# What is the charge of weak interaction?

That's right. As any other field theory, if you want to know how particles interact between them you have to look at lagrangian interaction terms. In the case of a gauge theory (and so also in the case of W,Z and γ interactions) the form of the interaction is "forced" by the requirement of gauge invariance.
The point is that when you write down the Yang-Mills lagrangian, the gauge fields that appear in the covariant derivative are all real fields. So at first they are all real, but in the case of Ws you have to do the transformation that I mentioned previously and so you have to substitute two real fields with a complex one and its conjugate.

But the Yang-Mill Lagrangian is invariant under the symmetry.Then to build the Yang-Mill Lagrangian we maybe need to apply the symmetry on the boson state?

Why the gauge field must real at first in Yang-Mill Lagrangian?

I mean why in covarian derivative the gauge field must be real?

That deduce from definition of Lie group.Is that correct?Then we can deduce that gluons are electric chargeless and so does photon.

I'll give you an example. In a YM theory you require the invariance of the lagrangian under local gauge transformations belonging to a certain Lie group. This transformation act, for example on fermion fields, as:

$$\psi(x)\rightarrow U(x)\psi(x)$$

where U(x) belongs to a particular representation of the considered Lie group. In order to have an invariant lagrangian under such transformations a covariant derivative is introduced. This derivative is written as:

$$D_\mu\psi(x)=(\partial_\mu + i\Gamma_mu(x))\psi(x)$$

where $\Gamma_\mu(x)$ are called connections. The connections are written as a real combination of the Lie group's generators:

$$\Gamma_\mu(x)=A^i_\mu(x)T^i$$

and A is the gauge field. So, yes, that's why gluons, photons and Zs are neutral.

The symmetry transformation is unita transformation?

Yes.

Why we choose unita symmetry SU(3),SU(2),U(1) but do not choose other Lie group?Is it required by Quantum Mechanics for Quantum Field Theory?

Unitary symmetry transformations are required by Quantum Mechanics because they preserve the scalar products, i.e. the observables.

Ok now I am confused. What does the fields being real or not have to with charge? And anyway aren't they all complex? SU(n) transformations are represented by complex matrices, so by doing local SU(n) transformations we can introduce arbitrary phase into the fields for any spacetime position we like.

Under SU(n) transformation the field operator changed like this:
U$\Psi$U$^{+}$,then the arbitray phase from U and U$^{+}$
is canceled each other.So if the field is real then it continues to be real after applying the unita transformation.

If you are talking to the electric charge, then its presence is conseguence of the invariance of the lagrangian under phase transformations of the form:

$$\phi(x)\rightarrow e^{iq\alpha}\phi(x)$$

Now the lagrangian for a real scalar field has, for example, the kinetic term written as $\frac{1}{2}\partial_\mu\phi(x)\partial^\mu\phi(x)$, that is obviously not invariant under the previous transformation.
On the other hand, if your field is complex then you have to consider also another phase transformation for the conjugate field:

$$\phi^*(x)\rightarrow e^{-iq\alpha}\phi^*(x)$$

In this case the kinetic term of the lagrangian is $\partial_\mu\phi(x)\partial^\mu\phi^*(x)$, and this is invariant.
So only complex field have a lagrangian invariant under phase transformation and so only them posses electric charge.

Hmm ok I guess. Is this really significant though? I mean we could design a gauge theory with fields that transformed under the real representations of the gauge groups, and those fields would still be charged under those gauge group, e.g. if instead of SU(2) doublets we chose to use triplet fields.

Actually I suppose the SU(2) gauge bosons themselves are an example, those are real fields but they still have weak charge.

I guess we cannot do this for U(1), although we could of course split our complex field into two real fields and use the 2x2 real matrix representation of U(1), but this is totally equivalent I think...

Last edited:
I'm not sure if I understood your point. However, I was strictly talking of gauge fields, i.e. the ones that appears in the connection. They are real by construction and so they are neutral (referring to electric charge).
In the previous post I just tried to explain why real fields are electrically neutral, and that's because electric charge arises from global invariance under phase transformations.

I do not understand why there is not repulsive force between same color quarks.I mean why we can not consider the representation 8+1=9 gluons,then the same color quarks can interact with each other by colorless gluons(this force is repulsive?).In general speaking why we do not consider a physics described by direct sum of many representations,because this physics clearly different from physics described by single representation.

Colorless gluons don't partecipate to the interactions. You can see this, for example, from the interaction term between gluons and quarks (that is completely determined by symmetry). In QCD lagrangian this term is:

$$g\bar{\psi}\gamma^\mu T^a\psi A_\mu^a$$

and leads to the following Feynman rule for the quarks-gluon vertex:

$$g\gamma_\mu T^a_{ij}$$, where i and j are the colors of the quarks (incoming and outcoming), since T is the generator of the fundamental representation of SU(3). If you want to exchange a colorless gluon then this means that quark's color remains unchanged. This means i=j and so you vertex become proportional to $T^a_{ii}=Tr(T^a)=0$. And so colorless gluons don't partecipate to QCD.

Then in 8 gluon representation,the interaction between same color quarks is repulsive force?
That explain why there are only color singlet hadrons,because the mass of colorful hadrons is very large?So there are both repulsive and attractive force in QCD?

Then in 8 gluon representation,the interaction between same color quarks is repulsive force?
That explain why there are only color singlet hadrons,because the mass of colorful hadrons is very large?So there are both repulsive and attractive force in QCD?
I don't understand your link between mass and color. However, in QCD there are both attractive and repulsive forces. For example, if you consider the interaction between two quarks, this is represented in group theory by $3\times3=\bar{3}+6$. If you do the right calculation it turns out that the resulting $\bar{3}$ representation is attractive, while the $6$ is repulsive.
If you consider, for example, a meson, which is a $q\bar{q}$ bound state then you have to consider $3\times\bar{3}=1+8$. In the same way, the octect turns out to be repulsive, while the singlet is attractive. So, as a meson is a bound quark-antiquark state (it means that the force has to be attractive), it has to be a color singlet.

I think inside colorful ''hadrons'' there are repulsive force,then it leads to higher energy hadrons,then we deduce that the hadrons have higher mass(unstable) ?

How can we calculate the values of ''charge''(electric charge,hypercharge.....)?Why we know the element particles(except quarks) have electric charge being +1 and -1?It seems that we can not use Standard Model to deduce the values of charge?

I don't think your point on the repulsive forces in hadrons is correct, but I'm not an expert.

How can we calculate the values of ''charge''(electric charge,hypercharge.....)?Why we know the element particles(except quarks) have electric charge being +1 and -1?It seems that we can not use Standard Model to deduce the values of charge?
I'm not really sure, but for the electric charge I suppose is an experimental issue. We knew about the charge of electrons before we built up the QED as a gauge theory. On the other hand, if you consider for example the weak isospin assigned to leptons (for example $I^W(\nu)=I^W(e^-)=1/2$ and $I^W_3(\nu)=-I^W_3(e^-)=1/2$) then I think this is an assumption that Glashow made in constructing his theory, because he was looking for something similar to Hesenberg's isospin symmetry in strong interactions.

Can we demonstrate that in representations 10,8,8 of color representation of baryons 3x3x3=10+8+8+1 there is exist two quarks in the baryon repel each other? Because this problem resolves the color singlet problem(?).

Probably yes, I have never done the calculation. However, the 8 representation is surely repulsive and the 1 is attractive. If the 10 is also repulsive this explains why you have barions as color singlets, because you want quarks to attract themselves in order to create boun states.