Identical Particles and Fock Space

The oscillator algebra from the previous page already knows how to count quanta. What it does not yet explain is why this is the correct language for identical particles. In ordinary one-particle quantum mechanics, a state vector describes one object. In QFT, the Hilbert space must describe any number of quanta, and those quanta usually have no permanent individual labels.

For bosons, exchanging two particles does not change the state. The $N$-particle wavefunction is symmetric,

$$\psi_N(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_N) = \psi_N(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_N),$$

and similarly in momentum space. The natural basis is therefore not a list of named particles, but a list of occupation numbers: how many particles occupy each one-particle state.

The bosonic Fock space is the direct sum of all symmetric $N$-particle sectors,

$$\mathcal F_B(\mathcal H_1) = \bigoplus_{N=0}^{\infty}\operatorname{Sym}^N\mathcal H_1.$$

Creation and annihilation operators move between adjacent sectors. They add or remove one particle in a chosen mode, and their commutation relations encode the symmetry factors that otherwise have to be inserted by hand.

The central reader’s trap is to carry over too much intuition from distinguishable particles. A bosonic Fock state is not a wavefunction for a collection of named objects. It is a vector whose components say which one-particle modes are occupied and how many times.

Identical bosons and symmetric states

Discrete one-particle basis

This page first uses an orthonormal discrete one-particle basis $|i\rangle$, $i=1,2,\ldots,K$. In a finite box, $i$ may be a momentum label $\mathbf p$. The bosonic creation and annihilation operators obey

$$[a_i,a_j^\dagger]=\delta_{ij}, \qquad [a_i,a_j]=0, \qquad [a_i^\dagger,a_j^\dagger]=0.$$

Continuum normalizations such as

$$[a(\mathbf p),a^\dagger(\mathbf q)] =(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q)$$

are introduced only after the finite-volume algebra is under control.

Imagine two particles and two available one-particle levels, labeled $1$ and $2$. If the particles were distinguishable, say particle $A$ and particle $B$, then the ordered assignments would be

$$(A:1,B:1), \qquad (A:1,B:2), \qquad (A:2,B:1), \qquad (A:2,B:2).$$

The two mixed assignments are different if the labels $A$ and $B$ are physical. For identical bosons, however, there is no physical experiment that tracks permanent labels $A$ and $B$. The states are instead

$$|2,0\rangle, \qquad |1,1\rangle, \qquad |0,2\rangle.$$

Here $|2,0\rangle$ means “two particles in level $1$ and none in level $2$,” while $|1,1\rangle$ means “one particle in each level.” The order in which we name the particles has disappeared.

For two one-particle levels, two distinguishable particles have four ordered assignments. Two identical bosons have three occupation states: $|2,0\rangle$, $|1,1\rangle$, and $|0,2\rangle$. The two mixed ordered assignments differ only by exchanging labels, so they represent the same bosonic state.

This example is also a warning about probabilities. Statements such as “the probability of a configuration is $1/3$” or “the probability is $1/4$” are not universal facts; they depend on which ensemble is being declared equally likely. The universal statement is the state-counting statement: bosonic states are labeled by occupations, not by ordered particle labels.

Let $\mathcal H_1$ be the one-particle Hilbert space. If particles carried physical labels, the $N$-particle Hilbert space would be the full tensor product

$$\mathcal H_1^{\otimes N} = \underbrace{\mathcal H_1\otimes\cdots\otimes\mathcal H_1}_{N\ \text{factors}}.$$

A coordinate-space vector in this tensor product has wavefunction

$$\psi_N(x_1,\ldots,x_N).$$

For identical bosons, exchanging any pair of arguments must leave the wavefunction unchanged:

$$\psi_N(x_{\pi(1)},\ldots,x_{\pi(N)}) = \psi_N(x_1,\ldots,x_N), \qquad \pi\in S_N.$$

Thus the bosonic $N$-particle sector is the symmetric subspace

$$\operatorname{Sym}^N\mathcal H_1 = \{\psi\in\mathcal H_1^{\otimes N}:P_\pi\psi=\psi\ \text{for all}\ \pi\in S_N\},$$

where $P_\pi$ permutes the tensor factors. The symmetrization operator is

$$\mathcal S_N = \frac{1}{N!}\sum_{\pi\in S_N}P_\pi.$$

It is a projection:

$$\mathcal S_N^2=\mathcal S_N, \qquad \mathcal S_N^\dagger=\mathcal S_N.$$

For example, if $|i\rangle$ and $|j\rangle$ are orthonormal one-particle states with $i\ne j$, the normalized two-boson state with one particle in each mode is

$$|i,j\rangle_B = \frac{1}{\sqrt2} \left(|i\rangle\otimes|j\rangle+|j\rangle\otimes|i\rangle\right).$$

If both particles are in the same state $i$, the state is simply

$$|i,i\rangle_B=|i\rangle\otimes|i\rangle.$$

There is no extra factor of $1/\sqrt2$ in this second formula, because $|i\rangle\otimes|i\rangle$ is already normalized.

Counting occupation states

Suppose there are $K$ one-particle levels and $N$ identical bosons. A bosonic state in the occupation basis is specified by nonnegative integers

$$(n_1,n_2,\ldots,n_K), \qquad n_1+n_2+\cdots+n_K=N.$$

The number of such occupation patterns is the stars-and-bars number

$$\dim\operatorname{Sym}^N\mathbb C^K = \binom{N+K-1}{N} = \binom{N+K-1}{K-1}.$$

For $N=2$,

$$\dim\operatorname{Sym}^2\mathbb C^K = \binom{K+1}{2} = \frac{K(K+1)}{2}.$$

This is the number of independent symmetric coefficients $T_{ij}=T_{ji}$ in a two-particle wavefunction expanded as

$$\psi_2=\sum_{i,j=1}^K T_{ij}|i\rangle\otimes|j\rangle, \qquad T_{ij}=T_{ji}.$$

The diagonal components $T_{ii}$ describe double occupation of one level. The off-diagonal components $T_{ij}$ with $i<j$ describe one particle in each of two different levels.

This replacement of ordered labels by occupation numbers is the first large simplification of many-particle quantum mechanics. Instead of carrying redundant information about which identical particle is which, the occupation basis keeps only the physical information.

Fock space

A fixed-$N$ Hilbert space is not enough for field theory. Interactions can create and annihilate quanta, and even when particle number happens to be conserved, the operator language is clearest when all $N$ sectors are present at once.

The bosonic Fock space over $\mathcal H_1$ is

$$\mathcal F_B(\mathcal H_1) = \mathbb C|0\rangle \oplus \mathcal H_1 \oplus \operatorname{Sym}^2\mathcal H_1 \oplus \operatorname{Sym}^3\mathcal H_1 \oplus\cdots.$$

The vector $|0\rangle$ is the vacuum, the unique state in the zero-particle sector. It is not the zero vector. It is a normalized physical state satisfying

$$\langle0|0\rangle=1, \qquad a_i|0\rangle=0.$$

A general Fock-space vector is a sequence

$$|\Psi\rangle = (\psi_0,\psi_1,\psi_2,\ldots), \qquad \psi_N\in\operatorname{Sym}^N\mathcal H_1.$$

The total particle-number operator is

$$\mathcal N=\sum_i a_i^\dagger a_i.$$

It acts diagonally:

$$\mathcal N\psi_N=N\psi_N.$$

A state with definite particle number lies entirely in one sector. A general Fock state may be a superposition of different particle numbers.

Creation and annihilation as maps between sectors

Creation and annihilation operators can be described in two closely related languages. The unnormalized symmetric-list notation makes the combinatorics visible: annihilation literally counts how many matching entries can be removed. The normalized occupation notation is better for matrix elements, probabilities, and Hamiltonian eigenstates. Moving between these two languages is one of the basic skills of QFT.

More explicitly, define

$$|p_1,\ldots,p_N\rangle_u = a_{p_1}^\dagger\cdots a_{p_N}^\dagger|0\rangle.$$

Because the creation operators commute, this state depends only on the multiset of occupied modes, not on the order of the displayed labels. If mode $q$ occurs $n_q$ times, then in a discrete orthonormal basis its squared norm is

$${}_u\langle p_1,\ldots,p_N|p_1,\ldots,p_N\rangle_u = \prod_q n_q!.$$

The subscript $u$ is therefore a reminder: these states are convenient algebraically, but are not generally normalized. Creation by a mode $p$ is written simply as

$$a_p^\dagger|p_1,\ldots,p_N\rangle_u = |p,p_1,\ldots,p_N\rangle_u.$$

Annihilation removes one occurrence of $p$ from the list and sums over all places it could have been found:

$$a_p|p_1,\ldots,p_N\rangle_u = \sum_{r=1}^N\delta_{p,p_r} |p_1,\ldots,\widehat{p_r},\ldots,p_N\rangle_u.$$

The hat means that the entry is omitted. For a single mode this reduces to

$$a^\dagger|N\rangle_u=|N+1\rangle_u, \qquad a|N\rangle_u=N|N-1\rangle_u.$$

Here the factor $N$ appears because there are $N$ identical quanta that could be removed.

The commutator follows immediately:

$$aa^\dagger|N\rangle_u=(N+1)|N\rangle_u, \qquad a^\dagger a|N\rangle_u=N|N\rangle_u,$$

so

$$[a,a^\dagger]|N\rangle_u=|N\rangle_u.$$

Since this holds for every $N$,

$$[a,a^\dagger]=1.$$

Normalized states are obtained by dividing out the factorials generated by repeated creation. For one mode,

$$|N\rangle = \frac{1}{\sqrt{N!}}(a^\dagger)^N|0\rangle.$$

Equivalently, if $|N\rangle_u=(a^\dagger)^N|0\rangle$, then

$$|N\rangle=\frac{1}{\sqrt{N!}}|N\rangle_u.$$

The same operator actions become

$$a^\dagger|N\rangle=\sqrt{N+1}\,|N+1\rangle, \qquad a|N\rangle=\sqrt N\,|N-1\rangle.$$

For many modes,

$$|n_1,n_2,\ldots\rangle = \prod_i\frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}}|0\rangle, \qquad n_i=0,1,2,\ldots,$$

where only finitely many $n_i$ are nonzero for a finite-particle state. Then

$$a_i^\dagger|n_1,\ldots,n_i,\ldots\rangle = \sqrt{n_i+1}\,|n_1,\ldots,n_i+1,\ldots\rangle,$$

and

$$a_i|n_1,\ldots,n_i,\ldots\rangle = \sqrt{n_i}\,|n_1,\ldots,n_i-1,\ldots\rangle.$$

The mode number operator is

$$N_i=a_i^\dagger a_i,$$

and

$$N_i|n_1,n_2,\ldots\rangle=n_i|n_1,n_2,\ldots\rangle.$$

For normalized bosonic occupation states, $a^\dagger$ raises the occupation number with coefficient $\sqrt{n+1}$, while $a$ lowers it with coefficient $\sqrt n$. In an unnormalized basis the same rules are $a^\dagger|n\rangle_u=|n+1\rangle_u$ and $a|n\rangle_u=n|n-1\rangle_u$.

The square-root coefficients are not arbitrary conventions. They are the unique coefficients compatible with normalized states and the commutator $[a_i,a_j^\dagger]=\delta_{ij}$.

Inner products and repeated momenta

When all momenta are distinct, a symmetrized $N$-particle state can be written as

$$|p_1,\ldots,p_N\rangle_B = \frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} |p_{\pi(1)}\rangle\otimes\cdots\otimes|p_{\pi(N)}\rangle.$$

This formula must be modified when some momenta are repeated, because the sum over permutations then repeats identical tensor products. The occupation-number formula avoids this trap automatically.

If the mode $q$ occurs $n_q$ times, the normalized occupation state is

$$|\{n_q\}\rangle = \prod_q\frac{(a_q^\dagger)^{n_q}}{\sqrt{n_q!}}|0\rangle.$$

In tensor-product notation, if $p_1,\ldots,p_N$ is a list with occupation numbers $n_q$, then the same normalized bosonic state can be written as

$$|\{n_q\}\rangle = \frac{1}{\sqrt{N!\prod_q n_q!}} \sum_{\pi\in S_N} |p_{\pi(1)}\rangle\otimes\cdots\otimes|p_{\pi(N)}\rangle.$$

The extra factor $\prod_q n_q!$ compensates for repeated terms in the permutation sum. Check the two extremes: if all $p_i$ are distinct, every $n_q$ is $0$ or $1$ and the coefficient reduces to $1/\sqrt{N!}$; if all $N$ particles occupy the same mode, the permutation sum contains $N!$ identical terms and the formula gives a single normalized tensor product. This is one of the places where occupation numbers are not merely convenient; they are safer than ordered particle labels.

Hamiltonian and wavefunctions in occupation language

Let the one-particle Hamiltonian have eigenstates $|i\rangle$ with energies $\varepsilon_i$:

$$h|i\rangle=\varepsilon_i|i\rangle.$$

The corresponding free many-boson Hamiltonian is

$$H=\sum_i\varepsilon_i a_i^\dagger a_i.$$

Acting on an occupation state,

$$H|n_1,n_2,\ldots\rangle = \left(\sum_i\varepsilon_i n_i\right)|n_1,n_2,\ldots\rangle.$$

In a relativistic momentum basis,

$$\varepsilon_i\rightarrow\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2},$$

and therefore

$$H=\sum_{\mathbf p}\omega_{\mathbf p}a_{\mathbf p}^\dagger a_{\mathbf p}.$$

A state with occupation numbers $n_{\mathbf p}$ has energy

$$E=\sum_{\mathbf p}n_{\mathbf p}\omega_{\mathbf p}.$$

This is the additive many-particle spectrum, now written without naming the particles.

A vector in the $N$-particle sector can also be represented by a coordinate-space wavefunction

$$\psi_N(\mathbf x_1,\ldots,\mathbf x_N) = \langle\mathbf x_1,\ldots,\mathbf x_N|\psi_N\rangle,$$

or by a momentum-space wavefunction

$$\widetilde\psi_N(\mathbf p_1,\ldots,\mathbf p_N) = \langle\mathbf p_1,\ldots,\mathbf p_N|\psi_N\rangle.$$

For bosons both are symmetric:

$$\psi_N(\mathbf x_{\pi(1)},\ldots,\mathbf x_{\pi(N)}) = \psi_N(\mathbf x_1,\ldots,\mathbf x_N),$$

and

$$\widetilde\psi_N(\mathbf p_{\pi(1)},\ldots,\mathbf p_{\pi(N)}) = \widetilde\psi_N(\mathbf p_1,\ldots,\mathbf p_N).$$

For example, the normalized two-boson state with one particle in one-particle wavefunction $f$ and one in $g$, with $\langle f|g\rangle=0$, has coordinate wavefunction

$$\psi_2(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \left(f(\mathbf x_1)g(\mathbf x_2)+g(\mathbf x_1)f(\mathbf x_2)\right).$$

The same state in Fock notation is

$$a_f^\dagger a_g^\dagger|0\rangle,$$

where

$$a_f^\dagger=\sum_i f_i a_i^\dagger, \qquad f_i=\langle i|f\rangle.$$

The operator notation hides the symmetrization because the commutation relation already performs it. This is the main practical advantage of second quantization: the many-particle symmetry is built into the algebra, so calculations can focus on operators rather than repeatedly symmetrizing wavefunctions.

A two-level check

Let the one-particle levels have energies $\varepsilon_1$ and $\varepsilon_2$. The two-boson sector has three normalized occupation states:

$$|2,0\rangle = \frac{(a_1^\dagger)^2}{\sqrt{2!}}|0\rangle,$$ $$|1,1\rangle = a_1^\dagger a_2^\dagger|0\rangle,$$

and

$$|0,2\rangle = \frac{(a_2^\dagger)^2}{\sqrt{2!}}|0\rangle.$$

The free Hamiltonian

$$H=\varepsilon_1 a_1^\dagger a_1 + \varepsilon_2 a_2^\dagger a_2$$

has eigenvalues

$$H|2,0\rangle=2\varepsilon_1|2,0\rangle,$$ $$H|1,1\rangle=(\varepsilon_1+\varepsilon_2)|1,1\rangle,$$

and

$$H|0,2\rangle=2\varepsilon_2|0,2\rangle.$$

Now act with $a_1$:

$$a_1|2,0\rangle=\sqrt2|1,0\rangle, \qquad a_1|1,1\rangle=|0,1\rangle, \qquad a_1|0,2\rangle=0.$$

The coefficient $\sqrt2$ is not a mysterious interaction. It is the square root of the number of identical quanta available in mode $1$.

Summary

Fock space replaces particle labels by mode occupations. For bosons,

$$\mathcal F_B(\mathcal H_1) = \bigoplus_{N=0}^{\infty}\operatorname{Sym}^N\mathcal H_1,$$

and the normalized occupation state is

$$|n_1,n_2,\ldots\rangle = \prod_i\frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}}|0\rangle.$$

Creation and annihilation act by

$$a_i^\dagger|\ldots,n_i,\ldots\rangle = \sqrt{n_i+1}\,|\ldots,n_i+1,\ldots\rangle,$$

and

$$a_i|\ldots,n_i,\ldots\rangle = \sqrt{n_i}\,|\ldots,n_i-1,\ldots\rangle.$$

The free Hamiltonian is diagonal in this basis:

$$H=\sum_i\varepsilon_i a_i^\dagger a_i, \qquad H|n_1,n_2,\ldots\rangle = \left(\sum_i n_i\varepsilon_i\right)|n_1,n_2,\ldots\rangle.$$

The physical lesson is sharper than the notation suggests. The statement “there are three bosons in mode $i$” is meaningful. The statement “this particular boson is particle number one” is not. Creation and annihilation operators are the linear maps that respect this fact.

Common pitfalls

Treating exchange as a physical motion

Exchange symmetry is not the statement that two particles physically travel around each other. It is a statement about how the state is represented when two identical arguments or labels are interchanged. For bosons, the state is invariant under this relabeling.

Confusing symmetric states with product states

The state

$$\frac{1}{\sqrt2} \left(|i\rangle\otimes|j\rangle+|j\rangle\otimes|i\rangle\right)$$

is not the same vector as $|i\rangle\otimes|j\rangle$ in the full tensor product. It lies in the symmetric subspace. The occupation notation $|1_i,1_j\rangle$ automatically means the symmetric state.

Forgetting normalization when particles occupy the same mode

The normalized state with two bosons in the same mode is

$$|2_i\rangle=\frac{(a_i^\dagger)^2}{\sqrt{2!}}|0\rangle.$$

Dropping the $\sqrt{2!}$ gives an unnormalized state. That may be harmless in a short algebraic derivation if one is consistent, but it gives wrong matrix elements if used as a normalized state.

Assuming Bose counting fixes probabilities by itself

Bose symmetry tells us which states are distinct. It does not by itself determine a probability distribution over those states. Probabilities require dynamics, a density matrix, or an ensemble assumption.

Mixing fixed-particle and Fock-space notation

A wavefunction $\psi_N(x_1,\ldots,x_N)$ lives in a fixed $N$-particle sector. A Fock vector contains all sectors at once. Operators like $a_i$ and $a_i^\dagger$ move between sectors, so they are not operators on a single fixed-$N$ Hilbert space alone.

Exercises

Exercise 1: counting bosonic states

Show that the number of ways to put $N$ identical bosons into $K$ one-particle levels is

$$\binom{N+K-1}{N}.$$

Check explicitly that for $N=2$ and $K=3$ the answer is $6$.

Solution

A bosonic occupation state is specified by nonnegative integers

$$(n_1,\ldots,n_K), \qquad n_1+\cdots+n_K=N.$$

This is the standard stars-and-bars problem. Represent the $N$ particles as stars and use $K-1$ bars to separate the $K$ levels. For example, with $K=4$,

$$\star\star|\ |\star|\star\star$$

represents $(2,0,1,2)$. There are $N$ stars and $K-1$ bars, so there are $N+K-1$ slots in total. Choosing the $N$ slots occupied by stars gives

$$\binom{N+K-1}{N}.$$

For $N=2$ and $K=3$,

$$\binom{2+3-1}{2}=\binom{4}{2}=6.$$

The six states are

$$(2,0,0),\quad (0,2,0),\quad (0,0,2), \quad (1,1,0),\quad (1,0,1),\quad (0,1,1).$$

Exercise 2: deriving the ladder coefficients

Starting from

$$|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle, \qquad [a,a^\dagger]=1, \qquad a|0\rangle=0,$$

show that

$$a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle, \qquad a|n\rangle=\sqrt n\,|n-1\rangle.$$

Solution

For creation,

$$a^\dagger|n\rangle = \frac{(a^\dagger)^{n+1}}{\sqrt{n!}}|0\rangle.$$

Since

$$|n+1\rangle = \frac{(a^\dagger)^{n+1}}{\sqrt{(n+1)!}}|0\rangle,$$

we have

$$a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle.$$

For annihilation, use

$$a(a^\dagger)^n=(a^\dagger)^n a+n(a^\dagger)^{n-1}.$$

Acting on the vacuum removes the first term because $a|0\rangle=0$, so

$$a|n\rangle = \frac{n(a^\dagger)^{n-1}}{\sqrt{n!}}|0\rangle.$$

Using

$$|n-1\rangle = \frac{(a^\dagger)^{n-1}}{\sqrt{(n-1)!}}|0\rangle,$$

we get

$$a|n\rangle = \frac{n\sqrt{(n-1)!}}{\sqrt{n!}}|n-1\rangle = \sqrt n\,|n-1\rangle.$$

Exercise 3: annihilation from an unnormalized symmetric list

Let

$$|p_1,p_2,p_3\rangle_u$$

be an unnormalized symmetric momentum state. Use

$$a_q|p_1,\ldots,p_N\rangle_u = \sum_{r=1}^N\delta_{q,p_r}|p_1,\ldots,\widehat{p_r},\ldots,p_N\rangle_u$$

to compute $a_q|p,p,q\rangle_u$ and $a_p|p,p,q\rangle_u$ for $p\ne q$.

Solution

First remove a particle of momentum $q$:

$$a_q|p,p,q\rangle_u = \delta_{q,p}|p,q\rangle_u + \delta_{q,p}|p,q\rangle_u + \delta_{q,q}|p,p\rangle_u.$$

Since $p\ne q$, this becomes

$$a_q|p,p,q\rangle_u=|p,p\rangle_u.$$

Now remove a particle of momentum $p$:

$$a_p|p,p,q\rangle_u = \delta_{p,p}|p,q\rangle_u + \delta_{p,p}|p,q\rangle_u + \delta_{p,q}|p,p\rangle_u.$$

Again using $p\ne q$,

$$a_p|p,p,q\rangle_u=2|p,q\rangle_u.$$

The factor $2$ appears because there are two identical entries with momentum $p$ that can be removed.

Exercise 4: two-boson wavefunction from operators

Let $f$ and $g$ be orthonormal one-particle wavefunctions. Define

$$a_f^\dagger=\sum_i f_i a_i^\dagger, \qquad a_g^\dagger=\sum_i g_i a_i^\dagger,$$

where $f_i=\langle i|f\rangle$ and $g_i=\langle i|g\rangle$. Show that the state

$$a_f^\dagger a_g^\dagger|0\rangle$$

corresponds to the normalized symmetric coordinate wavefunction

$$\psi_2(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \left(f(\mathbf x_1)g(\mathbf x_2)+g(\mathbf x_1)f(\mathbf x_2)\right).$$

Solution

In an unsymmetrized tensor product, the product state with one particle in $f$ and one in $g$ would be

$$|f\rangle\otimes|g\rangle.$$

For identical bosons we must project to the symmetric subspace:

$$\mathcal S_2(|f\rangle\otimes|g\rangle) = \frac12\left(|f\rangle\otimes|g\rangle+|g\rangle\otimes|f\rangle\right).$$

Because $f$ and $g$ are orthonormal, the norm of the numerator is

$$\langle f,g|f,g\rangle + \langle f,g|g,f\rangle + \langle g,f|f,g\rangle + \langle g,f|g,f\rangle =1+0+0+1=2.$$

Thus the normalized state is

$$\frac{1}{\sqrt2} \left(|f\rangle\otimes|g\rangle+|g\rangle\otimes|f\rangle\right).$$

Taking the coordinate-space matrix element gives

$$\psi_2(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \left(f(\mathbf x_1)g(\mathbf x_2)+g(\mathbf x_1)f(\mathbf x_2)\right).$$

The operator state $a_f^\dagger a_g^\dagger|0\rangle$ gives exactly this vector because the creation operators commute:

$$a_f^\dagger a_g^\dagger=a_g^\dagger a_f^\dagger.$$

So the ordering of creation is not a physical label; it only constructs the symmetric state.

References and further reading

• Sidney Coleman, Lectures of Sidney Coleman on Quantum Field Theory, Chapter 2.
• Mark Srednicki, Quantum Field Theory, Chapters 2–4.
• Steven Weinberg, The Quantum Theory of Fields, Volume I, Chapters 2, 4, and 5.
• Michael E. Peskin and Daniel V. Schroeder, An Introduction to Quantum Field Theory, Chapter 2.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Chapter 2.