Charged meson masses under strong magnetic fields: gauge invariance and Schwinger phases

We start by considering the electromagnetic field strength $F^{\mu\nu}$ associated with a general electromagnetic field ${\cal A}^{\mu}(x)$,

$\displaystyle F^{\mu\nu}=\partial^{\mu}\mathcal{A}^{\nu}-\partial^{\nu}\mathcal{A}^{\mu}\ .$

(II.1)

Throughout this work we use the Minkowski metric $g^{\mu\nu}=\mbox{diag}(1,-1,-1,-1)$, while for a space-time coordinate four-vector $x^{\mu}$ we adopt the notation $x^{\mu}=(t,\vec{x})$, with $\vec{x}=(x^{1},x^{2},x^{3})$. We also consider the covariant derivative ${\cal D}^{\mu}$ that appears in the field equations associated with an electrically charged particle,

$${\cal D}^{\mu}\ =\ \partial^{\mu}+i\,Q\mathcal{A}^{\mu}\ ,$$

(II.2)

where $Q$ is the particle electric charge. Now, under a gauge transformation $\Lambda(x)$ the electromagnetic field transforms as

$\displaystyle{\cal A}^{\mu}\ \rightarrow\ {\tilde{\cal A}}^{\mu}={\cal A}^{\mu}+\partial^{\mu}\Lambda\ .$

(II.3)

While the electromagnetic field strength $F^{\mu\nu}$ is invariant under this transformation, the operator ${\cal D}^{\mu}$ transforms in a covariant way, namely

$${\cal D}^{\mu}\ \rightarrow\ \tilde{\cal D}^{\mu}=e^{-iQ\Lambda(x)}\,{\cal D}^{\mu}\,e^{iQ\Lambda(x)}\ .$$

(II.4)

So far we have considered a general external electromagnetic field $\mathcal{A}^{\mu}(x)$. In what follows we concentrate on the case associated with a static and uniform magnetic field $\vec{B}$. The tensor $F^{\mu\nu}$ is given in this case by

$$F^{ij}=F_{ij}=-\epsilon_{ijk}B^{k}\ ,\qquad\qquad F^{0j}=0\ ,$$

(II.5)

with $i,j=1,2,3$, whereas the corresponding electromagnetic field can be written as

$\displaystyle{\cal A}^{\mu}(x)=\frac{1}{2}x_{\nu}F^{\nu\mu}+\partial^{\mu}\Psi(x)\ ,$

(II.6)

where $\Psi(x)$ is, in principle, an arbitrary function. For any form of this function one obtains a particular gauge. Without loosing generality, one can now choose the 3-axis to be parallel (or antiparallel) to the magnetic field, writing $\vec{B}=(0,0,B)$. In addition, one can take $\Psi(x)$ to be only a function of the spatial coordinates that are perpendicular to $\vec{B}$, i.e., $x^{1}$ and $x^{2}$. The reason is that only the components $F^{12}$ and $F^{21}$ of the field strength tensor are different from zero, which implies that only $\partial^{1}\mathcal{A}^{2}$ and $\partial^{2}\mathcal{A}^{1}$ are relevant. In what follows we adopt this coordinate choice.

For the considered situation, some commonly used gauges are

	Symmetric gauge (SG),	$\displaystyle\quad\Psi(x)=0\ ,$	$\displaystyle{\cal A}^{\mu}(x)=\left(0,-\frac{B}{2}\,x^{2},\frac{B}{2}\,x^{1},0\right)\ ;$		(II.7)
	Landau gauge 1 (LG1),	$\displaystyle\quad\Psi(x)=\frac{B}{2}\,x^{1}x^{2}\ ,$	$\displaystyle{\cal A}^{\mu}(x)=\left(0,-B\,x^{2},0,0\right)\ ;$		(II.8)
	Landau gauge 2 (LG2),	$\displaystyle\quad\Psi(x)=-\frac{B}{2}\,x^{1}x^{2}\ ,$	$\displaystyle{\cal A}^{\mu}(x)=\left(0,0,B\,x^{1},0\right)\ .$		(II.9)

In what follows, we refer to them as “standard gauges”.

According to the above introduced coordinate choice, given a four vector $V^{\mu}$ we find it convenient to distinguish between “parallel” components, $V^{0}$ and $V^{3}$, and “perpendicular” components, $V^{1}$ and $V^{2}$. Thus, we introduce the definitions

$$V_{\parallel}^{\mu}\equiv(V^{0},0,0,V^{3})\ ,\qquad\qquad V_{\perp}^{\mu}\equiv(0,V^{1},V^{2},0)\ .$$

(II.10)

In addition, we define the metric tensors

$$g_{\parallel}^{\mu\nu}=\text{diag}(1,0,0,-1)\ ,\qquad\qquad g_{\perp}^{\mu\nu}=\text{diag}(0,-1,-1,0)\ .$$

(II.11)

The scalar products of parallel and perpendicular vectors are thus given by

	$\displaystyle V_{\parallel}^{\mu}\,W_{\parallel\,\mu}$	$\displaystyle=$	$\displaystyle V_{\parallel}\cdot W_{\parallel}=V^{0}W^{0}-V^{3}W^{3}\ ,$
	$\displaystyle V^{\mu}_{\perp}\,W_{\perp\,\mu}$	$\displaystyle=$	$\displaystyle-\vec{V}_{\perp}\cdot\vec{W}_{\perp}=-(V^{1}W^{1}+V^{2}W^{2})\ ,$
	$\displaystyle V_{\parallel}^{\mu}\,W_{\perp\,\mu}$	$\displaystyle=$	$\displaystyle 0\ .$		(II.12)

We introduce here the so-called Schwinger phase, which will be a relevant quantity throughout this work. Given a particle P with electric charge $Q_{\mbox{{\scriptsize P}}}$, we denote the associated SP by $\Phi_{\mbox{{\scriptsize P}}}(x,y)$; its explicit form is Schwinger:1951nm

$$\Phi_{\mbox{{\scriptsize P}}}(x,y)=Q_{\mbox{{\scriptsize P}}}\int_{x}^{y}d\xi_{\mu}\left[{\cal A}^{\mu}(\xi)+\frac{1}{2}\,F^{\mu\nu}\,(\xi_{\nu}-y_{\nu})\right]\ ,$$

(II.13)

where $F^{\mu\nu}$ is assumed to be constant, and the integration is performed along an arbitrary path that connects $x$ with $y$. In general, the SP is found to be not invariant under either translations or gauge transformations. On the other hand, the integral in Eq. (II.13) is shown to be path independent; thus, it can be evaluated using a straight line path. In this way, using Eq. (II.6) one can obtain a closed expression for the SP associated to a static and uniform magnetic field in an arbitrary gauge. It reads

$\displaystyle\Phi_{\mbox{{\scriptsize P}}}(x,y)=\frac{Q_{\mbox{{\scriptsize P}}}}{2}\,x_{\mu}\,F^{\mu\nu}\,y_{\nu}-Q_{\mbox{{\scriptsize P}}}\big{[}\Psi(x)-\Psi(y)\big{]}\ .$

(II.14)

From Eqs. (II.3) and (II.6), it is seen that under a gauge transformation the SP transforms as

$$\Phi_{\mbox{{\scriptsize P}}}(x,y)\ \rightarrow\ {\tilde{\Phi}}_{\mbox{{\scriptsize P}}}(x,y)=\Phi_{\mbox{{\scriptsize P}}}(x,y)-Q_{\mbox{{\scriptsize P}}}\left[\Lambda(x)-\Lambda(y)\right]\ .$$

(II.15)

The expressions for the SP in the particular gauges introduced above can be now readily obtained from Eq. (II.14). We have

	SG:	$\displaystyle\Phi_{\mbox{{\scriptsize P}}}(x,y)=-\frac{Q_{\mbox{{\scriptsize P}}}B}{2}(x^{1}y^{2}-y^{1}x^{2})\ ;\hskip 42.67912pt$		(II.16)
	LG1:	$\displaystyle\Phi_{\mbox{{\scriptsize P}}}(x,y)=-\frac{Q_{\mbox{{\scriptsize P}}}B}{2}(x^{2}+y^{2})(x^{1}-y^{1})\ ;\hskip 42.67912pt$		(II.17)
	LG2:	$\displaystyle\Phi_{\mbox{{\scriptsize P}}}(x,y)=\frac{Q_{\mbox{{\scriptsize P}}}B}{2}(x^{1}+y^{1})(x^{2}-y^{2})\ .\hskip 42.67912pt$		(II.18)

It is worth noticing that in all cases the SP includes products that mix the coordinates of the points $x^{\mu}$ and $y^{\mu}$. Clearly, there is no way in which these combinations could be expressed in terms of the difference between a scalar function evaluated at $x^{\mu}$ and the same function evaluated at $y^{\mu}$. Therefore, it follows from Eq. (II.15) that if the SP does not vanish in a given gauge it will be nonvanishing in any gauge. This means that the SP cannot be “gauged away”.

Let us study now the propagators of charged particles. We start by considering the propagator of a spin zero meson, e.g. a charged pion, which we denote as $\Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)$ with ${\cal Q}=\pm 1$. The particle charge is then given by $Q_{\pi}={\cal Q}\,e$, where $e$ denotes the proton charge.

The equation that defines the meson propagator is

$\displaystyle\left({\cal D}^{\mu}\,{\cal D}_{\mu}+m_{\pi}^{2}\right)\Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)=-\,\delta^{(4)}(x-y)\ .$

(II.19)

If we perform now a gauge transformation, using Eq. (II.4) we get

$$\left(\tilde{\cal D}^{\mu}\,\tilde{\cal D}_{\mu}+m_{\pi}^{2}\right)e^{-iQ_{\pi}\Lambda(x)}\,\Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)=-\,e^{-iQ_{\pi}\Lambda(x)}\,\delta^{(4)}(x-y)\ ;$$

(II.20)

hence, the propagator has to transform according to

$$\Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)\ \rightarrow\ \tilde{\Delta}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)=e^{-iQ_{\pi}\Lambda(x)}\ \Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)\ e^{iQ_{\pi}\Lambda(y)}\ ,$$

(II.21)

which is the natural extension of a gauge covariant transformation for the case of a bilocal object. It is seen that the phase difference appearing in the transformed propagator is just the same quantity that appears in Eq. (II.15) for the gauge-transformed SP. Therefore, one can always write the meson propagator as

$$\Delta_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)=e^{i\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)}\ {\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)\ ,$$

(II.22)

where ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)$ is a gauge invariant function; the gauge dependence of the propagator is carried by the SP, which has a well defined expression.

Since we are dealing with a system subject to a static and uniform magnetic field, the invariance under translations in time and space, under rotations around any axis parallel to the magnetic field, and under boosts in directions parallel to the magnetic field, is expected to be preserved. Translations in time, as well as translations and boosts in the direction of $\vec{B}$, can be treated in the same way as in the case of a free particle, since they do not involve the axes 1 or 2. Thus, let us focus on the translations in the plane perpendicular to $\vec{B}$ and in the rotations around the $\vec{B}$ direction. Noticeably, the expected invariance seems to be at odds with the fact that the charged pion propagator is known to be not invariant under these transformations. The aim of the following discussion is to clarify this point and see how the invariance implies further constraints on the form of the propagator.

Let us first consider space translations in the perpendicular plane, i.e. a general transformation of the form $x^{\mu}\rightarrow x^{\prime\mu}=x^{\mu}+b^{\mu}_{\perp}$. From Eq. (II.6), under this transformation one has

$$\mathcal{A}^{\mu}(x)\ \rightarrow\ \mathcal{A}_{t}^{\mu}(x)\equiv\mathcal{A}^{\mu}(x^{\prime})=\mathcal{A}^{\mu}(x)-\frac{1}{2}\,F^{\mu\nu}\,b_{\perp\nu}+\partial^{\mu}\Psi\left(x^{\prime}\right)-\partial^{\mu}\Psi\left(x\right)\ .$$

(II.23)

It is rather easy to see that this is fully equivalent to a gauge transformation

$$\mathcal{A}_{\mu}(x)\rightarrow\tilde{\mathcal{A}}^{\mu}(x)=\mathcal{A}^{\mu}(x)+\partial^{\mu}\Lambda_{t}(x;b_{\perp})\ ,$$

(II.24)

with

$$\Lambda_{t}(x;b_{\perp})=\Psi(x^{\prime})-\Psi(x)-\frac{1}{2}\ x_{\mu}\,F^{\mu\nu}\,b_{\perp\,\nu}\ .$$

(II.25)

From Eq. (II.25) we can readily get the expressions of $\Lambda_{t}(x;b_{\perp})$ in the particular gauges introduced in the previous subsections. We have

	SG:	$\displaystyle\Lambda_{t}(x;b_{\perp})=-\frac{B}{2}\left(x^{1}b^{2}-x^{2}b^{1}\right)\ ;\hskip 42.67912pt$		(II.26)
	LG1:	$\displaystyle\Lambda_{t}(x;b_{\perp})=\frac{B}{2}\,b^{2}\left(2x^{1}+b^{1}\right)\ ;\hskip 42.67912pt$		(II.27)
	LG2:	$\displaystyle\Lambda_{t}(x;b_{\perp})=-\frac{B}{2}\,b^{1}\left(2x^{2}+b^{2}\right)\ .\hskip 42.67912pt$		(II.28)

A similar relation between the translation $x^{\mu}\rightarrow x^{\prime\mu}=x^{\mu}+b^{\mu}_{\perp}$ and the gauge transformation in Eqs. (II.24) and (II.25) can be obtained for the Schwinger phase. Under the translation, the SP transforms as

$$\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)\ \rightarrow\ \Phi_{\pi^{\scriptscriptstyle{\cal{Q}}},t}(x,y)\equiv\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x^{\prime},y^{\prime})=\frac{Q_{\pi}}{2}\ x^{\prime}_{\mu}\,F^{\mu\nu}\,y^{\prime}_{\nu}-Q_{\pi}\left[\Psi(x^{\prime})-\Psi(y^{\prime})\right]\ ,$$

(II.29)

whereas performing the corresponding gauge transformation one gets

$$\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)\ \rightarrow\ {\tilde{\Phi}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)=\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)-Q_{\pi}\left[\Lambda_{t}(x;b_{\perp})-\Lambda_{t}(y;b_{\perp})\right]\ .$$

(II.30)

Taking into account the form of $\Lambda_{t}(x;b_{\perp})$ in Eq. (II.25) we observe that $\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}},t}={\tilde{\Phi}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)$. Now we can turn back to Eq. (II.19), writing the propagator as in Eq. (II.22). From the above equations it is seen that under the considered translation the operator $\left({\cal D}^{\mu}\,{\cal D}_{\mu}+m_{\pi}^{2}\right)$ and the factor $\exp[i\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)]$ transform in the same way as under the gauge transformation $\Lambda_{t}(x,b_{\perp})$. Together with the requirement that Eq. (II.19) be translational invariant, this implies that the gauge invariant factor ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)$ has to be also translational invariant. Thus, we can write

$${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)={\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x-y)\ ,$$

(II.31)

and it is possible to obtain a Fourier transform ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(v_{\parallel},v_{\perp})$ that satisfies

$\displaystyle{\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x-y)=\,\int\frac{d^{4}v}{(2\pi)^{4}}\ e^{-i\,v\,(x-y)}\,{\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(v_{\parallel},v_{\perp})\ .$

(II.32)

Notice that in this expression we have made it explicit that one gets in general different dependences on parallel and perpendicular momenta.

We consider now a rotation of arbitrary angle $\alpha$ around an axis parallel to the magnetic field $\vec{B}$. The effect of such a rotation on an arbitrary vector $v^{\mu}$ acts only on the perpendicular component $v^{\mu}_{\perp}$. Choosing the 3-axis in the direction of $\vec{B}$, for a rotation matrix $R_{\hat{3}}(\alpha)$ we have $x^{\prime\mu}={\cal R}_{\,\,\nu}^{\mu}\ x^{\nu}$, with

$${\cal R}_{\,\,\nu}^{\mu}\equiv R_{\hat{3}}(\alpha)_{\,\,\nu}^{\mu}=\left(\begin{array}[]{cccc}1&0&0&\ 0\\ \ 0&\cos\alpha&-\sin\alpha&\ 0\\ 0&\sin\alpha&\cos\alpha&\ 0\\ 0&0&0&\ 1\end{array}\right)\ .$$

(II.33)

At this point it is important to specify if we adopt an active or a passive point of view for the rotation; here we adopt the passive point of view and define $\bar{x}^{\prime\mu}={\bar{\cal R}}_{\,\,\nu}^{\mu}\,x^{\nu}$, with ${\bar{\cal R}}\equiv R_{\hat{3}}(-\alpha)$. From Eq. (II.6), under the rotation $R_{\hat{3}}(\alpha)$ the electromagnetic field transforms as

$$\mathcal{A}^{\mu}(x)\ \rightarrow\ \mathcal{A}_{r}^{\mu}(x)={\cal R}_{\,\,\nu}^{\mu}\ \mathcal{A}^{\nu}\left(\bar{x}^{\prime}\right)=-\frac{1}{2}\ {\cal R}_{\,\,\tau}^{\mu}\,F^{\tau\delta}\,{\bar{\cal R}}_{\delta}^{\,\,\,\nu}\,x_{\nu}+{\cal R}_{\,\,\tau}^{\mu}\,\frac{\partial\Psi(z)}{\partial z_{\tau}}\Big{|}_{z\,=\,\bar{x}^{\prime}}\ .$$

(II.34)

Noting that ${\cal R}_{\,\,\tau}^{\mu}\,F^{\tau\delta}\,{\bar{\cal R}}_{\delta}^{\,\,\,\nu}=F^{\mu\nu}$, the result in Eq. (II.34) can be reinterpreted as a gauge transformation

$$\mathcal{A}^{\mu}(x)\ \rightarrow\ \tilde{\mathcal{A}}^{\mu}(x)=\mathcal{A}^{\mu}(x)+\partial^{\mu}\Lambda_{r}(x;\alpha)\ ,$$

(II.35)

where

$\displaystyle\Lambda_{r}(x,\alpha)=\Psi(\bar{x}^{\prime})-\Psi(x)\ .$

(II.36)

As in the case of the translations, the equivalence between the rotation $R_{\hat{3}}(\alpha)$ and the gauge transformation $\Lambda_{r}(x;\alpha)$ is also obtained for the SP, i.e., one gets $\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}},r}(x,y)={\tilde{\Phi}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)$. The explicit expressions for the function $\Lambda_{r}(x;\alpha)$ in the standard gauges read

	SG:	$\displaystyle\Lambda_{r}(x;\alpha)=0\ ;$		(II.37)
	LG1:	$\displaystyle\Lambda_{r}(x;\alpha)=-\frac{B}{2}\,\sin\alpha\,\Big{[}2x^{1}x^{2}\sin\alpha+\big{(}(x^{1})^{2}-(x^{2})^{2}\big{)}\cos\alpha\Big{]}\ ;$		(II.38)
	LG2:	$\displaystyle\Lambda_{r}(x;\alpha)=\frac{B}{2}\,\sin\alpha\,\Big{[}2x^{1}x^{2}\sin\alpha+\big{(}(x^{1})^{2}-(x^{2})^{2}\big{)}\cos\alpha\Big{]}\ .$		(II.39)

Turning back once again to Eq. (II.19), and writing the propagator as in Eq. (II.22), we observe that under a rotation $R_{\hat{3}}(\alpha)$ the operator $\left({\cal D}^{\mu}\,{\cal D}_{\mu}+m_{\pi}^{2}\right)$ and the factor $e^{i\Phi_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x,y)}$ transform in the same way as under the gauge transformation $\Lambda_{r}(x;\alpha)$. This equivalence, together with the requirement that Eq. (II.19) be invariant under $R_{\hat{3}}(\alpha)$ rotations, implies that ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(x-y)$ has to be invariant under these transformations. Since, in addition, the propagator has to be invariant under boosts along the 3-axis, one can conclude that the Fourier transform ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(v_{\parallel},v_{\perp})$ defined in Eq. (II.32) can depend only on the quantities $v_{\parallel}^{2}$ and $v_{\perp}^{2}$.

An entirely similar analysis can be performed for the case of spin 1/2 and spin 1 particles. Thus, the spin 1/2 fermion propagator $S_{f}(x,y)$ can be written as

$$S_{f}(x,y)=e^{i\Phi_{f}(x,y)}\,\bar{S}_{f}(x-y)\ ,$$

(II.40)

with

$$\bar{S}_{f}(x-y)=\,\int\frac{d^{4}v}{(2\pi)^{4}}\ e^{-i\,v(x-y)}\,\bar{S}_{f}(v_{\parallel},v_{\perp})\ .$$

(II.41)

Here the propagator is a matrix in Dirac space that involves products of the $\gamma^{\mu}$ Dirac matrices. Due to the invariance under rotations around the 3-axis (i.e., the $\vec{B}$ axis) and under boosts in that direction, it is easy to see that $\bar{S}_{f}(v_{\parallel},v_{\perp})$ has to be a function of $v_{\parallel}^{2}$, $v_{\perp}^{2}$, $\gamma_{\parallel}\cdot v_{\parallel}$ and $\vec{\gamma}_{\perp}\cdot\vec{v}_{\perp}$.

In the case of a charged vector meson propagator, for instance, a $\rho$ meson propagator $D_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(x,y)$, we can write

$$D_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(x,y)=e^{i\Phi_{\rho^{\scriptscriptstyle{\cal{Q}}}}(x,y)}\bar{D}_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(x-y)\ ,$$

(II.42)

with

$\displaystyle\bar{D}_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(x-y)=\,\int\frac{d^{4}v}{(2\pi)^{4}}\ e^{-i\,v(x-y)}\,\bar{D}_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(v_{\parallel},v_{\perp})\ .$

(II.43)

Similarly to the previous cases, invariance under rotations around the 3-axis and under boosts in that direction implies that $\bar{D}_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(v_{\parallel},v_{\perp})$ will be given by a linear combination of tensors of order two built from the tensors $g_{\parallel}^{\mu\nu}$, $g_{\perp}^{\mu\nu}$, $F^{\mu\nu}$ and the vectors $v_{\parallel}^{\mu}$, $v_{\perp}^{\mu}$, with coefficients given by functions that depend only on $v_{\parallel}^{2}$ and $v_{\perp}^{2}$.

Obviously, the above statements can be corroborated by carrying out detailed calculations of the propagators. This is sketched in Sec. IV, where the explicit forms of ${\bar{\Delta}}_{\pi^{\scriptscriptstyle{\cal{Q}}}}(v_{\parallel},v_{\perp})$, $\bar{S}_{f}(v_{\parallel},v_{\perp})$ and $\bar{D}_{\rho^{\scriptscriptstyle{\cal{Q}}}}^{\nu\gamma}(v_{\parallel},v_{\perp})$ are given.

In conclusion, we have seen that the Schwinger phase carries all gauge, translation and rotation non-invariance that is present in particle propagators. In fact, this should not be surprising, since the breakdown of translational and rotational symmetry is precisely produced by the gauge choice. When calculating a physical quantity we specify a particular gauge and use propagators that, in general, break both translational and rotational invariance; however, all these symmetries are simultaneously recovered in the final result.

Let us consider a charged scalar particle in a static and homogeneous magnetic field. We introduce the scalar functions ${{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$, solutions of the eigenvalue equation

$${\cal D}^{\mu}\,{\cal D}_{\mu}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\,=\,f_{\bar{q}}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ ,$$

(III.1)

where $Q$ is the particle electric charge, ${\cal D}^{\mu}$ is the covariant derivative defined in Eq. (II.2), and the corresponding electromagnetic field ${\cal A}^{\mu}(x)$ is of the form given by Eq. (II.6). In Eq. (III.1), $\bar{q}$ stands for a set of four labels that are needed to completely specify each eigenfunction. One can be more explicit and write the eigenvalue equation in the form

$\displaystyle\left[\partial^{\mu}\,\partial_{\mu}\,-\,Q\,\vec{B}\cdot\vec{L}\;+\,\frac{Q^{2}}{4}\Big{(}\vec{x}\times\vec{B}\Big{)}^{2}\right]e^{iQ\Psi(x)}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ =\ f_{\bar{q}}\,e^{iQ\Psi(x)}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ ,$

(III.2)

where $L^{k}=i\epsilon_{klm}\,x_{l}\,\partial_{m}$. From this equation it is seen that while the eigenfunctions ${\cal F}_{Q}(x,\bar{q})$ are gauge dependent, the eigenvalues $f_{\bar{q}}$ are not. As discussed in the previous section, the magnetic field can always be taken to lie along the 3-axis, and then $\Psi(x)$ can be assumed to depend only on the two spatial coordinates perpendicular to $\vec{B}$, $x^{1}$ and $x^{2}$. Consequently, as in the case of a free particle, the eigenvalues of the components of the four-momentum along the time direction, $q^{0}$, and the magnetic field direction, $q^{3}$, can be taken as two of the labels required to specify ${\cal F}_{Q}(x,\bar{q})$. On the other hand, as is well known, the eigenvalues of Eq. (III.1) are given by

$\displaystyle f_{\bar{q}}=-\left[(q^{0})^{2}-(2k+1)B_{Q}-(q^{3})^{2}\right]\ ,$

(III.3)

where $B_{Q}\equiv|QB|$, and $k$ is a non-negative integer, to be related with the so-called Landau level. This means that the eigenvalues depend only on three of the labels included in $\bar{q}$. There is a degeneracy, which arises, of course, as a consequence of gauge invariance; to fully specify the eigenfunctions, a fourth quantum number $\chi$ is required, i.e. one has $\bar{q}=(q^{0},k,\chi,q^{3})$.

Although it is not strictly necessary, the quantum number $\chi$ can be conveniently chosen according to the gauge in which the eigenvalue problem is analyzed Wakamatsu:2022pqo . In particular, since for the standard gauges SG, LG1 and LG2 one has unbroken continuous symmetries, in those cases it is natural to consider quantum numbers $\chi$ associated with the corresponding group generators. Usual choices are

	SG:	$\displaystyle\chi=\imath\ ,$	nonnegative integer, associated to $L^{3}$		(III.4)
	$\displaystyle\mbox{(eigenvalue of $L^{3}$: $m={\rm sign}(QB)(\imath-k)$)}\ ;$
	LG1:	$\displaystyle\chi=q^{1}\ ,$	$\displaystyle\mbox{real number, eigenvalue of}\ -i\frac{\partial\ }{\partial x^{1}}\ ;$		(III.5)
	LG2:	$\displaystyle\chi=q^{2}\ ,$	$\displaystyle\mbox{real number, eigenvalue of}\ -i\frac{\partial\ }{\partial x^{2}}\ .$		(III.6)

The explicit forms of the functions ${{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$ for the above standard gauges and quantum numbers $\chi$ are given in App. A. They are shown to satisfy the completeness and orthogonality relations

	$\displaystyle\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}\ {{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})^{\ast}\,{{\cal F}_{\scriptscriptstyle{Q}}}(y,\bar{q})$	$\displaystyle=$	$\displaystyle\delta^{(4)}{(x-y)}\ ,$		(III.15)
	$\displaystyle\int d^{4}x\ {{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q}^{\prime})^{\ast}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\ .$		(III.16)

Here we have introduced some shorthand notation whose explicit form depends on the chosen gauge. For SG we have

$$\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}\equiv\dfrac{1}{(2\pi)^{2}}\sum_{k,\imath=0}^{\infty}\int\frac{dq^{0}}{2\pi}\frac{dq^{3}}{2\pi}\ ,\quad\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\equiv(2\pi)^{4}\,\delta_{kk^{\prime}}\,\delta_{\imath\imath^{\prime}}\,\delta(q^{0}-q^{\prime 0})\,\delta(q^{3}-q^{\prime 3})\ ,$$

(III.17)

while for LG1 (LG2) we have

$$\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}\equiv\dfrac{1}{2\pi}\sum_{k=0}^{\infty}\int\frac{dq^{0}}{2\pi}\frac{dq^{i}}{2\pi}\frac{dq^{3}}{2\pi}\ ,\quad\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\equiv(2\pi)^{4}\,\delta_{kk^{\prime}}\,\delta(q^{0}-q^{\prime 0})\,\delta(q^{i}-q^{\prime i})\,\delta(q^{3}-q^{\prime 3})\ ,$$

(III.18)

where $i=1$ ($i=2$). For later use we also find it convenient to define $\breve{q}=(k,\chi,q^{3})$, $s=\mbox{sign}(QB)$ and

$$\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\{\bar{q}_{E}\}}=\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}2\pi\delta(q^{0}-E)\ .$$

(III.19)

It can be shown that the functions ${{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$ satisfy the useful relations Coppola:2018ygv

	$\displaystyle{\cal D}^{0}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle-iq^{0}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ ,$
	$\displaystyle\left({\cal D}^{1}\pm i{\cal D}^{2}\right){{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle(\mp s)\,\big{[}(2k+1\mp s)B_{Q}\big{]}^{1/2}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q}_{k\mp s})\ ,$
	$\displaystyle{\cal D}^{3}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle-iq^{3}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ ,$		(III.20)

where $\bar{q}_{k\pm s}=(q^{0},k\pm s,\chi,q^{3})$. We also notice that under a gauge transformation $\Lambda(x)$ the functions ${{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})$ [with $\bar{q}=(q^{0},k,\chi,q^{3})$] transform as

$\displaystyle{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ \rightarrow\ \tilde{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})=e^{-iQ\Lambda(x)}\,{{\cal F}_{\scriptscriptstyle{Q}}}(x,\bar{q})\ .$

(III.21)

Let us start by considering the gauged Klein-Gordon action for a point-like charged pion in the presence of a static and homogeneous magnetic field. We have

$${\cal S}_{KG}=-\,\int d^{4}x\ \pi^{\scriptscriptstyle{\cal{Q}}}(x)^{\ast}\left({\cal D}^{\mu}\,{\cal D}_{\mu}+m_{\pi}^{2}\right)\pi^{\scriptscriptstyle{\cal{Q}}}(x)\ ,$$

(III.22)

where, as in Sect. II.3, we have denoted the pion charge by $Q_{\pi}={\cal Q}\,e$, with ${\cal Q}=\pm 1$. From the action in Eq. (III.22) one gets the associated gauged Klein-Gordon equation, namely

$\displaystyle\left({\cal D}^{\mu}\,{\cal D}_{\mu}+m_{\pi}^{2}\right)\pi^{\scriptscriptstyle{\cal{Q}}}(x)=0\ .$

(III.23)

We notice that, taking into account Eq. (II.4), the gauge invariance of the gauged Klein-Gordon action requires that under a gauge transformation $\Lambda(x)$ the $\pi^{\scriptscriptstyle{\cal{Q}}}(x)$ field transform as

$$\pi^{\scriptscriptstyle{\cal{Q}}}(x)\ \rightarrow\ \tilde{\pi}^{\scriptscriptstyle{\cal{Q}}}(x)=e^{-iQ_{\pi}\Lambda(x)}\,\pi^{\scriptscriptstyle{\cal{Q}}}(x)\ .$$

(III.24)

Using the notation introduced in the previous subsection, the quantized charged pion field can be written as

$$\pi^{\scriptscriptstyle{\cal{Q}}}(x)\ =\ \pi^{-\scriptscriptstyle{\cal{Q}}}(x)^{\dagger}\ =\ \,\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\{\bar{q}_{E_{\pi}}\}}\frac{1}{2E_{\pi}}\Big{\{}a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})\;\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q})+a_{\pi}^{-\scriptscriptstyle{\cal{Q}}}(\breve{q})^{\dagger}\;\mathbb{F}^{\scriptscriptstyle{-\cal{Q}}}(x,\bar{q})^{\ast}\Big{\}}\ .$$

(III.25)

Here the pion energy is given by $E_{\pi}=\sqrt{m_{\pi}^{2}+(2k+1)B_{\pi}+(q^{3})^{2}}$, with $k\geq 0$, while the functions $\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q})$ are given by

$$\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q})={\cal F}_{Q_{\pi}}(x,\bar{q})\ .$$

(III.26)

According to Eqs. (III.15) and (III.16), they satisfy the relations

	$\displaystyle\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}\,\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q})^{\ast}\,\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(y,\bar{q})$	$\displaystyle=$	$\displaystyle\delta^{(4)}{(x-y)}\ ,$		(III.35)
	$\displaystyle\int d^{4}x\ \mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q})^{\ast}\,\mathbb{F}^{\scriptscriptstyle{\cal{Q}}}(x,\bar{q}^{\prime})$	$\displaystyle=$	$\displaystyle\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\ .$		(III.36)

On the other hand, the creation and annihilation operators in Eq. (III.25) satisfy the commutation relations

	$\displaystyle\left[\,a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})\,,\,a_{\pi}^{\pm\scriptscriptstyle{\cal{Q}}}(\breve{q}\,^{\prime})\,\right]$	$\displaystyle\ =\ \left[\,a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})^{\dagger}\,,\,a_{\pi}^{\pm\scriptscriptstyle{\cal{Q}}}(\breve{q}\,^{\prime})^{\dagger}\,\right]\ =\ \left[\,a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})\,,\,a_{\pi}^{-\scriptscriptstyle{\cal{Q}}}(\breve{q}\,^{\prime})^{\dagger}\,\right]\ =\ 0\ ,$
	$\displaystyle\left[\,a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})\,,\,a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q}\,^{\prime})^{\dagger}\,\right]$	$\displaystyle\ =\ 2E_{\pi}\,(2\pi)^{3}\,\delta_{kk^{\prime}}\,\delta_{\chi\chi^{\prime}}\,\delta(q^{3}-q^{\,\prime\,3})\ .$		(III.37)

Note that according to the above definitions the operators $a_{\pi}^{\scriptscriptstyle{\cal{Q}}}(\breve{q})$ and $a_{\pi}^{-\scriptscriptstyle{\cal{Q}}}(\breve{q})$ turn out to have different dimensions from the creation and annihilation operators that are usually defined in absence of the external magnetic field.

Let us consider the gauged Dirac action for a point-like quark of flavor $f$ in the presence of a static and homogeneous magnetic field. We express the quark charge as $Q_{f}={\cal Q}_{f}\,e$, with ${\cal Q}_{u}=2/3$, ${\cal Q}_{d}=-1/3$ for $f=u,d$. The gauged action is given by

$${\cal S}_{D}=\int d^{4}x\ \bar{\psi_{f}}(x)\left(i\,\hbox to0.0pt{/\hss}\!{\cal D}-m_{f}\right)\psi_{f}(x)\ ,$$

(III.38)

where, as usual, $\bar{\psi}_{f}=\psi_{f}^{\dagger}\,\gamma^{0}$ and $\hbox to0.0pt{/\hss}\!{\cal D}=\gamma_{\mu}{\cal D}^{\mu}$; the associated gauged Dirac equation reads

$\displaystyle\left(i\,\hbox to0.0pt{/\hss}\!{\cal D}-m_{f}\right)\psi_{f}(x)=0\ .$

(III.39)

In a similar way as in the case of the charged pion, gauge invariance of the gauged Dirac action requires that under a gauge transformation $\Lambda(x)$ the field $\psi_{f}(x)$ transform as

$$\psi_{f}(x)\ \rightarrow\ \tilde{\psi}_{f}(x)=e^{-iQ_{f}\Lambda(x)}\,\psi_{f}(x)\ .$$

(III.40)

The quantized quark fields are given by

$$\psi_{f}(x)\ =\ \operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\{\bar{q}_{E_{f}}\}}\ \sum_{a=1,2}\,\frac{1}{2E_{f}}\,\Big{\{}\,b_{f}(\breve{q},a)\,U_{f}(x,\bar{q},a)+d_{f}(\breve{q},a)^{\dagger}\,V_{f}(x,\bar{q},a)\Big{\}}\ ,$$

(III.41)

where the quark energy is given by $E_{f}=\sqrt{m_{f}^{2}+2kB_{f}+(q^{3})^{2}}$, with $k\geq 0$; for $k=0$, only the value $a=1$ in the sum over $a$ is allowed. Once again we use here the definitions $\bar{q}=(q^{0},k,\chi,q^{3})$, $\breve{q}=(k,\chi,q^{3})$, $B_{f}=|Q_{f}B|$ and $s=\mbox{sign}(Q_{f}B)$. The spinors $U_{f}$ and $V_{f}$ in Eq. (III.41) can be written as

	$\displaystyle U_{f}\left(x,\bar{q},a\right)\$	$\displaystyle=\ \mathbb{E}^{\scriptscriptstyle{{\cal Q}_{f}}}(x,\bar{q})\ u_{\scriptscriptstyle{{\cal Q}_{f}}}(k,q^{3},a)\ ,$
	$\displaystyle V_{f}\left(x,\bar{q},a\right)\$	$\displaystyle=\ \mathbb{\tilde{E}}^{\scriptscriptstyle{-{\cal Q}_{f}}}(x,\bar{q})\ v_{-\scriptscriptstyle{{\cal Q}_{f}}}(k,q^{3},a)\ ,$		(III.42)

where $\mathbb{E}^{\scriptscriptstyle{{\cal Q}_{f}}}(x,\bar{q})$ and $\mathbb{\tilde{E}}^{\scriptscriptstyle{-{\cal Q}_{f}}}(x,\bar{q})$ are Ritus functions Ritus:1978cj . Their explicit forms are

	$\displaystyle\mathbb{E}^{\scriptscriptstyle{{\cal Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle\sum_{\lambda=\pm}\ \Gamma^{\lambda}\,{\cal F}_{Q}(x,\bar{q}_{\lambda})\ ,$
	$\displaystyle\mathbb{\tilde{E}}^{\scriptscriptstyle{-{\cal Q}}}(x,\bar{q})$	$\displaystyle=$	$\displaystyle\sum_{\lambda=\pm}\ \Gamma^{\lambda}\,{\cal F}_{-Q}(x,\bar{q}_{-\lambda})^{\ast}\ ,$		(III.43)

where $\Gamma^{\pm}=(1\pm S^{3})/2$, $S^{3}=i\gamma^{1}\gamma^{2}$ being the 3-component of the spin operator in the spin one-half representation. We have also used the definition $\bar{q}_{\lambda}=(q^{0},k_{s\lambda},\chi,q^{3})$, with $k_{s\pm}\ =\ k-(1\mp s)/2$. The explicit form of the spinors $u_{\scriptscriptstyle{{\cal Q}_{f}}}\left(k,q^{3},a\right)$ and $v_{-\scriptscriptstyle{{\cal Q}_{f}}}\left(k,q^{3},a\right)$ in Eq. (III.42), as well as the anticommutation relations between the fermion creation and annihilation operators and some properties of the functions $\mathbb{E}^{\scriptscriptstyle{{\cal Q}}}(x,\bar{q})$ are given in App. B. Using these properties it is easy to show that the spinors $U_{f}$ and $V_{f}$ satisfy orthogonality and completeness relations, namely Coppola:2018ygv

	$\displaystyle\int d^{4}x\,\,\overline{U}_{f}(x,\bar{q},a)\,U_{f}(x,\bar{q}^{\prime},a^{\prime})$	$\displaystyle=2m_{f}\,\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\,\delta_{aa^{\prime}}\ ,$
	$\displaystyle\int d^{4}x\,\,\overline{V}_{f}(x,\bar{q},a)\,V_{f}(x,\bar{q}^{\prime},a^{\prime})$	$\displaystyle=-2m_{f}\,\hat{\delta}_{\bar{q}\bar{q}^{\prime}}\,\delta_{aa^{\prime}}\ ,$
	$\displaystyle\int d^{4}x\,\,\overline{V}_{f}(-x,\bar{q},a)\,U_{f}(x,\bar{q}^{\prime},a^{\prime})$	$\displaystyle=\int d^{4}x\,\,\overline{U}_{f}(x,\bar{q},a)\,V_{f}(-x,\bar{q}^{\prime},a^{\prime})=0$		(III.44)

and

$$\frac{1}{2m_{f}}\operatorname*{\mathchoice{\ooalign{$\displaystyle\sum$\cr$\displaystyle\int$\cr}}{\ooalign{\raisebox{0.14pt}{\scalebox{0.7}{$\textstyle\sum$}}\cr$\textstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}{\ooalign{\raisebox{0.2pt}{\scalebox{0.6}{$\scriptstyle\sum$}}\cr$\scriptstyle\int$\cr}}}_{\bar{q}}\sum_{a}\left[U_{f}(x,\bar{q},a)\,\overline{U}(x^{\prime},\bar{q},a)-V_{f}(-x,\bar{q},a)\,\overline{V}_{f}(-x^{\prime},\,\bar{q},a)\right]=\delta^{\left(4\right)}(x-x^{\prime})\ .$$

(III.45)

On the right hand side of this last equation, an identity in Dirac space is understood.