Return to Answer

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.

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.

To obtain the actual Dirac fields, $\mathfrak{h}$ needs to have the form $\mathfrak{h}=\mathfrak{k}\oplus\mathfrak{k}$, where this direct sum is orthogonal. These direct summands correspond 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 $$\psi(f)=\frac{1}{\sqrt{2}}(b(f)+c^*(f))\ ,\psi^*(f)=\psi(f)^*=\frac{1}{\sqrt{2}}(c(f)+b^*(f))\ ,\quad f\in\mathfrak{k}\ .$$ As before, we point that $f\mapsto\psi(f)$ and $f\mapsto\psi^*(f)$ are only real linear. 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 $$\{\psi(f),\psi(g)\}=0\ ,\,\{\psi(f),\psi^*(g)\}=\langle f,g\rangle\mathbf{1}\ .$$ We remark that, as before with the Majorana field operators, $b(f)$, $c(f)$ (and therefore also $a((f_1,f_2))=b(f_1)+c(f_2)$) can be recovered from $\psi(f)$ and $\psi^*(f)$ for each $f(,f_1,f_2)\in\mathfrak{k}$ in a way similar to the recovery of the bosonic annihilation operators from the free bosonic field operators - to wit, $$b(f)=\frac{1}{\sqrt{2}}(\psi(f)+i\psi(if))\ ,\,c(f)=\frac{1}{\sqrt{2}}(\psi^*(f)+i\psi^*(if))\ ,$$ hence $$b^*(f)=\frac{1}{\sqrt{2}}(\psi^*(f)-i\psi^*(if))\ ,\,c^*(f)=\frac{1}{\sqrt{2}}(\psi(f)-i\psi(if))\ .$$ 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.

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.

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 $$\psi_R(f)=\frac{1}{\sqrt{2}}(a^*(f)+a(f))\ .$$

To obtain the actual Dirac fields, $\mathfrak{h}$ needs to have the form $\mathfrak{h}=\mathfrak{k}\oplus\mathfrak{k}$, where this direct sum is orthogonal. These direct summands correspond 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 $$\psi(f)=\frac{1}{\sqrt{2}}(b(f)+c^*(f))\ ,\psi^*(f)=\psi(f)^*=\frac{1}{\sqrt{2}}(c(f)+b^*(f))\ ,\quad f\in\mathfrak{k}\ .$$ We point that $h\mapsto\psi_R(h)$, $f\mapsto\psi(f)$ and $f\mapsto\psi^*(f)$ are only real linear. 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 $$\{\psi(f),\psi(g)\}=0\ ,\,\{\psi(f),\psi^*(g)\}=\langle f,g\rangle\mathbf{1}\ .$$ We remark that $b(f)$, $c(f)$ (and therefore $a((f_1,f_2))=b(f_1)+c(f_2)$) can be recovered from $\psi(f)$ and $\psi^*(f)$ for each $f(,f_1,f_2)\in\mathfrak{k}$ in a way similar to the recovery of the bosonic annihilation operators from the free bosonic field operators - to wit, $$b(f)=\frac{1}{\sqrt{2}}(\psi(f)+i\psi(if))\ ,\,c(f)=\frac{1}{\sqrt{2}}(\psi^*(f)+i\psi^*(if))\ ,$$ hence $$b^*(f)=\frac{1}{\sqrt{2}}(\psi^*(f)-i\psi^*(if))\ ,\,c^*(f)=\frac{1}{\sqrt{2}}(\psi(f)-i\psi(if))\ .$$ 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.

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 $\mathfrak{h}=\mathfrak{k}\oplus\mathfrak{k}$, where this direct sum is orthogonal. These direct summands correspond 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 $$\psi(f)=\frac{1}{\sqrt{2}}(b(f)+c^*(f))\ ,\psi^*(f)=\psi(f)^*=\frac{1}{\sqrt{2}}(c(f)+b^*(f))\ ,\quad f\in\mathfrak{k}\ .$$ As before, we point that $f\mapsto\psi(f)$ and $f\mapsto\psi^*(f)$ are only real linear. 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 $$\{\psi(f),\psi(g)\}=0\ ,\,\{\psi(f),\psi^*(g)\}=\langle f,g\rangle\mathbf{1}\ .$$ We remark that, as before with the Majorana field operators, $b(f)$, $c(f)$ (and therefore also $a((f_1,f_2))=b(f_1)+c(f_2)$) can be recovered from $\psi(f)$ and $\psi^*(f)$ for each $f(,f_1,f_2)\in\mathfrak{k}$ in a way similar to the recovery of the bosonic annihilation operators from the free bosonic field operators - to wit, $$b(f)=\frac{1}{\sqrt{2}}(\psi(f)+i\psi(if))\ ,\,c(f)=\frac{1}{\sqrt{2}}(\psi^*(f)+i\psi^*(if))\ ,$$ hence $$b^*(f)=\frac{1}{\sqrt{2}}(\psi^*(f)-i\psi^*(if))\ ,\,c^*(f)=\frac{1}{\sqrt{2}}(\psi(f)-i\psi(if))\ .$$ 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.