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Any linear transformation wrought on the Lie algebra of a Lie group yields a valid Lie algebra as I think you understand (the Gell Mann matrices are actually $i$ times the skew-symmetric Lie algebra members), and your proposed $\lambda_3$ is a linear combination of the Gell Mann matrices. The basis comprising $i$ times the Gell Mann matrices does indeed span ...


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You appear confused by how spin is introduced in ordinary QM. It is rather ad hoc: Given a Hilbert space without spin degrees of freedom of a particle $\mathcal{H}_0$, and the spin $s$ of the particle, we take the total space of states of the particle to be $\mathcal{H}_0\otimes \mathcal{S}_s$, where $\mathcal{S}_s$ is a $2s+1$-dimensional complex Hilbert ...


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Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.


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Your final result looks right to me. Everything should be half-integers. A basic rule of combining two quantized angular momenta is that the quantum number of the resultant can be anywhere between the sum of the original quantum numbers and the absolute value of the difference of them, in integer steps. Consider $\ell_1$ = 1 combining with $\ell_2$=3. The ...


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$\newcommand{\ket}[1]{\left| #1 \right>}$ Note that $ l=0$ has only one state $m=0$. Therefore the tensor product of $l=1$ and $l=0$ can be written as: $$ (l=1)\otimes (l=0) = \left\{ \begin{array} &\ket{l= 1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \end{array} \right\}=(l=1) $$ As ...


1

Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system. For 2 particles with total angular momentum eigenvalues $ l_1,l_2 $, such that for example $ ...


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Can these two pictures be connected in some way? Yes, that's why the Wikipedia spinor article features a picture of a Möbius strip: GNUFDL image by Slawekb, see Wikipedia The Mobius strip also features in the Mathspages Dirac's belt article where you can read that it's "reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be ...


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I) Perhaps it is helpful to point out that even if the physical system $S$ has no rotational symmetry (e.g. if the system $S$ is a 3D an-isotropic harmonic oscillator), then the Lie group $G=SO(3)$ of rotations still has a group action $G \times S \to S$ on the system. See also e.g. this Phys.SE post. In particular the Hilbert space ${\cal H}$ of the system ...


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A spinor has two components that describes its transformation under a lorentz transformation, classified by two indices $\Psi_{a\dot{b}}$, the first transforms under the left part of the lorentz group $SO(1,3)\approx SU(2)_L \otimes SU(2)_R$ (in more detail $\Psi_{a\dot{b}}\rightarrow L(\Lambda)_a^{a'} R(\Lambda)_\dot{b}^{\dot{b}'}\Psi_{a'\dot{b}'}$ where ...


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I'm not very strong in group theory, so could someone please explain in simple terms what it means for the left handed parts to transform as triplets and right handed parts to transform as singlets? How would you go about writing down such terms in a Lagrangian? Groups are abstract. They have elements that can be "multiplied" and they have other ...



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