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There is a general expression for the matrix representation of $N$ Grassmann numbers (called the Clifford-Jordan-Wigner representation), and it is intimately related to the matrix representation of the Euclidean $\gamma$-matrices in $D=2N$ dimensions ($\gamma$-matrices in $D=2N+1$ and Lorentzian $\gamma$-matrices are simple to find from these). I will simply ...

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There is no need of starting with $SU(2)$ symmetry and then extending it to $SU(2)_L\otimes SU(2)_R$. In fact, the moment you write down the Lagrangian, the symmetry is by default $U(2)_L\otimes U(2)_R$ with $U(2)_L$ acting on the left handed quark doublet $(u , d)_L$ and $U(2)_R$ acting on the right handed quark doublet $(u , d)_R$, which can be extended ...

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Question 1 First of all, when you discuss chiral symmetry spontaneous breaking, you need to assume pure QCD theory. QCD lagrangian with $u-,d-,s-$quarks (they have relatively small masses in compare with $b, t, c$-quarks) has the form $$\tag 1 L_{QCD} = \bar{q}_{i}i\gamma_{\mu}D^{\mu}_{ij}q_{j} - \frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - ... 1 The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense,$$ \frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0, $$than it commutes with S-operator:$$ [\hat{Q}, \hat{S}] = 0 $$So that these two operators can be diagonalized simulatenously: in ... 1 There is a long and formal way, and also an easy and dirty way. I will tell you the easy option. The algebra tells you that [\delta_Q (\epsilon_1), \delta_Q (\epsilon_2)] = \delta_{P}(\xi^\mu_3) where \epsilon is your SUSY parameter and \xi^\mu_3 = \bar\epsilon_1 \gamma^\mu \epsilon_2 is your translation parameter. Now, the only Lorentz vector that ... 0 Comments to the question (v3): The point group in question is the chiral tetrahedral symmetry group T of order 12, i.e. the symmetry group of the tetrahedron. Problem 3.1(c) confusingly talks about a 2-dimensional irreducible representation E, which is in fact the reducible sum of a 1-dimensional representation and its complex conjugate ... 1 Since you know about SU(2) characters, this is a doable exercise. Let me remind you that the character of spin-j is$$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}. The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, ...

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Short answers Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows ...

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This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...

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