# Rigorous construction of fermionic field theory?

In section X.7 of Reed & Simon's book 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.

• Just a remark: Fermionic QFTs are easier to construct than Bosonic ones. May 27, 2021 at 14:29

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).

• Pedro, thanks for the answer. Although I'm not much familiar with $C^{*}$-algebras, I got the idea and appreciate the details and quality of your exposure. I know Bratelli & Robinson's book, so I might need to learn a little bit of $C^{*}$-algebra to fully understand your reasoning. In any case, your answer will be an important guide for me. Also, this discussion led me to think about another topic, and I should post another question soon. May 26, 2021 at 21:14
• You can look at Volume 1 of Bratteli-Robinson for an introduction to C*-algebras. Another nice book on the subject is G. Murphy's C${}^*\!$-Algebras and Operator Theory, but BR1 is more than enough to follow Volume 2. May 26, 2021 at 21:22
• I will take a look at both references! Thanks! May 26, 2021 at 22:03
• Pedro, by the way, you probably know this but there is a book by E. de Faria and W. de Mello called "Mathematical Aspects of Quantum Field Theory" where the quantization of Dirac fields is briefly discussed. But, there, the authors introduce Grassmann variables instead and I don't understand where the quantization is actually discussed. Anyways, I thought the quantization using Grassmann variables was used only when one is trying to quantize it via path integrals. If possible, could you comment on that? May 27, 2021 at 14:43
• Pedro, could it be that in the Dirac construction, as it is written, the anticommutator $\left\{\psi(f),\psi^*(g)\right\}=\text{Re}\langle f , g \rangle \textbf{1}$? Instead, in order to get the inner product on th RHS of the anticommutator, may be one can take the conjugate (pre)Hilbert space $\overline{\mathfrak{k}}$ in the second slot, I mean $\mathfrak{h}=\mathfrak{k} \oplus \overline{\mathfrak{k}}$ ? Also this would make $f \mapsto \psi(f)$ anti-linear. Nov 9, 2022 at 13:51