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Your reasoning is correct. The Euler angles are not the components of $\vec{\theta}$. Here is what $\vec{\theta}$ is. Let $\vec{\theta}=(\theta_1,\theta_2,\theta_3)=\theta \hat{n}$. Let's derive the 3x3 matrix (ie: the group elements $R(\vec{\theta})$) for rotating an object by $\theta$ radians about an arbitrary direction specified by the unit vector $\... 1 $$\epsilon = i\sigma_2 = -\epsilon^{-1}, \\ \leadsto \epsilon \sigma_i \epsilon^ {-1}= -\sigma_i^*,$$ the conjugate representation. As a similarity transformation it leads to $$\epsilon \sigma_i^n \epsilon^ {-1}= (-\sigma_i^*)^n, \leadsto \\ f(\epsilon \sigma_i \epsilon^ {-1})= f(-\sigma_i^*),$$ for any function f regular at the origin, including the ... 1 take the Taylor series for this rotation matrix: $$R_x(\theta_1)=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \theta_{{1}} \right) &-\sin \left( \theta_{{1}} \right) \\ 0&\sin \left( \theta_{{1}} \right) &\cos \left( \theta_{{1}} \right) \end {array} \right]$$ you obtain $$R_x(\theta_1)=\left[ \begin {array}{ccc} (1)&0&... 0 The error is in the professor Rychlewski's paper, i.e. the last two permutations are incorrect (wrong). The correct Hooke's tensor symmetrization permutations are as follows: (1234),(1243),(2134),(2143),(3412),(4312),(3421),(4321) 3 The multiplicity of the J_3 eigenvalues is always 1 in every irreducible representation of \mathrm{SU}(2). Irreducible representations of semisimple Lie algebras can be obtained as (quotients of) Verma modules. By the Poincaré–Birkhoff–Witt theorem we know that a basis of this space, for the case of \mathrm{SU}(2) is$$ (J_-)^{k_-}(J_+)^{k_+}(J_3)^{... 2 For any pair of Lie groups$G$and$H$, they will necessarily have associated Lie algebras,$\mathfrak{g}$and$\mathfrak{h}$. Note that it is a general property from Lie group theory that the Lie algebra of some Lie group is isomorphic to the tangent space at the identity of the Lie group, as viewed as a manifold. Now, in general, it may be the case that ... 1 The moonshine phenomenon is a deep subject involving conformal field theory and string theory that is based on the observation of some relationships between the representation theory of finite groups of "big size"(but finite) and modular forms (obtained as partition functions of some CFTs). References: Miranda Cheng - Progress on moonshine TASI ... 1 In the study of equilibrium configurations and vibrational properties of molecules, clusters and solids, finite groups representations play a key role to extract all consequences of point symmetries of the average atomic positions. Also electronic states in molecules, clusters and solids provide a different example of applied representation group theory. ... 0 This is a long comment : Physics is about observing nature and using mathematical models to fit the observed data and also choose models that predict new observations. It is an ongoing interactive process, data, model, data, correction or new models, data ..... The eightfold way is the history of how quarks were found and a physicist should understand how, ... 3 The reflection isn't unitary. The action of unitaries on the density matrix is by conjugation, which is trivial for the scalar matrices, giving a reduction $$U(2) \to U(2)/U(1) = SO(3)$$ meaning there are only rotations, not reflections. 2 Perhaps this will help. Let$t^a$be the generators of the Lie algebra is some representation. For example, this could be the adjoint representation as you're asking about and each$t^a$will be a$3\times 3$matrix. Then any field$\phi$valued in the Lie algebra may be written generically as$\phi=\phi_a t^a$. This is a common trick employed when dealing ... 2 It's actually self-explanatory, but you have made a hash of terminology and illustrations to concoct a mystery. You are talking about the same object, really. Your unitary V s are group elements, exponentials of elements L in the Lie algebra (in which you also take ψ to be: in your setup, it is traceless hermitian). The dimensionality of the matrices you ... 2 From$U^\dagger U = 1$, you get$U^\dagger = U^{-1}$; then$UU^\dagger = UU^{-1}= 1$. You don't have to worry about anything;$\det U = 1$, this matrix is invertible. 4 Your intuition is correct. One way to understand this is as follows. Let$\mathcal{C}$be the space of all gauge connections$A$and let$G$be the gauge group. Then the space$\hat{\mathcal{C}} = \mathcal{C} /G$can be understood as the space of gauge inequivalent connections. To make this construction explicit, we can think of the action of$G$on a single ... 0 Well it is because the way we defined what is called "symmetry". According to Weinberg's QFT Page 50, A symmetry transformation is a change in our point of view that does not change the results of possible experiments.If an observer$O$sees a system in a state represented by a ray$\mathscr{R}$or$\mathscr{R}_{1}$or$\mathscr{R}_{2} \ldots,$... 1 As Emmy said, the trace in your lagrangian does not necessarily vanish. We can take for example $$(T^a)_{jk} = -\varepsilon_{ajk}$$ for our generators which corresponds to the choice in the wikipedia article. Then $$(T^aT^b)_{jk} = \varepsilon_{ajl}\varepsilon_{blk} = \delta_{ak}\delta_{jb} - \delta_{ab}\delta_{jk}$$ so that $$tr(T^aT^b) = - 2\delta_{ab} \... 4 The product of two matrices which have zeroes on the diagonal does not always have zeroes on the diagonal, so \mathrm{tr}\{T^aT^b\} \neq 0 for some a and b 6 What is a particle? Before we worry about how to classify particles, we should try to be clear about what "particle" means. Ideally, we would like the definition to have these features: Particles can be localized. Particles can be counted. The vacuum state has none of them. Even in flat spacetime, this wish-list already has a problem. In ... 4 There is a systematic way to define spinor representations in any number of dimensions. The key idea is the following. The SO(d-1,1) algebra can be written$$i[ \Sigma^{\mu\nu}, \Sigma^{\sigma\rho}]=\eta^{\nu\sigma} \Sigma^{\mu\rho} + \eta^{\mu\rho}\Sigma^{\nu\sigma} - \eta^{\nu\rho} \Sigma^{\mu\sigma} -\eta^{\mu\sigma} \Sigma^{\nu\rho}$$where \eta^{\mu\... 4 As an experimental physicist my vote goes to :"because these groups happen to give a theory that agrees with experiments" The SU(2) represented spin, and then isotopic spin. Then the plethora of new resonances was organised in the eightfold away in SU(3) representations. It is the flavor SU(3) representation. And then the strong force came in in ... 6 Spinors are projective representations of the SO(m,n), i.e. they provide representations up to some phases:$$ U(g)U(h)=e^{i\theta_{g,h}}U(gh) $$These phases turns out to represent the group of closed loops in the SO(m,n), also known as the fundamental group of SO(m,n). Loops in the SO(m,n) manifold from 0 to 2\pi are not continuously connected ... 1 The reason, why \eta^{'} meson is significnatly heavier, that than the mesons in the octet is rather nontrivial and a interesting story, known as \eta-\eta^{'} puzzle. It is resolved by 't Hooft instanton mechanism https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.37.8, whose ​\frac{1}{N} realization is also known as the Witten–Veneziano ... 1 It's possible you're overthinking this -- the step is not too complicated. When you prove that$$A R(g)\left| \psi \right\rangle= \lambda R(g) \left| \psi \right\rangle$$for \left| \psi \right\rangle a \lambda-eigenvalue of A, you're proving that R(g)\left| \psi \right\rangle is a \lambda-eigenvalue of A for every g\in G. This directly ... 3 Let me make some statements which may help organize things a bit. Explicit forms for the Lorentz transformations on the states are not necessary to see what's going on with the little group, and may obfuscate things a bit. Let P^\mu be our 4-momentum operator and suppose that our states are written |p,\sigma\rangle where \sigma is any index needed for ... 0 You have not shown your work, so it is hard to tell how you are applying your formulae, and plugging in to get your answers. The Dynkin indices (a,b,c,d) of an SU(5) irrep Λ give you the dimensionality$$ N(a,b,c,d) =(1+a)(1+b)(1+c)(1+d)\\ \times \left(1+\frac{a+b}{2}\right) \left(1+\frac{b+c}{2}\right) \left(1+\frac{c+d}{2}\right) \left(1+\frac{a+b+c}{3}... 0 (Quadratic) Casimir elements can be represented in terms of elements of the (universal enveloping algebra of the) associated Lie algebra. For SU($n$) this Lie algebra is$\frak{su}(n)$. This Lie algebra can be represented in terms of the generators of the group by $$[T_a,T_b] = f_{ab}^{\;\;c}T_c$$ Then a (quadratic) Casimir$C$is one for which$$[T_a,C]=0$...