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.
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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