- #1
LAHLH
- 405
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Hi,
I'm just looking at the stuff on left and right handed spinor fields in Srednicki. Srednciki distinguishes fields in the left rep from those in the right rep by putting a dot over them. Since hermitian conjugation swaps the two SU(2) algebras, the hermitian conj of a left spinor is a right spinor etc. e.g. \left[\psi_a(x)\right]^{\dag}=\psi^{\dag}_{\dot{a}}(x)
Under a Lorentz transformation we have, U(\Lambda)^{-1}\psi^{\dag}_{\dot{a}}(x)U(\Lambda)=R^{\dot{b}}_{\dot{a}}(\Lambda)\psi^{\dag}_{\dot{b}}(\Lambda^{-1}x).
Under infinitesimal transformations we have, R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}. R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}.
From this one finds,
\left[\psi^{\dag}_{\dot{a}}(0),M^{\mu\nu}\right]=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)Now I want to take the Hermitian conjugate of this to get Srednicki's 34.16. I presume this conjugate is w.r.t. the a,b, internal indices not the spacetime indices,
\psi^{\dag}_{\dot{a}}(0)M^{\mu\nu}-M^{\mu\nu}\psi^{\dag}_{\dot{a}}(0)=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)
OK, so taking the h.c.:
(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)\right]^{\dag}
(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\psi_{b}(0)\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\right]^{\dag}
But given that the M's have no a,b type indices I don't really see what this would mean.
Could anyone point me in the right direction on how to derive this 34.16, thanks
I'm just looking at the stuff on left and right handed spinor fields in Srednicki. Srednciki distinguishes fields in the left rep from those in the right rep by putting a dot over them. Since hermitian conjugation swaps the two SU(2) algebras, the hermitian conj of a left spinor is a right spinor etc. e.g. \left[\psi_a(x)\right]^{\dag}=\psi^{\dag}_{\dot{a}}(x)
Under a Lorentz transformation we have, U(\Lambda)^{-1}\psi^{\dag}_{\dot{a}}(x)U(\Lambda)=R^{\dot{b}}_{\dot{a}}(\Lambda)\psi^{\dag}_{\dot{b}}(\Lambda^{-1}x).
Under infinitesimal transformations we have, R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}. R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}.
From this one finds,
\left[\psi^{\dag}_{\dot{a}}(0),M^{\mu\nu}\right]=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)Now I want to take the Hermitian conjugate of this to get Srednicki's 34.16. I presume this conjugate is w.r.t. the a,b, internal indices not the spacetime indices,
\psi^{\dag}_{\dot{a}}(0)M^{\mu\nu}-M^{\mu\nu}\psi^{\dag}_{\dot{a}}(0)=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)
OK, so taking the h.c.:
(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)\right]^{\dag}
(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\psi_{b}(0)\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\right]^{\dag}
But given that the M's have no a,b type indices I don't really see what this would mean.
Could anyone point me in the right direction on how to derive this 34.16, thanks
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