In section X.7 of [Reed & Simon's book](https://www.amazon.com.br/II-Fourier-Analysis-Self-Adjointness-2/dp/0125850026) there is a nice rigorous construction of the free scalar field theory which applies to the Klein-Gordon field.


**Question:** Are there references which discuss, in analogous fashion, the construction of (rigorous) fermionic field theories? At the moment, I'm more interested in the Dirac field theory because it includes the spin and two kinds of creation and annihilation operators.



Rigorous construction of fermionic field theory?
There is the construction of the C${}^\*\!$-algebra of canonical anticommutation relations (CAR's), which is actually somewhat easier than the construction of free bosonic fields: given any complex pre-Hilbert space $\mathfrak{h}$, which may be thought of as our "one-particle" space, define the unital \*-algebra $\text{CAR}\_0(\mathfrak{h})$ given by the generators $a(f),a^\*(f)=a(f)^\*$, $f\in\mathfrak{h}$ (respectively, the *annihilation* and *creation* operators for the particle with wave function $f$), and the following relations:


* The map $f\mapsto a(f)$ is *anti-linear* (hence the map $f\mapsto a^\*(f)$ is *linear*);
* $\{a(f),a(g)\}=0$, where $\{A,B\}=AB+BA$ is the *anti-commutator* of two elements $A,B$ in an associative algebra (hence $\{a^\*(f),a^\*(g)\}=\{a(f),a(g)\}^\*=0$);
* $\{a(f),a^\*(g)\}=\langle f,g\rangle\mathbf{1}$, where we assume that the scalar product $\langle\cdot,\cdot\rangle$ of $\mathfrak{h}$ is anti-linear in the first variable and linear in the second variable.


Notice that the CAR's imply that $$(a^\*(f)a(f))^2=a^\*(f)\{a(f),a^\*(f)\}a(f)=\|f\|^2a^\*(f)a(f)\ ,\quad f\in\mathfrak{h}$$ hence if there is a C${}^\*\!$-norm $\|\cdot\|$ on $\text{CAR}\_0(\mathfrak{h})$, we must have $$\|(a^\*(f)a(f))^2\|=\|a^\*(f)a(f)\|^2=\|f\|^2\|a^\*(f)a(f)\|\ ,$$ therefore $\|a(f)\|=\|a^\*(f)\|=\|f\|$ for all $f\in\mathfrak{h}$. In other words, unlike for bosonic fields, fermionic creation and annihilation operators are necessarily *bounded*, thanks to the Pauli exclusion principle encoded in the CAR's. To show that such a C${}^\*\!$-norm actually exists, notice that there is a nontrivial \*-representation of $\text{CAR}\_0(\mathfrak{h})$ in the fermionic (i.e. anti-symmetric) Fock space $\mathfrak{F}\_-(\mathfrak{h})$ generated by $\mathfrak{h}$: $$\mathfrak{F}\_-(\mathfrak{h})=\bigoplus^\infty\_{n=0}\wedge^n\mathfrak{h}\ ,$$ where $\wedge^0\mathfrak{h}=\mathbb{C}$, $\wedge^1\mathfrak{h}=\mathfrak{h}$, and $\wedge^n\mathfrak{h}$ for $n>1$ is the vector space generated by the $n$-fold wedge (i.e. anti-symmetrized tensor) product of elements of $\mathfrak{h}$: $$f\_1\wedge\cdots\wedge f\_n=\frac{1}{n!}\sum\_{\sigma\in\mathbb{S}\_n}\epsilon\_\sigma f\_{\sigma(1)}\otimes\cdots\otimes f\_{\sigma(n)}\ ,$$ where $\mathbb{S}\_n$ is the group of permutations of $n$ elements and $\epsilon\_\sigma$ ( = $(-1)^{inv(\sigma)}$, $inv(\sigma)=$number of inversions of $\sigma\in\mathbb{S}\_n$) is the sign of the permutation $\sigma$, endowed with the scalar product $$\langle f\_1\wedge\cdots\wedge f\_n,g\_1\wedge\cdots\wedge g\_n\rangle=\det[\langle f\_i,g\_j\rangle]\ .$$ The direct sum is assumed to be orthogonal. The action of $a(f),a^\*(f)$ on $\mathfrak{F}\_-(\mathfrak{h})$ is given by $$a(f)\lambda=0\ ,\,a^\*(f)\lambda=\lambda f\ ,\quad\lambda\in\mathbb{C}\ ,$$ $$a(f)f\_1\wedge\cdots\wedge f\_n=\frac{1}{\sqrt{n}}\langle f,f\_1\rangle f\_2\wedge\cdots\wedge\cdots\wedge f\_n\ ,$$ $$a^\*(f)f\_1\wedge\cdots\wedge f\_n=\sqrt{n+1}f\wedge f\_1\wedge\cdots\wedge f\_n\ ,\quad f\_1,\ldots,f\_n\in\mathfrak{h}\ .$$ It is easy to verify that this defines a \*-representation of $\text{CAR}\_0(\mathfrak{h})$ by bounded linear operators on $\mathfrak{F}\_-(\mathfrak{h})$ (boundedness is guaranteed by the CAR's as shown above). This ensures that the completion of $\text{CAR}\_0(\mathfrak{h})$ w.r.t. the maximal C${}^\*\!$-norm $\|\cdot\|$ is a C${}^\*\!$-algebra $\mathfrak{A}$ acting on the Fock Hilbert space $\overline{\mathfrak{F}\_-(\mathfrak{h})}$, called the *CAR algebra* associated to $\mathfrak{h}$. As written, one is only able to define *Majorana* (i.e. Hermitian) fermionic field operators $$\psi\_R(f)=\frac{1}{\sqrt{2}}(a^\*(f)+a(f))\ ,$$ from which $a(f)$ and $a^\*(f)$ may be recovered through the formula $$a(f)=\frac{1}{\sqrt{2}}(\psi\_R(f)+i\psi\_R(if))\ ,\,a^\*(f)=\frac{1}{\sqrt{2}}(\psi\_R(f)-i\psi\_R(if))\ ,\quad f\in\mathfrak{h}\ .$$ We remark that the map $f\mapsto\psi\_R(f)$ is only *real* linear.


To obtain the actual Dirac fields, $\mathfrak{h}$ needs to have the form $\mathfrak{h}=\mathfrak{k}\oplus\bar{\mathfrak{k}}$, where this direct sum is *orthogonal* and $\bar{\mathfrak{k}}$ is equal to $\mathfrak{k}$ as a *real* vector space, but with the *complex* scalar multiplication equaling that of $\mathfrak{k}$ composed with complex conjugation of the scalar factor. In other words, we have that for all $f,g\in\mathfrak{k}$, $\alpha\in\mathbb{C}$ (**edit - November 11th 2022**, see Alan Garbarz's comment below) $$(f,g)+(f',g')=(f+f',g+g')\ ,\,\alpha(f,g)=(\alpha f,\bar{\alpha}g)$$ and $$\langle (f,f'),(g,g')\rangle=\langle f,g\rangle+\overline{\langle f',g'\rangle}=\langle f,g\rangle+\langle g',f'\rangle\ .$$ The above definition for the scalar product of $\mathfrak{h}$ from that of $\mathfrak{k}$ guarantees sesquilinearity as well as the mutual orthogonality of both direct summands w.r.t. each other. These correspond respectively to the particle and antiparticle sectors. Defining $$b(f)=a((f,0))\ ,\,c(f)=a((0,f))\ ,\quad f\in\mathfrak{k}\ ,$$ one can define the *Dirac field operators* $$\psi(f)=\frac{1}{\sqrt{2}}(b(f)+c^\*(f))\ ,\psi^\*(f)=\psi(f)^\*=\frac{1}{\sqrt{2}}(c(f)+b^\*(f))\ ,\quad f\in\mathfrak{k}\ .$$ Now, thanks to the definition of $\bar{\mathfrak{k}}$, we have that $f\mapsto c(f)$ is actually *linear*, therefore $f\mapsto\psi(f)$ is *antilinear* and $f\mapsto\psi^\*(f)$ is *(complex) linear*, contrary to the case of Majorana field operators. The above yields the *CAR's in Dirac form* $$\{b(f),b(g)\}=\{b(f),c(g)\}=\{c(f),c(g)\}=\{b(f),c^\*(g)\}=\{c(f),b^\*(g)\}=0\ ,$$ $$\{b(f),b^\*(g)\}=\{c^\*(f),c(g)\}=\langle f,g\rangle\mathbf{1}$$ for all $f,g\in\mathfrak{k}$, hence $$\{\psi(f),\psi(g)\}=0\ ,\,\{\psi(f),\psi^\*(g)\}=\langle f,g\rangle\mathbf{1}\ .$$ All that is left is the construction of the one-particle pre-Hilbert space $\mathfrak{k}$, which can be identified with the space of initial data for positive-energy solutions of the Dirac equation.


We point that a similar procedure to that used to obtain Dirac fields can be employed for the free *complex* (bosonic) scalar field, which also enjoys a similar concept of particle and antiparticle sectors.


Apart from the last section on Dirac fields, all of the above may be found e.g. in Section 5.2 of the book by Ola Bratteli and Derek W. Robinson, *Operator Algebras and Quantum Statistical Mechanics 2. Equilibrium States, Models in Quantum Statistical Mechanics* (2nd. ed., Springer-Verlag, 1997).


