All Questions
39
questions
3
votes
1
answer
58
views
Why semi-simple and compact Gauge Group in YM Theory? [duplicate]
I'm studying the Yang-Mills theory, with the Action:
$$
S=-\frac{1}{2}\int\mathrm{tr}_{\rho}(\mathcal{F}\wedge\star\mathcal{F})
$$
where $\mathcal{F}:=\mathrm{d} \mathcal{A}+\frac{1}{2}[\mathcal{A},\...
- 77
0
votes
1
answer
95
views
In QFT when performing path integral, why don’t we divide it by the volume of Poincaré group, as what we did for gauge group?
When performing path integral in gauge theory, we naively want to compute
$$
Z = \int DA \exp(iS[A])
$$
But we noticed, that because the action is the same for gauge equivalent conditions, we should ...
- 99
3
votes
1
answer
403
views
What is the "volume of the gauge group"?
I often see the term "volume of the gauge group" and I am not clear on what this is referring to. For example, in the second volume of Weinberg (page 22), he says
...the volume of the gauge ...
- 3,350
1
vote
0
answers
46
views
Group factors in scalar-gauge box diagram
So, I'm currently writing my Thesis, which involves one-loop beta functions of a general $SU(N)$ for scalars and fermions fields, Yukawa coupling and one scalar self-coupling.
To this moment I was ...
- 21
3
votes
1
answer
130
views
Why does this infinitesimal transformation tell us the boson vectors carry charge?
I'm studying from "Cheng, Li - Gauge theory of elementary particle physics", and in section 8.1, where it talks about non-Abelian gauge symmetry and the theory you can construct imposing ...
- 693
4
votes
2
answers
430
views
Why are there no instantons in the gauge group $U(1)$?
I am working my way through Srednicki's QFT book and am in chapter 93. Near the end, Srednicki says "If the gauge group is $U(1)$, there are no instantons, and hence no vacuum angle."
I'm ...
- 143
2
votes
1
answer
294
views
Why standard model uses Lie groups like $SU(2)$ and not $SL(2,\mathbb{C})$?
First of all, the question is written in section $2)$. Also, I known that the $SU(2)$ group do not appears "alone" in standard model, rather, inside the Glashow-Salam-Weinberg model.
1) ...
- 3,085
2
votes
1
answer
599
views
Nonlinear symmetry realization: what is it for and caveats?
I have several doubts regarding the nonlinear realization of a spontaneously broken symmetry and hope they are appropriate to be grouped, and I appreciate any insights.
Consider the group breaking ...
- 1,742
0
votes
0
answers
168
views
$SU(3)$ and $SU(3)\times SU(2)\times U(1)$ Symmetry Breaking [duplicate]
For my master's project I was doing spontaneous symmetry breaking in which I covered U(1), SU(2), SU(2)×U(1) symmetry breaking. My supervisor has said that for the project this much is enough. But now ...
0
votes
0
answers
63
views
Gauge Groups in Nature [duplicate]
It is well known that the relevant gauge groups appearing in particle physics are $U(1)$, $SU(2)$, and $SU(3)$. In many respects these are among the simplest Lie groups, such as in the dimension of ...
2
votes
2
answers
256
views
Why do we demand $SU(2)$ and $SU(3)$ gauge invariance when we construct the standard model?
If one tries to verify the construction of the standard model, one has to find a Lagrangian that is invariant under $U(1)\times SU(2) \times SU(3)$. While it seems kind of logic that the Lagrangian ...
- 437
5
votes
0
answers
62
views
Infinite dimensional gauge theories [duplicate]
Gauge theories depend on finite-dimensional symmetry groups like $SU(2)$. Is it possible to construct sensible gauge theories (or at least something similar in spirit) in QFT based on infinite ...
- 1,471
9
votes
1
answer
369
views
Why are gauge anomalies characterised by the non-triviality of $\pi_5(\mathcal G)$?
The folklore in 4-dimensional gauge theories is that the existence of potential gauge anomalies from the triangle diagrams that need to be cancelled are characterised by the non-triviality of the ...
- 8,492
5
votes
1
answer
188
views
Is this measure employed in the Faddeev-Popov procedure related to the Haar measure?
In the Faddeev-Popov procedure one defines the Faddeev-Popov determinant through the formula $$\int {\mathcal{D}\alpha \ } \delta\big[G(A^\alpha)\big]\Delta[A]=1,\tag{1}$$
where $G(A^\alpha)$ is the ...
- 36.4k
3
votes
1
answer
306
views
Most general Gauge Lie group in a Yang-Mills theory
Mathematicians have done a complete classification of all possible Lie groups. Is there a set of conditions that allows us to identify which Lie groups from the classification can possibly act as a ...
- 1,182