It is possible to write down a Feynman diagram as below for production of a Higgs boson.

I am slightly confused because the fermion-fermion-$Z$ vertex has a factor of $c_V - c_A \gamma^5$, but the $HZZ$ vertex does not have this factor. When the matrix element is squared, because there is only one factor of $c_V - c_A \gamma^5$ there will be an overall imaginary contribution to the sum, but the matrix element squared must be a real number. What happens to this contribution?

Another example is the tree-level diagram in the Weinberg model where $e^- e^+$ annihilates to $W^+ W^-$ via an intermediate photon. The amplitude contains a factor of the form $a + b \gamma^5$, but would this not produce an imaginary contribution when the amplitude is squared?

Edit: The answer might be that when you take the trace a $(1 - \gamma^5 / 2)$ term in the trace gives you a term like $i \epsilon^{\alpha \mu \beta \nu} p'_{\alpha} p_{\beta}$. This is non-zero when dotted with another term of the same form when squaring the matrix element, but vanishes when summed over on its own.

Edit: Did some reading of papers from the 1980s where they deal with stuff like this and it seems like the part with the $\gamma_{\mu} \gamma_5$ contribution to a vertex is UV divergent, so in the end the contributions cancel out by necessity for the Weinberg model to work. If not, the $U(1)$ coupling constants for left-handed and right-handed electrons would have to be renormalized separately, which isn't possible because interaction terms between the Higgs boson and electrons would not be gauge invariant.

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1 Answer 1


the matrix element squared must be a real number

No, it doesn't, but the modulus of the matrix element squared, $|\mathcal{M}|^2 = \mathcal{M}^\dagger\mathcal{M}$, does :)

  • $\begingroup$ Yes thanks, I think I got confused for some reason. $\endgroup$
    – Tom
    Feb 23 at 15:00

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