5.2 Classical Dirac field
A Dirac field is defined as a Dirac spinor-valued function \(\Psi : \bbR ^d\to \bbC ^N\,,~x\mapsto \Psi (x)\) on the Minkowski spacetime \(\bbR ^d\). The transformation law of a Dirac field under active Poincaré transformations is then given by
\(\seteqnumber{0}{5.}{37}\)\begin{flalign} \label {eqn:DiracfieldLorentz} T_{(\omega ,b)} : \Psi (x)~\longmapsto ~T_{(\omega ,b)}\Psi (x) \,=\, e^{\frac {\ii }{2}\omega _{\mu \nu } S^{\nu \mu }}\,\Psi \big (e^{-\omega }(x-b)\big )\quad , \end{flalign} where \(S^{\nu \mu }\) are the Lorentz generators in the Dirac spinor representation (5.8). Using the Dirac adjoint from (5.20), we introduce the Dirac action functional
\begin{equation} \label {eqn:Diracaction} S_{\mathrm {Dirac}}^{}[\Psi ,\overline {\Psi }]\,:=\, -\int _{\bbR ^d} \overline {\Psi }\,\big (\slashed {\partial } + m \big )\,\Psi \,\dd x\,:=\, -\int _{\bbR ^d} \overline {\Psi }\,\big (\gamma ^\mu \,\partial _\mu + m \big )\,\Psi \,\dd x\quad , \end{equation}
where \(m>0\) is a positive parameter that will be interpreted later as the mass. (To simplify this section, we consider only the massive Dirac field because the massless case \(m=0\) is slightly more subtle.) Here and in the following we shall use the convenient Feynman slash notation
\begin{equation} \slashed {B} \,:=\, \gamma ^\mu \,B_\mu \end{equation}
to denote the contraction of a covector \(B_\mu \) with the gamma-matrices \(\gamma ^\mu \). Using (5.22) and (5.24), one easily checks that the Dirac action is invariant under the Poincaré transformations given in (5.38). Using also the anti-Hermiticity properties (5.16), one shows that the Dirac action is, as required, real-valued
\(\seteqnumber{0}{5.}{40}\)\begin{flalign} \nn S_{\mathrm {Dirac}}^{}[\Psi ,\overline {\Psi }]^\dagger \,&=\, -\int _{\bbR ^d} \Big (\Psi ^\dagger \,(\ii \gamma ^0)\,\gamma ^\mu \,\partial _\mu \Psi + m \,\Psi ^\dagger \,(\ii \gamma ^0)\,\Psi \Big )^\dagger \,\dd x\\ \nn \,&=\, -\int _{\bbR ^d} \Big (-\partial _\mu \Psi ^\dagger \,(\ii \gamma ^0)\,\gamma ^\mu \,\Psi + m \,\Psi ^\dagger \,(\ii \gamma ^0)\,\Psi \Big )\,\dd x\\ \,&=\, -\int _{\bbR ^d} \Big ( \Psi ^\dagger \,(\ii \gamma ^0)\,\gamma ^\mu \,\partial _\mu \Psi + m \,\Psi ^\dagger \,(\ii \gamma ^0)\,\Psi \Big )\,\dd x = S_{\mathrm {Dirac}}^{}[\Psi ,\overline {\Psi }]\quad , \end{flalign} where in the third step we have integrated by parts.
The Euler-Lagrange equations associated with the Dirac action are given by
Note that the second equation is simply the Dirac adjoint of the first equation
\(\seteqnumber{0}{5.}{42}\)\begin{flalign} 0 = \overline {(\gamma ^{\mu }\partial _\mu + m)\Psi } = \partial _\mu \Psi ^\dagger \,(\gamma ^\mu )^\dagger \,(\ii \gamma ^0) + m\, \overline {\Psi } = -\partial _\mu \overline {\Psi }\,\gamma ^\mu + m\,\overline {\Psi }\quad . \end{flalign} The equation \((\slashed {\partial } + m)\Psi =0\) is the famous Dirac equation, which was originally discovered by Dirac in his search for a “square root” of the Klein-Gordon operator \(-\partial ^2 + m^2\). Indeed, multiplying both sides of the Dirac equation by the “wrong sign Dirac operator” \((-\slashed {\partial } + m)\), we observe with some Clifford algebra (5.7) that
\(\seteqnumber{0}{5.}{43}\)\begin{flalign} \nn 0 &= (-\slashed {\partial } + m) (\slashed {\partial } + m) \Psi = -\slashed {\partial }^2\,\Psi +m^2\,\Psi =-\gamma ^\nu \gamma ^\mu \, \partial _\nu \partial _\mu \Psi + m^2 \,\Psi \\ &= -\frac {1}{2}\{\gamma ^\nu ,\gamma ^\mu \} \, \partial _\nu \partial _\mu \Psi + m^2 \,\Psi = -\eta ^{\nu \mu }\partial _\nu \partial _\mu \Psi + m^2\,\Psi = (-\partial ^2 +m^2)\Psi \end{flalign} gives the Klein-Gordon equation. This implies in particular that every solution \(\Psi \) of the Dirac equation satisfies automatically the Klein-Gordon equation \((-\partial ^2+m^2)\Psi =0\) too.
Finding solutions of the Dirac equation is more involved than in the case of the Klein-Gordon equation, because the Dirac spinor \(\Psi (x)\in \bbC ^N\) is a multicomponent field and the Dirac equation (5.42) imposes nontrivial conditions among the individual components. Since the Dirac equation is linear, we can write any solution \(\Psi (x)\) as a superposition of the following two types of plane waves
\(\seteqnumber{0}{5.}{44}\)\begin{flalign} \Psi (x) = u(\mathbf {k}) \,e^{\ii \,k\,x}\quad ,\qquad \Psi (x) = v(\mathbf {k})\,e^{-\ii \,k\,x}\quad , \end{flalign} where \(k = (\omega _{\mathbf {k}},\mathbf {k})\in \bbR ^d\) is a relativistic on-shell momentum with positive energy \(\omega _{\mathbf {k}}=\sqrt {\mathbf {k}^2 +m^2}\). Note that the on-shell condition follows from the fact that \(\Psi (x)\) satisfies the Klein-Gordon equation, which for plane waves enforces \(k^2 + m^2 =0\). Inserting these plane waves into the Dirac equation, we find that the spinor polarizations \(u(\mathbf {k}),v(\mathbf {k})\in \bbC ^N\) must satisfy the conditions
\(\seteqnumber{0}{5.}{45}\)\begin{flalign} (\ii \,\slashed {k} +m)\,u(\mathbf {k}) =0\quad ,\qquad (-\ii \,\slashed {k} +m)\,v(\mathbf {k}) =0\quad . \end{flalign} To understand these equations better, let us introduce the linear maps
\begin{equation} P^\pm _{} (\mathbf {k}) \,:=\,\frac {1}{2m}\,\big (\pm \ii \,\slashed {k} + m\big ) \end{equation}
on the space of Dirac spinors. The spinor polarization equations then read equivalently as
\(\seteqnumber{0}{5.}{47}\)\begin{flalign} \label {eqn:spinorpolarizationconditions} P^+_{} (\mathbf {k}) \,u(\mathbf {k}) = 0\quad ,\qquad P^-_{} (\mathbf {k})\,v(\mathbf {k}) =0\quad . \end{flalign} By direct inspection, one observes that these linear maps satisfy the following identities
\(\seteqnumber{0}{5.}{48}\)\begin{flalign} P^+_{} (\mathbf {k}) + P^-_{} (\mathbf {k}) \,=\, 1~~,\quad P^\pm _{} (\mathbf {k}) \,P^\pm _{} (\mathbf {k}) \,=\, P^\pm _{} (\mathbf {k})~~,\quad P^\pm _{} (\mathbf {k})\,P^\mp _{} (\mathbf {k})\,=\,0\quad , \end{flalign} which means that they are complementary projectors. This allows us to decompose the space of Dirac spinors into a direct sum
\(\seteqnumber{0}{5.}{49}\)\begin{flalign} \label {eqn:spinordecomposition} \bbC ^{N} = \Imm \big (P^-_{} (\mathbf {k})\big )\oplus \Imm \big (P^+_{} (\mathbf {k})\big ) =\Ker \big (P^+_{} (\mathbf {k})\big )\oplus \Ker \big (P^-_{} (\mathbf {k})\big )\quad , \end{flalign} where we use that
\(\seteqnumber{0}{5.}{50}\)\begin{flalign} \Imm \big (P^\pm _{} (\mathbf {k})\big ) \,=\,\Ker \big (P^\mp _{} (\mathbf {k})\big ) \quad . \end{flalign}
-
Lemma 5.2. The decomposition (5.50) of the space of Dirac spinors is orthogonal with respect to the Dirac inner product. Furthermore, the Dirac inner product is positive definite on the first summand \(\Ker \big (P^+_{} (\mathbf {k})\big )\) and it is negative definite on the second summand \(\Ker \big (P^-_{} (\mathbf {k})\big )\). In formulas, all this means that, for any nonzero \(u(\mathbf {k})\in \Ker \big (P^+_{} (\mathbf {k})\big )\) and \(v(\mathbf {k})\in \Ker \big (P^-_{} (\mathbf {k})\big )\), we have
\(\seteqnumber{0}{5.}{51}\)\begin{flalign} \overline {u(\mathbf {k})} \,v(\mathbf {k}) \,= \,0 \,=\, \overline {v(\mathbf {k})} \,u(\mathbf {k}) ~~,\quad \overline {u(\mathbf {k})} \,u(\mathbf {k})\, >\, 0~~,\quad \overline {v(\mathbf {k})} \,v(\mathbf {k})\, < \,0\quad . \end{flalign}
-
Proof. Let us first note that the Dirac adjoint of the projector \(P^{\pm }(\mathbf {k})\) is given by
\(\seteqnumber{0}{5.}{52}\)\begin{flalign} \overline {P^{\pm }(\mathbf {k})\,\psi } = \psi ^\dagger \,\big (P^{\pm }(\mathbf {k}) \big )^\dagger \,(\ii \gamma ^0) =\psi ^\dagger \,(\ii \gamma ^0)\, P^{\pm }(\mathbf {k}) =\overline {\psi }\, P^{\pm }(\mathbf {k})\quad . \end{flalign} Using (5.50), we can write \(u(\mathbf {k}) = P^-_{} (\mathbf {k})\,\psi \) and \(v(\mathbf {k})= P^+_{} (\mathbf {k})\,\tilde {\psi }\) for some spinors \(\psi ,\tilde {\psi }\in \bbC ^N\), from which we deduce that
\(\seteqnumber{1}{5.54}{0}\)\begin{flalign} \overline {u(\mathbf {k})} \,v(\mathbf {k}) = \overline {P^-_{} (\mathbf {k}) \psi } \,P^{+}_{}(\mathbf {k})\tilde {\psi } = \overline {\psi }\,P^-_{} (\mathbf {k})\,P^+_{} (\mathbf {k})\,\tilde {\psi } =0 \end{flalign} and similarly
\(\seteqnumber{1}{5.54}{1}\)\begin{flalign} \overline {v(\mathbf {k})} \,u(\mathbf {k}) = \overline {P^+_{} (\mathbf {k}) \tilde {\psi }} \,P^{-}_{}(\mathbf {k})\psi = \overline {\tilde {\psi }}\,P^+_{} (\mathbf {k})\,P^-_{} (\mathbf {k})\,\psi =0\quad . \end{flalign} To prove the statement about the definiteness of the Dirac inner product, we use that the Dirac inner product is Lorentz invariant. This means that we can transform, without loss of generality, to the rest frame where \(\mathbf {k} = \mathbf {0}\) and consequently \(k = (m,\mathbf {0})\). The projectors in the rest frame simplify to \(P^\pm _{} (\mathbf {0})= \frac {1}{2}\big (1 \mp \ii \,\gamma ^0\big )\) and they enjoy the following useful properties
\(\seteqnumber{0}{5.}{54}\)\begin{flalign} \big (P^\pm _{} (\mathbf {0})\big )^\dagger = P^\pm _{} (\mathbf {0})~~,\quad (\ii \gamma ^0)\,P^\pm _{}(\mathbf {0}) = \mp \,P^\pm _{}(\mathbf {0})\quad . \end{flalign} Writing again \(u(\mathbf {0}) = P^-_{} (\mathbf {0})\,\psi \) and \(v(\mathbf {0}) = P^+_{} (\mathbf {0})\,\tilde {\psi }\) for some spinors \(\psi ,\tilde {\psi }\in \bbC ^N\), we find for the Dirac inner products
\(\seteqnumber{1}{5.56}{0}\)\begin{flalign} \overline {u(\mathbf {0})} \,u(\mathbf {0}) = \psi ^\dagger \,(\ii \gamma ^0)\, P^-_{}(\mathbf {0})\,P^-_{} (\mathbf {0})\,\psi = \big (P^-_{}(\mathbf {0})\,\psi \big )^\dagger \,\big (P^-_{}(\mathbf {0})\,\psi \big ) = \big \| P^-_{}(\mathbf {0})\,\psi \big \|^2 >0 \end{flalign} and similarly
\(\seteqnumber{1}{5.56}{1}\)\begin{flalign} \overline {v(\mathbf {0})} \,v(\mathbf {0}) = \tilde {\psi }^\dagger \,(\ii \gamma ^0)\, P^+_{}(\mathbf {0})\,P^+_{} (\mathbf {0})\,\tilde {\psi } = -\big (P^+_{}(\mathbf {0})\,\tilde {\psi }\big )^\dagger \,\big (P^+_{}(\mathbf {0})\,\tilde {\psi }\big ) =-\big \| P^+_{}(\mathbf {0})\,\tilde {\psi }\big \|^2<0\quad , \end{flalign} where in the last steps we recognized the standard norm \(\| \cdot \|\) on \(\bbC ^N\). □
The relevance of this lemma is that it allows us to choose vector space bases
\(\seteqnumber{1}{5.57}{0}\)\begin{flalign} \Big \{u^s(\mathbf {k})\in \Ker \big (P^+_{} (\mathbf {k})\big ) \Big \}\quad ,\qquad \Big \{v^s(\mathbf {k})\in \Ker \big (P^-_{} (\mathbf {k})\big )\Big \} \end{flalign} that are orthonormal, with respect to the indefinite Dirac inner product, according to
\begin{flalign} \overline {u^s(\mathbf {k})}\, u^r(\mathbf {k}) \,&=\,2m\,\delta ^{sr}~~,\quad \\ \overline {v^s(\mathbf {k})}\, v^r(\mathbf {k}) \,&=\,-2m\,\delta ^{sr}~~,\quad \\ \overline {u^s(\mathbf {k})}\, v^r(\mathbf {k}) \,&=\, 0\,=\,\overline {v^s(\mathbf {k})}\, u^r(\mathbf {k}) \quad . \end{flalign}
The projectors \(P^\pm _{} (\mathbf {k})\) can be written in terms of these bases as
\(\seteqnumber{1}{5.58}{0}\)\begin{flalign} P^{+}_{}(\mathbf {k}) = -\frac {1}{2m} \sum _{s} v^s(\mathbf {k})\,\overline {v^s(\mathbf {k})}\quad ,\qquad P^{-}_{}(\mathbf {k}) = \frac {1}{2m} \sum _{s} u^s(\mathbf {k})\,\overline {u^s(\mathbf {k})}\quad , \end{flalign} which, upon rearrangement, yields the so-called spin sum identities
\begin{flalign} \sum _{s} u^s(\mathbf {k})\,\overline {u^s(\mathbf {k})} \,&=\, \big ({-}\ii \,\slashed {k} + m\big )~~,\quad \\ \sum _{s} v^s(\mathbf {k})\,\overline {v^s(\mathbf {k})} \,&=\, - \big (\ii \,\slashed {k} +m\big )\quad . \end{flalign}
-
Remark 5.3. One can derive further spinor polarization identities that will facilitate our calculations with Dirac spinors in the next sections. These identities arise from the observation that taking the ordinary adjoint of the projector \(P^{\pm }_{}(\mathbf {k})\) gives
\(\seteqnumber{0}{5.}{58}\)\begin{flalign} \big (P^{\pm }_{}(\mathbf {k})\big )^\dagger = \frac {1}{2m}\big (\mp \ii \,\gamma ^0\omega _\mathbf {k} \pm \ii \,\gamma ^i\,k_i +m\big )^\dagger = \frac {1}{2m}\big (\mp \ii \,\gamma ^0\omega _\mathbf {k} \mp \ii \,\gamma ^i\,k_i +m\big ) = P^{\pm }_{}(-\mathbf {k})\quad , \end{flalign} where \(-\mathbf {k}\) is the oppositely pointing spatial momentum. The conditions (5.48) for the spinor polarizations then translate for the ordinary adjoint spinors to
\(\seteqnumber{0}{5.}{59}\)\begin{flalign} u^s(\mathbf {k})^\dagger \,P^+_{}(-\mathbf {k}) \,=\, 0\quad ,\qquad v^s(\mathbf {k})^\dagger \,P^-_{}(-\mathbf {k}) \,=\, 0\quad . \end{flalign} Using also \(P^+_{}(-\mathbf {k})\,v^r(-\mathbf {k})=v^r(-\mathbf {k})\) and \(P^-_{}(-\mathbf {k})\,u^r(-\mathbf {k})=u^r(-\mathbf {k})\), we obtain the orthogonality conditions
\(\seteqnumber{1}{5.61}{0}\)\begin{flalign} u^s(\mathbf {k})^\dagger \,v^r(-\mathbf {k}) &= u^s(\mathbf {k})^\dagger \,P^+_{}(-\mathbf {k}) \,v^r(-\mathbf {k})=0\quad ,\\ v^s(\mathbf {k})^\dagger \,u^r(-\mathbf {k}) &= v^s(\mathbf {k})^\dagger \,P^-_{}(-\mathbf {k}) \,u^r(-\mathbf {k})=0 \end{flalign} for oppositely pointing spatial momenta. We will also need formulas for the standard inner products \(u^s(\mathbf {k})^\dagger \,u^r(\mathbf {k})\) and \(v^s(\mathbf {k})^\dagger \,v^r(\mathbf {k})\) without Dirac adjoints. These can be obtained using our projectors via the following trick
\(\seteqnumber{0}{5.}{61}\)\begin{flalign} u^s(\mathbf {k})^\dagger \,u^r(\mathbf {k}) = \frac {1}{2}\,u^s(\mathbf {k})^\dagger \,\big (P^-_{}(\mathbf {k}) + P^{-}_{}(-\mathbf {k})\big )\,u^r(\mathbf {k})=\frac {\omega _{\mathbf {k}}}{2m}\, \overline {u^s(\mathbf {k})}\,u^r(\mathbf {k}) + \frac {1}{2}\,u^s(\mathbf {k})^\dagger \,u^r(\mathbf {k})\quad , \end{flalign} which after rearrangement and using that \(\overline {u^s(\mathbf {k})}\,u^r(\mathbf {k})=2m\,\delta ^{sr}\) yields
\(\seteqnumber{0}{5.}{62}\)\begin{flalign} u^s(\mathbf {k})^\dagger \,u^r(\mathbf {k}) = 2\omega _{\mathbf {k}}\,\delta ^{sr} \quad . \end{flalign} Similarly, for the \(v\)’s we use
\(\seteqnumber{0}{5.}{63}\)\begin{flalign} v^s(\mathbf {k})^\dagger \,v^r(\mathbf {k}) = \frac {1}{2}\,v^s(\mathbf {k})^\dagger \,\big (P^+_{}(\mathbf {k}) + P^{+}_{}(-\mathbf {k})\big )\,v^r(\mathbf {k})= - \frac {\omega _{\mathbf {k}}}{2m}\, \overline {v^s(\mathbf {k})}\,v^r(\mathbf {k}) + \frac {1}{2}\,v^s(\mathbf {k})^\dagger \,v^r(\mathbf {k})\quad , \end{flalign} which yields
\(\seteqnumber{0}{5.}{64}\)\begin{flalign} v^s(\mathbf {k})^\dagger \,v^r(\mathbf {k}) = 2\omega _{\mathbf {k}} \,\delta ^{sr}\quad . \end{flalign} Note that, in contrast to the Dirac inner product (5.57), the standard inner product is, of course, positive definite.
To conclude this section, we shall discuss briefly the Hamiltonian and conserved Noether charges of the Dirac field. More details can be found in the literature, see in particular the textbooks listed in Section 1.3.
Hamiltonian: To derive the Hamiltonian, let us take any splitting of the coordinates \(x = (t,\mathbf {x})\in \bbR ^d\) of Minkowski spacetime into time and space coordinates. This allows us to write the Dirac action (5.39) as
\(\seteqnumber{0}{5.}{65}\)\begin{flalign} S_{\mathrm {Dirac}} = \ii \,\int _{\bbR ^d} \Big (\Psi ^\dagger \,\dot {\Psi } - \Psi ^\dagger \,\gamma ^0 \gamma ^i\,\partial _i \Psi - m\,\Psi ^\dagger \,\gamma ^0\,\Psi \Big )\,\dd x\quad , \end{flalign} where by \(\dot {\Psi } = \partial _0 \Psi = \frac {\partial }{\partial t}\Psi \) we denote again the time derivative. The canonical momenta are
\(\seteqnumber{0}{5.}{66}\)\begin{flalign} \pi _{\Psi } = \frac {\partial \LL }{\partial \dot {\Psi }} = \ii \,\Psi ^\dagger \quad ,\qquad \pi _{\Psi ^\dagger } = \frac {\partial \LL }{\partial {\dot {\Psi }}^\dagger } = 0\quad , \end{flalign} which gives the Hamiltonian
\(\seteqnumber{0}{5.}{67}\)\begin{flalign} \label {eqn:DiracHamiltonian} H \,=\,\ii \int _{\bbR ^{d-1}}\Big (\Psi ^\dagger \,\gamma ^0\,\gamma ^i\,\partial _i\Psi + m\,\Psi ^\dagger \,\gamma ^0\,\Psi \Big )\,\dd \mathbf {x}\quad . \end{flalign} Here and in what follows we simply write \(\ii \,\Psi ^\dagger \) instead of \(\pi _{\Psi }\) for the canonical momentum, which is the usual convention in the literature.
Conserved charges: By construction, the Dirac action (5.39) is invariant under the Poincaré transformations given in (5.38). Furthermore, in analogy to the complex scalar field from Example 2.5, it has an internal continuous symmetry given by the constant complex phase rotations \(T_{\alpha }\Psi (x)= e^{-\ii \,\alpha }\,\Psi (x)\) and \(T_{\alpha }\overline {\Psi }(x)= e^{\ii \,\alpha }\,\overline {\Psi }(x)\). Noether’s Theorem 2.6 then yields, via a routine calculation that you should try on your own, conserved on-shell currents and charges that are associated to these symmetries. The conserved charges associated with spacetime translations read as
\(\seteqnumber{0}{5.}{68}\)\begin{flalign} \label {eqn:DiracPmu} P^{\mu } \,=\, -\ii \int _{\bbR ^{d-1}} \Psi ^\dagger \, \partial ^\mu \Psi \,\dd \mathbf {x}\quad . \end{flalign} Rewriting the Dirac equation \((\gamma ^\mu \partial _\mu +m)\Psi =0\) as \(\partial _0 \Psi = \gamma ^0\gamma ^i\partial _i\Psi +m\,\gamma ^0\Psi \), one checks that the \(\mu =0\) component of these conserved charges agrees with the Hamiltonian (5.68),
\(\seteqnumber{0}{5.}{69}\)\begin{flalign} P^0 = \ii \int _{\bbR ^{d-1}} \Psi ^\dagger \, \partial _0 \Psi \,\dd \mathbf {x} = \ii \int _{\bbR ^{d-1}}\Big (\Psi ^\dagger \,\gamma ^0\,\gamma ^i\,\partial _i\Psi + m\,\Psi ^\dagger \,\gamma ^0\,\Psi \Big )\,\dd \mathbf {x} = H\quad . \end{flalign} The conserved charges associated with Lorentz transformations read as
\(\seteqnumber{0}{5.}{70}\)\begin{flalign} \label {eqn:DiracLmunu} L^{\mu \nu } \,=\, -\ii \int _{\bbR ^{d-1}} \Big (\Psi ^\dagger \big (x^\mu \,\partial ^\nu - x^\nu \,\partial ^\mu \big )\Psi -\ii \,\Psi ^\dagger S^{\mu \nu }\Psi \Big )~\dd \mathbf {x}\quad . \end{flalign} Comparing this with our results for the Klein-Gordon field in Example 2.10, we note that the second term involving the Lorentz generator \(S^{\mu \nu }\) from (5.9) is a new feature. This term captures the internal relativistic angular momentum, a.k.a. the spin, of the Dirac field. The conserved charge associated with complex phase rotations reads as
\(\seteqnumber{0}{5.}{71}\)\begin{flalign} \label {eqn:DiracQ} Q = \int _{\bbR ^{d-1}}\Psi ^\dagger \,\Psi \,\dd \mathbf {x}\quad . \end{flalign}