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As of today, nobody knows how to quantise a classical theory with first class constraints. Given a classical Lagrangian, one may attempt to formulate a quantum theory using Dirac's algorithm, where the canonical brackets are replaced by (anti)commutators. This algorithm becomes meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A_{ij}\sim\int \mathrm d\varphi \ f_{ij}(\varphi)\ \mathrm e^{iS[\varphi]} $$

Attempts for formalise the integral above have failed. In practice, one discretises the space of field configurations. Of course, if the integral has a gauge symmetry, the discretised integral diverges by the usual arguments, which means that the continuum limit is meaningless: not even in the discrete case does the integral converge. Therefore, Feynman's approach cannot work either.

Even in the most pragmatical sense, the functional integral is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the functional-integral through Hori's formula, $$ Z[J]\sim \mathrm e^{iS_\mathrm{int}[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.

As of today, nobody knows how to canonically quantise a classical theory with gauge symmetries. The standard approach (Dirac's algorithm) where one replaces the canonical brackets by (anti)commutators is meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this. In practice, in order to formulate a consistent theory in the canonical formalism one must first eliminate the gauge symmetries, either by turning them into (second class) constraints or by more elaborate methods.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A\sim\int a(\varphi)\ \mathrm e^{iS[\varphi]}\ \mathrm d\varphi $$

Attempts for formalise the integral above in as much generality as needed have failed. A possible approach, to discretise the space of field configurations, has two possible outcomes: the lattice formulation either breaks gauge invariance (in which case we have essentially fixed the gauge by means of the regularisation), or it doesn't (in which case the integral diverges, inasmuch as we are integrating over $\mathbb R^n$ a function that does not decay in some directions). In either case, we see that a naïve implementation of Feynman's approach cannot work either.

Even in the most pragmatical sense, the quantum theory is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the theory through its Feynman rules (formally speaking, through Hori's formula), $$ Z[J]\sim \mathrm e^{iS_\mathrm{int}[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

None of these approaches seems to work. The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.

As of today, nobody knows how to quantise a classical theory with first class constraints. Given a classical Lagrangian, one may attempt to formulate a quantum theory using Dirac's algorithm, where the canonical brackets are replaced by (anti)commutators. This algorithm becomes meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A_{ij}\sim\int \mathrm d\varphi \ f_{ij}(\varphi)\ \mathrm e^{iS[\varphi]} $$

Attempts for formalise the integral above have failed. In practice, one discretises the space of field configurations. Of course, if the integral has a gauge symmetry, the discretised integral diverges by the usual arguments, which means that the continuum limit is meaningless: not even in the discrete case does the integral converge. Therefore, Feynman's approach cannot work either.

Even in the most pragmatical sense, the functional integral is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the functional-integral through Hori's formula, $$ Z[J]\sim \mathrm e^{iS_\mathrm{int}[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.

As of today, nobody knows how to quantise a classical theory with first class constraints. Given a classical Lagrangian, one may attempt to formulate a quantum theory using Dirac's algorithm, where the canonical brackets are replaced by (anti)commutators. This algorithm becomes meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A_{ij}\sim\int \mathrm d\varphi \ f_{ij}(\varphi)\ \mathrm e^{iS[\varphi]} $$

Attempts for formalise the integral above have failed. In practice, one discretises the space of field configurations. Of course, if the integral has a gauge symmetry, the discretised integral diverges by the usual arguments, which means that the continuum limit is meaningless: not even in the discrete case does the integral converge. Therefore, Feynman's approach cannot work either.

Even in the most pragmatical sense, the functional integral is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the functional-integral through Hori's formula, $$ Z[J]\sim \mathrm e^{iS_\mathrm{int}[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.

As of today, nobody knows how to quantise a classical theory with first class constraints. Given a classical Lagrangian, one may attempt to formulate a quantum theory using Dirac's algorithm, where the canonical brackets are replaced by (anti)commutators. This algorithm becomes meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A_{ij}\sim\int \mathrm d\varphi \ f_{ij}(\varphi)\ \mathrm e^{iS[\varphi]} $$

Attempts for formalise the integral above have failed. In practice, one discretises the space of field configurations. Of course, if the integral has a gauge symmetry, the discretised integral diverges by the usual arguments, which means that the continuum limit is meaningless: not even in the discrete case does the integral converge. Therefore, Feynman's approach cannot work either.

Even in the most pragmatical sense, the functional integral is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the functional-integral through Hori's formula, $$ Z[J]\sim \mathrm e^{-iS[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.

As of today, nobody knows how to quantise a classical theory with first class constraints. Given a classical Lagrangian, one may attempt to formulate a quantum theory using Dirac's algorithm, where the canonical brackets are replaced by (anti)commutators. This algorithm becomes meaningless if the symplectic form is degenerate. See Quantization of Gauge Systems, by Marc Henneaux & Claudio Teitelboim for a full discussion of this.

A second, more direct approach is to follow Feynman's quantisation, where we postulate that the matrix elements can be calculated from a functional integral, $$ A_{ij}\sim\int \mathrm d\varphi \ f_{ij}(\varphi)\ \mathrm e^{iS[\varphi]} $$

Attempts for formalise the integral above have failed. In practice, one discretises the space of field configurations. Of course, if the integral has a gauge symmetry, the discretised integral diverges by the usual arguments, which means that the continuum limit is meaningless: not even in the discrete case does the integral converge. Therefore, Feynman's approach cannot work either.

Even in the most pragmatical sense, the functional integral is ill-defined in the presence of gauge symmetries: if we convene to sidestep all the formal manipulations and define the functional-integral through Hori's formula, $$ Z[J]\sim \mathrm e^{iS_\mathrm{int}[\delta]}\mathrm e^{-\frac i2 J\cdot \Delta\cdot J} $$ where $\Delta$ is the inverse of the quadratic part of the Lagrangian, the programme fails, because $$ \mathcal L_0\equiv\frac 14 F^2 $$ is not invertible.

The problem can be traced back to the representations of the Poincaré Group. One may show, using the properties of the Poincaré group but nothing about Lagrangians or path integrals, that the propagator of an arbitrary vector field is $$ \Delta(p)=\frac{-1+pp^t/m^2}{p^2-m^2}-\frac{pp^t/m^2}{p^2-\xi m^2} $$ where $m$ is the mass of the spin $j=1$ particles created by the vector field, and $\xi\equiv m^2/m_L^2$, where $m_L$ is the mass of the spin $j=0$ particles created by the vector field.

It's easy to check that the limits $\xi\to\infty$ and $m\to 0$ are both separately well-defined, but you cannot take both limits at the same time. This means that you cannot have, at the same time, a vector field that creates massless spin $j=1$ particles and no longitudinal states. So you must either

• use massive particles, as in the Proca Lagrangian,
• accept that there can be negative norm states, as in $R_\xi$ QED,
• or that the field that creates particles is not a vector, as in QED in the Coulomb gauge.

In the first case the term $\frac 12 m^2 A^2$, and in the second case the term $\frac 12\xi^{-1}(\partial\cdot A)^2$, breaks the gauge invariance of the Lagrangian. In the third case the gauge is fixed by a constraint. In neither of these cases is the Lagrangian gauge invariant.