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Since this question looks like homework we will be somewhat brief. OP's notes are apparently describing the symmetry of the corresponding Young diagram for each $SU(3)$ irrep. Each box corresponds to an index. Roughly speaking, indices in same row (column) are symmetric (antisymmetric), respectively. Examples: A single box $[~~]$ corresponds to the ...

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I think I have found a rigorous answer (http://en.wikipedia.org/wiki/Representation_theory_of_SU%282%29). This is the complexified algebra of SU(2) $$[J_z,J_\pm]=\pm J_{\pm}\\ [J_+,J_-]= J_z$$ The complex Lie algebra (i.e. the complexification of the Lie algebra) doesn't affect the representation theory. So both the real $[J_a,J_b]=i\varepsilon_{abc}J_c$ ...

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For $d=3$ the group theoretic meaning of total angular momentum is that it is the Casimir operator of $SO(3)$. For $SO(d)$ where $d>3$ you have more than one Casimir operator, so it's not clear what you mean by "total angular momentum" In particular the number of Casimir operators is $[d]/2$, where $[d]=d$ or $d-1$ depending whether $d$ is even or odd.

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The book where the derivation is described sufficiently pedagogically is Ballentine's Quantum mechanics - A Modern development, chapter 3. I am going to give a sketch of the 30-page chapter. (Beware, I suppress vector notation) Transformations of the quantum state are expressible as unitary transformations. The first order expansion of a unitary ...

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Say you had didn't know about quantum mechanics and no idea what a spinor is. You're given an "operator" that rotates things. One of the most basic assumptions you will make is that a rotation of $2\pi$ changes nothing. This is really quite reasonable. As @Phoenix87 said, we can identify $SU(2)$ has having a Lie algebra isomorphic to that of $SO(3)$. We ...

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Different Lie groups can have the same (up to isomorphism) Lie algebra. This is the case of, say, $SO(3)$ and $SU(2)$, the latter being the universal 2-cover of the former. When you are given a Lie algebra $\mathfrak g$ and you want to integrate it to a Lie group $G$ having $\mathfrak g$ as a Lie algebra, you will end up with a simply connected group. Hence, ...

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Well this is pretty similar to the calculations I have done to find the spectra of the quantum geometric volume operator in Loop Quantum Gravity. Given that I don't think that you will be able to find a closed analytical expression for this summation. I would be reasonably straightforward to write a numerical routine to calculate this. Here is the link to ...

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You are right that that the symmetry breaking breaks all three symmetries of $SU(2)$. Thus the $SU(2)$ generators give you three goldstone bosons in the theory with broken symmetry. However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. ...

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The transformation of a quark field under a group require you have to choose a representation of that group. It happen that the fundamental representation and the anti fundamental (bar) of $SU(N)$ with $N>2$ are inequivalent in the sense that there no non singular matrix independent of the representation chosen that allow us to make a change of basis and ...

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The Lie algebra of a non-Abelian Lie group can't be commutative, or the group itself would have to be Abelian. What you need to check is, e.g., that the only matrices that commute with any other matrix of $SU(3)$ is a multiple of the identity matrix, as required by Schur's lemma.

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The way to get from exponential form to the most general Lorentz transformation can be an extremely tedious computation to carry out. In principle there is nothing else to do but to explicitly compute the exponential of a given matrix. I would suggest you try to fix every generator and get an idea of the transformation it is generating. For example, fix the ...

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I. To obtain the matrix form you quote from the infinitesimal transformations: Start from the Lorentz transformation as usually written per component, e.g. in the wikipedia article. Write the infinitesimal form of these transformations. E.g. for a boost in the $x^1$ direction you have $$\tag{1-1} x'^0 = x^0 + \beta x^1 + \mathcal{O}(\beta^2),$$ $$... 2 We are not entirely sure what OP's question (v4) is asking, but here are some comments: I) The Dirac belt trick demonstrates that the Lie group SO(3) of 3D rotations is doubly connected,$$ \pi_1(SO(3))~=~\mathbb{Z}_2. $$II) As for the title question Are spinors somehow connected to spacetime? one answer could be: Yes, in the sense that the mere ... 0 You might equally well ask, "How does the physical belt in the Dirac trick sense the topology?" This question is, when you think about it, no less mysterious than yours. The answer, by experiment, is that it simply does. And ultimately, if something transforms "compatibly" with the Lorentz group, or with SO(3), then there is really only a one-bit question ... 2 (Strong) SU(2)_F isospin is a global u\leftrightarrow d flavor symmetry of the strong force (but not a symmetry of the EM force, and hence only an approximate symmetry). The SU(2)_F sits inside an approximate global SU(3)_F flavor symmetry of the u, d and s quark, cf. e.g. this Wikipedia page. (Strong) isospin is different from the weak ... 3 The Wilson loop observables inside 3d Chern-Simons gauge field theory are secretly themselves the quantization of a 1d field theory in terms of coadjoint orbits. This possibly still surprising-sounding statement was hinted at already on p. 22 of the seminal Edward Witten, Quantum Field Theory and the Jones Polynomial Commun. Math. Phys. 121 (3) (1989) ... 2 I think it is easier to compute direct products when you write the matrices in component form; basically, you just have to multiply each element of the first matrix by the whole second matrix:$$ \mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} A_{11} \mathbf{B} & \cdots & A_{1n} \mathbf{B} \\ \vdots & \ddots & \vdots \\ A_{n1} \mathbf{B} & ...

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Each Pauli matrix has two non-zero elements. Therefore, direct product of Pauli matrices will have four non-zero elements. Your answer, unfortunately, has eight.

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