In mathematics, the special unitary group of degree n, denoted SU(n), is the Lie group of n × n unitary matrices with determinant 1.
The more general unitary matrices may have complex determinants with absolute value 1, rather than real 1 in the special case.
The group operation is matrix multiplication. The special unitary group is a subgroup of the unitary group U(n), consisting of all n×n unitary matrices. As a compact classical group, U(n) is the group that preserves the standard inner product on
C
n
{\displaystyle \mathbb {C} ^{n}}
. It is itself a subgroup of the general linear group,
SU
(
n
)
⊂
U
(
n
)
⊂
GL
(
n
,
C
)
{\displaystyle \operatorname {SU} (n)\subset \operatorname {U} (n)\subset \operatorname {GL} (n,\mathbb {C} )}
.
The SU(n) groups find wide application in the Standard Model of particle physics, especially SU(2) in the electroweak interaction and SU(3) in quantum chromodynamics.The simplest case, SU(1), is the trivial group, having only a single element. The group SU(2) is isomorphic to the group of quaternions of norm 1, and is thus diffeomorphic to the 3-sphere. Since unit quaternions can be used to represent rotations in 3-dimensional space (up to sign), there is a surjective homomorphism from SU(2) to the rotation group SO(3) whose kernel is {+I, −I}. SU(2) is also identical to one of the symmetry groups of spinors, Spin(3), that enables a spinor presentation of rotations.
I know people have looked into what it would mean if photons had a mass. But what would it mean if gluons had a mass? ie. if there was a small violation of SU(3) symmetry.
In other words, how do we know (experimentally) there is SU(3) symmetry?
Hi,
I'm currently reading a book on particle physics, which tells me this about SU(3):
"...The generators may be taken to be any 3x3-1=8 linearly independent traceless matrices. Since it possible to have only two of these traceless matrices diagonal, this is the maximum number of commuting...
Hi,
1: I just want to ask that what does SU(3) x SU(2) x U(1) means?
and when a lagrangian is invariant under SU(3) x SU(2) x U(1)
what does that mean?
Does it mean that lagrangian L is invariant under SU(3) and then SU(2) and
then U(1)? so if one wants to check SU(3) invariance of L...
For SU(2) rotational invariance there is a clear interpretation of what all the generators mean. They correspond to operators whose eigenvalues are some observable. The noether charges correspond to the x,y and z components of angular momentum. The casimir is just the total angular momentum...
Dear PFers,
I am looking for a way to visualize SU(3). I have heard from a friend
(who heard it as a rumor) that like SU(2) can be
visualized with a (Dirac) belt, also SU(3) can be visualized, but
with three belts, because SU(3) has three independent copies of
SU(2) as subgroups.
I...
The 3-dimensional harmonic oscillator has SU(3) symmetry. This is stated in many papers. It seems to be due to the spherical symmetry of the system. (After all, the idea of a 3d harmonic oscillator is that a mass is attached to the origin with a spring, and that the mass can move in 3...
Hi all,
I need to construct the Casimir op. of group SU(3).
I have these relations;
T2=\sum C_{i}_{j}T_{i}T_{j} i,j=1,2...,8 ...eq1
[Ti , Tj]= \sum f_{i,j,k} T_{k} ...eq2
[T2 , Ti]=[\sum C_{i}_{j}T_{i}T_{j} , Ts]=\sum C_{i}_{j}T_{i}[T_{j}, T_{s}] + \sum C_{i}_{j}[T_{i}...
Hi there!
As most people already might know, we can decompose the 27 dimensional representation for the baryons under SU(3) flavour symmetry as 27 = 10 + 8 + 8 + 1. I can find a lot of information about the particles that lie in the decuplet and in the octet, but nothing about which particle...
How are gluons related to the generators of SU(3),
the Gell-Mann matrices?
I do not understand how the structure constants f and d
describe how, for example, a red-antigreen gluon transforms
into a red-antiblue and a blue-antigreen one.
Do the f or the d factors describe the three-gluon...
Hi all!
I am currently preparing for an oral exam in quantum field theory and particle physics and I have some problems with the SU(3)-Hadron Multipletts and the relation to the Gell-Mann-Nishijima equation: First, for SU(2) Multipletts you take your Casimir-Operator J, some commuting...
Can anyone please explain to me how quarks are the fundamental representation of SU(3)?
Why is a proton exactly uud and not another combination of quarks?
What is a multiplet?
Thank you for answers :)
Can someone help me? It is correct at all to make the conection between the SU(3) symmetry and zeros of (radial) wave function? To make more clear: can I say that the fact that radial wave function has only the simple zeros automatically excludes the existence of SU(3) symmetry for given quantum...
This springs from section 15.1.3 of Superstring Theory (Vol 2) by GS&W (should anyone have that to hand).
K is a compact 6 dimensional space, thus it's holonomy group is a subgroup of SO(6). Fine. \eta is covariantly constant on K (comes from SUSY constraints). Thus need subgroup of SO(6)...
I find it awkward that quarks are in fundamental representation of SU(3) while gluons are in adjoint representation of SU(3). Is there a reason as to why this is the case? Why aren't they in the same representation or in the current specific representation?
I'm trying to learn some theory of inflation, and recently read a paper by Andrei Linde from 1982 where he suggested a scenario where inflation is driven by the phase transition SU(5) -> SU(3) x SU(2) x U(1). Now SU(5) is known to not be a good candidate as an extension of the standard model...
Hello,
I have a question with respect to the decomposition in irreducible representations of antiquark - antiquark ( SU(3) color ).
In the case of quark - quark what you have is a triplet with an antitriplet and what you obtain is an antitriplet and a sextet, and from the Young tables...
The more I read about group theory and SU(3), the stronger is my suspicion that they are very similar to the principle of directional invariance, which is based on the ideas of describing a dynamic one-dimensional cube and its eight properties.
Maybe I am wrong and too engross in my own ideas...