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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, ...


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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 ...


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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 ...


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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 ...


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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\...


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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 ...


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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 ...


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So I would point out that, of course, you can always define whatever you want. But there is a good reason why linear representations are so important, so that's what I'll talk about. Suppose we are dealing with a Lagrangian theory and we have some symmetry thereof. Then we know by Noether's theorem that there will be a conserved current, and hence a ...


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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 ...


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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}...


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(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$...


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I think it's easiest to understand this from the bottom up. Suppose you have a vector in some Hilbert space which describes a state where there is just one particle in a pure momentum state. This particle is described by a momentum vector and has some internal spin degrees of freedom. If you act on this vector with elements of the Poincare group, you'll ...


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You don't build the spin $1$ matrices from the spin $\frac 12$ matrices. You rather repeat the whole procedure, which you learned with $2\times 2$ matrices for spin $\frac 12$. But now you do it with $3\times 3$ matrices for spin $1$. As you already know from spin $\frac 12$ the 3 matrices are not unique. So there is much freedom in choosing a possible set ...


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You don't really derive higher-spin matrices from lower-spin ones. Rather, they are derived from the algebra of spin operators and from how they act oon the chosen spin basis. The basis chosen is usually the basis of eigenvectors of operator $S_z$, and are often denoted $|j,m\rangle$, $2j\in\mathbb{N}$, $m\in\{-j,-j+1\dots,j\}$. Operator $S_z$ act on them as ...


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Hints : REFERENCE : Total spin of two spin- 1/2 particles. $=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$ We could derive the spin-1 matrices adding two spin-1/2. But this addition is not so easy as you may expect since ...


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I am not sure I understand your problem. The topology is continuously connected rotations around the identity. Let's fix to standard physics notation. You want, e.g., a rotation $U|11\rangle=|10\rangle$ $$ \begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix} = U \begin{pmatrix} 1 \\ 0\\ 0 \end{pmatrix}. $$ Look at this as a Cartesian transformation (instead of ...


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In general, for any spin, you can always construct the raising and lowering operators $S_\pm = S_x \pm i S_y$ which take a state with azimuthal angular momentum quantum number $m$ and change it to $m \pm 1$ (or annihilate the state if $|m|=s$, where the eigenvalue of $S^2$ is $s(s+1)$). So, to answer the first part of your question, just use $S_\pm$. The ...


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This is what's happening. You have $\vartheta^{\gamma\delta}$ parameters for the $\text{SO}(1,3)$; they constitute an antisimmetric matrix in their $\gamma,\delta$ indexes, so that the actual free parameters are the usual $6$ for the Lorentz transformation. That said consider that an infinitesimal Lorentz transformation will be $$ {\Lambda^\alpha}_\beta \...


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I suspect you are unfamiliar with the (mercifully) loose language of physicists. Strictly matrix realizations of a Lie algebra, and hence group, are termed Representations, like (3.18); but anything else, including (3.16) and its SU(2) antecedent, are just termed Realizations: versatile maps (linear, in this case) which satisfy the Lie Algebra (3.17). In ...


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One can compute the the weights by filling in the numbers 1,2,3 according to the rule for semi-standard tableaux (not decreasing along the rows, strictly increasing down the columns). Each of the eight possible tableaux gives the eigenvalues of $\lambda_3$ (the number of 1's minus the number of 2's) and $\lambda_8$ (number of 1's plus number of 2's minus ...


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