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When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Yes. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is ...

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Tricritical Ising model belongs to the family of minimal models ($M(5,4)$). There are several different coset constructions that represent them, one of them is the following: $M(m+1,m)=SU(2)_{m-2} \times SU(2)_1/SU(2)_{m-1}$

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The way you've written them, those are the spin operators in the $\hat{S}_z$ eigenbasis for a spin-1/2 particle. The two $\hat{S}_z$ eigenstates are spin up (written as $|+\rangle$ or $\uparrow$) and spin down ($|-\rangle$ or $\downarrow$), which can be written as $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ in the $\hat{S}_z$ ...

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These are mainly conventions. Conventionally, the kets $|+\rangle$ and $|-\rangle$ are taken to be eigenkets of the z-spin operator with, respectively, z-spin of $+\hbar/2$ and -$\hbar/2$. S_x and S_y are chosen such that they obey the canonical commutation relations for angular momenta $$[S_i,S_j]=i\epsilon_{ijk}S_k$$ E.g., $$[S_x,S_y]=iS_z$$ and so ...

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Here is my answer to some of your questions - this is based purely on my understanding of these concepts and could be wrong. (1) Whenever Lorentz transformations are a symmetry of any quantum system, they must necessarily be represented by unitary linear transformations on the quantum Hilbert space of the system. Operators representing Lorentz boosts on a ...

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The relevant Lie group isomorphism reads $$\tag{1a} U(2)~\cong~[U(1)\times SU(2)]/\mathbb{Z}_2.$$ In detail, the Lie group isomorphism (1a) is given by $$U(2)~\ni~ g\quad\mapsto\quad$$ $$\left(\sqrt{\det g}, \frac{g}{\sqrt{\det g}}\right) ~\sim~ \left(-\sqrt{\det g}, -\frac{g}{\sqrt{\det g}}\right)$$ $$\tag{1b}~\in ~[U(1)\times SU(2)]/\mathbb{Z}_2.$$ ...

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This is frequently good enough, but in your specific case its actually much easier to show this holds exactly in the finite case. I won't do this for you, but note that this is a global symmetry, i.e. alpha has no dependence on x. At most, you might also need to use the Baker-Campbell-Hausdorff formula.

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I'll use a notation that probably you know. I'll denote group representations by Dynkin labels. Consider the following: \begin{equation*} [1,0]_3 = [1]_2 q^1 + [0]_2q^0 \ , \end{equation*} and \begin{equation*} [0,1]_3 = [1]_2 q^{-1} + [0]_2q^0 \ . \end{equation*} Where $q$ is the $U(1)$-charge of the representation. By Dynkin diagrams the $SU(3)$ algebra ...

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The mistake is that $W^a_\mu [\tau_a, \tau_b] W^{\mu b} = 0$ does not imply $[\tau_a, \tau_b] = 0$, this is true for any antisymmetric object $A_{ab}$. In your example, you can see this in the following way: Let $A_{ab}$ be antisymmetric: $A_{ab} = -A_{ba}$. Then: \begin{align} W^a_\mu A_{ab} W^{\mu b} &= W^b_\mu A_{ba} W^{\mu a} \qquad ... 1 No, one can only conclude that the symmetrization[ \tau_a,\tau_b] + (a\leftrightarrow b)~=~0$$is zero, which is indeed true because the commutator is antisymmetric. 1 \newcommand{\ket}[1]{\lvert #1 \rangle}There is no "reason" you can use the ladder operators. Rather, they are the reason angular momentum is quantized in integer steps. You can define them, there's no inconsistency, so you can use them, and using them leads you to conclude that the angular momentum is raised/lowered in integer steps by them, in the way ... 2 I think I have found a rigorous answer (http://en.wikipedia.org/wiki/Representation_theory_of_SU%282%29). This is the complexified algebra of SU(2)$$[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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