# 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).

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