# Anti-commutation relation for quantized fields

• A
• The Tortoise-Man
The Tortoise-Man
Could somebody elaborate following statement from wikipedia in detail on interplay between the "choice" of anti- or commutation relation for quantized fields and the the associated statistics which the field satisfies before get quantized:

We impose an anticommutator relation (as opposed to a commutation relation as we do for the bosonic field) in order to make the operators compatible with Fermi–Dirac statistics.

Very roughly the story with second quantization is one start with a field ( as honest "function" in variables ##x_1, x_2,...##, so before running the quantization machinery) ##\psi(x_1,x_2,..., x_n)## and - depending on how the sign of the field changes ( symmetrically or antisym) when we permute the arguments ##x_i## - the field is called bosonic or fermionic, more less "by definition".

And my question is how this fits in the picture that if we have a fermionic ( resp bosonic) field given, then the "only right" bracket relation in the quatization procedure for this field must be the be anti-commutating (resp commutating) one?

In other words, what would run into troubles if we not set the bracket relation exactly that way? What does that exactly mean ( as definition) for an operator to be "compatible" with certain given statistic/ distribution?

The Tortoise-Man said:
Very roughly the story with second quantization is one start with a field ( as honest "function" in variables , so before running the quantization machinery)
"Before running the quantization machinery" it makes no sense to talk about what you are asking about. The commutation or anticommutation relations are things that only come into play after "running the quantization machinery", at which point the quantum fields are operators (or more precisely operator-valued functions on spacetime).

In other words, the commutation or anticommutation relations are operator relations in QFT. Saying that two operators commute (or anticommute) is not the same as saying that a function of two variables takes the same value (or the same magnitude but the opposite sign) if you swap the arguments.

The Tortoise-Man said:
What does that exactly mean ( as definition) for an operator to be "compatible" with certain given statistic/ distribution?
It means that the field operators produce the appropriate statistics when applied to states. For example, applying the same fermionic creation operator twice to the vacuum state vanishes, since no two fermions can be in the same state.

vanhees71 and topsquark
Let's discuss non-relativistic QFT (or non-relativistic QT in "2nd quantization"). The relation between quantum fields and states is as follows. You start with some basis of the single-particle Hilbert space, e.g., the position, spin eigenbasis ##|\vec{x},\sigma \rangle##, where ##\vec{x} \in \mathbb{R}^3## and ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}## for particles with spin ##s##, where ##s \in \{0,1/2,1,3/2,\ldots \}##.

Now you define field operators ##\hat{\psi}(\vec{x},\sigma)## (working in the Schrödinger picture of time evolution). The Hilbert space describes many-particle systems of indistinguishable particles with an unspecified number of particles, the socalled Fock space, and the meaning of ##\hat{\psi}(\vec{x},\sigma)## is that when applied to such a many-particle state you take out a particle with spin ##\sigma## at the position ##\vec{x}## ("annihilation operator") and ##\hat{\psi}^{\dagger}(\vec{x},\sigma)## adds a particle with spin ##\sigma## at the place ##\vec{x}## ("creation operator").

The creation and annihilation operators obey commutator or anti-commutator relations,
$$[\hat{\psi}(\vec{x},\sigma),\hat{\psi}(\vec{x}',\sigma')]_{\pm}=0, \quad [\hat{\psi}(\vec{x},\sigma),\hat{\psi}^{\dagger}(\vec{x}',\sigma')]_{\pm}=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma \sigma'}.$$
For the lower sign (bosons) these are the commutation relations for creation and annihilation operators of independent harmonic oscillators. The case of the upper sign (fermions) is very similar. You can show that you can build a basis of the Fock state by defining "the vacuum", ##|\Omega \rangle##, that describes the case that no particle is present, and thus obeys
$$\hat{\psi}(\vec{x},\sigma) |\Omega \rangle=0$$
for all ##\vec{x}## and ##\sigma##.

From this you can build the Fock-space basis by repeated application of creation operators. Obviously the order you apply these creation operators doesn't matter, because for bosons you get always the same basis vector since the creation operators commute and for fermions you get the same basisvector up to a sign, because the creation operators anticommute. For fermions you can't have more than 1 particle at the same place, because due to the anticommutation relations ##[\hat{\psi}(\vec{x},\sigma)]^2=0##. Thus the anticommutation relations lead automatically to the Pauli exclusion principle.

The relation to the 1st-quantization formalism is that there you deal with states with a definite number, ##N##, of indistinguishable particles, described by state vectors ##|\Psi \rangle##, which can be represented uniquely by the corresponding wave functions
$$\Psi(\vec{x}_1,\sigma_1;\ldots,\vec{x}_N,\sigma_N) = \langle \vec{x}_1,\sigma_1;\ldots;\vec{x}_N,\sigma_N|\Psi \rangle.$$
In the 2nd-quantization formalism you simply have
$$|\vec{x}_1,\sigma_1;\ldots;\vec{x}_N,\sigma_N \rangle=\hat{\psi}^{\dagger}(\vec{x}_1,\sigma_1) \cdots \hat{\psi}^{\dagger}(\vec{x}_N,\sigma_N) |\Omega \rangle.$$
For bosons, interchanging any pair of one-particle arguments ##(\vec{x}_j,\sigma_j)## and ##\vec{x}_k,\sigma_k)## doesn't change this basis vector at all, and the corresponding ##N##-particle wave function is thus symmetric under reordering of its arguments, showing that the commutation relations for bosonic field operators lead to the description of bosons as defined in the 1st-quantization formulation. For fermions a change of the ordering implies a sign according to the sign of the permutation of the arguments, i.e., interchaning any pair of arguments leads to a sign change of the basis state. That's still the same quantum state, because overall phase factors are irrelevant for physical quantities, but now the corresponding wave function in the 1st-quantization formalism are antisymmetric under such an exchanges of arguments, i.e., now we describe fermions.

The 1st- and 2nd-quantization formalism are completely equivalent if the Hamiltonian preserves particle particle numbers. In the 2nd-quantization formalism that means that ##\hat{H}## commutes with the particle-number operator,
$$\hat{N}=\int_{\mathbb{R}}^3 \sum_{\sigma} \hat{\psi}^{\dagger}(\vec{x},\sigma) \hat{\psi}(\vec{x},\sigma).$$
This means that, when you have prepared a state with a definite number of particles, the time evolution doesn't change this number of particles, i.e., the entire dynamics is such that everything is described within the subspace of the Fock space with this fixed particle number, and the entire physics for such a system is exactly the same as described in the 1st-quantization formalism with a fixed number of indistinguishable bosons or fermions. In this case the advantage of the 2nd-quantization formalism compared to the 1st-quantization formalism is that you can express everything in terms of field operators, which have to Bose symmetrization or Fermion antisymmetrization of the states built in from the very beginning.

PeterDonis said:
"Before running the quantization machinery" it makes no sense to talk about what you are asking about. The commutation or anticommutation relations are things that only come into play after "running the quantization machinery", at which point the quantum fields are operators (or more precisely operator-valued functions on spacetime).

In other words, the commutation or anticommutation relations are operator relations in QFT. Saying that two operators commute (or anticommute) is not the same as saying that a function of two variables takes the same value (or the same magnitude but the opposite sign) if you swap the arguments.It means that the field operators produce the appropriate statistics when applied to states. For example, applying the same fermionic creation operator twice to the vacuum state vanishes, since no two fermions can be in the same state.

That's not completely what I intended to say, sorry for confused formulation. What I meant/wanted to know, was the following: we use @vanhees71's setting, so say we start with some basis of the single-particle Hilbert space, e.g., the position, spin eigenbasis ##|\vec{x},\sigma \rangle##, where ##\vec{x} \in \mathbb{R}^3## and ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}## for particles with a given spin ##s## (here the values of ##s## more or less says which spin irrep we choose .)

So - at this stage - this is the given initial data with which we assume to start. Now we want to define field operator ##\hat{\psi}(\vec{x},\sigma)## for this particles. To this operator we have to associate creation & annihilation operators and need at this stage to know which bracket relation we should choose: the anti- or commutation relation.

And my question is essentially how do we "know" at this stage (from this given data) which bracket relation the creation & annihilation operators for our desired field operator ought satisfy? And from "where" do we "extract" this information?

(The part in the question with "many particle wave function" ##\psi(x_1,x_2,..., x_n)## in the OP is misleadingly formulated and should be ignored; I assumed when I prepared this question that this wave function should be considered as a part of the initial datum, but that's not the case)

A guess: Is it basically the case that all we need to know in order to decide "which bracket relation" for our field we have to choose, lies completely in the assumption (which we consider as "given" datum) if the spin ##s## of our given particle sort has half integer value (=fermions) or integer (=boson)?

And then we refer to the spin statistic theorem (which we consider at this stage a big black box) that in order to make our field operator "consistent" there is only one reasonable choice, namely if ##s## half integer (=fermion), then the bracket relation must be set to be anti-commutating, and if ##s## is integer (=boson), then the bracket relation must be set to be commutating?

Are my reasonings correct now? Ie, to put in a nutshell: The choice for bracket relation for the field operator we are going to constuct depends purely on the value of spin ##s## and why this is the case is the (rather deep) essence of spin statistic theorem?

Last edited:
vanhees71
The Tortoise-Man said:
The choice for bracket relation for the field operator we are going to constuct depends purely on the value of spin ##s## and why this is the case is the (rather deep) essence of spin statistic theorem?
I would say this is correct, yes.

vanhees71
Also, you wrote

It means that the field operators produce the appropriate statistics when applied to states. For example, applying the same fermionic creation operator twice to the vacuum state vanishes, since no two fermions can be in the same state.

This part I not completely understand. The statistic itself refers formally to the states themself (=the states of the Fock space), more precisely the "statistic" tells us - in simple words- if we pick some state, how big is the probability that the system is in this picked state, right?

Therefore, I not understand your statement, how field operators can produce the appropriate statistics. The field operator acts on the states, but the statistic inself "weights" the states with appropriate probabilities, right? Therefore I'm not sure how to interpret the phrase that a field operator "produce" certain statstic, when it acts on states. Could you elaborate a bit this point? (Formally: How following association precislely works: Say we have already build up our field operator. How does it "produce" a statistic?

The Tortoise-Man said:
the "statistic" tells us - in simple words- if we pick some state, how big is the probability that the system is in this picked state, right?
No.

The Tortoise-Man said:
The field operator acts on the states
Yes.

The Tortoise-Man said:
but the statistic inself "weights" the states with appropriate probabilities, right?
No.

The Tortoise-Man said:
I'm not sure how to interpret the phrase that a field operator "produce" certain statstic, when it acts on states. Could you elaborate a bit this point?
What do operators do when they act on states? They produce other states.

For example, a creation operator ##a^\dagger## acting on the vacuum state, ##a^\dagger \ket{0}##, produces a Fock state where the particular 1-particle state the creation operator is creating is occupied, and no others. We might write this state as ##\ket{a}##.

If the creation operator is for a fermion state, then applying the same creation operator twice to the vacuum, ##a^\dagger a^\dagger \ket{0}##, must vanish, because that operation, physically, means trying to produce two fermions in the same state, which is impossible. We could write this as ##a^\dagger \ket{a} = 0##.

Now consider two different fermionic creation operators, ##a^\dagger## and ##b^\dagger##. The fact that they anticommute means that swapping them flips the sign, so, for example, applying them both to the vacuum state, we have ##a^\dagger b^\dagger \ket{0} = - b^\dagger a^\dagger \ket{0}##. Which is precisely what we mean when we say that the state changes sign if we exchange two fermions.

The corresponding reasoning for bosons should be simple to work through given the above.

Note, btw, that the above also shows what we mean when we say states have certain statistics. We could write the states produced by the two different fermionic creation operators above as follows: ##a^\dagger b^\dagger \ket{0} = \ket{a, b}## and ##b^\dagger a^\dagger \ket{0} = \ket{b, a}##. And then we would write ##\ket{a, b} = - \ket{b, a}##, to express, once again, that the state changes sign if you exchange two fermions.

vanhees71
The Tortoise-Man said:
That's not completely what I intended to say, sorry for confused formulation. What I meant/wanted to know, was the following: we use @vanhees71's setting, so say we start with some basis of the single-particle Hilbert space, e.g., the position, spin eigenbasis ##|\vec{x},\sigma \rangle##, where ##\vec{x} \in \mathbb{R}^3## and ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}## for particles with a given spin ##s## (here the values of ##s## more or less says which spin irrep we choose .)

So - at this stage - this is the given initial data with which we assume to start. Now we want to define field operator ##\hat{\psi}(\vec{x},\sigma)## for this particles. To this operator we have to associate creation & annihilation operators and need at this stage to know which bracket relation we should choose: the anti- or commutation relation.

And my question is essentially how do we "know" at this stage (from this given data) which bracket relation the creation & annihilation operators for our desired field operator ought satisfy? And from "where" do we "extract" this information?

(The part in the question with "many particle wave function" ##\psi(x_1,x_2,..., x_n)## in the OP is misleadingly formulated and should be ignored; I assumed when I prepared this question that this wave function should be considered as a part of the initial datum, but that's not the case)

A guess: Is it basically the case that all we need to know in order to decide "which bracket relation" for our field we have to choose, lies completely in the assumption (which we consider as "given" datum) if the spin ##s## of our given particle sort has half integer value (=fermions) or integer (=boson)?

And then we refer to the spin statistic theorem (which we consider at this stage a big black box) that in order to make our field operator "consistent" there is only one reasonable choice, namely if ##s## half integer (=fermion), then the bracket relation must be set to be anti-commutating, and if ##s## is integer (=boson), then the bracket relation must be set to be commutating?

Are my reasonings correct now? Ie, to put in a nutshell: The choice for bracket relation for the field operator we are going to constuct depends purely on the value of spin ##s## and why this is the case is the (rather deep) essence of spin statistic theorem?
One should know, that the spin-statistics theorem comes from relativistic QFT. I haven't ever seen a convincing proof within non-relativistic QT, and indeed nothing goes wrong when "2nd-quantizing" a spin-0 Schrödinger field as fermions. The only point is that no spin-0 fermion has ever been found in Nature (and also no integer-spin fermions too). That's why I think in non-relativistic QFT the answer to the question, whether to quantize the fields as fermions or bosons is just an empirical input, and the observation that integer-spin particles are bosons and half-inter-spin particles are fermions is just an observational fact. That this "spin-statistics relation" can be proven from the basic assumptions of local relativistic QFT (microcausality for local observable-operators and boundedness of energy from below), is one of the great achievements of this theory.

the "statistic" tells us - in simple words- if we pick some state, how big is the probability that the system is in this picked state, right?

PeterDonis said:
No.
But what is then in this context a "statistic" precisely? How should it formally be correctly defined? Up to now I thought about it propability theoretically as a distribution on certain given space/set (=the "sample space"). Can the notion of a "statistic" in the QFT context we discussed up to now expressed in this probability theoretic language. Over which underlying "sample space" the considered statistic is here defined?

Starting slowly: What is precisely wrong in my quoted statement above?

Last edited:
It's not so clear what you mean by "pick a state". Quantum theory describes probabilities for the outcome of measurements on a quantum system prepared in a quantum state.

The Tortoise-Man said:
what is then in this context a "statistic" precisely?
The word "statistics" in this context is used more for historical reasons than anything else. "Bose-Einstein statistics" in this context just means that quantum field operators commute, and states stay the same under exchange. "Fermi-Dirac statistics" in this context just means that quantum field operators anticommute, and states change sign under exchange. The "spin-statistics theorem" just shows how QFT requires integer spin fields to be bosonic in the sense just stated, and half-integer spin fields to be fermionic in the sense just stated.

The historical reason why "statistics" is used in this context is that, when you are analyzing the specific case of a large number of particles in a gas in thermal equilibrium, the formula for the probability of finding a particle with a particular energy changes depending on whether the particles are bosons or fermions, and both formulas are different from the classical Maxwell-Boltzmann formula that was discovered in the 19th century when classical statistical mechanics was developed, and which leads to the "ultraviolet catastrophe" that led Planck in 1900 to first propose the quantum hypothesis.

The Tortoise-Man
vanhees71 said:
It's not so clear what you mean by "pick a state". Quantum theory describes probabilities for the outcome of measurements on a quantum system prepared in a quantum state.

Yes, that was based on my attept to interpret the notion of "statistic" here in this context in the stochatistic language , falsely assuming (compare with @PeterDonis' answer #11) that here a "statistic" can be also regarded directly as probabilistic distribution on certain "sample space" ##\Omega##. And therefore by "pick a state" formulation I thought that a base vector in Fock space (= a "state") could be identified as certain event in the underlying "sample space" - regarding a "statistic" in stochastic language.

And there - in stochastic setting -you can literally say "pick an element or an subset" of the underlying sample space ##\Omega## and the "statistic" gives it's probability.

But so far I understand now based on Peter Donis' explanations in #No 11 in QFT a "statistic" in this context primary means really the "datum" encoding the behavior & properties of the field operators (eg which bracket rule is choosen), so in light of this definition of a "statistic" the phrase "pick a state" of course not make any sense.

But (a guess): is it maybe possible if we start with this quantum theortical definition of a "statistic" (= as set of rules for operators), and "construct" from this datum canonically a "statistic" in stochastic language (compare with this; it seems there is an intime link between these two notions of the term "statistic")

#EDIT: This seems exactly to be the loosesly connection between the two notions of "statistic" which @PeterDonis explains in second part of #No 11

Last edited:
The Tortoise-Man said:
compare with this; it seems there is an intime link between these two notions of the term "statistic"
What specifically in that article gives you this impression?

Unfortunately it's a bit more complicated than that. One must be careful with what represents a state. It is sufficient to discuss pure states. These are uniquely determined not by the Hilbert-space vectors but by the corresponding projection operators, ##\hat{\rho}_\psi = |\psi \rangle \langle \psi|## with ##|\psi \rangle## a Hilbert-space vector normalized to ##1##. Now ##\hat{\rho}_{\psi}## doesn't change if one uses ##|\psi' \rangle =\exp(\mathrm{i} \varphi) |\psi \rangle## with ##\varphi \in \mathbb{R}##, i.e., in this case you get the same pure state ##\hat{\rho}_{\psi'} = \exp(\mathrm{i} \varphi) |\psi \rangle \langle \psi|\exp(-\mathrm{i} \varphi)=|\psi \rangle \langle \psi|=\hat{\rho}_{\psi}##.

Let's now first concentrate on the simple case of only two indistinguishable particles. If ##\mathcal{H}## is the single-particle Hilbert space then the two-particle space must be a subspace of the product space, ##\mathcal{H}_2=\mathcal{H} \otimes \mathcal{H}##. Now you can define the exchange of the two particles ##\hat{P}## by its action on an arbitrary product basis,
$$\hat{P} |\xi_1,\xi_2 \rangle=|\xi_2,\xi_1 \rangle.$$
This implies that ##\hat{P}## is unitary and ##\hat{P}^2=\hat{1}## (which implies that ##\hat{P}## is also self-adjoint).

Now we have to come to the indistinguishability of the particles. This means that our theory must be built such that there is no observable on the two particles that can distinguish the particles, and this implies that for all observable-operators ##\hat{A}## we must have
$$\hat{P} \hat{A} \hat{P}^{\dagger}=\hat{A} \; \Rightarrow \; [\hat{A},\hat{P}]=0.$$
For each observable-operator there must thus be a common eigensystem of ##\hat{A}## and ##\hat{P}##. The possible eigenvalues of ##\hat{P}## are obviously ##\pm 1##.

Since we define (elementary) particles such that the observable algebra is represented by irrducible representations it follows from the fact that ##\hat{P}## commutes with all observable-operators, ##\hat{P}## must be proportional to the unit operator. Thus for each sort of particles the correct two-particle Hilbert space for two indistinguishable particles can only be either the space, where ##\hat{P}=+\hat{1}## (bosons), or ##\hat{P}=-\hat{1}## (fermions).

That means that the entire observable algebra must be realized on either the bosonic Fock space,
##\mathcal{H}_2^+##, which is spanned not by all product vectors of single-particle basis vectors but only by the symmetrized ones,
$$|\xi_1,\xi_2 \rangle_+=\frac{1}{\sqrt{2}} (|\xi_1 \rangle \otimes \xi_2 \rangle + |\xi_2 \rangle \otimes |\xi_1 \rangle.$$

In the analogous way you argue that the fermionic Fock space is spanned by the corresponding antisymmetrized products.

If you now have more than 2 indistinguishable particles, the argument is not as simple since now you have to consider all kinds of unitary representations of the permutation groups of ##N## particles with ##N \geq 3##, and there the symmetries under arbitrary reorderings of the individual particles within an ##N##-particle state are not given only by the trivial representation, i.e., all ##\hat{P}##-operators being simply ##\hat{1}## or ##\hat{P}=\sign(P) \hat{1}##, where ##\sign(P)## is the signum of the permutation, i.e., it's +1 if you need an even number of pair interchanges to get the permutation ##P## and (-1) if you need an odd number of such pair interchanges but also non-trivial other representations.

There are, however, topological arguments in connection with the representation of the rotation group in 3 dimensions, which exclude any such non-trivial realizations of the symmetry under particle interchanges, i.e., the realization of the indistinguishability of particles in many-body systems, i.e., also for more than ##2## particles, there can be only bosons or fermions, i.e., totally symmetric or antisymmetric Hilbert space under particle interchanges.

Only for 2D systems, there can be more complicated realizations of the indistinguishability, i.e., you can also have what's called "anyons", and indeed such beasts have been found as "quasiparticles" in condensed-matter systems which are effectively retricting the particles to live in a 2D space (like graphene).

For the topological argument see

https://arxiv.org/abs/1610.09260

## What is the anti-commutation relation for quantized fields?

The anti-commutation relation for quantized fields is a mathematical expression that specifies how fermionic field operators behave when they are interchanged. For two fermionic field operators, ψ(x) and ψ(y), the anti-commutation relation is given by {ψ(x), ψ(y)} = ψ(x)ψ(y) + ψ(y)ψ(x) = 0, where { , } denotes the anti-commutator. This implies that swapping the order of two fermionic operators results in a sign change.

## Why do fermionic fields obey anti-commutation relations?

Fermionic fields obey anti-commutation relations because of the Pauli exclusion principle, which states that no two fermions can occupy the same quantum state simultaneously. The anti-commutation relations naturally enforce this principle by ensuring that the wavefunction of a system of fermions changes sign when two fermions are exchanged, thus becoming zero if they are in the same state.

## How do anti-commutation relations differ from commutation relations?

Anti-commutation relations apply to fermionic fields and involve the sum of the product of two operators in both orders, resulting in a sign change when the operators are swapped. Commutation relations, on the other hand, apply to bosonic fields and involve the difference of the product of two operators in both orders. For example, for bosonic operators φ(x) and φ(y), the commutation relation is [φ(x), φ(y)] = φ(x)φ(y) - φ(y)φ(x). In contrast, for fermionic operators ψ(x) and ψ(y), the anti-commutation relation is {ψ(x), ψ(y)} = ψ(x)ψ(y) + ψ(y)ψ(x).

## What role do anti-commutation relations play in quantum field theory?

Anti-commutation relations are crucial in quantum field theory for describing the behavior of fermions, such as electrons, quarks, and neutrinos. They ensure that the quantized fields representing these particles obey the Pauli exclusion principle. These relations are fundamental to the construction of Fermi-Dirac statistics and the formulation of the Standard Model of particle physics, which describes the interactions of elementary particles.

## Can anti-commutation relations be applied to bosonic fields?

No, anti-commutation relations are specific to fermionic fields. Bosonic fields, which describe particles like photons and gluons, obey commutation relations instead. This difference arises from the intrinsic properties of fermions and bosons: fermions have half-integer spins and follow the Pauli exclusion principle, while bosons have integer spins and can occupy the same quantum state without restriction. Therefore, the mathematical framework for

Replies
18
Views
4K
Replies
13
Views
3K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
10
Views
1K
Replies
67
Views
10K