Lorentz transformation of Weyl fields In the Srednicki's textbook, Chapter 35, the author states (Equation 35.28):
$$
U(\Lambda)^{-1}[\psi^\dagger \bar\sigma^\mu \chi ] U(\Lambda) = \Lambda^\mu_{\,\,\nu} [\psi^\dagger \bar\sigma^\nu \chi ]. \tag{35.28}
$$
When I tried to derive this myself I got $\Lambda^{-1}$ instead of $\Lambda$ on the right hand side.
My attempt goes as follows:
First take Equation 35.9 that is simply the invariance of the sigma element:
$$
\sigma^{\rho}_{a\dot{a}} = \Lambda^\rho_{\, \, \tau} \, L(\Lambda)^{\,\,b}_{a} \, R(\Lambda)^{\,\,\dot{b}}_{\dot{a}} \, \sigma^{\tau}_{b\dot{b}} 
$$
Now using the definition $\bar\sigma^{\rho \dot{a} a} = \varepsilon^{ab} \varepsilon^{\dot{a}\dot{b}}\sigma^\rho_{b\dot{b}}$ and multiplying the quation above by $\Lambda^{-1}$ and $\varepsilon$'s we can get:
$$
(\Lambda^{-1})^{\mu}_{\,\, \rho} \bar\sigma^{\rho \dot{a} a}  = L(\Lambda)^{ab} R(\Lambda)^{\dot{a}\dot{b}} \sigma^\mu_{b \dot{b}} 
$$
This can be used to show that:
$$
U(\Lambda)^{-1}[\psi^\dagger \bar\sigma^\mu \chi ] U(\Lambda) = U(\Lambda)^{-1}\psi^\dagger_\dot{a} U(\Lambda)  \bar\sigma^{\mu \dot{a} a} U(\Lambda)^{-1}\chi_a U(\Lambda) = R(\Lambda)_\dot{a}^{\,\,\dot{d}} \psi^\dagger_\dot{d} \bar\sigma^{\mu \dot{a} c} L(\Lambda)_{c}^{\,\,e}\chi_e = \\
 = \psi^\dagger_\dot{d} [L(\Lambda)^{ce} R(\Lambda)^{\dot{a}\dot{d}} \sigma^{\mu}_{c\dot{a}}] \chi_e = \psi^\dagger_\dot{d} [ (\Lambda^{-1})^{\mu}_{\,\, \rho} \bar\sigma^{\rho \dot{d} e} ] \chi_e = 
(\Lambda^{-1})^{\mu}_{\,\, \rho} \psi^\dagger_\dot{d} \bar\sigma^{\rho \dot{d} e} \chi_e
$$
where I first used the fatct that $\sigma$ is invariant and then lowering/raising indices. What am I doing wrong here?
 A: I don't have Sredicki to hand so I don't fully understand your notation, but the isssue of matrices re inverse matrices in transformation equations  is part of a general argument that is often messed up. Lets take the transformation of sigma  matrices as an example. It is claimed that for   any rotation matrix $R \in {\rm SO}(3)$ there is a $U(R)\in {\rm SU}(2)$ such that
$$
U^{-1} (R)\sigma ^\mu U(R) = {R^\mu}_\nu \sigma^\nu.
$$
and $U(R_1)U(R_2)= U(R_1R_2)$ so $U(R)$ is representation of the rotation group.
To see if this expression has   the inverses in the correct place,
and should not be  for example
$$
U(R)\sigma ^\mu U^{-1}(R) \stackrel{?}{=} {R^\mu}_\nu \sigma^\nu,
$$
or
$$
U^{-1} (R)\sigma ^\mu U(R) \stackrel{?}{=} {[R^{-1}]^\mu}_\nu \sigma^\nu
$$
lets try doing this twice. We have
$$
U^{-1}(R_2)U^{-1} (R_1) \sigma^\mu U(R_1)U(R_2)= U^{-1}(R_2)\{{[R_1]^\mu}_\nu \sigma^\nu \}U(R_2)\\
=  {[R_1]^\mu}_\nu \{U^{-1}(R_2)\sigma^\nu U(R_2)\}\\
= {[R_1]^\mu}_\nu{[R_2]^\nu}_\lambda \sigma^\lambda.
$$
Notice that the $U$'s pass through the ${[R_1]^{\mu}}_\nu$  because these are just a set of numbers and $U$ is a linear map.
Setting $U(R_1)U(R_2) = U(R_1R_2)$ we get
$$
U^{-1}(R_1R_2) \sigma^\mu U(R_1R_2)= {[R_1R_2]^\mu}_\lambda \sigma^\lambda,
$$
which is consistent. If we accepted one of the the equations with the "?" we would have
$U(R_1)U(R_2)= U(R_2R_1)$ and so  not a representation.
I suspect that  interchanging the order of indices on the $R(\Lambda)$ and $L(\Lambda)$ is the problem. Isn't it true that   ${L(\Lambda)^\alpha}_\beta =  {L(\Lambda^{-1})_\beta}^\alpha$ because  $\chi^\alpha \chi_\alpha$ is an invariant? Your first invariance eq has the second indices on $R$ and $L$ contracted  to those on $\sigma$ and in the last line  you have the first indices  contracted to the $\sigma$. I think the $\Lambda^{-1}$  comes from the need to transpose the indices in equating your antipenultimate expression  to the penultimate one.
