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I'm a bit confused about this following issue concerning representations of $SU(2)$. Denote by 1 the 1-dimensional representation of the group $SU(2)$ (=the spin 0). … But since 1 and 3 both represent the same group $SU(2)$, so does their direct sum (a reducible representation). It follows that 1+3 is a representation of both $SU(2)$ and $SU(2)\times SU(2)$. …

If the symmetry group were one single SU(2), I would understand that $M$ transforms as $$ M \rightarrow g^\dagger M g. $$ Generalizing to $SU(2)_L \times SU(2)_R$, I would guess that $M$ transforms as … My further question is, supposing the field operator transforms in the fundamental representation, how does it transform under $SU(2) \times SU(2)$? …

What is the normalizer of $SU(2)\times SU(2)$ in $SU(4)$ or how would I find it? … Reason for the question: with 2 qubits, if I was interested in conjugation of 2-qubit gates with generic $SU(2)$ elements, instead of with elements of the Pauli group, what 2-qubit gates would be preserved …

The question is regarding SU(2) group and SU(2) algebra. … The SU(2) group can be generated by exponentiating the generators of SU(2) algebra $X_a$ as $exp(i t_a X_a )$ with $t_a$ being three parameters. …

What's the difference between the $j = 1/2$ representation of $SU(2)$ and the $(j,j') = ( 1/2 , 1 )$ representation of $SU(2)\times SU(2)$? …

Like my title, what is the difference between $SU(2)_L$ and $SU(2)_R$? I know they transform differently, but I don't know how to write down the transformation. …

In other words, How do I construct the $SU(2)$ representation of the Lorentz Group using the fact that $SU(2)\times SU(2) \sim SO(3,1)$? … Here is some background information: Zee has shown that the algebra of the Lorentz group is formed from two separate $SU(2)$ algebras [$SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$] because the Lorentz …

Following from this question and the links within, I have a couple of questions about the use of $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ for the classification of finite real restricted Lorentz algebra … $ and $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$? …

My current understanding is that $SU(2)$ rotates a single spin 1/2 particle, and $SU(2) \times SU(2)$ rotates both particles (but not necessarily with the same axis and angles). … This was confusing to me because I previously thought that two spin 1/2 degrees of freedom led to $SU(2) \times SU(2)$ symmetry. …

If $e^{i\sigma_j\phi_j}$ were to be a $SU(2)$ matrix then the $\phi_j$ ought to be real. … Therefore this is no longer a proper $SU(2)$ theory. What is going on? …

Therefore my symmetry group is $SU(2)_L \times SU(2)_R \times U(1)_Y$. … In my understanding: $SU(2)_L$ has 3 generators, all of which are broken $SU(2)_R$ has 3 generators, but only $T^3_R$ is broken. …

Let us call the generators of $su(2)$ in the spin $A$ or spin $B$ representation $J^A_i$ and $J^B_i$ respectively. What are the generators of $su(2) \oplus su(2)$ in the $(A, B)$ representation ? … And how do they act on the intertwiners $u_{ab}$ and $v_{ab}$, for a field in the representation $(A, B)$ of $SL(2, \mathbb{C})$ with particles in the representation of $ISO(2)$ (or its appropriate cover …

The author then said that the symmetry of this Lagrangian is $SU(2)\times SU(2)$. However, I thought it was just $SU(2)$ or $U(1)\times U(1)$? … I then googled this and found that the Lorentz group is isomorphic to $SU(2) \times SU(2)$, which I guess is one explanation. …

One way to classify Lorentz representations is to consider the Lie algebra isomorphic of Lorentz group to $SU(2)\times SU(2)$. So that we can classify it by two integers $(j_1,j_2)$. …

Following the approach of Weinberg's book to discuss the chiral symmetry, at a certain point he says If the $\rm SU(2)\times SU(2)$ symmetry is exact and unbroken, then this would require any one-hadron … (2)$ subgroup. …