I know that a cut boson propagator is replaced with the mass shell delta function. But what happens when you cut a fermion propagator? Do you just replace the denominator with a mass shell delta function, and do nothing to the numerator? Why? If so, it's a bit peculiar because the numerator actually reduce the degree of singularity and might alter the behavior of the poles in the complex plane.
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It's correct that you only replace the denominators $1/(p^2-m^2+i\epsilon)$ by $-2\pi i \delta(p^2-m^2)$ in the propagators to compute the discontinuities. The fermionic propagators must first be rewritten so that they contain the denominator I just mentioned. You're right that the numerator isn't affected in the Cutkosky rules. In some formal sense, you could also write the discontinuity of the fermionic propagators as $2\pi i \delta(p_\mu \gamma^\mu -m)$ but whenever there is a confusion, it just means the same thing indicated by the procedure in the previous paragraph. Indeed, the fermionic propagators only have a "simple pole" near the mass shell: that's related to the fact that the propagators are the "inverse differential operators" and the equations for fermions are first-order rather than second-order as they are for bosons. This different "degree of divergence" near the mass shell is reflected by the Cutkosky rules. However, just to be sure and avoid a potential misunderstanding that may be implicitly included in your question, the numerators don't really "annihilate" the delta-function. The function $x\,\delta(x)$ vanishes because $x=0$ at the only point where the delta-function has a support: $f(x)\delta(x) = f(0)\delta(x)$. However, $x\,\delta(x)$ is not what we see in the fermionic cutting rules. Instead of $x$, the numerator is is $p_\mu \gamma^\mu - m$ or something like that. This numerator "formally vanishes" when it acts on a solution of the Dirac equation. However, in the cutting rules, you compute the whole matrix and not just its action on a specific spinor. And the matrix doesn't vanish on the mass shell: only two of its four eigenvalues are set to zero in the combination. So you still get a nonzero result. |
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