# Gupta-Bleuler and Lorenz Gauge: I don't understand the principle behind Gupta-Bleuler

I would like to make the link between the Gupta-Bleuler Lagrangian and the Lorenz Gauge for Electromagnetism because everything is not clear to me.

I am looking for a simple explanation without too many references if possible.

So, from what I have understood :

We have $$\mathcal{L}_0=-\frac 14 F_{\mu \nu} F^{\mu \nu}$$

The problem with this Lagrangian is that we have $\Pi_0=0$: the momentum associated to $A_0$ is $0$.

Also, the field equations are:

$$\Box A_{\mu}-\partial_\mu \partial^{\nu} A_\nu=0$$

And there is no propagator associated with this equation (no Green function).

Two problems here. The momentum associated to $A_0$ is $0$ and there is no propagator associated with the equation of motion.

The "trick" we do is to add a term $\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$ to the Lagrangian.

When we do it, the Lagrangian is now $$\mathcal{L}=-\frac 14 F_{\mu \nu} F^{\mu \nu}+\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$$

Then, we can show if we take $\xi=1$, and do a few integrations by parts, that this new Lagrangian is equivalent to $\mathcal{L}=\partial_\mu A_\nu \partial^\mu A^\nu$ and thus, will give the equations of motion :

$$\Box A_{\mu} = 0$$

Then, we can quantize everything as the Klein Gordon field.

$$A_\mu=\int d\widetilde{k}\left(A_\mu(k)^+ e^{ikx} + A_\mu(k)^- e^{-ikx}\right)$$

And we implement the Lorenz Gauge by saying $\partial_\mu A^{\mu, +} |\psi_\text{Phys} \rangle=0$ where $|\psi_\text{Phys} \rangle$ are "physical" states of our Hilbert space. (we define them like this).

My questions:

I don't understand the global spirit of this. For me we modified the dynamics because we modified the Lagrangian. I think everything relies upon the fact that we work with an equivalent Lagrangian as the one we started with but I don't find where this is true? Why can we modify our Lagrangian like this?

I would like the most simple answer even if it is not very rigorous, I just would like to get the average Idea of the thing.

The sketch of the philosophy:

1. You modified the Lagrangian by introducing the $\xi$ term. This shouldn't really bother you because the Lagrangian is not measurable, so there is, in principle, nothing wrong about modifying it. But the Lagrangian does tell you about the underlying theory, so it is understandable that you don't like the gauge-fixing term.

2. What is measurable is the $S$-matrix. To define this matrix you have to specify what states we are allowed to scatter off each other. This is where the Gupta-Bleuler condition comes in. The in- and out-states are restricted to be those that satisfy $\langle a|\partial\cdot A|b\rangle\equiv 0$. You set up the $S$-matrix by evolving these states in time, $$S_{1\cdots n;1'\cdots n'}\equiv \langle a_1\dots a_n;\text{in}|b_{1'}\dots b_{n'};\text{out}\rangle$$

3. Using the Ward-Takahashi identity, you can prove that $S$ is actually independent of $\xi$. This is a non-trivial statement because $S$ is calculated using $\xi$-dependent objects. This way, we conclude that the gauge-fixing term has not really affected the theory, and its introduction has been nothing but a formal tool. It would be nice to be able to set up the $S$ matrix directly, without the need to use non-physical auxiliary objects. This is not really possible, as one can learn from several no-go theorem (that we shall not discuss here).

4. Formally, you can eliminate the unwanted term $\frac{1}{2\xi}(\partial\cdot A)^2$ by taking the so-called unitary limit, $\xi\to\infty$. As $S$ is $\xi$ independent, you may want to argue that you have taken the unitary limit, thus getting rid of the unwanted term, while at the same time taking $\xi=1$, thus simplifying the evaluation of $S$. The 't Hooft-Feynman gauge is more convenient from the practical point of view, and the unitary one is more convincing from the Lagrangian point of view. Gauge invariance is essential if these different POVs are to coexist.

5. This formalism is admittedly ad-hoc. Its ingredients can be motivated but not in a completely convincing manner. Moreover, it is not really possible to extend it to other gauge theories. In this sense, Gupta-Bleuler is best thought of as a formalism that is proved to work a posteriori, but that should not be pursued beyond what an historical analysis is concerned. Modern methods -- being more general and systematic -- are preferred (cf. the BV-BRST formalism; for the equivalence to Gupta-Bleuler in the abelian case, see this PSE post).

Note: the recent paper 1801.06311 argues that there is an equivalent formalism that rests on a different philosophy: one may take the gauge-fixed Lagrangian as fundamental, and forget that it comes from a gauge-invariant theory. In this scenario, the Hamiltonian is in principle not positive-definite. By insisting that $H\ge0$ we essentially get the Gupta-Bleuler condition, so the latter may be understood as the requirement for the theory either to contain only physical states, or for it to contain only positive-energy states. The two points of view are equivalent.

• Ad 5.: It might be worth to mention the names of the more general formalisms: The Dirac-Bergmann recipe and BRST(-BV) quantization. – ACuriousMind Dec 5 '17 at 10:48
• @ACuriousMind In fact, I actually was referring to Bohr-Sommerfeld quantization ;-P (but yeah, good point). – AccidentalFourierTransform Dec 5 '17 at 12:26
• About the BRST formalism... I know this is covered in non-abelian gauge theory. Does this mean if I wait until QCD, the gauge-fixing term will be justified clearly? – doublefelix Jan 26 '20 at 22:10