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consider the photon in QED and the corresponding EOM of its Green's functional in k-space: $$(k^\mu k^\nu-k^2g^{\mu\nu})\Delta_{\nu\rho}(k)=i\delta^\mu_\rho.$$

Now, I understand that $U^{\mu\nu}(k):=k^\mu k^\nu-k^2g^{\mu\nu}$ is not injective, since $U^{\mu\nu}k_\nu=0$ and thus $\det U=0$. That is why $U$ is not invertible.

In the literature I read that gauge fixing solves this problem. Using the $R_\xi$ gauges, one then obtains a new $U^{\prime\mu\nu}=(1-\xi^{-1})k^\mu k^\nu-k^2g^{\mu\nu}$. It follows that $U^{\prime\mu\nu}k_\nu=-\xi^{-1}k^2k^\mu$ and thus $k_\nu$ does not have the eigenvalue zero anymore.

  1. How can we be sure that there aren't any other vanishing eigenvalues? Why don't we diagonalise the operator?

Also, I remember that in scalar field theory we solved the invertibility problem by analytical continuation and then Feynman-shifting the poles away from the real axis: $p^2-m^2 \mapsto p^2-m^2+i\epsilon.$

We can do the same here, can't we? If we write $U^{\prime\mu\nu}=k^\mu k^\nu-k^2g^{\mu\nu}+i\epsilon$, then we arrive at $U^{\prime\mu\nu}k_\nu=i\epsilon\neq0$ for $\epsilon>0$.

  1. Why do we need gauge fixing to make $U$ invertible? Why isn't it sufficient to analytically continue the operator and then Feynman-shift its poles, as we do in scalar field theory?
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  • $\begingroup$ You probably wanted to say "How can we be sure that there aren't any other eigenvectors to eigenvector 0?", and the answer is that we are not. The point is that there are $\xi$ values for which there are none. So you do your computation knowing that there is a $\xi$ value for which you can invert the operator, and find "miraculously" in the end that physical observables are independent of $\xi$. I do not understand the part "Why don't we diagonalise the operator?", why should we? What would that buy us? $\endgroup$
    – user178876
    Commented Jun 18, 2019 at 2:06
  • $\begingroup$ But how do we know that it's invertible for some $\xi$ without calculating the eigenvalues? In diagonalising the operator, we can find out, if there is a $\xi$, such that none of the eigenvalues vanish. Then the determinant does not vanish either and hence the operator is invertible. I mean, that is the point. To make it invertible, isn't it? Also, why isn't it enough to Feynman-shifti the poles to deal with the singularities in the reals? $\endgroup$
    – Thomas Wening
    Commented Jun 18, 2019 at 6:11
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    $\begingroup$ You can just work out the eigenvalues of your matrix $U$ and they are $-k^2$, $-k^2$, $-k^2$ and $-\frac{k^2}{\xi}$. Thus, for any finite value of $\xi$, the matrix is invertible. $\endgroup$
    – Prahar
    Commented Mar 27, 2020 at 5:13

3 Answers 3

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  1. That is indeed a good exercise that any serious student of QFT should do at least once. In momentum space, the trick is to decompose vectors in components parallel and perpendicular to $k^{\mu}$.

  2. Unlike for scalar theory, in the case of gauge theories, the $i\epsilon$-prescription alone is not enough to render the path integral well-defined without gauge-fixing. (Don't forget that at the end of the day we should take the limit $\epsilon\to 0^+$.) See also e.g. this & this Phys.SE posts.

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    $\begingroup$ Unfortunately, neither one of the posts answers my questions. I am aware that gauge invariance means that we have not yet divided out the equivalence classes but how exactly does this manifest itself in the non-invertibility of the differential operator in the EOM? Many books state that the invertibility is a direct consequence of the gauge invariance. But looking at the EOM, how so explicitly? $\endgroup$
    – Thomas Wening
    Commented Jun 18, 2019 at 10:27
  • $\begingroup$ @ThomasWening: The implication (gauge invariance) $\Rightarrow$ (non-invertibility) heuristically goes like this: By gauge invariance, $A_\mu=\partial_\mu \phi$ is equivalent to $A_\mu=0$, and in momentum space, that's just $A_\mu\sim k_\mu$, hence $U\cdot k=0$. In other words, pure gauge modes don't propagate. If you gauge-fix the Lagrangean in a way that there is no residual gauge freedom, the argument does not apply anymore and you'd expect an invertible operator. (Recall that the linearised Lagrangean is essentially $A\cdot U \cdot A$.) $\endgroup$
    – Toffomat
    Commented Jun 18, 2019 at 11:03
  • $\begingroup$ @Toffomat You mean $A_\mu\mapsto A_\mu+\partial_\mu\Phi$, right? $\endgroup$
    – Thomas Wening
    Commented Jun 19, 2019 at 19:26
  • $\begingroup$ @ThomasWening Right. Hence, a field that ist globally a gradient, $A_\mu=\partial_\mu\phi$, should not give different results than a zero field; in particular, $F_{\mu\nu}F^{\mu\nu}"="A_\mu U^{\mu \nu} A_\nu=0$, where the equality is up to total derivatives and possibly a minus sign. $\endgroup$
    – Toffomat
    Commented Jun 20, 2019 at 7:43
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In covariant notation Maxwell's equations can be written as $\partial_\mu ( \partial^\mu A^\nu - \partial^\nu A^\mu) =-j^\nu / \epsilon_0 $. This equation does not fix a one on one relation between field an source so it cannot be inverted. The cause is gauge invariance. To find the Green's function you need to invert the equation of motion. This is not possible for Maxwell's equations because they are gauge invariant, so physicists "fix the gauge" and then argue that the final results are independent of the particular choice they made.
For a valid non gauge invariant theory of electromagnetism see my paper at https://arxiv.org/abs/physics/0106078.

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  • $\begingroup$ I do understand that. You have several fields, that map to the same action. But what does that have to do with calculating the Green's functional? After all, the field $A_\mu$ appears nowhere in the defining equation $U^{\mu\nu}\Delta_{\nu\rho}=\delta^\mu_\rho$. This is, as far as I understand, a matrix equation, that should be solvable by means of linear algebra. $\endgroup$
    – Thomas Wening
    Commented Jun 19, 2019 at 19:24
  • $\begingroup$ So, I have done some further calculations. If one starts with the EOM with source and gauge breaking term, one ends up with $(g^{\mu\nu}\square-(1-\epsilon^{-1})\partial^{\mu\nu})A_\mu=j^\nu$. We see that for $\epsilon\rightarrow\infty$ the two fields $A_\mu,A^\prime_\mu:=A_\mu+\partial_\mu\alpha$ have the same image under $g^{\mu\nu}\square-\partial^{\mu\nu}$. This is the usual result of gauge invariance, okay. So I guess the actual question amounts to: What equation defines the Green's functional? The EOM $U^{\mu\nu}A_\mu=j^\nu$ or $U^{\mu\nu}\Delta_{\nu\rho}(x)=\delta^\mu_\rho\delta(x)$? $\endgroup$
    – Thomas Wening
    Commented Jun 19, 2019 at 21:14
  • $\begingroup$ To find the Green's function you need to invert the equation of motion. This is not possible for Maxwell's equations because they are gauge invariant, so physicists "fix the gauge" and then argue that the final results are independent of the particular choice they made. $\endgroup$
    – my2cts
    Commented Jun 19, 2019 at 22:01
  • $\begingroup$ Great, this makes it crystal clear. Then my error was thinking that $U^{\mu\nu}:=g^{\mu\nu}\square-\partial^{\mu\nu}$ can be inverted from the equation $U^{\mu\nu}\Delta_{\nu\rho}(x-y) =\delta^\mu_\rho\delta(x-y)$. If you put this explicitly into an answer, I'll mark it as the answer to this thread. $\endgroup$
    – Thomas Wening
    Commented Jun 20, 2019 at 8:06
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The gauge invariance implies that the $k$-combination acts like a projection operation. In fact, there are two projection operations $$ P_1 = \frac{k^{\mu}k^{\nu}}{k^2} $$ and $$ P_2 = \frac{k^{\mu}k^{\nu}}{k^2}-g^{\mu\nu} . $$ A projection operator cannot be inverted because of its null space. However, if one can restrict operation to the region onto which the projection process is projected then it acts just like an identity operator which is trivially inverted. So the propagator is then given by the combination of these two projection operators combined by an arbitrary constant, which becomes tantamount to gauge fixing.

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