5.3 Canonical quantization of the Dirac field
The quantization of the Dirac field can be achieved by promoting the fixed-time Dirac field \(\Psi (\mathbf {x}) = \Psi (t_0,\mathbf {x})\) and its canonical momentum \(\pi _{\Psi }(\mathbf {x}) = \ii \,\Psi ^\dagger (\mathbf {x}) =\ii \, \Psi ^\dagger (t_0,\mathbf {x})\) to operators. However, in contrast to the Klein-Gordon field discussed in Section 3.1, one can not use equal-time commutation relations as in (3.1) to obtain a consistent quantization of the Dirac field. The reason for this is rather deep and it is rooted in the so-called spin-statistics theorem of QFT. Without going too much into the details, which lie beyond the scope of this module, the main insight from this theorem is that integer spin fields must be quantized by using commutators and that half-integer spin fields must be quantized by using anticommutators. Recall that the anticommutator of two operators \(A\) and \(B\) is defined by \(\{A,B\}:= A\,B + B\,A\). Since the Dirac field is spin \(\frac {1}{2}\), and hence in particular of half-integer spin, it makes perfectly sense in the light of the spin-statistics theorem to start instead of (3.1) with the following equal-time anticommutation relations
\begin{flalign} \big \{\Psi _\alpha (\mathbf {x}),\Psi _\beta (\mathbf {y})\big \} \,&=\,0\, =\, \big \{\Psi ^\dagger _\alpha (\mathbf {x}),\Psi ^\dagger _\beta (\mathbf {y})\big \} \quad ,\\[4pt] \qquad \big \{\Psi _\alpha (\mathbf {x}),\Psi ^\dagger _\beta (\mathbf {y})\big \}\,&=\,\delta _{\alpha \beta }\,\delta (\mathbf {x}-\mathbf {y})\quad , \end{flalign}
where the indices \(\alpha ,\beta \) are used to denote the components of Dirac spinors. Note that there is no factor of \(\ii \) on the right-hand side of the nontrivial anticommutation relation, because the canonical momentum differs from \(\Psi ^\dagger \) by a factor of \(\ii \), namely \(\pi _{\Psi }(\mathbf {x}) = \ii \,\Psi ^\dagger (\mathbf {x})\). To obtain the Heisenberg picture Dirac field operator \(\Psi (x)\), which is an operator on spacetime \(x\in \bbR ^d\), one has to solve Heisenberg’s equation
\(\seteqnumber{0}{5.}{73}\)\begin{flalign} \frac {\partial }{\partial t} \Psi (x) \,=\,\ii \,\big [H,\Psi (x)\big ]\quad , \end{flalign} subject to the initial condition \(\Psi (0,\mathbf {x})=\Psi (\mathbf {x})\), where \(H\) denotes the Hamiltonian operator that is obtained by quantizing the Dirac Hamiltonian (5.68). (Normal ordering of the Hamiltonian operator \(H\) is inessential at this point, because under the commutator we have that \([\noor {H},-]=[H,-]\).) Note that, in contrast to the anticommutators \(\{- ,-\}\) in (5.73), the Heisenberg equation involves the usual commutator \([- , -]\). Working out this commutator by using (5.68) and (5.73), one finds that
\(\seteqnumber{0}{5.}{74}\)\begin{flalign} \frac {\partial }{\partial t} \Psi (x) \,=\,\gamma ^0\,\gamma ^i\,\partial _i\Psi (x) + m\,\gamma ^0\,\Psi (x)\quad \Longleftrightarrow \quad (\slashed {\partial } +m)\Psi (x) =0\quad , \end{flalign} i.e. the Heisenberg picture Dirac field operator \(\Psi (x)\) satisfies the Dirac equation.
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Remark 5.4. The interplay between commutators and anticommutators is best understood by using the concept of a graded commutator. For this we declare both the Dirac field operator \(\Psi (\mathbf {x})\) and its adjoint \(\Psi ^\dagger (\mathbf {x})\) to be odd operators, which we write in terms of the \(\bbZ _2\)-parities \(\vert \Psi (\mathbf {x})\vert =1\in \bbZ _2\) and \(\vert \Psi ^\dagger (\mathbf {x})\vert =1\in \bbZ _2\). The identity operator \(1\) is declared to be even, written as \(\vert 1\vert =0\in \bbZ _2\). This concept of odd and even can be extended to products of operators: Given two operators \(A\) and \(B\) with parities \(\vert A\vert \) and \(\vert B\vert \), then their product \(A\,B\) has parity \(\vert A\,B\vert :=(\vert A\vert + \vert B\vert ) \mod 2\in \bbZ _2\). For example, the product \(\Psi ^\dagger \,\Psi \) has parity \(\vert \Psi ^\dagger \Psi \vert = (1 + 1 )\mod 2 = 0\), hence it is an even operator. More generally, even monomials of the \(\Psi \)’s and \(\Psi ^\dagger \)’s define even operators and odd monomials are odd operators. The Hamiltonian operator associated with (5.68) is even because it is quadratic. The graded commutator of two operators \(A\) and \(B\) with parities \(\vert A\vert \) and \(\vert B\vert \) is defined as
\(\seteqnumber{0}{5.}{75}\)\begin{flalign} \big [A,B\big ]_{\mathrm {gr}}^{} := A\,B - (-1)^{\vert A\vert \,\vert B\vert }\,B\,A\quad . \end{flalign} Note that if either \(A\) or \(B\) (or both) are even, then this is a commutator, while when both \(A\) and \(B\) are odd, the graded commutator gives the anticommutator. This means that we can write the equal-time anticommutation relations (5.73) equivalently in terms of the graded commutator as
\(\seteqnumber{1}{5.77}{0}\)\begin{flalign} \big [\Psi _\alpha (\mathbf {x}),\Psi _\beta (\mathbf {y})\big ]_{\mathrm {gr}}^{} \,&=\,0\, =\, \big [\Psi ^\dagger _\alpha (\mathbf {x}),\Psi ^\dagger _\beta (\mathbf {y})\big ]_{\mathrm {gr}}^{} \,=\,0\quad ,\\[4pt] \qquad \big [\Psi _\alpha (\mathbf {x}),\Psi ^\dagger _\beta (\mathbf {y})\big ]_{\mathrm {gr}}^{}\,&=\,\delta _{\alpha \beta }\,\delta (\mathbf {x}-\mathbf {y})\quad \end{flalign} and Heisenberg’s equation as
\(\seteqnumber{0}{5.}{77}\)\begin{flalign} \frac {\partial }{\partial t} \Psi (x) \,=\,\ii \,\big [H,\Psi (x)\big ]_{\mathrm {gr}}\quad . \end{flalign} The computation of the right-hand side of Heisenberg’s equation is then easy by using the following identities for the graded commutator
\(\seteqnumber{1}{5.79}{0}\)\begin{flalign} \big [A,B\big ]_{\mathrm {gr}}^{} \,&=\, - (-1)^{\vert A\vert \,\vert B\vert }\,\big [B,A\big ]_{\mathrm {gr}}^{}\quad ,\\ \big [A,B\,C\big ]_{\mathrm {gr}}^{}\,&=\,\big [A,B\big ]_{\mathrm {gr}}^{}\,C + (-1)^{\vert A\vert \,\vert B\vert }\, B\,\big [A,C\big ]_{\mathrm {gr}}^{}\quad ,\\ \big [A\,B,C\big ]_{\mathrm {gr}}^{}\,&=\,A\,\big [B,C\big ]_{\mathrm {gr}}^{} + (-1)^{\vert B\vert \,\vert C\vert }\, \big [A,C\big ]_{\mathrm {gr}}^{}\,B\quad , \end{flalign} which I recommend you to confirm on your own. The graded commutator also satisfies the graded Jacobi identity
\(\seteqnumber{0}{5.}{79}\)\begin{flalign} (-1)^{\vert A\vert \,\vert C\vert }\,\big [A,[B,C]_{\mathrm {gr}}^{}\big ]_{\mathrm {gr}}^{}+ (-1)^{\vert B\vert \,\vert A\vert }\,\big [B,[C,A]_{\mathrm {gr}}^{}\big ]_{\mathrm {gr}}^{}+ (-1)^{\vert C\vert \,\vert B\vert }\,\big [C,[A,B]_{\mathrm {gr}}^{}\big ]_{\mathrm {gr}}^{}=0\quad , \end{flalign} which we however do not need at the moment.
To solve Heisenberg’s equation for the Dirac field operator, which as we have seen above is equivalent to the Dirac equation, we can use our knowledge from Section 5.2 of all plane wave solutions. Using the Dirac spinor bases from (5.57), we can write
\begin{equation} \Psi (x) = \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2 \omega _{\mathbf {k}}}}\sum _s \Big (a_s(\mathbf {k})\,u^s(\mathbf {k}) \,e^{\ii \,k\,x} + b_s^\dagger (\mathbf {k})\,v^s(\mathbf {k}) \,e^{-\ii \,k\,x} \Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \end{equation}
where \(k = (\omega _{\mathbf {k}},\mathbf {k})\in \bbR ^d\) is the relativistic on-shell Fourier momentum. The Dirac adjoint of this expression is given by
\begin{equation} \overline {\Psi }(x) = \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2 \omega _{\mathbf {k}}}}\sum _s \Big (b_s(\mathbf {k})\,\overline {v^s(\mathbf {k})} \,e^{\ii \,k\,x} + a_s^\dagger (\mathbf {k})\,\overline {u^s(\mathbf {k})} \,e^{-\ii \,k\,x} \Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{equation}
Comparing with the case of the real Klein-Gordon quantum field (3.20), the following remarks are in order:
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• In (5.81) there is a sum over all spinor polarizations and consequently there are spinor polarization labels \(s\) on the annihilation and creation operators. For a scalar field (3.20) this sum is trivial because \(\Phi \) has only a single component.
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• In (5.81) there are two different sets of annihilation and creation operators given by \(a_s(\mathbf {k})\), \(a^{\dagger }_s(\mathbf {k})\) and \(b_s(\mathbf {k})\), \(b^{\dagger }_s(\mathbf {k})\). This is due to the fact that \(\Psi (x)\) is the quantization of a complex field. A similar feature is present for the complex Klein-Gordon field, where instead of the operator \(a^\dagger (\mathbf {k})\) in (3.20) one has a \(b^\dagger (\mathbf {k})\). The Hermiticity condition \(\Phi (x)^\dagger = \Phi (x)\) for the real Klein-Gordon field forces the two sets of operators to coincide, i.e. \(b^\dagger (\mathbf {k})=a^\dagger (\mathbf {k})\) and \(b(\mathbf {k})=a(\mathbf {k})\).
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• As a consequence of the equal-time anticommutation relations (5.73), the annihilation and creation operators in (5.81) have to satisfy anticommutation relations too. The nontrivial ones are given explicitly by
\(\seteqnumber{1}{5.82}{0}\)\begin{equation} \big \{a_s(\mathbf {k}),a_r^\dagger (\mathbf {q})\big \} \,=\, \big \{b_s(\mathbf {k}),b_r^\dagger (\mathbf {q})\big \} \,=\,\delta _{sr}\,(2\pi )^{d-1}\,\delta (\mathbf {k}-\mathbf {q})\quad , \end{equation}
while all other anticommutators are zero, i.e.
\(\seteqnumber{1}{5.82}{1}\)\begin{flalign} \big \{a_s(\mathbf {k}),a_r(\mathbf {q})\big \} &= \big \{a_s(\mathbf {k}),b_r(\mathbf {q})\big \}=\big \{a_s(\mathbf {k}),b_r^\dagger (\mathbf {q})\big \}=0\quad ,\\ \big \{a^\dagger _s(\mathbf {k}),a^\dagger _r(\mathbf {q})\big \} &= \big \{a^\dagger _s(\mathbf {k}),b_r(\mathbf {q})\big \}=\big \{a^\dagger _s(\mathbf {k}),b_r^\dagger (\mathbf {q})\big \}=0\quad ,\\ \big \{b_s(\mathbf {k}),b_r(\mathbf {q})\big \} &=\big \{b_s^\dagger (\mathbf {k}),b_r^\dagger (\mathbf {q})\big \}=0\quad . \end{flalign}
To check that (5.81) together with the anticommutation relations (5.82) really implies the equal-time anticommutation relations (5.73), one has to remember and use the spin sum identities from (5.58).
Similarly to the case of the quantum Klein-Gordon field in Section 3.3, we can build a Hilbert space \(\HH \) by introducing a vacuum state \(\ket {0}\in \HH \) that is characterized by the property of being normalized \(\braket {0}{0}=1\) and being annihilated by all annihilation operators, i.e.
\begin{equation} a_s(\mathbf {k})\ket {0} = 0~~,\quad b_s(\mathbf {k})\ket {0} =0\quad . \end{equation}
Multiparticle states are then obtained by acting on \(\ket {0}\) with suitably normalized creation operators. Since there are two sets of such creation operators, namely \(a_s^\dagger (\mathbf {k})\) and \(b_s^{\dagger }(\mathbf {k})\), which further depend on a choice of spinor polarization \(s\), such multiparticle states have more labels than only a relativistic on-shell momentum \(k=(\omega _{\mathbf {k}},\mathbf {k})\in \bbR ^d\). We denote such states by
\begin{flalign} \nn &\big \vert (k_1,s_1,+),\dots ,(k_l,s_l,+),(k_{l+1},s_{l+1},-),\dots ,(k_n,s_n,-)\big \rangle \,:=\,\\[4pt] &\qquad ~\qquad \sqrt {2\omega _{\mathbf {k}_1}}\cdots \sqrt {2\omega _{\mathbf {k}_n}}~a_{s_1}^{\dagger }(\mathbf {k}_1)\cdots a_{s_l}^{\dagger }(\mathbf {k}_l)\, b_{s_{l+1}}^{\dagger }(\mathbf {k}_{l+1})\cdots b_{s_n}^{\dagger }(\mathbf {k}_n)\ket {0}\quad ,\label {eqn:Diracmultistate} \end{flalign}
where we attach a \(\pm \) sign to distinguish between \(a\)-type and \(b\)-type particles. As a direct consequence of the anticommutation relations (5.82), we observe that exchanging any neighboring pairs of labels in such states gives a minus sign, hence the particles associated to the Dirac quantum field describe fermions.
To study further properties of these particles, and in particular to figure out what’s the difference between \(a\)-type and \(b\)-type particles, we use the conserved charges that we have computed in (5.69), (5.71) and (5.72). Using (5.81) and the additional spinor polarization identities from Remark 5.3, one can express these conserved charges in terms of annihilation and creation operators. For the relativistic momentum (5.69) one then finds
\(\seteqnumber{0}{5.}{84}\)\begin{flalign} P^{\mu } = \sum _s \int _{\bbR ^{d-1}} k^\mu \,\Big (a^\dagger _s(\mathbf {k})\,a_s(\mathbf {k}) - b_s(\mathbf {k})\,b^\dagger _s(\mathbf {k})\Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{flalign} The normal ordering of products of odd operators can be defined similarly as in Definition 3.1, with the additional rule that exchanging the order of two odd operators gives a minus sign. With this the normal ordering of \(P^\mu \) reads as
\(\seteqnumber{0}{5.}{85}\)\begin{flalign} \noor {P^{\mu }} = \sum _s \int _{\bbR ^{d-1}} k^\mu \,\Big (a^\dagger _s(\mathbf {k})\,a_s(\mathbf {k}) + b_s^\dagger (\mathbf {k})\,b_s(\mathbf {k})\Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{flalign} Acting with this operator on single particle states gives
\(\seteqnumber{0}{5.}{86}\)\begin{flalign} \noor {P^\mu }\ket {k,s,\pm } = k^\mu \,\ket {k,s,\pm }\quad , \end{flalign} hence the label \(k=(\omega _{\mathbf {k}},\mathbf {k})\) has the physical interpretation of the on-shell momentum of a particle with mass \(m\). Note that both \(a\)-type and \(b\)-type particles have positive energies and identical mass \(m\). The normal ordered operator associated with the relativistic angular momentum (5.71) is more complicated to work out, hence we shall skip the details. Even without looking into the details, one sees that due to the second term in (5.71), which captures the intrinsic angular momentum a.k.a. spin \(S^{\mu \nu }\), the single particle states \(\ket {k,s,\pm }\) describe spin \(\frac {1}{2}\) particles. The difference between \(a\)-type and \(b\)-type particles lies in their electric charge: The operator associated with (5.72) is given in terms of annihilation and creation operators by
\(\seteqnumber{0}{5.}{87}\)\begin{flalign} Q = \sum _s \int _{\bbR ^{d-1}} \Big (a^\dagger _s(\mathbf {k})\,a_s(\mathbf {k}) + b_s(\mathbf {k})\,b_s^\dagger (\mathbf {k})\Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \end{flalign} which after normal ordering (performed as above with minus signs when exchanging odd operators) becomes
\(\seteqnumber{0}{5.}{88}\)\begin{flalign} \noor {Q} = \sum _s \int _{\bbR ^{d-1}} \Big (a^\dagger _s(\mathbf {k})\,a_s(\mathbf {k}) -b^\dagger _s(\mathbf {k})\,b_s(\mathbf {k})\Big )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{flalign} Acting on single particle states gives
\(\seteqnumber{0}{5.}{89}\)\begin{flalign} \label {eqn:chargeeigenstates} \noor {Q}\ket {k,s,\pm } = \pm \, \ket {k,s,\pm }\quad , \end{flalign} which means that \(a\)-type particles have positive charge and \(b\)-type particles have negative charge.
Summing up, we have seen that the quantization of the Dirac field gives two types of fermionic spin \(\frac {1}{2}\) particles, which have identical mass \(m\) but differ by a sign in their electric charge. Such pairs are typically called particle/antiparticle pairs and a physical example is the electron/positron pair. To avoid any misconceptions, I would like to stress that the existence of antiparticles has nothing to do with the fermionic spin \(\frac {1}{2}\) structure of the Dirac field, but rather with the fact that Dirac spinors are complex-valued. As already indicated in the itemization above, when studying a complex Klein-Gordon field, one also finds two independent types of annihilation and creation operators, \(a\)-type and \(b\)-type, that give rise to particle/antiparticle pairs. The Hermiticity condition \(\Phi (x)^\dagger = \Phi (x)\) for the real Klein-Gordon field forces the \(b\)-type operators to coincide with the \(a\)-type operators (compare with (3.20)). This is why people often say that “the particles associated with a real field are their own antiparticles”.