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7

The Hamiltonian for the hydrogen atom $$H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of ...

2

It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf

0

A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$. For ...

2

How about just testing the two different cases? I.e. if $\mu\not=0$ then the LHS becomes $$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i$$ while the RHS becomes $$(\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}).$$ For $\mu=0$, the case ...

7

Defining a Lie algebra by the commutation relations $[T^a,T^b]=if^{abc} T^c$, the adjoint representation is defined by $(T_{adj}^a)^{bc}= if^{abc}$. Now it turns, that in the special case of $so(3)=su(2)$, you have $f^{abc} = \epsilon^{abc}$, where $\epsilon^{abc}$ is the totally antisymmetric tensor. So, your representation is the adjoint representation. ...

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 ...

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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 ...

2

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 ...

3

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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Theories with fundamental quarks which experience spontaneous chiral symmetry breaking: $$SU_L(N_f) \times SU_R(N_f) \rightarrow SU_A(N_f)$$ ($N_f$ is the number of flavors) (This is the observed approximate symmetry breaking in nature where the pions are the approximate Goldstone bosons). In contrast, theories with adjoint quarks experience the chiral ...

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I am not sure if I know the correct answer (as I am a student my self), but I will try (and if I am wrong, someone please correct me). The first thing that took me some time to figure out is what they mean by adjoint representation. In Georgi's book he defines the adjoint representation of a generator as: [T_i]_{jk} \equiv -if_{ijk} ...

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