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Rotation of Pauli Vectors with $SU(2)$ reproduces the $SO(3)$ matrix. but do all $SU(2)$ matrices reproduces $SO(3)$?

Hint: a real 3-vector $\vec r=(x,y,z)^T$ is mapped, in your complex doublet SU(2) illustration to "Pauli vectors", $$ \vec r \mapsto \vec r\cdot \vec \sigma, $$ in the fundamental of the su(...
Cosmas Zachos's user avatar
1 vote

Subgroup of Lorentz Group Generated by Boosts

Let us consider a closed subgroup of the (proper orthochronous) Lorentz group which includes all boost transformations. (We are actually interested in the smallest such subgroup, but what follows does ...
Valter Moretti's user avatar
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Abelian vs non-abelian discrete symmetries in neutrino physics

I am going to give you a partial response that hopefully will help you clear some concepts about symmetries in particle physics. I will not focus on the papers that you are reading. Part 1: Discrete ...
Gabriel Ybarra Marcaida's user avatar
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Do Lie Groups need to satisfy the “addition property”?

You can disappear into this problem to any bottomless depth you like, or at least to the depth that you need as the crucial step to solve one of the great problems of 20th Century Mathematics, namely ...
Selene Routley's user avatar
3 votes

Why do physicists refer to irreducible representations as "charges" or "charge sectors"?

If a physical system has a continuous symmetry, a Lie group acts on it, cf. e.g. this Phys.SE post. The symmetry generating elements are often realized as certain quantities/operators in the theory ...
Qmechanic's user avatar
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3 votes

Why use the double cover instead of the universal cover of $SO(2)$?

Generally speaking, in quantum physics of course we are interested in the largest "universal covering group" of a given classical symmetry group. In most cases this is a double cover, ...
printf's user avatar
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4 votes

Why use the double cover instead of the universal cover of $SO(2)$?

$Spin(2)$ is, by definition, the double-cover of $SO(2)$. We don't use the notation $Spin(2)$ for the universal cover for a very simple reason: the universal cover already has a name: $\mathbb R$. It ...
AccidentalFourierTransform's user avatar
2 votes

Why do representations of $SU(2)$ correspond to angular momentum eigenstates?

Let us take for granted that the group $SO(3)$ encodes how spatial rotations in three dimensional Euclidean space work. Actually, we must also fix the scalar in front of the commutation relation of ...
Silly Goose's user avatar
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1 vote

Measurable effects of the global structure of the SM

The gauge group of the Standard Model is the quotient by $\mathbb Z_6$. Larger groups may be physically relevant, but it would take the discovery of new particles not in representations of the ...
benrg's user avatar
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6 votes

Why do representations of $SU(2)$ correspond to angular momentum eigenstates?

$\mathrm{SU}(2)$ is the so-called universal cover of $\mathrm{SO}(3)$, which is the group of rotations. The representations of $\mathrm{SU}(2)$ are in one-to-one correspondence with the projective ...
Níckolas Alves's user avatar
1 vote
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Wigner $ D $ matrix equivalent for cyclic symmetry

$\newcommand{\ket}[1]{\left|#1\right\rangle}$Ok now that I think about it more, maybe the answer is obvious. The Hilbert space for the $ C_5 $ symmetrized problem is indeed $ 8 $ dimensional but a ...
Ian Gershon Teixeira's user avatar
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Wigner $ D $ matrix equivalent for cyclic symmetry

You don't need to think of representations of the rotation group SU(2) as necessarily in the D-matrix spherical basis, but you may. When you tensor n such, you get a reducible $2^n \times 2^n$ matrix....
Cosmas Zachos's user avatar
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Complete positive (CP) divisible in Markovian and non-Markovian

This Physics Reports article by Chruściński (arXiv) is a nice starting point to understand divisibility, non-Markovianity, and the like. Section 7 is devoted to time-dependent generators & ...
Frederik vom Ende's user avatar
2 votes
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Are projective representiations of a Lie group a representation of the semi-direct product of the group with $U(1)$ if the norm is preserved?

If they were, then it would always be possible to obtain a true representation by restricting to $\mathbf 1\in \mathrm{U}(1)$, wouldn't it? Let $G$ be a group and $V$ be a complex vector space. It is ...
J. Murray's user avatar
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How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group?

An easy way to do this is to prove $\Lambda \in \rm O(3,1)$ preserves the sign of $v^0$ for a timelike vector $v \in \mathbb R^4$ iff $\Lambda \in \rm O^+(3,1) = \{\Lambda \in \mathrm O(3,1) : \Lambda^...
jajaperson's user avatar
1 vote

Why semi-simple and compact Gauge Group in YM Theory?

The compact criterion comes from the fact that the kinetic term showing up in the lagrangian is actually, strictly defined via a associative bilinear form (often the killing form). But for the theory ...
mika's user avatar
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