# Why is the spinor field anti-commutator not made gauge invariant?

When we introduce minimal coupling for the Dirac spinor field, we introduce terms into the Lagrangian, by the substitution $i\frac{\partial}{\partial x^\mu}\mapsto i\frac{\partial}{\partial x^\mu}+eA_\mu$, so that the Lagrangian is invariant under arbitrary changes of phase at every point of space-time, $$\psi(x)\mapsto e^{-ieG(x)}\psi(x),\qquad A_\mu\mapsto A_\mu+\partial_\mu G(x),$$ where $G(x)$ is a scalar function. We don't make this substitution, however, in the anti-commutator $$\left\{\psi_\xi(x),\overline{\psi_{\xi'}(x')}\right\}=\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}i\Delta(x-x'),$$ which, it seems, is therefore not invariant under the transformation $\psi(x)\mapsto e^{-ieG(x)}\psi(x)$, $\psi(x')\mapsto e^{-ieG(x')}\psi(x')$. Presumably the path-integral formalism doesn't care about this, but does this have consequences in subsequent derivations in the canonical approach?

I expect that in fussing about this I'm missing something very obvious. Am I? What is it?

EDIT: So it appears from the Answers, for which Thanks, that the anti-commutator given is valid for both space-like and time-like separation for the free Dirac field, and it's valid for space-like separation in QED, but it's nae valid for time-like separation in QED. Put somewhat loosely, this is OK in the canonical formalism because this anti-commutator is valid for the phase space. I'd rather like to know what a valid expression for the anti-commutator is at time-like separation in QED, but since that would, in a sense, be a solution of the theory I guess I'll have to whistle for it. I'll be grateful for Comments if this is an obviously obtuse reading of the Answers, though I ask that you consider that I would like a manifestly covariant version (so to speak, perhaps obscurely) of the canonical formalism before leaping in.

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$e^{-ieG(x)}\cdot e^{ieG(x)}=1$ ? – Raskolnikov May 2 '11 at 15:15
x' in one of the factors, I suppose. – Peter Morgan May 2 '11 at 15:24
Is $\Delta$ the Feynman propagator? – Raskolnikov May 2 '11 at 15:25
No, I would make a point of writing the Feynman propagator as $\Delta_F(x)$. The Fourier transform of this propagator is $\tilde\Delta(k)=i\,2\pi\delta(k^2-m^2)\epsilon(k_0)$. – Peter Morgan May 2 '11 at 15:31
Sorry, that's $i\tilde\Delta(k)=2\pi\delta(k^2-m^2)\epsilon(k_0)$. – Peter Morgan May 2 '11 at 15:41

Because $\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}i\Delta(x-x')$ is a symbolic expression for a given analytical right-hand side. It is written so for convenience (not yet calculated) but it is a specific expression like $\delta(x-x')$ or $\delta(x-x')'$. It should not acquire any "gauge extension" by definition. This expression does not contain a "particle momentum".

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But the left hand side would be multiplied by $e^{ie(G(x')-G(x)}$ under a gauge transformation? Or not? – Peter Morgan May 2 '11 at 15:25
Yes but for $x'\neq x$ the RDS is zero so you must only consider the case $x'= x$. – Vladimir Kalitvianski May 2 '11 at 15:29
$\Delta(x-x')$ is zero for $x\not=x'$? It has a Bessel function form for $(x-x')^2\not=0$. Language getting in the way, RDS=RHS, right hand side? – Peter Morgan May 2 '11 at 15:37
Yes, it should have been RHS, my typo, sorry. – Vladimir Kalitvianski May 2 '11 at 15:40
But you stick by the $\Delta(x)=0$ when $x\not=0$?? The explicit form, I think, is $i\Delta(x)=\frac{m\theta(x^2)}{8\pi\sqrt{x^2}}\left[Y_1(m\sqrt{x^2})+i\epsilon(‌​x_0)J_1(m\sqrt{x^2})\right]+\frac{m\theta(-x^2)}{4\pi^2\sqrt{-x^2}}K_1(m\sqrt{-x^‌​2})-\frac{i}{4\pi}\epsilon(x_0)\delta(x^2)$. – Peter Morgan May 2 '11 at 16:10

Dear Peter, there is no freedom to "make" the (anti)commutator of your fields whatever you want, so there is no freedom to "make" it gauge-invariant, either.

The canonical momentum is obtained as the derivative of the Lagrangian density with respect to the time-derivative of the canonical coordinate, and the (anti)commutator of these two is just the Kronecker delta (multiplied by the delta-function). The remaining (anti)commutators are defined to vanish.

It's clear that the cubic coupling of the Dirac field to the electromagnetic field, $\bar\psi A_\mu \gamma^\mu \psi$, doesn't contain any time derivatives, so it also doesn't modify the canonical momenta.

It follows that the anticommutators of $\psi$ and $\bar \psi$ - which are just a Kronecker delta times a delta function (arising from the $\bar\psi p_\mu \gamma^\mu \psi$ kinetic term in the Lagrangian) - are completely unaffected by this coupling to the electromagnetic field - which is needed for gauge invariance of the action. And indeed, $$\{ \psi_a(x), \psi^\dagger_b(x') \} = \delta_{ab} \delta^{(d-1)} (x-x').$$ The equation above transforms trivially under the gauge transformation. The left-hand side gets multiplied by $$e^{-ieG(x)+ieG(x')}$$ and you may add the same factor to the right hand side as well because the phase above is equal to one for $x=x'$, and it's the only point where it matters because of the $\delta(x-x')$ factor.

For some reasons, you wrote the anticommutator as the Dirac differential operator acting on the Dirac propagator (although I am confused about your method to deal with the Dirac spinor indices - it's probably wrong), but the propagator is nothing else than the inverse Dirac differential operator (acting on a delta function), so the Dirac operators cancel and you're left with the delta function.

The previous sentence is meant to be an explicit elegant proof, but even if you decided to write the position representation of the propagator in terms of Bessel's functions, you should be able to see that the right hand side of your complicated form of the anticommutator will be just a delta function. Note that you only want to evaluate the commutator at equal times, $t=t'$, and as you have already established - as far as your comments under Vladimir's answer go - the function vanishes for spacelike separations.

You only need to be careful about the behavior of the right hand side as a distribution for very small spatial separations of $x$ and $x'$ and for $t=t'$. That's why the non-vanishing of the function for timelike separations would be irrelevant for the anticommutator. However, despite the irrelevance, it's still true that when you act on the Dirac propagator with the Dirac differential operator, you get a delta-function that vanishes everywhere in the spacetime except for $x=x'$ - localized in all spacetime dimensions. That's the function whose spacetime Fourier transform is proportional to $1$.

But don't get me wrong: the anticommutator does not vanish for time-like separations.

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Thanks Luboš, Useful after my education at Vladimir's hands. The way I wrote the anticommutator is eq. 3-170 of Itzykson&Zuber, but I've been getting increasingly unhappy at their failure to distinguish conjugation and raising of spinor components, so I'll take this as requiring me to move on. I think I've been missing important issues for a long time, which is reflected in this misapprehension. I'm very struck that this expression (if it were done right) is not valid for time-like separated $x$ and $x'$. The canonical formalism is not manifestly covariant, which ought not to be surprising. – Peter Morgan May 2 '11 at 18:14
š, do you always use the notation you use in your Answer for spinor indices? I suspect you never make mistakes with the various factors of -1, conjugation, etc., but I prefer to use notations that help me to keep control of them, wherever possible, 'cause otherwise I lose some of 'em. – Peter Morgan May 2 '11 at 18:37
Dear Peter, oh, I take my complaint back. Your $\Delta$ is just the normal Klein-Gordon propagator which is proportional to the unit matrix in the Dirac spinor indices, so nothing is missing in your formula. – Luboš Motl May 2 '11 at 19:23
Otherwise there are lots of $i$'s and minus signs and $\gamma_0$ one may get wrong. For example, in my anticommutator, $\bar\psi$ should have been $\psi^\dagger$ actually - because this is how the special 0-direction of spacetime, corresponding to the slice of spacetime, is acknowledged. I will fix it. Then it's clear that the right hand side must be real and positive - because the left-hand side is positively definite, you know, $\psi \psi^\dagger+\psi ^\dagger\psi$, for the right components. – Luboš Motl May 2 '11 at 19:25