What is the charge of weak interaction?

[ndung200790]

The strong interaction has color ''charge'',the electromagnetic interaction has electric charge,the gravity interaction has mass ''charge'',then what is the charge in weak interaction?

Each interaction corresponds to a symmetry.Symmetry SU(3) for strong interaction,
SU(2)xU(1) for electroweak interaction,then what is the symmetry for gravity interaction?

Why we know the gluons in QCD carry color charge,or other saying when quarks absorb or emit gluon their color must be changed?

 

[Einj]

Actually I don't know if there is a name for the "weak charge". However, in electromagnetism, electric charge is the coupling constant used in the lagrangian in front of the $U(1)_{e.m.}$ generator. On the other hand, usually in the SM electromagnetic gauge symmetry is unified with weak symmetry under the $SU(2)\times U(1)$ gauge symmetry. If we want to interpret "charge" in the same way of e.m. symmetry, then in this new symmetry you have two different charges, the one referred to $U(1)$ group which is called "weak hypercharge" $Y=Q-I_3^W$ (where the last term is the third component of the weak isospin), and the one associated to the $SU(2)$ symmetry, which I usually call as "g" but I don't know if it has a proper name.

As far as you second question is concerned, we aren't able to assign a symmetry to gravitational interaction because we haven't reach the complete grand unified theory! :tongue:

 

[ndung200790]

Then how about the color charge and coupling constant in QCD?

 

[Einj]

I'm sorry, I completely forgot to answer that question! :tongue: However, while electrodynamics is an abelian theory (i.e. the generetors of the symmetry group commute with each other), QCD, but also electroweak theory, is a non-abelian one. This means that in the lagrangian of the theory you can find terms of interaction of the gauge fields between them. For example, in QCD you can have a vertex with three or four gluons interacting. This situation obviously can be possibile only if the gauge fields (gluons, Ws, ecc) carry the charge of the interaction themselves.

 

[ndung200790]

In QCD (for example),do the color charge and the coupling constant ''correspond'' with each other?

 

[mfb]

What do you mean with "correspond"?
Do mass and the gravitational constant "correspond" with/to (?) each other?

 

[Einj]

The concept of coupling constant and of charge are not exacty the same, but they are closely related. In the case of electrodynamics they coincide (because QED is abelian). In the case, for example, of QCD, you have a coupling constant usually denoted with g (or something like that) and which "identify" the intensity of the interaction. The term "charge", on the other side, is usually related to the way of transforming of the fields under gauge transformations and so involves both the coupling constant and the generators of the symmetry. In QCD, for example the "color charge", should involve the coupling constant and the Gell-Mann matrices for quarks or the matrices of the adjoint representation for the gluons.
However you can find some explanation on the difference between charge and coupling here: http://en.wikipedia.org/wiki/Color_charge.
I hope I was useful

 

[ndung200790]

Why are gluons to be bicolor(two color) carriers while quarks are single color carrier?

 

[Einj]

In my opinion you can see it in two possible ways. The physical one is that if, for example, you have a G quark that emits a gluon and become a R quark, then the gluon must have a coloro-anticolor propriety in order to conserv color itself.
On the other hand, you can consider a mathematical motivation. When you talk about a certain "color" you are referring to one of the three possible indices of the matrices of the fundamental ($3$) representation of SU(3). The gluons belongs to the adjoint representation ($8$), which is obtained by a tensor product $3\times\bar{3}=8+1$ and so is obtained "mixing" color and anticolor togheter.

 

[ndung200790]

Thank you very much for your usefull teaching!

 

[ndung200790]

How does QFT make the relation between ''charge'' and coupling constant?

 

[Einj]

It's in the way of transforming of the fields. For example, in QCD, quark fields transform with the fundamental representation of SU(3) which is given by $\lambda_a/2$, where $\lambda_a$ are the 8 Gell-Mann matrices, i.e. the generators of SU(3). So, the law of tranformation of quark fields is:

$$\psi(x)\rightarrow exp\left[i g_s \frac{\lambda_a}{2}\omega_a(x)\right]\psi(x)$$

thus, as you can see, the coupling constant $g_s$ and the generators of the symmetry are related via this relation.

However, if you want to have a better explanation you can read Mandl-Shaw "Quantum Field Theory". It's quite essential and a good book.

 

[kurros]

The strong interaction has color ''charge'',the electromagnetic interaction has electric charge,the gravity interaction has mass ''charge'',then what is the charge in weak interaction?

I always find this an interesting conversation. I think the naming conventions are very confusing on this topic. Einj already explained the story but I just wanted to say that I do not really know why we talk about the SU(2) charges differently than SU(3).

Someone correct me if I am wrong (which I am pretty sure some of it is because I have forgotten a lot of group theory), but this is how it seems to me:

For SU(3), the quarks are in the fundamental representation, and so we have eigenvectors r=(1,0,0), g=(0,1,0), b=(0,0,1) (although I get confused about what happens in SU(3) since there are 2 Cartan generators, what do the eigenvalues for these generators tell us?)

Likewise in SU(2) the (left handed) leptons are in the fundamental representation, with eigenvectors up=(1,0), down=(0,1) (here with only one Cartan generator we can map these straight to the eigenvalues of T3, so we can talk about leptons having +1/2 or -1/2 3rd component of weak isospin and be describing the corresponding eigenvectors)

So when we talk about colour charge we really mean that in colour space a particle is one of these colour eigenvectors. Why do we not do the same thing for weak charge? Weak charge is just being one of the "up" or "down" eigenvectors in weak isospin space. I suppose we do, it is just that the names are not very catchy.

But let us go a little further. As Einj explained the gluons are in the adjoint representation, and because of the way the representation decomposes into the $3$ and $3\times\bar{3}$ representations then we can talk about them having colour and anti-colour. Why do we not do the same thing for weak vector bosons, i.e. talk about them in terms of the decomposition $2\times2=3+1$ (actually I would also like to know why it is $2\times2$ not $2\times\bar{2}$, although I suppose it is just that the fundamental and anti-fundamental representations are equivalent for SU(2)? There is a sign difference for one of the matrices, but I guess that doesn't matter?)

Example: We should have 3 weak vector bosons, with up.up, up.down+down.up, down.down charges.

Although I guess we do, since we think of this as "like spin" and so up.up is really like 1/2+1/2=1, down.down is -1/2-1/2=-1, and the neutral combination is 0, so we just talk about the eigenvalues of T3 in the adjoint representation.

Anyway I guess that was slightly rambling, but the point was I think we could easily talk about SU(2) and SU(3) charges in the same kind of language, if we wanted to.

 

[ndung200790]

I think we have decomposition 2x$\bar{2}$=3+1 intead 2x2=3+1 in weak interaction because the flavor of fermions can be changed,then the weak bosons carry flavor and antiflavor.

 

[ndung200790]

If we can say about ''charge'' in weak interaction why we can not say about potential concept and about ''normal'' force(as electromagnetic force).But I hear that weak interaction is an ''exotic'' force.

 

[Einj]

For SU(3), the quarks are in the fundamental representation, and so we have eigenvectors r=(1,0,0), g=(0,1,0), b=(0,0,1) (although I get confused about what happens in SU(3) since there are 2 Cartan generators, what do the eigenvalues for these generators tell us?)

Firstly, r,b and g are not eigenvectors for all the generators, as they are not all diagonal. In fact, in the case of SU(3) you have two Cartan generators, i.e. two of the eight generators which commute with each other. This means that this particular generators (which are $T_3$ and $T_8$) are both diagonal and so r,b and g are eigenvectors of them.
Actually, I don't know the meaning of their eigenvalues for the color representation (if they have one), but if you consider instead the SU(3) flavor symmetry (which is an approximate one), the eigenvalues of the Cartan generators correspond to the third component of weak isospin and to the hypercharge of the states if we identify them with the following quark: u=(100), d=(010), s=(001).

But let us go a little further. As Einj explained the gluons are in the adjoint representation, and because of the way the representation decomposes into the 3 and 3×3ˉ representations then we can talk about them having colour and anti-colour. Why do we not do the same thing for weak vector bosons, i.e. talk about them in terms of the decomposition 2×2=3+1 (actually I would also like to know why it is 2×2 not 2×2ˉ, although I suppose it is just that the fundamental and anti-fundamental representations are equivalent for SU(2)? There is a sign difference for one of the matrices, but I guess that doesn't matter?)

I'm not really sure, but I think your reasoning is valid. According to what I've studied, gauge fields always transform according to the stucture constants of the group, i.e. with the adjoint representation. So the situation of vector bosons for SU(2) should be similar to the situation of gluons. But, as I said, I'm not really sure and, as kurros said, the situation is a little confusing. But, however is just a matter of names

As far as the decomposition of SU(2) is considered, in the case of this particular group the antifundamental representation doesn't exist. One usually say that $2$ representation of SU(2) is pseudo-real, which mean that is equivalent (but not really equal) to the $\bar{2}$ as you can obtain the second one from the first one via a transformation of equivalence involving $\sigma_2$ Pauli's matrix.

If we can say about ''charge'' in weak interaction why we can not say about potential concept and about ''normal'' force(as electromagnetic force).But I hear that weak interaction is an ''exotic'' force.

In QFT one does not talk about "forces" anymore, because a force is a concept related to an action at a distance, which is unacceptable in a relativistic framework. Instead, one usually talks about "interactions", which is clearer because we are talking about local action between particles.

 

[ndung200790]

In SU(2)xU(1) model,why we know W carry electric charge but Z and photon are electric chargeless(meaning why we know W field are complex field and the others are real field)?

 

[kurros]

In SU(2)xU(1) model,why we know W carry electric charge but Z and photon are electric chargeless(meaning why we know W field are complex field and the others are real field)?

I'm not sure what you are talking about re: complex vs real fields, but the simplest answer to your other question is the formula relating electric charge and weak isospin which Einj already gave: $Y=Q-I_3^W$. Weak vector bosons have hypercharge zero, so their isospin eigenvalue (of the diagonal generator) is just straight equal to their electric charge. This is a post-electroweak symmetry breaking thing though; the unbroken SU(2) gauge bosons do not have electric charge, indeed nothing has electric charge back then. For a more complete answer, i.e. where this formula comes from, you have to dig into how the electroweak symmetry breaking works.

 

[ndung200790]

Then the formula Y=Q-I$^{W}_{3}$ can be aplyed both for fermions and bosons?
And how about photon?(W,Z have ''isospin'' 1,0-1 so electric charge also 1,0,-1)

 

[kurros]

Then the formula Y=Q-I$^{W}_{3}$ can be aplyed both for fermions and bosons?
And how about photon?(W,Z have ''isospin'' 1,0-1 so electric charge also 1,0,-1)

I have always assumed so and it appears to work out. Although I just realized there is a factor of 1/2 missing, it should be $Y/2=Q-I^{W}_{3}$, but I suppose this is just a matter of the definition of weak hypercharge and probably it is different in different books.

As for photons, well they have $Y=I^{W}_{3}=Q=0$, so there is no problem.

 

[ndung200790]

Regorously speaking,the charge of photon is zero because of U(1) symmetry where there is nothing changed in ''color''(electric charge of fermions) as in QCD(SU(3))?Because how we know formula Y=Q-I$^{W}_{3}$ can be aplyed to photon?What is Y for photon?

 

[ndung200790]

At the moment,I am be able understand your teaching.The answer consist in that we can use different representations.We can apply the symmetry transformation(SU(2)xU(1)) on fermion states or boson states.Then the formula can be used both for fermions and bosons.
Is that correct?

 

[Einj]

I'm not really sure if applying this formula to gauge bosons is possible. I have always used it only for the fermion weak multiplets, mainly because you need it in order to know how fermions transform under the $U(1)_Y$ symmetry of the electroweak theory.

However, as far as your question about W and Z charges is concerned, the answer lays in the mathematical methods we use to manage gauge bosons from gauge symmetry. In fact, when you first write the lagrangian (gauge invariant) of you system you don't have directly W and Z fields. You start with other four gauge fields. These fields are usually written as $W_1^\mu (x) \; , \; W^\mu_2 (x) \; , \; W_2^\mu (x) \; , \; B^\mu(x)$. One usually redifine two of them in the following way:

$$W_\mu(x)=\frac{W_{1\mu}-iW_{2\mu}}{\sqrt{2}}$$

and

$$W_\mu^\dagger(x)=\frac{W_{1\mu}+iW_{2\mu}}{\sqrt{2}}$$

Now, when you write the lagrangian you find out that these last two fields are already mass eigenstate, while the fields $W_{3\mu}(x)\;,\;B_\mu(x)$ are not. So you introduce the following fields:

\begin{eqnarray*}
W_{3\mu}(x)=\cos \theta_W Z_\mu(x)+\sin \theta_W A_\mu(x) \\
B_\mu(x)=-\sin \theta_W Z_\mu(x)+\cos \theta_W A_\mu(x)\\
\end{eqnarray*}

These last two fields are both mass eigenstate but, while the W fields are complex these one (which are Z and photon fields) are real and so they are referred to neutral particles.

 

[ndung200790]

I think that when we consider the interaction between W,Z , photon with themself we must apply the symmetry transformation on the boson states(the process of building the interaction Lagrangian of this interaction).Is that correct?

 

[ndung200790]

Sorry,the interaction Lagrangian between the bosons is made from Yang-Mill Lagrangian.
Then why we know the field of W are complex and those for Z and photon are real?

 

[Einj]

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.

 

[ndung200790]

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?

 

[ndung200790]

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

 

[ndung200790]

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

 

[ndung200790]

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

 

[Einj]

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.

 

[ndung200790]

The symmetry transformation is unita transformation?

 

[Einj]

Yes.

 

[ndung200790]

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?

 

[Einj]

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