1 Introduction
Relativistically invariant field theories feature[1-4] conserved quantities, which can be calculated either from Noether's theorem[2-3] or, in the case of free fields, from careful analysis of the behaviour of the studied field under the transformations of the Poincaré group. The expressions of these conserved quantities, the so-called Poincaré generators, in terms of the ladder operators for the quantised field are easily computed for the free scalar field (which describes spin $0$ bosons) and are found in many textbooks (see Refs. [3] and [5] for instance). On the other hand, for the case of the massless vector field of electrodynamics their expression has, as far as we know, only been given by Białynicki-Birula and Bial ynicka-Birula.[1,6] The derivation, sketched in Ref. [1], is conducted through Noether's theorem, and makes use of the energy-momentum tensor of the electromagnetic field. In this paper we derive these Poincaré generators using a more direct approach based on Wigner's theorem and the study of the behaviour of the four-potential under Poincaré transformations. Our approach also allows for an investigation of the role of gauge fixing and gauge freedom in Lorentz transformations, and highlights the relevance of the discussions of this topic found in Refs. [2] and [4]. An important prerequisite to the derivation is a thorough study of the polarisation vectors of electrodynamics. A central object in that respect is the so-called connection on the light cone, a quantity introduced in Ref. [7] and later shown[8] to be relevant to the construction of Poincaré generators for massless particles. In Sec. 2 we briefly recall some properties of the Lorentz group. Section 3 is then devoted to establishing a priori the commutators of the generators of Poincaré transformations with the four-potential field operator. It is then the goal of Sec. 4, the central section of this paper, to build the Poincaré generators so that they will fulfill the commutation relations given in Sec. 3. Section 5 is a brief summary of our work.
2 The Lorentz Group
We remember the definition of the Lorentz group: a transformation $\Lambda$ is part of the Lorentz group if and only if
$ (\Lambda^\top\eta\Lambda)_{\mu\nu}=\eta_{\mu\nu}\,, $
where $\eta_{\mu\nu}=\mathrm{diag}\left(+1,-1,-1,-1\right)$ is the flat Minkowski metric. We write, around the identity transformation within the Lorentz group, $\Lambda^{\mu\nu}=\eta^{\mu\nu}+\omega^{\mu\nu}$. It is easy to show that Eq. (1) demands that $\omega^{\mu\nu}=-\omega^{\nu\mu}$. Such antisymmetric $\left(4\times4\right)$ matrices have six independent components and can be expanded[9] over the basis
$ (M^{\rho\sigma})^{\mu\nu}=\eta^{\rho\mu} \eta^{\sigma\nu}-\eta^{\rho\nu}\eta^{\sigma\mu}\,, $
as
$ \omega^{\mu\nu}=\frac{1}{2}\Omega_{\rho\sigma} (M^{\rho\sigma})^{\mu\nu}\,. $
The tensor $\Omega_{\rho\sigma}$ is antisymmetric. Its six independent components are six parameters, which label the infinitesimal transformation $\Lambda^{\mu\nu}$: they contain information about rotation angles and boost velocities.
3 Wigner-Bargmann Analysis of the Poincaré Generators
$ P(m\leftrightarrow n)=\left|\langle m\!\mid\!n\rangle\right|^2\,, $
be independent of the inertial frame of reference in which they are computed. Wigner's theorem[11] states that to satisfy this requirement, only two types of transformations of the Hilbert space states are possible, namely, linear and unitary transformations, or antilinear and antiunitary transformations. Since we shall focus on the proper ($\mathrm{Det}\Lambda=+1$) orthochronous ($\Lambda_0^{\,0}\geq1$) Lorentz group, we want the identity transformation to be included: thus the topological structure of the Lorentz group[2,10] allows to conclude that the only acceptable transformations are linear and unitary.
For the theory to be complete, operators must be included in this formalism. Indeed, it is the very definition of a linear operator that its action on a state of the Hilbert space yields another state. Accordingly, Wigner's theorem should be extended to operators. Start with the quantity $\langle\varphi\!\mid\hat{O}\mid\!\psi\rangle$. Since $\hat{O}\mid\!\psi\rangle$ is a state, then this quantity is a scalar product and its square modulus, which is a probability, should be invariant under Poincaré transformations. These transformations combine Lorentz transformations (that is, rotations and pure boosts) $\Lambda^\mu_{\,\nu}$ with spacetime translations $a^\mu$. To enforce this, we should ask that since the state vectors transform as
$ \mid\psi\rangle\rightarrow\hat{U} (\Lambda,a)\mid\psi\rangle\,, $
where $\hat{U}(\Lambda,a)$ is a linear unitary operator, operators should transform as
$ \hat{O}\rightarrow\hat{U}(\Lambda,a) \hat{O}\hat{U}^\dagger(\Lambda,a)\,. $
Indeed, if this is the case, then
$$\langle\varphi\!\mid\hat{O}\!\mid\psi\rangle\rightarrow \langle\varphi\!\mid\hat{U}^\dagger\left(\Lambda,a\right) \hat{U}\left(\Lambda,a\right)\hat{O}\hat{U}^\dagger \left(\Lambda,a\right)\hat{U}\left(\Lambda,a\right)\mid\!\psi\rangle =\langle\varphi\!\mid\hat{O}\!\mid\psi\rangle\,, $$
and the probability is conserved under Poincaré transformations. According to Eq. (6), $\hat{U}(\Lambda,a) \hat{A}^\mu\left(x\right)\hat{U}^\dagger(\Lambda,a)$ is the Poincaré-transformed field operator. Since the field operator is a vector, we ask that[11]
$ \qquad\quad {\hat{U}}^{†}\left(\Lambda,a\right)\hat{A}^\nu\left(x\right) \hat{U}\left(\Lambda,a\right)= \Lambda^\mu_{\,\nu}\hat{A}^\mu(\Lambda^{-1} (x-a)) =\hat{A}^{\prime\mu}\left(x\right)\,. $
where $\hat{A}^{\prime\mu}(x')=\Lambda^\mu_{\,\nu} \hat{A}^\nu(x)$ and $x^{\prime\mu}=\Lambda^\mu_{\,\nu}x^\nu+a^\mu$. Wigner-Bargmann analysis of the Poincaré group tells us[12] that there is a one-to-one correspondence between this behaviour under Poincaré transformations and the fact that vector fields obey the Proca equations, or, in the massless case at hand, the Maxwell equations. The transformation law (7) allows us to work out the explicit expression for the unitary operators $\hat{U}(\Lambda,a)$. These can be written
$ \hat{U}(\Lambda,a)=\mathrm{e}^{\mathrm{i}a^\mu\hat{P}_\mu +({\mathrm{i}}/{2})\Omega^{\mu\nu}\hat{J}_{\mu\nu}}\,, $
$ \hat{U}^\dagger(\Lambda,a) =\mathrm{e}^ {-\mathrm{i}a^\mu\hat{P}_\mu-({\mathrm{i}}/{2}) \Omega^{\mu\nu}\hat{J}_{\mu\nu}}\,, $
where $\hat{P}_\mu$ and $\hat{J}_{\mu\nu}$, which are Hermitian operators,[2] are the generators of the Poincaré transformations. $\hat{P}_\mu$ generates space-time translations, while $\hat{J}_{\mu\nu}$ generates Lorentz transformations, that is, rotations and pure Lorentz boosts. As mentioned in Sec. 2, $\Omega^{\mu\nu}$ parametrises the Lorentz transformation realised by $\hat{U}(\Lambda,a)$. The other parameter $a^\mu$ parametrises the translation. Now, assume that $a^\mu$ and $\Omega^{\mu\nu}$ are infinitesimal parameters. To first order in these parameters the rightmost expression in Eq. (7) reads
$ \left(\delta^\mu_{\,\nu}+\omega^\mu_{\,\nu}\right)\hat{A}^\nu \left(\Lambda^{-1} (x-a)\right) =\hat{A}^\mu\left(x\right)+\omega^\mu_{\,\nu}\hat{A}^\nu \left(x\right)-[x^\nu-\left(\Lambda^{-1} (x-a)\right)^\nu]\partial_\nu\hat{A}^\mu(x) \\ \mbox{(first order)} =\hat{A}^\mu\left(x\right)+\omega^\mu_{\,\nu}\hat{A}^\nu \left(x\right)-\left(a^\rho+\omega^\rho_{\,\sigma}x^\sigma\right)\partial_\rho\hat{A}^\mu\left(x\right)\,. $
To first order in $a^\mu$ and $\Omega^{\mu\nu}$ the leftmost expression in Eq. (7) reads
$ {\hat{U}}^{†}\left(\Lambda,a\right)\hat{A}^\mu(x)\hat{U}\left(\Lambda,a\right) =\hat{A}^\mu\left(x\right)-\mathrm{i}a_\nu[\hat{P}^\nu,\hat{A}^\mu(x)] -\frac{\mathrm{i}}{2}\Omega_{\rho\sigma}[\hat{J}^{\rho\sigma},\hat{A}^\mu(x)]\,. $
Now all there is to do is equate Eq. (9) with Eq. (10). Use Eqs. (2) and (3) to find[13]
$ [\hat{P}^\nu,\hat{A}^\mu\left(x\right)]=-\mathrm{i}\, \partial^\nu\hat{A}^\mu\left(x\right), $
$ [\hat{J}^{\rho\sigma},\hat{A}^\mu\left(x\right)] =\mathrm{i}\, [\left(x^\rho\partial^\sigma-x^\sigma\partial^\rho\right)\hat{A}^\mu \left(x\right)+(\eta^{\rho\mu}\hat{A}^\sigma\left(x\right) -\eta^{\sigma\mu}\hat{A}^\rho\left(x\right))]\,. $
The commutator (11b) of the field with the generators $\hat{J}^{\rho\sigma}$ of Lorentz transformations will be our beacon for the rest of the paper. The Lorentz generators should be built so that they fulfil this identity. We should also note that to completely determine the Poincaré generators, it should be asked that their action upon the vacuum state of the theory yield zero. Indeed, it is a natural requirement[10] that the vacuum state of the theory be invariant under Poincaré transformations. Accordingly $\hat{U}\left(\Lambda,a\right)\mid\!0\rangle=\mid\!0\rangle$, which, to first order, yields
$ \hat{P}_\mu\mid\!0\rangle =0\,, $
$ \hat{J}_{\mu\nu}\mid\!0\rangle =0\,. $
This allows us to build the Poincaré generators. The translation generators $\hat{P}_\mu$ are easy enough to determine: use Eqs. (11) and (12) along with, either, in the Coulomb case, Eqs. (17) and (21) or, in the Lorenz case, Eqs. (40) and (45) to get
$ \hat{P}_\mu=\sum_\lambda\int\tilde{\mathrm{d}k}\,k_\mu\, \hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right) \hat{a}_{\left(\lambda\right)}\left({k}\right)\,, $
with the usual creation $\hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right)$ and annihilation $\hat{a}_{\left(\lambda\right)}\left({k}\right)$ operators. On the other hand, the generators $\hat{J}_{\mu\nu}$ of Lorentz transformations (rotations and boosts) are difficult to build. This is the subject of Subsecs. 4.2 and 4.3.
4 Lorentz Generators for the Massless Vector Field
4.1 Gauge Fixing
As with most difficulties encountered in the theory of the free electromagnetic field, the delicate nature of the derivation of the Lorentz generators comes from the fact that photons are particles of helicity $\pm1$ described by a four-vector. It is a well known problem that the four components of $A^\mu$ carry redundant information and that irrelevant degrees of freedom must thus be dealt with by the procedure of gauge fixing. Remember that the Wigner-Bargmann equation for the potential vector of electrodynamics is[2,12]
$ \partial^\nu\partial_\nu A_\mu-\partial^\nu\partial_\mu A_\nu=0\,. $
Since photons can have two distinct polarisations, it is expected that $A^\mu$ carries only two degrees of freedom. The Maxwell equations (14) restrain the number of degrees of freedom to no fewer than three. Gauge fixing can then be used to restrain, in certain cases, one further degree of freedom.
$\bullet$ It readily does if we choose the Coulomb gauge condition $A^0=0,\,\nabla\cdot{A}=0$, which kills two degrees of freedom. In that case the Maxwell equations (14) are reduced to $\partial^\mu\partial_\mu A_i=0$, which is simply the zero mass Klein-Gordon equation and does not kill any further degree of freedom. Of course the Coulomb gauge condition is not invariant under Lorentz boosts. This is further discussed at the end of SubSec. 4.2.
$\bullet$ On the other hand, if we choose the Lorenz gauge condition $\partial^\mu A_\mu=0$, this kills one degree of freedom and the Maxwell equations read, again, $\partial^\mu\partial_\mu A_\nu=0$. Thus the total number of degrees of freedom is three. This is problematic, but, as is well known, can be solved when quantising the field through the Gupta-Bleuler[14-15] procedure.
We introduce here, for future reference,
$ \tilde{\mathrm{d}k}\equiv\frac{\mathrm{d}^4k} {\left(2\pi\right)^4}2\pi\,\delta\left(k_0^2 -{k}^2\right)\theta\left(k_0\right)\,, $
$ \sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k} \equiv\sum_{\lambda=\pm}\int\tilde{\mathrm{d}k}\,. $
4.2 Coulomb Gauge
In the Coulomb gauge the wave equation (14) is solved by[16]
$ \hat{A}^i \left(x\right) = \sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k} [\hat{a}_{\left(\lambda\right)}\left({k}\right) \epsilon_{\left(\lambda\right)}^i\left({k}\right) \mathrm{e}^{-\mathrm{i}k^\mu x_\mu}+\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)\epsilon_{\left(\lambda\right)}^{*i}\left({k}\right) \mathrm{e}^{\mathrm{i}k^\mu x_\mu}]\,. $
Here the polarisation vectors $ {\epsilon}_{\left(\lambda\right)} \left({k}\right)$ are linear combinations of the unit spherical vectors
$ {e}_1\left({k}\right) = \cos\theta\cos {e}_x + \cos\theta\sin\varphi {e}_y - \sin\theta {e}_z\,,\\ {e}_2\left({k}\right) = -\sin\varphi {e}_x + \cos\varphi {e}_y\,,\\ {e}_3\left({k}\right) = \sin\theta\cos\varphi {e}_x + \sin\theta\sin\varphi {e}_y + \cos\theta {e}_z\,. $
namely[17]
$ {\epsilon}_{\left(\lambda\right)}\left({k}\right) \equiv\frac{\mathrm{e}^{-\mathrm{i}\lambda\,\chi\left(\mathbf{k}\right)}} {\sqrt{2}}\left( {e}_1\left({k}\right)+\mathrm{i}\lambda {e}_2 \left({k}\right)\right)\,, $
where $\lambda=\pm1$ and $\chi$ is an arbitrary function of $\theta$ and $\varphi$. Of high importance are the following identities:
$ {\epsilon}_{\left(\lambda\right)}^*\left({k}\right) \cdot{\epsilon}_{\left(\kappa\right)}\left({k}\right)=\delta_{\lambda\kappa}\,, $
$ {\epsilon}_{\left(\lambda\right)}^*\left({k}\right) \times{\epsilon}_{\left(\kappa\right)}\left({k}\right)=\mathrm{i}\, \delta_{\lambda\kappa}\,\lambda\,\frac{{k}}{\left|\left|{k}\right|\right|}\,, $
$ \sum_{\lambda=\pm}\epsilon_{\left(\lambda\right)}^i\left({k}\right) \epsilon_{\left(\lambda\right)}^{*j}\left({k}\right)=-\eta^{ij}-\frac{k^ik^j}{{k}^2}\,, $
which follow readily from Eq. (19). The commutation relation for the ladder operators is[16]
$ [\hat{a}_{\left(\kappa\right)}\left({k}\right), \hat{a}_{\left(\lambda\right)}^\dagger\left( {q}\right)] =2\left|\left| {k}\right|\right|\left(2\pi\right)^3\delta \left({k}- {q}\right)\delta_{\kappa\lambda}\,. $
We are now ready to look for the Lorentz generators $\hat{J}^{\mu\nu}$. Let us focus directly on
$ \hat{J}_i \equiv-\frac{1}{2}\,\epsilon_{0ijk}\hat{J}^{jk}\,, $
$ \hat{K}_i \equiv-\hat{J}_{0i}\,, $
the respective generators for rotations and boosts. According to Eqs. (11b) and (22) we should build these operators such that they verify
$ [\hat{J}_i,\hat{A}^p\left(x\right)] = -\frac{\mathrm{i}}{2}\,\epsilon_{0ijk}[2x^j\partial^k\hat{A}^p\left(x\right) +(\eta^{jp}\hat{A}^k\left(x\right)-\eta^{kp}\hat{A}^j \left(x\right))]\,, $
$ [\hat{K}_i,\hat{A}^p\left(x\right)] = -\mathrm{i}\,[\left(x_0\partial_i-x_i\partial_0\right) \hat{A}^p\left(x\right) +(\delta_0^{\,p}\hat{A}_i\left(x\right)-\delta_i^{\,p} \hat{A}_0\left(x\right))]\,. $
Using Eqs. (20b) and (20c) we can compute the second terms on the right-hand side of both equalities in Eq. (23). This yields
$ \frac{\mathrm{i}}{2}\epsilon_{0ijk}[\eta^{jp} \epsilon_{\left(\lambda\right)}^k\left({k}\right) -\eta^{kp}\epsilon_{\left(\lambda\right)}^j\left({k}\right)] =\frac{\lambda}{\left|\left| {k}\right|\right|} (\epsilon_{\left(\lambda\right)}^p\left({k}\right)k_i +\epsilon_{\left(-\lambda\right)i}\left({k}\right)k^p)\,, $
$ \mathrm{i}\left[\delta_0^{\,p}\epsilon_{\left(\lambda\right)i} \left({k}\right)-\delta_i^{\,p}\epsilon_{\left(\lambda\right)0} \left({k}\right)\right] =0\,. $
Using what we know of the easier case of the scalar field,[3] we introduce the naive guess consisting of
$ \hat{J}_i^{\mathrm{naive}}=\mathrm{i}\,\epsilon_{0ijk}\sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k}\,\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)k^j\frac{\partial}{\partial k_k}\hat{a}_{\left(\lambda\right)} \left({k}\right)\,, $
$ \hat{K}_i^{\mathrm{naive}} =\mathrm{i}\sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k}\,\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)\left|\left| {k}\right|\right|\frac{\partial} {\partial k^i}\hat{a}_{\left(\lambda\right)}\left({k}\right)\,, $
knowing that it cannot work, but merely as a starting point. Going further requires to study the behaviour of the polarisation vectors (19) under differentiation with respect to $ {k}$. Following Ref. [1] we write $ {\epsilon}\left({k}\right)\equiv {\epsilon}_{\left(+1\right)} \left({k}\right)= {\epsilon}_{\left(-1\right)}^*\left({k}\right)$. From Eq. (20) we can deduce that
$ \partial_{ {k}i} {\epsilon}\left({k}\right)= -\mathrm{i}\alpha_i\left({k}\right) {\epsilon}\left({k}\right) -\epsilon_i\left({k}\right)\frac{ {k}}{ {k}^2}\,, $
$ \alpha_i\left({k}\right)\equiv\mathrm{i} {\epsilon}^* \left({k}\right)\cdot\partial_{ {k}i} {\epsilon} \left({k}\right)\,, $
is not directly constrained, but we can easily compute its curl through Eq. (20), which reads
$ \nabla_{ {k}}\times {\alpha}\left({k}\right) =-\frac{ {k}}{\left|\left| {k}\right|\right|^3}\,. $
Let us treat the rotation generators $\hat{J}_i$ first. The commutators of the corresponding naive generator (25a) with the field reads
$ [\hat{J}_i^{\mathrm{naive}},\hat{A}^p\left(x\right)] =-{i}\epsilon_{0ijk}x^j\partial^k\hat{A}^p\left(x\right) +\sum_{\lambda=\pm} \int \; \tilde{\mathrm{d}k}\,\lambda \Big[ \epsilon_{0ijk}\,k^j\alpha^k \left({k}\right)\epsilon_{\left(\lambda\right)}^p\left({k}\right) -\epsilon_{\left(\lambda\right)i}\left({k}\right)\frac{k^p}{\left|\left| {k}\right|\right|}\Big]\hat{a}_{\left(\lambda\right)}\left({k}\right) \mathrm{e}^{-\mathrm{i}k^\mu x_\mu} +\sum_{\lambda=\pm} \int \; \tilde{\mathrm{d}k}\, \lambda[\epsilon_{0ijk}\,k^j\alpha^k \left({k}\right)\epsilon_{\left(\lambda\right)}^{*p}\left({k}\right) -\epsilon_{\left(\lambda\right)i}^*\left({k}\right)\frac{k^p}{\left|\left| {k}\right|\right|}]\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)\mathrm{e}^{\mathrm{i}k^\mu x_\mu}\,. $
Accordingly we can guess, upon comparing Eq. (24a) with Eq. (29), the correct expression for the generators. Introducing the covariant derivative $D_{\mathbf{k}\left(\lambda\right)}^j$ on the light cone[1] as
$ D_{ {k}\left(\lambda\right)}^j=\frac{\partial}{\partial k_j} +\mathrm{i}\lambda\alpha^j\left({k}\right)\,, $
yields
$ \hat{J}_i=\sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k}\,\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)\Big(\mathrm{i}\,\epsilon_{0ijk}k^jD_{ {k}\left(\lambda\right)}^k +\lambda\frac{k_i}{\left|\left| {k}\right|\right|}\Big) \hat{a}_{\left(\lambda\right)}\left({k}\right)\,, $
which is in agreement with the result obtained in Ref. [1] by computing the energy-momentum tensor of the electromagnetic field. Things are not as simple for the boost generators $\hat{K}_i$. The commutators of the corresponding naive generator (25b) with the field read
$ [\hat{K}_i^{\mathrm{naive}},\hat{A}^p\left(x\right)] =-\mathrm{i}\left(x_0\partial_i-x_i\partial_0\right)\hat{A}^p\left(x\right) +\mathrm{i}\sum_{\lambda=\pm} \int\;\tilde{\mathrm{d}k}\,\left|\left| {k}\right|\right| \Big(-\mathrm{i}\lambda\alpha_i\left({k}\right)\epsilon_{\left(\lambda\right)}^p \left({k}\right)-\epsilon_{\left(\lambda\right)i} ({k}) \frac{k^p} { {k}^2}\Big)\hat{a}_{\left(\lambda\right)}\left({k}\right)\mathrm{e}^{-\mathrm{i}k^\mu x_\mu} +\;\mathrm{i}\sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k}\,\left|\left| {k}\right|\right| \Big(\mathrm{i}\lambda\alpha_i\left({k}\right)\epsilon_{\left(\lambda\right)}^{*p} \left({k}\right)-\epsilon_{\left(\lambda\right)i}^*\left({k}\right) \frac{k^p}{ {k}^2}\Big)\hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right) \mathrm{e}^{\mathrm{i}k^\mu x_\mu}\,. $
Compare this last equality with Eqs. (23b) and (24b). To build the correct $\hat{K}_i$ from $\hat{K}_i^{\mathrm{naive}}$, we would need to add to the latter a term the commutator of which with the vector potential is minus the two integrals on the right-hand side of Eq. (32). As far as we know, this is not possible. Let us, however, consider the boost generator given in Ref. [1], namely
$ \hat{\mathcal{K}}_i=\mathrm{i}c\sum_{\lambda=\pm} \int\;\tilde{\mathrm{d}k}\,\left|\left| {k}\right|\right|\hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right)\, D_{ {k}\left(\lambda\right)i}\,\hat{a}_{\left(\lambda\right)}\left({k}\right)\,. $
A different font is used here to emphasize the fact that Eq. (33) does not fulfill Eq. (23b). The corresponding commutator reads[13]
$ [\hat{\mathcal{K}}_i,\hat{A}^p\left(x'\right)]=-\mathrm{i}\, [\left(x_0\partial_i-x_i\partial_0\right)\hat{A}^p\left(x\right) +(\delta_0^{\,p}\hat{A}^i\left(x\right)-\delta_i^{\,p}\hat{A}_0 \left(x\right))]+\hat{G}_i^{\,p}\left(x\right)\,, $
where
$ \hat{G}_i^{\,p}\left(x\right)\equiv-\mathrm{i}\sum\sum_{\lambda=\pm} \int \;\tilde{\mathrm{d}k}\frac{k^p}{\left|\left| {k}\right|\right|} \big[\hat{a}_{\left(\lambda\right)}\left({k}\right) \epsilon_{\left(\lambda\right)i}\left({k}\right) \mathrm{e}^{-\mathrm{i}k^\mu x_\mu}+\hat{a}_{\left(\lambda\right)}^\dagger \left({k}\right)\epsilon_{\left(\lambda\right)i}^*\left({k}\right) \mathrm{e}^{\mathrm{i}k^\mu x_\mu}\big]\,. $
This term is best understood when investigating the transformation properties of the Coulomb gauge condition under a pure Lorentz boost. Compute, from Eq. (7), the three-divergence of the vector field in a frame of reference linked to that in which the Coulomb condition is imposed through a pure (infinitesimal) Lorentz boost of parameters $\Omega_{0j}$. If we accept that Eq. (33) is the boost generator, then we write
$ \partial'_i\hat{A}^{\prime i}(x')=\left(\delta_i^{\,\nu} +\omega_i^{\,\nu}\right)\partial_\nu\{\hat{A}^i\left(x'\right) -\mathrm{i}\,\Omega_{0q}[\hat{\mathcal{K}}^q,\hat{A}^i\left(x'\right)]\} =(\delta_i^{\,\nu}+\omega_i^{\,\nu})\partial_\nu \{\hat{A}^i\left(x'\right)-\mathrm{i}\,\Omega_{0q}\hat{G}^{qi} \left(x\right)-\mathrm{i}\,\Omega_{0q}(-\mathrm{i}\, [(x^0\partial^q-x^q\partial^0)\hat{A}^i\left(x\right) +(\eta^{0i}\hat{A}^q\left(x\right)-\eta^{qi}\hat{A}^0 \left(x\right))])\}\,. $
An analysis similar to that used in the derivation of Eq. (9) yields, with use of Eq. (35) in the last step,
$ \partial'_i\hat{A}^{\prime i}\left(x'\right) =\partial_i\hat{A}^i\left(x\right) -\Omega_{0q}\partial^0\hat{A}^q\left(x\right)-\mathrm{i}\,\Omega_{0q}\partial_i\hat{G}^{qi}\left(x\right)\\ =\partial_i\hat{A}^i\left(x\right)\,. $
Hence if we use the "wrong'' generator (33), we find that the Coulomb gauge condition remains valid after a boost. This result calls for the following dichotomy: either we choose the vector potential to transform as a Lorentz vector (see Eq. (7)), and hence to obey the Coulomb gauge condition in one frame of reference only. In that case, no explicit form for the boost generator can be written. At best, one can use $\hat{\mathcal{K}}_i$ and discard $\hat{G}^{pi}$ when writing the action of Lorentz boosts on $\hat{A}^i$, so that Eq. (23b) will be fulfilled. Or, we choose the vector potential to transform as a Lorentz vector only up to an extra term $\hat{G}^{pi}$ that corrects for the fact that the Coulomb gauge condition is not invariant under boosts. The latter choice is made in Refs. [2] and [4], and can be explained thusly: since the physical observable of free quantum electrodynamics is the gauge-invariant Faraday tensor $\hat{F}^{\mu\nu}=\partial^\mu\hat{A}^\nu -\partial^\nu\hat{A}^\mu$, we can relax Eq. (7) to
$ \hat{U}\left(\Lambda,a\right)\hat{A}^\mu\left(x\right)\hat{U}^\dagger \left(\Lambda,a\right)=\Lambda^\mu_{\,\nu}\hat{A}^\mu\left(\Lambda x+a\right) +\partial^{\mu}\hat{\lambda}\left(x\right)\,, $
without affecting the fact that $\hat{F}^{\mu\nu}$ transforms as a second-rank tensor under Lorentz transformations. Choosing the Coulomb gauge and the generator (33) amounts to choosing Eq. (38) with $\partial^i\hat{\lambda}\left(x\right) =\mathrm{i}\Omega_{0q}\hat{G}^{qi}\left(x\right)$. Hence we find[13]
$ \hat{\lambda}\left(t, {x}\right)=\frac{\Omega_{0q}}{4\pi c}\int \frac{\mathrm{d} {y}}{\left|\left| {x}- {y}\right|\right|} \partial_t\hat{A}^q\left(t, {y}\right)\,, $
as can be shown from Eq. (35) and some elementary distribution theory (also see Ref. [4] (exercise 2 of chapter 14)).
4.3 Lorenz Gauge
In the Lorenz gauge the wave equation (14) is solved by
$ \hat{A}_\mu\left(x\right)=\sum_{\lambda=0\to3} \int\tilde{\mathrm{d}k} [\hat{a}_{\left(\lambda\right)}\left({k}\right) \epsilon_{\left(\lambda\right)\mu}\left({k}\right)\mathrm{e}^{-\mathrm{i}k^\nu x_\nu} +\hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right)\epsilon_{\left(\lambda\right) \mu}^*\left({k}\right)\mathrm{e}^{\mathrm{i}k^\nu x_\nu}]\,. $
Here the polarisation vectors $\epsilon_{\left(\lambda\right)\mu}\left({k}\right)$ obey
$ \epsilon_{\left(\kappa\right)}^{*\mu}\left({k}\right)\epsilon_{\left(\lambda\right) \mu}\left({k}\right) =\eta_{\kappa\lambda}\,, $
$ \epsilon_{\left(1\right)}^\mu\left({k}\right)k_\mu=\epsilon_{\left(2\right)}^\mu \left({k}\right)k_\mu =0\,, $
and provide[3] a closure relation for the Minkowski metric
$ \sum_{\lambda=0}^3\eta_{\lambda\lambda}\epsilon_{\left(\lambda\right) \mu}^*\left({k}\right)\epsilon_{\left(\lambda\right)}^\nu \left({k}\right)=\delta_\mu^{\,\nu}\,. $
Since $k^\mu k_\mu=0$ it is automatically true that
$ \epsilon_{\left(0\right)}^{*\mu}\left({k}\right)k_\mu= -\epsilon_{\left(3\right)}^{*\mu}\left({k}\right)k_\mu\equiv\sigma\left({k}\right)\,. $
Also note from Eq. (43) that for
$\epsilon_{\left(0\right)}^\mu\left({k}\right)$ and $\epsilon_{\left(3\right)}^\mu \left({k}\right)$ to be four-vectors, $\sigma\left({k}\right)$ must be a scalar. Hence it can only be a function of $k^2=k_0^2- {k}^2$ which in the free case considered here is always zero. Hence $\sigma\left({k}\right)\equiv\sigma$ is a constant and we can rewrite Eq. (43) as[13]
$ k^\mu=\sigma(\epsilon_{\left(0\right)}^\mu \left({k}\right)+\epsilon_{\left(3\right)}^\mu \left({k}\right))\,. $
This time the commutation relation for the ladder operators is[16]
$ [\hat{a}_{\left(\kappa\right)}\left({k}\right), \hat{a}_{\left(\lambda\right)}^\dagger \left( {q}\right)]=-2k_0\left(2\pi\right)^3\delta \left({k}- {q}\right)\eta_{\kappa\lambda}\,. $
Similarly to the Coulomb case, we define
$ \epsilon_{\left(\pm\right)}^\mu\left({k}\right)\equiv\frac{\mathrm{e}^{-\mathrm{i} \lambda\,\chi\left({k}\right)}}{\sqrt{2}} (\epsilon_{\left(1\right)}^\mu\left({k}\right)\pm\mathrm{i}\epsilon_{\left(2\right)}^\mu \left({k}\right))\,, $
and write $\epsilon^\mu\left({k}\right)\equiv\epsilon_{\left(+\right)}^{*\mu} \left({k}\right)=\epsilon_{\left(-\right)}^\mu\left({k}\right)$. We also define the corresponding relation for the ladder operators
$ \hat{a}_{\left(\pm\right)}\left({k}\right)\equiv\frac{\mathrm{e}^{\mathrm{i}\lambda\, \chi\left({k}\right)}}{\sqrt{2}}\left(\hat{a}_{\left(1\right)}\left({k}\right) \mp\mathrm{i}\hat{a}_{\left(2\right)}\left({k}\right)\right)\,, $
which obey the same commutation relations as Eq. (45). From Eq. (41) and (46) we get
$ \epsilon^{*\mu}\left({k}\right)\epsilon_\mu\left({k}\right) =-1\,, $
$ \epsilon^\mu\left({k}\right)\epsilon_\mu\left({k}\right) =0\,, $
$ \epsilon^{\mu}\left({k}\right)k_{\mu} =0\,. $
Keeping in mind that, on the light cone, one can write the $ {k}$-space derivative
$ \partial^i_{ {k}}k_\mu=\delta^i_{\,\mu}-\delta^0_{\,\mu}\frac{k^i}{k^0}\,, $
we use Eqs. (48a) and (48b) to find the equivalent relation to Eq. (26), namely
$ \partial^i_{ {k}}\epsilon_\mu\left({k}\right)=-\mathrm{i}\,\alpha_X^i \left({k}\right)\epsilon_\mu\left({k}\right)+\alpha_{\left(0\right)}^i \left({k}\right)\epsilon_{\left(0\right)}^\mu\left({k}\right) +\alpha_{\left(3\right)}^i\left({k}\right)\epsilon_{\left(3\right)}^\mu\left({k}\right)\,, $
where the real vector $ {\alpha}_X\left({k}\right)$ is not directly constrained (one can compute its curl but the result is less eloquent than Eq. (28)) and the coefficients $\alpha_{\left(0/3\right)}^i\left({k}\right)$ obey
$ \sigma^*(\alpha_{\left(0\right)}^i\left({k}\right) -\alpha_{\left(3\right)}^i \left({k}\right))+\epsilon^i\left({k}\right) -\epsilon^0\left({k}\right)\frac{k^i}{k^0}=0\,, $
which is proved from Eq. (48c).
Remember that we are in the Lorenz gauge, for which $\partial^\mu A_\mu=0$ should be fulfilled. For this to be the case, Gupta-Bleuler quantisation tells us[3,5,13,16] that, for a given wave vector $k^\mu$, the component of $A^\mu$ along $\epsilon_{\left(0\right)}^\mu\left({k}\right) -\epsilon_{\left(3\right)}^\mu\left({k}\right)$ must be discarded, while the component along $\epsilon_{\left(0\right)}^\mu\left({k}\right) +\epsilon_{\left(3\right)}^\mu\left({k}\right)$, that is, according to Eq. (44), along $k^\mu$, is a gauge term and may be ignored (as can be shown,[5,13,16] it does not contribute to the physical variables, which are bilinear in the ladder operators, such as the Hamiltonian of the electromagnetic field). We shall, however, see that ignoring this latter component leads, under Poincaré transformations, to the apparition of gauge terms similar to the one which we obtained (see Eq. (35)) in the Coulomb gauge.
With the help of Eqs. (42) and (44) we rewrite the second term on the right-hand side of Eq. (11b) as
$ \mathrm{i}\,\eta^{\rho\mu}\epsilon^\sigma\left({k}\right)=-\mathrm{i}\, \epsilon^{*\rho}\left({k}\right)\epsilon^\mu\left({k}\right) \epsilon^\sigma\left({k}\right)+\frac{\mathrm{i}}{2}\Bigl[\frac{k^\rho}{\sigma^*} (\epsilon_{\left(0\right)}^\mu\left({k}\right)-\epsilon_{\left(3\right)}^\mu \left({k}\right))+(\epsilon_{\left(0\right)}^{*\rho}\left({k}\right) -\epsilon_{\left(3\right)}^{*\rho}\left({k}\right))\frac{k^\mu}{\sigma}\Bigr] \epsilon^\sigma\left({k}\right)\,. $
As was done in the Coulomb gauge, we start from the Lorentz generators for the scalar field
$ \hat{J}_{ij}^{\mathrm{naive}} =-\mathrm{i}\sum_{\lambda=\pm} \int\tilde{\mathrm{d}k}\, \hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right)\Bigl(k^i\frac{\partial} {\partial k_j}-k^j\frac{\partial}{\partial k_i}\Bigr)\hat{a}_{\left(\lambda\right)} \left({k}\right)\,, $
$ \hat{J}_{0i}^{\mathrm{naive}} =-\mathrm{i}\sum_{\lambda=\pm} \int\tilde{\mathrm{d}k}\, \hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right)\left|\left| {k}\right|\right|\frac{\partial}{\partial k^i}\hat{a}_{\left(\lambda\right)} \left({k}\right)\,, $
and modify them appropriately. With the use of Eqs. (50) and (51) we find that
$ \qquad\qquad \hat{\mathcal{J}}^{ij}=-\mathrm{i}\sum_{\lambda=\pm} \int\tilde{\mathrm{d}k}\, \hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right) \Bigl[\Bigl(k^i\Bigl(\frac{\partial}{\partial k_j}+\mathrm{i} \lambda\alpha_X^j\left({k}\right)\Bigr)-k^j\Bigl(\frac{\partial}{\partial k_i} +\mathrm{i}\lambda\alpha_X^i\left({k}\right)\Bigr)\Bigr) +\lambda(\epsilon_{\left(\lambda\right)}^{*i}\left({k}\right) \epsilon_{\left(\lambda\right)}^j\left({k}\right) -\epsilon_{\left(\lambda\right)}^i \left({k}\right)\epsilon_{\left(\lambda\right)}^{*j} \left({k}\right))\Bigr]\hat{a}_{\left(\lambda\right)}\left({k}\right)\,, $
$ \hat{\mathcal{J}}^{0i}=-\mathrm{i}\sum_{\lambda=\pm} \int\tilde{\mathrm{d}k}\, \hat{a}_{\left(\lambda\right)}^\dagger\left({k}\right) \Bigl[\left|\left| {k}\right|\right|\Bigl(\frac{\partial} {\partial k_i}+\mathrm{i}\lambda\alpha_X^i\left({k}\right) \Bigr)+\lambda\left(\epsilon_{\left(\lambda\right)}^{*0}\left({k}\right )\epsilon_{\left(\lambda\right)}^i\left({k}\right) -\epsilon_{\left(\lambda\right)}^0\left({k}\right) \epsilon_{\left(\lambda\right)}^{*i}\left({k}\right)\right)\Bigr] \hat{a}_{\left(\lambda\right)}\left({k}\right)\,, $
have the same features as the Coulomb gauge-boost generator (33), in the sense that discrepancies between the commutators $\left[\hat{\mathcal{J}}^{\rho\sigma}, \hat{A}^\mu\left(x\right)\right]$ computed from Eq. (54) and the expected result Eq. (52) are terms proportional to $k^\mu$:
$ [\hat{\mathcal{J}}^{\rho\sigma},\hat{A}^\mu\left(x\right)]=\mathrm{i}\, [\left(x^\rho\partial^\sigma-x^\sigma\partial^\rho\right)\hat{A}^\mu\left(x\right) -(\eta^{\rho\mu}\hat{A}^\sigma\left(x\right)-\eta^{\sigma\mu} \hat{A}^\rho(x))]+\hat{Q}^{\rho\sigma\mu}\left(x\right)\,, $
with
$ \qquad \hat{Q}^{\rho\sigma\mu}\left(x\right)=\Bigl[\,\mathrm{i}\!\int\tilde{\mathrm{d}k}\, \hat{a}_{ \left(-\right)}\left({k}\right)\mathrm{e}^{-\mathrm{i}k^\nu x_\nu} \left\{ \left[ k^\rho(\alpha_{\left(0\right)}^\sigma +\alpha_{\left(3\right)}^\sigma)+(\epsilon_{\left(0\right)}^{*\rho} \left({k}\right)-\epsilon_{\left(3\right)}^{*\rho}\left({k}\right)) \epsilon_{\left(-\right)}^\sigma\right]-\left[\rho\leftrightarrow\sigma\right]\right\} \frac{k^\mu}{\sigma} + \; \mathrm{i}\!\int \!\tilde{\mathrm{d}k}\,\hat{a}_{\left(+\right)} \left({k}\right)\mathrm{e}^{\mathrm{i}k^\nu x_\nu} \left\{\left[k^\rho(\alpha_{\left(0\right)}^{*\sigma}\left({k}\right) +\alpha_{\left(3\right)}^{*\sigma}\left({k}\right)) +(\epsilon_{\left(0\right)}^\rho\left({k}\right) -\epsilon_{\left(3\right)}^\rho\left({k}\right)) \epsilon_{\left(+\right)}^{\sigma}\left({k}\right)\right] \!-\left[\rho\leftrightarrow\sigma\right]\right\}\frac{k^\mu}{\sigma^*}\Bigr]-\mathrm{h.c.}\,, $
where it is understood that $\alpha_{\left(0\right)}^0\left({k}\right)$ and $\alpha_{\left(3\right)}^0\left({k}\right)$, which had not yet been introduced, vanish. In regard to Eq. (50) this can be understood as a notational way to indicate that polarisation vectors do not depend on the magnitude of the wave vector. In the Gupta-Bleuler formalism, terms such as Eq. (56) can be ignored. Indeed, this quantity is proportional to $k^\mu$, which, as discussed above, means that it is proportional to $\epsilon_{\left(0\right)}^\mu\left({k}\right)+\epsilon_{\left(3\right)}^\mu \left({k}\right)$. The effect of a Lorentz transformation on the vector potential is given by Eq. (10), which means that $\hat{Q}^{\rho\sigma\mu}$ brings to $\hat{A}^\mu$ this extra term proportional to $k^\mu$. As such, $\hat{Q}^{\rho\sigma\mu}$ is a gauge term. Indeed, if one performs a computation equivalent to the one that led to Eq. (37), this term is found to have no influence on the compatibility of Lorentz transformations with the Lorenz gauge condition. This is so because $\partial_\mu\hat{Q}^{\rho\sigma\mu}\left(x\right)=0$.
As was the case for the Coulomb gauge, keeping only the transverse degrees of freedom in the Lorenz gauge amounts to relaxing Eq. (7) to Eq. (38) with $\partial^\mu\hat{\lambda}\left(x\right)=-\mathrm{i}\Omega_{\rho\sigma} \hat{Q}^{\rho\sigma\mu}\left(x\right)$. There is no easy solution to this equation that would be equivalent to Eq. (39).
It is a good idea to check whether the Lorentz generators (54) have an appropriate action on the longitudinal component of $\hat{A}^\mu$, that is, the one which, for a given wave vector $k^\mu$, is parallel to $k^\mu$. One can show with use of Eq. (49) that the action of Eq. (54) on the longitudinal component of $\hat{A}^\mu$ is consistent with Eq. (11b) up to terms, which do not affect the validity of the Lorenz gauge condition (in other words, up to terms the four-divergence of which (w.r.t. $x^\mu$) vanishes).
Notice that, compared to the boost generator given in Ref. [1], which we showed was relevant in the Coulomb gauge, the Lorenz gauge boost generator (54b) features an extra term (the one which is displayed on the second line of Eq. (54b). This term clearly cancels if one switches over to the Coulomb gauge, where the polarisation four-vectors become three-vectors with a time component equal to zero.
5 Summary and Discussion
In this paper we have shown how a careful analysis of the behaviour of polarisation vectors under $ {k}$-space differentiation allows, in the two standard gauges of electrodynamics, to derive the explicit expression for the generators of Lorentz transformations in terms of the photon creation and annihilation operators.
In the Coulomb gauge we have retrieved Białynicki-Birula and Białynicka-Birula's result[1,6] for the rotation generators, and discussed their boost generator (33) in relation to Coulomb gauge fixing. As discussed at the end of SubSec. 4.2, the boost generator (33) has the very pleasant effect of correcting for the incompatibility of the Coulomb gauge condition with Lorentz boosts. Its action on the vector potential yields the extra term in the transformation law (7) that is accepted in Refs. [2] and [4] when quantising electrodynamics (in both these references the emphasis is on the Coulomb gauge). Note that an argument for the use of the Coulomb gauge has been given[7] in terms of the differential geometry of the $ {k}$-space light cone.
In the Lorenz gauge, similar results (54) are obtained for the rotation and boost generators, which clearly reduce to those obtained (see Eqs. (31) and (33)) earlier if one switches to the Coulomb gauge. It is also shown that in the framework of Gupta-Bleuler quantisation, the longitudinal part of the vector potential can be consistently ignored at the cost of, yet again, introducing an extra term in the transformation law (7) of the vector potential.