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What are the Feynman rules for the spinor-helicity formalism?

Physics Asked on May 11, 2021

If we do not work with helicity amplitudes, there are Feynman rules for the external legs of a Feynman diagram, i.e. $u_s(k),overline{v}_s(k),epsilon_r(k)$ for an incoming fermion, antifermion and gauge boson, respectively, and $overline{u}_s(k),v_s(k),epsilon^ast_r(k)$ for their outgoing pendants.

Are these correct rules for external states in the spinor-helicity formalism?

Particle & helicity rule for incoming rule for outgoing
L-fermion $u_L(k)=k]$ $overline{u}_L(k)=langle k$
R-fermion $u_R(k)=krangle$ $overline{u}_R(k)=[k$
L-antifermion $overline{v}_L(k)=langle k$ $v_L(k)=k]$
R-antifermion $overline{v}_R(k)=[k$ $v_R(k)=krangle$
(+)-gauge boson $epsilon^mu_+(k,r)=-frac{1}{sqrt{2}}frac{[rgamma^mu krangle}{[rk]}$ $epsilon^{muast}_+(k,r)=frac{1}{sqrt{2}}frac{[kgamma^mu rrangle}{langle rkrangle}$
(-)-gauge boson $epsilon^mu_-(k,r)=frac{1}{sqrt{2}}frac{[kgamma^mu rrangle}{langle rkrangle}$ $epsilon^{muast}_-(k,r)=-frac{1}{sqrt{2}}frac{[rgamma^mu krangle}{[rk]}$

Here, $r$ shall be an arbitrary reference momentum with $r^2=0,rkneq0$, and of course all fermions shall be massless.

One Answer

The table I have given turns out to be wrong for the antifermions. As I have explained in this answer, a left-chiral antifermion $overline{v}_L=(v_R)^daggergamma_0$ is described by the Dirac-conjugate of a right-chiral Dirac-spinor $v_R$. That means, $v_R(k)=k]leftrightarrowoverline{v}_L(k)=[k$ and conversely $v_L(k)=krangleleftrightarrowoverline{v}_R(k)=langle k$.

All this can now be put into a neat table.

Particle & helicity rule for incoming rule for outgoing
L-fermion $u_L(k)=k]$ $overline{u}_L(k)=langle k$
R-fermion $u_R(k)=krangle$ $overline{u}_R(k)=[k$
L-antifermion $overline{v}_L(k)=[k$ $v_L(k)=krangle$
R-antifermion $overline{v}_R(k)=langle k$ $v_R(k)=k]$
(+)-gauge boson $epsilon^mu_+(k,r)=-frac{1}{sqrt{2}}frac{[rgamma^mu krangle}{[rk]}$ $epsilon^{muast}_+(k,r)=frac{1}{sqrt{2}}frac{[kgamma^mu rrangle}{langle rkrangle}$
(-)-gauge boson $epsilon^mu_-(k,r)=frac{1}{sqrt{2}}frac{[kgamma^mu rrangle}{langle rkrangle}$ $epsilon^{muast}_-(k,r)=-frac{1}{sqrt{2}}frac{[rgamma^mu krangle}{[rk]}$

Correct answer by Thomas Wening on May 11, 2021

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