Why Photon Momentum Drops Out of QED Propagator

In summary: In QED, the vacuum expectation value is the product of the two time-ordered products: the Schr\"dinger equation and the Born equation. The Ward-Takahashi identity says that the two equations are equivalent, and so the vacuum expectation value is just the Born equation with all the terms replaced by their canonical values. The canonical values of k_\mu are zero, so the Ward-Takahashi identity says that the Fourier-transform of k_\mu is also zero.
  • #1
noether21
5
0
Since the 4-vector A couples to the conserved current j^\mu = \psi-bar gamma^\mu \psi in QED, k_\mu Fourier-transform(j^\mu) = 0. The Fourier-transform of j is a mess. Is there an easy way to see why terms containing photon momentum k_\mu drop out of the photon propagator in practical QED calculations?
 
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  • #2
You can find discussion of this point in section 8.5 in Weinberg's "The quantum theory of fields" vol. 1. The idea is the following:

The true interaction Hamiltonian in QED is

[tex] V(t) =-\int d^3x j(x,t)A(x,t) -\frac{1}{2}\int d^3x d^3y \frac{j^0(x,t)j^0(y,t)}{4 \pi|x-y|} [/tex]

and the true photon propagator is

[tex] D_{\mu \nu}(p) = \frac{-i}{(2 \pi)^4} \frac{P_{\mu \nu}(q)}{q^2} [/tex]

However, results for the S-matrix would not change if you simplify both the interaction and the propagator. In the interaction operator you drop the (current)x(current) term

[tex] V'(t) =-\int d^3x j(x,t)A(x,t) [/tex]

and in the photon propagator you replace the momentum-dependent function [tex] P_{\mu \nu}(q)[/tex] simply by [tex] g_{\mu \nu}[/tex]

[tex] D'_{\mu \nu}(p) = \frac{-i}{(2 \pi)^4} \frac{g_{\mu \nu}}{q^2} [/tex]

Weinberg does not prove that this trick works at all orders, but apparently it works.
 
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  • #3
Thanks for the pointer to Weinberg's book.

Weinberg's discussion also seems to merely make the claim without offering any proof;
he refers to Feynman's 1949 paper (section 8). Feynman has an argument to show
that the divergence of the amplitude of a process that emits a photon (real/virtual) vanishes,
implying that the dot product of $k$ and the Fourier transform of the amplitude vanishes. His
argument appears plausible (even if a bit sketchy) and doesn't appear to invoke current conservation to show that the divergence vanishes. If Feynman's argument implies that the $k^\mu k^\nu$ term can be ignored in the photon propagator, it's not obvious as to why/how. The $k^\mu k^\nu$ term does not appear in the propagator in Feynman gauge, but that is not the same as disregarding the term in other gauges. Perhaps there is some simple way to see that the Fourier-transform of the current occurs dotted to the $k_\mu$ term in all Feynman diagrams containing QED vertices?
 
  • #4
It's also discussed in Zee chapter II.7
As usual, you need external legs to be on shell.
 
  • #5
I think what you have is a special case of the Ward-Takahashi identity, which is most easily proven in the path integral formalism. If I recall, the identity says that classical equations of motion hold true inside the vacuum expectation value of time-ordered products.
 

1. What is the QED propagator?

The QED propagator is a mathematical formula used in quantum electrodynamics (QED) to describe the behavior of particles, specifically photons, in an interaction. It represents the probability amplitude for a photon to travel between two points in space and time.

2. Why does photon momentum drop out of the QED propagator?

The QED propagator is a complex-valued function that involves the momentum of the particles involved in the interaction. In the case of photons, the momentum is related to their energy and frequency through the equation E=hf. However, in the QED propagator, the momentum of the photon cancels out due to the mathematical structure of the formula. This is known as gauge invariance and is a fundamental principle in quantum field theory.

3. Does this mean that photons have no momentum?

No, photons do have momentum as demonstrated by their ability to transfer energy and momentum when interacting with particles. However, in the context of the QED propagator, the momentum is not explicitly included in the formula due to gauge invariance.

4. How does this affect our understanding of the behavior of photons?

The fact that photon momentum drops out of the QED propagator does not have a significant impact on our understanding of their behavior. It is simply a mathematical consequence of the theory and does not change the experimental results or predictions of QED.

5. Are there any other theories or models that can explain the behavior of photons without dropping out their momentum?

There are alternative theories to QED that attempt to explain the behavior of particles, including photons, without relying on gauge invariance. However, QED has been extensively tested and is considered the most accurate and reliable theory for describing the behavior of photons and other particles at the quantum level.

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