6.2 Canonical quantization à la Gupta-Bleuler

In what follows we will set the current \(j^\mu =0\) to zero and consider the free electromagnetic potential. In the next chapter we will see how coupling \(A_\mu \) to a charged Dirac field provides a quantum field theoretic model for a current.

The aim of this section is to discuss the quantization of the electromagnetic potential \(A_\mu \), which as we will see in a moment is more complicated than in our previous cases due to the presence of gauge symmetries. Let us first note that ignoring the gauge symmetries in the quantization procedure is not an option. Indeed, starting from the usual action (6.11) with \(j^\mu =0\), one computes the canonical momenta

\begin{flalign} \Pi ^\mu = \frac {\partial \LL }{\partial \dot {A}_\mu } = - F^{0\mu }\quad . \end{flalign} But wait: \(\Pi ^0 = -F^{00} =0\) due to antisymmetry of \(F^{\mu \nu } = -F^{\nu \mu }\), so the pair \((\Pi ^0,A_0)\) is not a good pair of canonically conjugate variables. Indeed, it is impossible to demand canonical Poisson bracket relations of the form \(\{A_0(\mathbf {x}),\Pi ^0(\mathbf {y})\} = \delta (\mathbf {x}-\mathbf {y})\), or upon quantization canonical commutation relations of the form \([A_0(\mathbf {x}),\Pi ^0(\mathbf {y})]=\ii \,\delta (\mathbf {x}-\mathbf {y})\), since \(\Pi ^0=0\) is identically zero. That this issue is indeed related to gauge symmetry can be seen quite easily: As discussed in the previous section, we can use gauge transformations to fix \(A_0=0\) in temporal gauge, so this variable is not dynamical and hence it doesn’t require a canonical momentum. Things get much better when we start from the action (6.18) that enforces Lorenz gauge fixing by a Lagrange multiplier. The canonical momenta for this action are

\begin{flalign} \Pi ^\mu = -F^{0\mu } - \xi \,\eta ^{0\mu }\,(\partial _\rho A^\rho )\quad , \end{flalign} so we see that for \(\xi \neq 0\) also \(A_0\) gets a nonvanishing canonical momentum \(\Pi ^0\).

One could now plough through the canonical quantization of this system, but we shall rather follow a simpler approach that starts from the equivalent action (6.22) where Lorenz gauge is imposed as a constraint by hand. The canonical momenta for the action (6.22) are very simple and given by

\begin{flalign} \Pi ^\mu = \dot {A}^\mu \quad . \end{flalign} That’s nice and looks pretty much like what one gets for a Klein-Gordon field, see Example 2.13. The Hamiltonian associated to the action (6.22) is given by

\begin{flalign} \label {eqn:HamiltonianMW} H\big [\Pi ,A\big ] \,=\,\int _{\bbR ^{d-1}}\frac {1}{2}\, \Big (\Pi ^\mu \,\Pi _\mu + \partial ^i A^\mu \,\partial _i A_\mu \Big ) \,\dd \mathbf {x}\quad . \end{flalign} Superficially, this looks pretty much like the Klein-Gordon Hamiltonian (2.58) with zero mass \(m^2=0\), but there is one key difference: The potential \(A_\mu \) is a multicomponent field, with \(\mu = 0,1,\dots , d-1\). Splitting into the time-like component \(0\) and the spatial components \(j\), the Hamiltonian reads as

\begin{flalign} H\big [\Pi ,A\big ] \,=\,\int _{\bbR ^{d-1}}\frac {1}{2}\, \bigg (-(\Pi ^0)^2 - (\nabla A_0)^2 + \sum _{j=1}^{d-1}\Big ((\Pi ^j)^2 + (\nabla A_j)^2\Big )\bigg ) \,\dd \mathbf {x}\quad , \end{flalign} where we use that \(\eta _{00} = \eta ^{00}=-1\) in the Minkowski metric. So each \((\Pi ^j,A_j)\) contributes to the Hamiltonian in exactly the same way as a massless Klein-Gordon field \((\Pi ,\Phi )\) would do, however \((\Pi ^0,A_0)\) has the wrong sign and contributes negatively to the Hamiltonian. This looks like a disaster! The Hamiltonian/energy of the system is unbounded from below, which is a potential source of unphysical features such as instability of the system, nonexistence of a ground state/vacuum, etc. The good news is that this disaster can be prevented by dealing correctly with the constraint \(\partial _\mu A^\mu =0\) that comes together with the action (6.22) and its residual gauge symmetries (6.24). Indeed, at the classical level, we can use the residual gauge symmetries to fix \(\Pi ^0=0\) and \(A_0=0\), hence the negative contribution to the Hamiltonian can be avoided. We will see later in this section that an analogous conclusion holds true for the quantized theory.

After all these words of warning, let us now finally carry out the quantization of the electromagnetic potential \(A_\mu \) and see what this gives. We follow the approach of Gupta and Bleuler, which consists of carrying out the following steps:

• 1. We start with quantizing the canonical variables \((\Pi ^\mu ,A_\mu )\) and the Hamiltonian (6.28) that is associated with the action in (6.22). The constraint \(\partial _\mu A^\mu =0\) and the residual gauge symmetries (6.24) are ignored at this point.

• 2. We implement the Lorenz gauge fixing constraint at the level of the QFT and analyze the quantum analog of the residual gauge symmetries.

Carrying out the first step is completely analogous to what we have done for the Klein-Gordon field in Sections 3.1 and 3.3, hence we can be relatively fast at this point. Quantization is achieved by promoting \(A_\mu \) and \(\Pi ^\mu \) to operators that satisfy the equal-time commutation relations

To obtain the Heisenberg picture field operators \(A_\mu (x)\) and \(\Pi ^\mu (x)\), one has to solve Heisenberg’s equations

\begin{flalign} \frac {\partial }{\partial t} A_\mu (x) \,=\,\ii \,\big [H,A_\mu (x)\big ]\quad ,\qquad \frac {\partial }{\partial t} \Pi ^\mu (x) \,=\,\ii \,\big [H,\Pi ^\mu (x)\big ]\quad , \end{flalign} subject to the initial conditions \(A_\mu (0,\mathbf {x})=A_\mu (\mathbf {x})\) and \(\Pi ^\mu (0,\mathbf {x})=\Pi ^\mu (\mathbf {x})\), where \(H\) denotes the Hamiltonian operator that is obtained by quantizing (6.28). (Normal ordering of the Hamiltonian operator \(H\) is inessential at this point, because under the commutator we have that \([\noor {H},-]=[H,-]\).) Introducing as in Section 3.1 annihilation and creation operators, one finds

and that \(\Pi ^\mu (x) = \dot {A}^\mu (x)\) is given by a time derivative, hence it contains no new information. A few remarks are in order:

• • The relativistic momentum \(k = (\vert \mathbf {k}\vert , \mathbf {k})\in \bbR ^d\) satisfies the massless energy-momentum relation (or on-shell condition)

  \(\seteqnumber{0}{6.}{32}\)

  \begin{equation} k^2 = \eta _{\mu \nu }\, k^\mu \,k^\nu =0 \end{equation}

  because there is no mass term in the Hamiltonian (6.28) (as well as in the action (6.22) and its Euler-Lagrange equation (6.23)). For comparison with the formulas from Section 3.1, use that the energy \(\omega _{\mathbf {k}} = \sqrt {\mathbf {k}^2 +m^2}\) is given in the massless case by the Euclidean norm \(\vert \mathbf {k}\vert \) of the spatial momentum.

• • There is a sum over a basis of polarization covectors \(\{\epsilon ^\lambda (\mathbf {k}) \in \bbR ^d: \lambda =0,\dots ,d-1\}\), which arises because \(A_\mu \) is a multicomponent field. (We have seen a similar sum over spinor polarizations in the description of the Dirac quantum field (5.81).) A possible choice of basis would be \(\epsilon ^\lambda _\mu (\mathbf {k}) = \delta ^\lambda _\mu \), but this choice isn’t the best for our discussion of the Hilbert space below. A better choice is given by taking a basis of polarization covectors that is adapted to time-like, transversal and longitudinal polarizations. In more detail, splitting covectors into their \(0\) component and spatial components, we choose

  \(\seteqnumber{1}{6.34}{0}\)

  \begin{flalign} \epsilon ^0(\mathbf {k}) \,=\,\begin{pmatrix} 1\\\mathbf {0} \end {pmatrix}\,\in \,\bbR ^d~~,\quad \epsilon ^j(\mathbf {k}) \,=\, \begin{pmatrix} 0 \\ \boldsymbol {\epsilon }^j(\mathbf {k}) \end {pmatrix} \,\in \,\bbR ^d\quad , \end{flalign} such that

  \(\seteqnumber{1}{6.34}{1}\)

  \begin{flalign} \boldsymbol {\epsilon }^j(\mathbf {k})\cdot \mathbf {k} &= 0 \qquad \text {for }j=1,\dots ,d-2\quad ,\\ \boldsymbol {\epsilon }^{d-1}(\mathbf {k}) &= \frac {\mathbf {k}}{\vert \mathbf {k}\vert }\quad . \end{flalign} This means that \(\epsilon ^0(\mathbf {k})\in \bbR ^d\) is a time-like polarization, \(\epsilon ^1(\mathbf {k}),\dots ,\epsilon ^{d-2}(\mathbf {k}) \in \bbR ^d \) are transversal spatial polarizations and \(\epsilon ^{d-1}(\mathbf {k})\in \bbR ^d\) is a longitudinal spatial polarization. We choose these polarization covectors to be orthonormal with respect to the inverse Minkowski metric, i.e.

  \(\seteqnumber{1}{6.35}{0}\)

  \begin{flalign} \epsilon ^{\lambda }_\mu (\mathbf {k})\,\epsilon ^{\lambda ^\prime }_\nu (\mathbf {k})\, \eta ^{\mu \nu }= \eta ^{\lambda \lambda ^\prime }\quad , \end{flalign} from which the following polarization sum identity follows

  \(\seteqnumber{1}{6.35}{1}\)

  \begin{flalign} \label {eqn:photonpolarizationsum} \sum _{\lambda = 0}^{d-1} \eta _{\lambda \lambda }\, \epsilon ^\lambda _\mu (\mathbf {k})\,\epsilon ^{\lambda }_\nu (\mathbf {k}) = \eta _{\mu \nu }\quad . \end{flalign} (These are the covector analogs of the spinor formulas (5.57) and (5.58) from our discussion of the Dirac field.)

• • The annihilation and creation operators carry a polarization label \(\lambda =0,1,\dots , d-1\) and they satisfy the following commutation relations

  \(\seteqnumber{0}{6.}{35}\)

  \begin{equation} \label {eqn:AnnCrephoton} \big [a_{\lambda }(\mathbf {k}),a_{\lambda ^\prime }(\mathbf {q})\big ] \,=\,0\,=\, \big [a_{\lambda }^\dagger (\mathbf {k}),a_{\lambda ^\prime }^\dagger (\mathbf {q})\big ]\quad ,\qquad \big [a_{\lambda }(\mathbf {k}),a_{\lambda ^\prime }^\dagger (\mathbf {q})\big ] \,=\,\eta _{\lambda \lambda ^\prime }\, (2\pi )^{d-1}\,\delta (\mathbf {k}-\mathbf {q}) \end{equation}

  involving the Minkowski metric \(\eta _{\lambda \lambda ^\prime }\). Note that the annihilation and creation operators associated with the time-like polarization \(\lambda =0\) have the “wrong sign”, which will cause some issues later that however can be resolved by taking into account the constraint \(\partial _\mu A^\mu =0\) and the residual gauge symmetries (6.24).

To build a Hilbert space \(\HH \), we proceed in complete analogy to our previous examples and introduce 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

Multiparticle states are then obtained by acting on \(\ket {0}\) with suitably normalized creation operators, leading to

Each particle is labeled by a pair \((k,\lambda )\) consisting of a relativistic momentum \(k= (\vert \mathbf {k}\vert ,\mathbf {k})\in \bbR ^d\) that satisfies the massless on-shell condition \(k^2=0\) and a covector polarization index \(\lambda =0,1,\dots ,d-1\). This looks so far quite good and it seems to indicate that the quantum particles associated with the electromagnetic potential \(A_\mu \), which are often called photons, are massless particles with nontrivial spin. Furthermore, from the commutation relations (6.36), we see that the multiparticle states are symmetric under the exchange of any of its labels, hence our photons are bosons as they should be. After a more careful look however, we recognize the following huge issue:

• !!! The states corresponding to photons with a time-like polarization \(\lambda =0\) have negative norm, as one can see by the following calculation

  \(\seteqnumber{0}{6.}{38}\)

  \begin{flalign} \nn \braket {q,\lambda ^\prime }{k,\lambda } \,&=\, \sqrt {2\,\vert \mathbf {q}\vert }\, \sqrt {2\,\vert \mathbf {k}\vert }\, \expect {0}{a_{\lambda ^\prime }(\mathbf {q})\,a^\dagger _{\lambda }(\mathbf {k})}{0}\\ \nn \,&=\,\sqrt {2\,\vert \mathbf {q}\vert }\, \sqrt {2\,\vert \mathbf {k}\vert }\, \expect {0}{\big [a_{\lambda ^\prime }(\mathbf {q}),a^\dagger _{\lambda }(\mathbf {k})\big ]}{0}\\ \,&=\,\eta _{\lambda ^\prime \lambda }\,2\, \vert \mathbf {k}\vert \,(2\pi )^{d-1} \,\delta (\mathbf {q}-\mathbf {k}) \quad . \end{flalign} This issue arises from the “wrong sign” commutation relation for time-like polarizations in (6.36). Negative norm states are an absolute no-go in quantum theory, because the norm of a state is interpreted as a probability which can never be negative.

The good news is that such negative norm states are just an artifact of the gauge symmetry of electromagnetism and that they can be removed by taking into account the Lorenz gauge fixing constraint and its residual gauge symmetries. This brings us to item 2. in the itemization above. Let us start thinking about how to implement the constraint \(\partial _\mu A^\mu =0\) at the level of the QFT. A first attempt would be to demand \(\partial _\mu A^\mu =0\) as an identity for operators. But that’s too naive because, as a consequence of (6.32) and the nontrivial commutation relations (6.36), one finds a nontrivial commutator \([\partial _\mu A^\mu (x),A_{\nu }(y)] \neq 0\) which is incompatible with implementing \(\partial _\mu A^\mu =0\). A smarter way is to implement the constraint at the level of the Hilbert space \(\HH \) by selecting a subclass of states \(\ket {\psi _{\mathrm {phys}}}\) one calls the physical states. This idea is called the Gupta-Bleuler method, named after the two physicists Gupta and Bleuler. The operator that selects the physical states is inspired by the Lorenz gauge fixing and it is constructed as follows: Let us denote by

\begin{flalign} A^{(+)}_\mu (x) \,= \,\int _{\bbR ^{d-1}}\frac {1}{\sqrt {2\,\vert \mathbf {k}\vert }}\, \sum _{\lambda =0}^{d-1} a_\lambda (\mathbf {k})\,\epsilon ^\lambda _{\mu }(\mathbf {k})\,e^{\ii \,k\,x} ~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}} \end{flalign} the annihilating part of the Heisenberg picture operator (6.32) and consider the operator

\begin{flalign} \nn \partial _\mu A^{(+)\mu }(x)\,&=\,\partial ^\mu A^{(+)}_{\mu }(x) \,=\, \int _{\bbR ^{d-1}}\frac {1}{\sqrt {2\,\vert \mathbf {k}\vert }}\, \sum _{\lambda =0}^{d-1} a_\lambda (\mathbf {k})\,\ii \,k^\mu \,\epsilon ^\lambda _{\mu }(\mathbf {k})\,e^{\ii \,k\,x} ~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ \,&=\,\ii \, \int _{\bbR ^{d-1}} \sqrt {\frac {\vert \mathbf {k}\vert }{2}}~ \Big ( a_0(\mathbf {k}) + a_{d-1}(\mathbf {k})\Big )\,e^{\ii \,k\,x} ~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \label {eqn:GuptaBleuleroperator} \end{flalign} where in the last equality we have used our particular choice of basis for the polarization covectors in (6.34). The Gupta-Bleuler condition that selects physical states is then given by

The collection of all physical states defines a vector subspace \(\HH _{\mathrm {phys}}\subset \HH \) that is called the physical Hilbert space. As a consequence of the Gupta-Bleuler condition (6.42), we find that all matrix elements

\begin{flalign} \expect {\phi _{\mathrm {phys}}}{\partial _\mu A^\mu (x)}{\psi _{\mathrm {phys}}} \,=\, \braket {\phi _{\mathrm {phys}}}{\partial _\mu A^{(+)\mu }(x)\,\psi _{\mathrm {phys}}} + \braket {\partial _\mu A^{(+)\mu }(x)\,\phi _{\mathrm {phys}}}{\psi _{\mathrm {phys}}} \,=\, 0 \end{flalign} of the operator \(\partial _\mu A^\mu \) between physical states vanish. Hence, the constraint gets implemented not as an operator identity \(\partial _\mu A^\mu =0\), which as we have seen above would be inconsistent, but in a weaker fashion at the level of matrix elements between physical states.

Note that the vacuum state \(\ket {0}\) satisfies the Gupta-Bleuler condition \(\partial _\mu A^{(+)\mu }(x)\ket {0}=0\), i.e. it is an element of the physical Hilbert space \(\HH _{\mathrm {phys}}\), because (6.41) consists only of annihilation operators that annihilate the vacuum. So the zeroth order sanity check is passed. It is instructive to characterize also the physical \(1\)-particle states. Since (6.41) only contains the annihilation operators for the time-like polarization \(\lambda =0\) and the longitudinal polarization \(\lambda =d-1\), we find that all transversally polarized states are physical, i.e.

\begin{flalign} \label {eqn:transversal} \partial _\mu A^{(+)\mu }(x)\ket {k,j} \,=\,0\qquad \text {for all }j=1,\dots ,d-2\quad . \end{flalign} This is nice as it matches the physical expectation that the polarization of photons is transversal. On the other hand, the time-like polarized state \(\ket {k,0}\) and the longitudinal one \(\ket {k,d-1}\) do not satisfy the Gupta-Bleuler condition, but one checks that the superposition

\begin{flalign} \ket {k,\mathrm {sp}} \,:=\, \ket {k,0} + \ket {k,d-1} \end{flalign} does satisfy this condition

\begin{flalign} \partial _\mu A^{(+)\mu }(x) \ket {k,\mathrm {sp}} \,=\,0\quad . \end{flalign} The label “sp” refers to spurious, which will be explained below. Summing up, we see that out of the \(d\) polarization degrees of freedom (per fixed momentum \(k\)) only \(d-1\) satisfy the Gupta-Bleuler condition for physical states. These are the \(d-2\) transversal polarizations (6.44) and \(1\) particular superposition (6.45) of the time-like and longitudinal polarizations.

To understand why I called the states (6.45) spurious, let’s look at their properties. These states are not only orthogonal to all transversally polarized states, i.e.

\begin{flalign} \braket {q,j}{k,\mathrm {sp}} \,=\, 0\qquad \text {for all }j=1,\dots ,d-2\quad , \end{flalign} but they are also orthogonal to themselves

\begin{flalign} \nn \braket {q,\mathrm {sp}}{k,\mathrm {sp}}\,&=\, \sqrt {2\,\vert \mathbf {q}\vert }\, \sqrt {2\,\vert \mathbf {k}\vert }~\expect {0}{\big (a_{0}(\mathbf {q}) + a_{d-1}(\mathbf {q}) \big ) \, \big (a^\dagger _{0}(\mathbf {k}) + a^\dagger _{d-1}(\mathbf {k}) \big ) }{0}\\ \,&=\,2\,\vert \mathbf {k}\vert ~ \big (\eta _{00} + \eta _{(d-1)(d-1)}\big )\, (2\pi )^{d-1}\,\delta (\mathbf {q}-\mathbf {k}) \,=\,0\quad . \end{flalign} This is again a consequence of the “wrong sign” commutation relations for time-like polarizations. So the spurious states do not interfere with all other physical states, and not even with themselves. Furthermore, one can show that the spurious states do not contribute to the expectation value of any physical quantity, such as energy, momentum, spin, etc. Let us show this for the energy/Hamiltonian: The Hamiltonian operator obtained from the normal ordered quantization of (6.28) reads as

\begin{flalign} \noor {H} \,=\, \int _{\bbR ^{d-1}} \vert \mathbf {k}\vert ~\sum _{\lambda =0}^{d-1} \eta _{\lambda \lambda } \, a_\lambda ^\dagger (\mathbf {k})\,a_\lambda (\mathbf {k})~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \end{flalign} from which one checks that

\begin{flalign} \expect {q,\mathrm {sp}}{\noor {H}}{k,\mathrm {sp}} \,=\,0\quad . \end{flalign} All of this implies that the spurious states do not contribute to physics, which means that we can take a quotient of the Hilbert space of physical states \(\HH _{\mathrm {phys}}\) by identifying all spurious states with \(0\), i.e.

\begin{flalign} \ket {\psi _{\mathrm {phys}}}\,\sim \, \ket {\psi _{\mathrm {phys}}} + \ket {\mathrm {spurious}}\quad . \end{flalign} This equivalence relation can be understood as the quantum analog of the residual gauge symmetries (6.24), see e.g. Exercise 7.2 in the textbook by Greiner/Reinhardt. At the level of the \(1\)-particles states, we are then left with \(d-2\) degrees of freedom (per fixed momentum \(k\)), which are described by the transversally polarized photon states \(\ket {k,j}\), for \(j=1,\dots , d-2\).