The Klein Gordon propagator is given (I use Peskin and Schroeder's conventions, if it matters...),

\begin{equation} \frac{ i }{ p ^2 - m ^2 + i \epsilon } \end{equation} The photon propagator (using Feynman gauge) is \begin{equation} \frac{ - i \eta^{\mu\nu}}{ k ^2 + i \epsilon } \end{equation} The time-like component of the photon field propagates with a different sign then the scalar field and the spatial components propagate with the same sign.

Is there a physical significance to the difference in sign between the two or it is just a consequence of our conventions?


1 Answer 1


Yes, there is a physical significance. The longitudinal mode $A^0$ is pure gauge, it does not propagate (in other words, the equation of motion for $A^0$ is a constraint [Gauss Law], not an equation of motion and it's canonical momenta is identically 0 , meaning we cannot impose canonical commutation relations on it). Some of the spatial modes do propagate, so they should have the same sign propagator as the scalars. This is really the easiest way to remember the sign of the photon propagator. The wrong sign in the $A^0$ propagator is at the heart of many problems in QFT. It marks the conflict between the positive definite norm of quantum mechanics and the indefinite norm on Minkowski space that we are forced to deal with because we want a manifestly local Lorentz invariant formulation of the theory.

  • $\begingroup$ Fascinating! Thanks for the answer. How come $A^0$ doesn't propagate? Is this always true or does it depend the gauge that you work in? $\endgroup$
    – JeffDror
    Feb 24, 2014 at 1:30
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    $\begingroup$ could you elaborate how exactly the sign difference implies the equation for $A^0$ is a constraint? $\endgroup$
    – Jia Yiyang
    Feb 24, 2014 at 1:40
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    $\begingroup$ It is not always true that A0 does not propagate. This is a gauge dependent statement. It is true that not all components of A are independent. This is easy to see because a 4-vector has four components but we know that a massless particle only has two degrees of freedom (the two transverse polarizations). However, which components we treat as independent depends on the choice of gauge. $\endgroup$
    – Heterotic
    Feb 24, 2014 at 15:48
  • $\begingroup$ Consider the Lagrangian of the theory. There are no time derivatives of A^0. They would need to be of the form F_00. This means that a propagating A_0 field does not change the action, its canonical momenta is identically 0. I agree that in different gauges different linear combinations of the gauge field components will not propagate, but the propagator in the question is not the propagator in these gauges either. Moreover, question of A_0 is not limited to gauge theory, this longitudinal mode still needs to decouple in the massive theory. Gauge theory requires you to remove an additional dof $\endgroup$
    – Dan
    Feb 24, 2014 at 18:57
  • $\begingroup$ @Dan: Could you kindly answer my question in the comment? I'm aware that in a vector theory, one can derive(from lagrangian) that equation for $A^0$ contains no time derivative, one can also derive there is a sign change in the propagator. I've always treated these two facts as two separate consequences of the vector theory, so I'd like to know how can you derive one from the other. $\endgroup$
    – Jia Yiyang
    Feb 26, 2014 at 11:56

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