exclusion principle: two identical fermions cannot occupy the same state. This rule does not hold for bosons.

In quantum field theory, a state or a wavefunction is described by field operators operating on some basic state called the vacuum. In order for the operators to project out the symmetric or antisymmetric component of the creating wavefunction, they must have the appropriate commutation law. The operator

${\displaystyle \iint \psi (x,y)\phi (x)\phi (y)\,dx\,dy}$

(with ${\displaystyle \phi }$ an operator and ${\displaystyle \psi (x,y)}$ a numerical function with complex values) creates a two-particle state with wavefunction ${\displaystyle \psi (x,y)}$, and depending on the commutation properties of the fields, either only the antisymmetric parts or the symmetric parts matter.

Let us assume that ${\displaystyle x\neq y}$ and the two operators take place at the same time; more generally, they may have spacelike separation, as is explained hereafter.

If the fields commute, meaning that the following holds:

${\displaystyle \phi (x)\phi (y)=\phi (y)\phi (x),}$

then only the symmetric part of ${\displaystyle \psi }$ contributes, so that ${\displaystyle \psi (x,y)=\psi (y,x)}$, and the field will create bosonic particles.

On the other hand, if the fields anti-commute, meaning that ${\displaystyle \phi }$ has the property that

${\displaystyle \phi (x)\phi (y)=-\phi (y)\phi (x),}$

then only the antisymmetric part of ${\displaystyle \psi }$ contributes, so that ${\displaystyle \psi (x,y)=-\psi (y,x)}$, and the particles will be fermionic.

Naively, neither has anything to do with the spin, which determines the rotation properties of the particles, not the exchange properties.

Spin–statistics relation

The spin–statistics relation was first formulated in 1939 by Markus Fierz^[3] and was rederived in a more systematic way by Wolfgang Pauli.^[4] Fierz and Pauli argued their result by enumerating all free field theories subject to the requirement that there be quadratic forms for locally commuting observables including a positive-definite energy density. A more conceptual argument was provided by Julian Schwinger in 1950. Richard Feynman gave a demonstration by demanding unitarity for scattering as an external potential is varied,^[5] which when translated to field language is a condition on the quadratic operator that couples to the potential.^[6]

Theorem statement

The theorem states that:

• The wave function of a system of identical integer-spin particles has the same value when the positions of any two particles are swapped. Particles with wave functions symmetric under exchange are called bosons.
• The wave function of a system of identical half-integer–spin particles changes sign when two particles are swapped. Particles with wave functions antisymmetric under exchange are called fermions.

In other words, the spin–statistics theorem states that integer-spin particles are bosons, while half-integer–spin particles are fermions.

Suggestive bogus argument

Consider the two-field operator product

${\displaystyle R(\pi )\phi (x)\phi (-x),}$

where R is the matrix that rotates the spin polarization of the field by 180 degrees when one does a 180-degree rotation around some particular axis. The components of ${\displaystyle \phi }$ are not shown in this notation. ${\displaystyle \phi }$ has many components, and the matrix R mixes them up with one another.

In a non-relativistic theory, this product can be interpreted as annihilating two particles at positions ${\displaystyle x}$ and ${\displaystyle -x}$ with polarizations that are rotated by ${\displaystyle \pi }$ relative to each other. Now rotate this configuration by ${\displaystyle \pi }$ around the origin. Under this rotation, the two points ${\displaystyle x}$ and ${\displaystyle -x}$ switch places, and the two field polarizations are additionally rotated by a ${\displaystyle \pi }$. So we get

${\displaystyle R(2\pi )\phi (-x)R(\pi )\phi (x),}$

which for integer spin is equal to

${\displaystyle \phi (-x)R(\pi )\phi (x)}$

and for half-integer spin is equal to

${\displaystyle -\phi (-x)R(\pi )\phi (x)}$

(proved at Spin (physics) § Rotations). Both the operators ${\displaystyle \pm \phi (-x)R(\pi )\phi (x)}$ still annihilate two particles at ${\displaystyle x}$ and ${\displaystyle -x}$. Hence we claim to have shown that, with respect to particle states:

${\displaystyle R(\pi )\phi (x)\phi (-x)={\begin{cases}\phi (-x)R(\pi )\phi (x)&{\text{ for integral spins}},\\-\phi (-x)R(\pi )\phi (x)&{\text{ for half-integral spins}}.\end{cases}}}$

So exchanging the order of two appropriately polarized operator insertions into the vacuum can be done by a rotation, at the cost of a sign in the half-integer case.

This argument by itself does not prove anything like the spin–statistics relation. To see why, consider a nonrelativistic spin-0 field described by a free Schrödinger equation. Such a field can be anticommuting or commuting. To see where it fails, consider that a nonrelativistic spin-0 field has no polarization, so that the product above is simply:

${\displaystyle \phi (-x)\phi (x).}$

In the nonrelativistic theory, this product annihilates two particles at ${\displaystyle x}$ and ${\displaystyle -x}$, and has zero expectation value in any state of definite particle number. In order to have a nonzero matrix element, this operator product must be between states with two more particles on the right than on the left:

${\displaystyle \langle 0|\phi (-x)\phi (x)|\psi \rangle .}$

Performing the rotation, all that we learn is that rotating the 2-particle state ${\displaystyle |\psi \rangle }$ gives the same sign as changing the operator order. This gives no additional information, so this argument does not prove anything.

Why the bogus argument fails

To prove the spin–statistics theorem, it is necessary to use relativity, as is obvious from the consistency of the nonrelativistic spinless fermion, and the nonrelativistic spinning bosons. There are claims in the literature of proofs of the spin–statistics theorem that do not require relativity,^[7][8] but they are not proofs of a theorem, as the counterexamples show, rather they are arguments for why spin–statistics is "natural", while wrong-statistics is "unnatural". In relativity, the connection is required.

In relativity, there are no local fields that are pure creation operators or annihilation operators. Every local field both creates particles and annihilates the corresponding antiparticle. This means that in relativity, the product of the free real spin-0 field has a nonzero vacuum expectation value, because in addition to creating particles which are not annihilated and annihilating particles which are not subsequently created, it also includes a part that creates and annihilates "virtual" particles whose existence enters into interaction calculations – but never as scattering matrix indices or asymptotic states.

${\displaystyle G(x)=\langle 0|\phi (-x)\phi (x)|0\rangle .}$

And now the heuristic argument can be used to see that ${\displaystyle G(x)}$ is equal to ${\displaystyle G(-x)}$, which tells us that the fields cannot be anti-commuting.

Fermionic fields

The spin–statistics theorem implies that half-integer–spin particles are subject to the Pauli exclusion principle, while integer-spin particles are not. Only one fermion can occupy a given quantum state at any time, while the number of bosons that can occupy a quantum state is not restricted. The basic building blocks of matter such as protons, neutrons, and electrons are fermions. Particles such as the photon, which mediate forces between matter particles, are bosons.

The Fermi–Dirac distribution describing fermions leads to interesting properties. Since only one fermion can occupy a given quantum state, the lowest single-particle energy level for spin-1/2 fermions contains at most two particles, with the spins of the particles oppositely aligned. Thus, even at absolute zero, a system of more than two fermions in this case still has a significant amount of energy. As a result, such a fermionic system exerts an outward pressure. Even at non-zero temperatures, such a pressure can exist. This degeneracy pressure is responsible for keeping certain massive stars from collapsing due to gravity. See white dwarf, neutron star, and black hole.

Bosonic fields

There are a couple of interesting phenomena arising from the two types of statistics. The Bose–Einstein distribution which describes bosons leads to Bose–Einstein condensation. Below a certain temperature, most of the particles in a bosonic system will occupy the ground state (the state of lowest energy). Unusual properties such as superfluidity can result.

Ghost fields

In quantum field theory, ghost fields are fields that have negative kinetic energy and are used to cancel out unphysical degrees of freedom in the theory. They are called ghost fields because they do not correspond to any physical particles. Ghost fields are used in the quantization of gauge theories such as quantum chromodynamics and electroweak theory.

Ghost fields do not obey the spin–statistics relation. See Klein transformation on how to patch up a loophole in the theorem.