Gauge invariance in a non-abelian theory SU(2)xU(1)xU(1)

manfromearth
Messages
5
Reaction score
0
Homework Statement
I'm given a gauge group G=SU(2)xU(1)xU(1) and matter fields in the following representations of G:

##\Psi## in the ##2_{(-1,0)}##
##\Phi_1## in the ##1_{(0,-1)}##
##\Phi_2## in the ##2_{(-2,0)}##
##\chi_L## in the ##2_{(-1,-1)}##
##\chi_R## in the ##1_{(1,-2)}##

where ##\chi_{L / R}## are Weyl chiral spinors, ##\Psi## is a Dirac spinor and ##\Phi_{1 / 2}## are complex scalar.
This notations means that, for example ##\Psi## is a doublet of the SU(2) factor, transforms with the first U(1) factor but is a scalar under the second U(1).

I'm asked to find a combinations of fields which is bilinear Lorentz invariant in ##\chi_{L / R}##, and invariant under gauge group G and that mixes Weyl spinors with the complex scalars.
Relevant Equations
let's denote ##U## a transformation of the SU(2) term,
##V## the ones of the first U(1) and ##T## the ones of the second U(1).
Then the transformations for the fields must be:

##\Psi \rightarrow UV \Psi ##
##\Phi_1 \rightarrow T \Phi_1##
##\Phi_2 \rightarrow UV \Phi_1##
##\chi_L \rightarrow UVT \chi_L##
##\chi_R \rightarrow VT \chi_R##
I believe what is asked is impossible. Here is why.
The U(1) factors are abelian, so V and T commute with each other and with U, so i can just try to build a term containing and even number of T-s,V-s and U-s.
From the transformation laws we see that a bilinear term in the Weyl fermions must have at least 2T-s, 2V-s and 1U in a transformation under G. Si I must add include other fields to have at least another U, but this is impossible without introducing other V-s or T-s.
 
Physics news on Phys.org


Gauge invariance is a fundamental principle in modern physics, and it is crucial for the consistency and validity of any theory. In a non-abelian theory, such as SU(2)xU(1)xU(1), gauge invariance plays an even more important role, as it allows for the description of non-abelian gauge fields and their interactions with matter fields.

In this context, gauge invariance refers to the invariance of the physical predictions of a theory under a local transformation of the gauge fields. This means that the mathematical description of the theory should remain unchanged under such transformations, and only physical observables should be affected.

In the case of SU(2)xU(1)xU(1), the gauge transformations involve the SU(2) and U(1) gauge fields, which are non-abelian. This means that the gauge fields themselves are transformed under these local transformations, and their transformation laws are more complex compared to abelian gauge theories.

Now, coming to the statement that gauge invariance in this theory is impossible, it is important to understand that gauge invariance is a necessary condition for any theory to be consistent and meaningful. It cannot be simply discarded or ignored.

In this particular case, the statement suggests that it is impossible to construct a gauge-invariant term in the Lagrangian that involves an even number of SU(2) and U(1) gauge fields. However, this is not true. In fact, there are many gauge-invariant terms that can be constructed using an even number of these gauge fields, and they are crucial for the consistency of the theory.

It is true that in a non-abelian theory, the gauge transformations of the matter fields are more complex, and they may involve other gauge fields (such as the U(1) gauge fields in this case). However, this does not make gauge invariance impossible. It simply means that the gauge transformations of the matter fields must also be taken into account in order to maintain gauge invariance.

In conclusion, gauge invariance in a non-abelian theory such as SU(2)xU(1)xU(1) is not impossible, and it is a fundamental principle that must be satisfied for the consistency and validity of the theory. The statement that it is impossible is not correct and may stem from a misunderstanding of the role of gauge invariance in non-abelian theories.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top