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4

The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1. In general, you need to know the Lie ...


3

Apologies for not producing a most general answer for arbitrary Lie groups, (which you might tease with great effort out of WP ), but only a trail-map for your particular (charmed!) problem. I call it charmed because it should remind you of the Lorentz group, with a,b,c parameterizing Kx,Ky,Kz boosts and d,e,f the three J rotation angles. Decent treatments ...


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The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle $$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...


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The spin $s$ of a particle characterizes how the rotation generators act on it. In $D$ dimensions, you represent the little group $SO(D-1)$ for massive particles and $SO(D-2)$ for massless ones. In fact, you really need to consider its universal cover $\textrm{Spin}(n)$ which happen to be just its double cover. Now, you can define the spin to be the largest ...


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This is the prototypical example of a superselection rule. The operator $U(2\pi)$ commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation ...


2

Let $\phi : \mathbb{R}^4\to V$ be a field with (complex) target vector space $V$, transforming in a finite-dimensional projective representation $\rho_\text{fin} : \mathrm{SO}(1,3)\to\mathrm{U}(V)$. As it is a field, the representation of the translations $\mathbb{R}^4$ on $V$ is the trivial one, since the field transforms as $\phi(x)\overset{x\mapsto ...


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Anthony Zee just came out with Group Theory in a Nutshell for Physicists - covers most of what a undergrad physics student needs including finite groups and representations, except Young diagrams.


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Your $SU(3)\otimes SU(3)={\bf 1}\oplus {\bf 8}$ above is a chimaeric typo from hell. OK, I'll just give you the self-evident answers, but they would be meaningless junk numbers if you failed to reproduce them directly on the basis of your SU(3) text or the WP article which explains the rules and the Dynkin representation notation, D(p,q), which connects to ...


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1) Universal covering groups are groups with the property of being simply connected. Each algebra has a unique covering group. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde ...


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Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down. The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging ...


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To a given Lie algebra $\mathfrak g$ there is a unique group $\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way ...


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Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3). For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), ...



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