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I was reading about how to choose divergent diagrams in QED by using the concept of Superficial degree of divergence. We have an empirical relation $$ D= 4-E_b -\frac{3}{2}E_f $$ where $E_b$ is number of external photons and $E_f$ is number of external fermions. There are some values for $(E_b,E_f) $ which mathematically satisfy this relation but are not taken into account because of some physical conditions. There is one such concept I didn't understand.

Conservation of Charge demands that $E_f$ can not be odd number. As I understand,we have to apply conservation of charge at each vertex; but how does it control number of external lines?

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  • $\begingroup$ if you have two electrons in, you cannot have only one out. $\endgroup$ – AccidentalFourierTransform Apr 12 '16 at 13:37
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I assume you are familiar with basic QFT concepts. The notation is the usual one in Peskin and Schroeder.

The resulting amplitude of any process must be a scalar. A Lorentz invariant, indeed. Following the Feynman rules, briefly (meaning I am not being rigorous with formal notation, I don't think it's necessary to understand your question):

  • Incoming fermion $u(p)$

  • Outcoming fermion $\bar{u}(p)$

  • Incoming antifermion $\bar{v}(p)$

  • Outcoming antifermion $v(p)$

To form a scalar quantity you have to combine a "bar"-ed one with one without the bar.

Consequences:

Fermion number must be even, as they have to be contracted by pairs.

More consequences:

You cannot violate charge conservation, for example, if you have an incoming electron ($u(p)$) you can:

  • Contract it with an incoming positron ($\bar{v}(p)$) to form a scalar ($\bar{v}(p)u(p)$)

  • Contract it with an outcoming electron ($\bar{u}(p)$) to form a scalar ($\bar{u}(p)u(p)$)

You cannot contract it with an outcoming positron, as $v(p)u(p)$ is not a scalar.

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