Gauge Invariance (QED)

1. Mar 8, 2015

PhyAmateur

My book says that in this case $$e^+e^- \rightarrow \gamma \gamma$$ gauge invariance requires that $$k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0=k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})$$ Please see attachment. My question is how does this statement hold?

Last edited: Mar 8, 2015
2. Mar 9, 2015

vanhees71

This is a Ward-Takahashi identity. It comes from current conservation, $\partial_{\mu} j^{\mu}=0$, which is a necessary condition for gauge invariance.

3. Mar 9, 2015

PhyAmateur

But the author insisted that these conditions are met although the quantities in the equations k1ν(Aμν+A~μν)=0=k2μ(Aμν+A~μν) each separately are all different from zero. Why would he say that if it is already a consequence of ward identity? @vanhees71

4. Mar 10, 2015

vanhees71

The Ward-Takahashi identities refer to the proper vertex function and is valid order by order in $\hbar$ (number of loops), i.e., it's valid only for the complete set of diagrams of a given order.

5. Mar 10, 2015

PhyAmateur

How does this have to do with my question? I can't relate.

6. Mar 10, 2015

vanhees71

It has all to do with your last question, why it doesn't apply to a single diagram but only to the sum of the two diagrams relevant for the N-point function at the given loop order (which here is tree-level).

7. Mar 10, 2015

Avodyne

If you write out the expressions for $A^{\mu\nu}$ and $\tilde{A}^{\mu\nu}$ that you get from computing these diagrams, you will find that $k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0$ and $k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0$. So, while this can be predicted from gauge invariance, it is also the result of doing the explicit calculation.

8. Mar 10, 2015

PhyAmateur

Are you telling me that Ward Takashi holds for the amplitude and not necessarily for each of the Feynman diagrams whose sum is the amplitude? @vanhees71

9. Mar 11, 2015

vanhees71

Yes, that's it. The same is true for other symmetries, like charge conjugation symmetry, which implies that n-photon vertices with an odd number of photons must vanish. This also holds true only for the sum at a given loop order. Take, e.g., the one-loop triangle diagrams. You need to add both contibutions (which are different only by the orientation of the electron-positron loop making up the triangle).