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3

First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is : $\varphi^i \to U^i{}_j \varphi^j$. By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$. Now, looking at the last expression with $U^\dagger$, one ...

3

Ok, I think there is a mistake here: A general tensor $\varphi^i$ transforms as: $$\varphi^i\rightarrow U^i_{\phantom{1}j}\varphi^j$$ whereas $\varphi_i$ transforms as: $$\varphi_i\rightarrow (U^\boldsymbol{\ast})_i^{\phantom{1}j}\varphi_j$$ Where did you find these equations? The unitary matrix element in the second line should not be a complex ...

3

First, to check the decomposition of a product of representations, you may use, as noticed by user26143, the tool Form Interfact to Lie. Choose Tensor product decomposition, then choose $A_1$ for $SU(2)$, or $A_2$ for $SU(3)$,click sur "Proceed", type your representation, and click on "Start" to have the decomposition. The name of the representations in this ...

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Because $SU(2)$ is not the same as $SU(3)$ ? The closest analog to your $SU(3)$ case would be two doublets: $\mathbf{2} \otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3}$, as you already know :) Afaik, $SU(3)$ has two independent $SU(2)$ subgroups, i.e., it has two "$L^2$" operators. You can still do Clebch-Gordan-style coefficients calculations but it ...

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It is proven in Tong's QFT script http://www.damtp.cam.ac.uk/user/dt281/qft.html section 4.1. in a quite nice fashion.

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I struggled with this one as well and once I found I have written it in LaTeX which I will copy here below. Do note that I am using slightly different conventions than P&S, however it should still work out the same. \begin{aligned} S^{\mu \nu} & = - \frac{i}{4}[\gamma^\mu,\gamma^\nu] \\& = - \frac{i}{4}(\gamma^\mu \gamma^\nu - ...

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Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get ...

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Note that if you lower an index of the Kronecker delta, it becomes the metric: $\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$ And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$. Then, the metric terms cancel and you neglect cuadratic terms. That should be enough to solve it.

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