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6

Now the question is that why is the emphasis on saying an "operator" instead of simply a matrix. A matrix is a representation of an operator expressed in a particular basis. Consider the operation $T$ which mirrors the 2D plane about the line $x=y$. If we construct the obvious basis vectors $\hat{x}$ and $\hat{y}$, then $T$ is represented as $$[T]_{xy} ... 2 The same thing happens in 3d you can have an operator of rotation of a quarter turn about a particular vector. Or you can have a matrix. But the latter requires a choice of basis. The former does not. In quantum mechanics, you often want to pick the basis after you know the operator and if you wrote every operator as a diagonal matrix with real numbers ... 4 Given the orbital angular momentum operator L on the "spatial space" \mathcal{H}_1 and the spin angular momentum operator S on the "spin space" \mathcal{H}_2, we have the total angular momentum operator on the combined space \mathcal{H}_1\otimes\mathcal{H}_2 given by$$ J = L \otimes \mathbf{1} + \mathbf{1} \otimes S$$where \mathbf{1} is the ... 1 I'll sketch how you do it. In a general Lie group \mathfrak{G} setting, the mapping \mathrm{Ad} is clearly a homomorphism, since the action of the image of \gamma_1\,\gamma_2 under \mathrm{Ad} is$$X\mapsto \gamma_1\,\gamma_2\,X\,(\gamma_1\,\gamma_2)^{-1} = \gamma_1\,\gamma_2\,X\,\gamma_2^{-1}\,\gamma_1^{-1} = ...

1

A more physical construction: Let $R_3(\theta, \bf{n})$ be the matrix of a rotation of angle $\theta$ around axis $\bf{n}$ in $\mathbb{R}^3$. Then if $\hat{J}_i$, $i=1,2,3$ are corresponding SO(3) generators, $$\left[\hat{J}_i, \hat{J}_j \right] = i \epsilon_{ijk} \hat{J}_k$$ we have $$R_3(\theta, {\bf n} ) = exp\left(-i\;\theta \;n^i \hat{J}_i\right) ... 1 Given an element \phi of SU(2), let the first row of \phi be (P,Q) where P and Q are complex numbers. Let q(\phi) be the quaternion P+Qj. Now for any (x,y,z)\in {\mathbb R}^3, consider the quaternion q(\phi)(xi+yj+zk)q(\phi)^{-1} = (ai+bj+ck). The map (x,y,z)\mapsto (a,b,c) is a rotation of {\mathbb R}^3 and hence an element of ... 1 The OPE coefficients respect the symmetries. For example, consider the OPE$$ \mathcal{O}^i(x) \times \mathcal{O}^j(0)=\sum_{k} C^{ij}_{k}\left[|x|^{\Delta_k-\Delta_i-\Delta_j}\mathcal{O}^k(0)+\mathrm{descendants}\right] $$where for the time being I am suppressing the spin index. Then the C^{ij}_{k} transform as$$ C^{ij}_{k}\rightarrow U^{i}_a U^{i}_b ...

0

I interpret the question as follows. Let $SU(3)$ be the group of complex $3\times 3$ matrices with $U^\dagger U=I$ and $\det U=1$ acting as linear operators in $\mathbb C^3$. Is there a subspace $M \subset \mathbb C^3$ with $M \neq \{0\}$, $M \neq \mathbb C^3$ such that $U(M) \subset M$ for every $U \in SU(3)$? In other words, is the natural action of ...

1

Actually there are analogies or generalisations of results which reduce to Noether's theorems under usual cases and which do hold for discrete (and not necesarily discretised) symmetries (including CPT-like symmetries) For example see: Anthony C L Ashton (2008) Conservation Laws and Non-Lie Symmetries for Linear PDEs, Journal of Nonlinear Mathematical ...

1

I wanted to complement the answers above. For (1) $so(4) = so(3) \times so(3)$, one $so(3)$ is from the geometric 3D symmetry of the Hamiltonian, and the other $so(3)$ is from the potential term of $\frac{k}{r}$. For (2). the second $so(3)$ symmetry is a dynamic symmetry and only holds when potential term is inversely proportional to $r$. One has to do ...

1

This answer outlines how the defining matrix representation of the symplectic group Sp(2m,R) is ray optics, whilst the infinite-dimensional unitary rep of Sp(2m,R) carried on the space of wavefunctions is diffractive optics in the Fresnel approximation. The outline is for Sp(2,R) (cylindrical lenses) but the generalization to Sp(2m,R) is reasonably ...

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The Weyl system $\exp\left[\frac{i}{\hbar} Q \hat{p}\right]$ and $\exp\left[\frac{i}{\hbar}P\hat{q}\right]$ comprise two "presentation" elements of the Heisenberg group. To the extent $\hat{p}$ is a derivative with respect to position q, Q is the shift amount q in any function of it is translated by the action of $\exp\left[\frac{i}{\hbar} Q ... 2 I'm as confused as you by the boxed equation. At best the author is making that all-too-common mistake of reordering the expressions in a transitive equals relation, making the equation nonsensical when read left to right. However, it is not quite a tautology to prove what I think this is trying to prove: If$E$and$\psi$satisfy$H \psi = E \psi$, ... 5 I) The main point is that we usually only consider tensor products$V \otimes W$of vector spaces$V$,$W$; but groups (say$G$,$H$) are often not vector spaces. If we only consider tensor products of vector spaces, then the object$G \otimes H$is nonsense, mathematically speaking. With further assumptions on the groups$G$and$H\$, it is sometimes ...

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