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  Conserved topological charge for d=3 Yang-Mills. G=U(2)

+ 4 like - 0 dislike
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Consider a pure Yang-Mills lagrangian density $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}_aF^a_{\mu\nu}$$ with gauge group $U(2)$.

Take the generators for $U(2)$ to be $t_0$, $t_i \ i=1,...,3$ with commutation relations given by $$[t_0,t_i]=0$$ $$[t_i,t_j]=i\epsilon_{ijk}t_k$$ In particular $t_0$ is the generator of the $\mathfrak{u(1)}$ factor in the expansion $\mathfrak{u(2)}\simeq \mathfrak{u(1)}\times \mathfrak{su(2)}$ and $t_i$ are the generators of the Lie Algebra $\mathfrak{su(2)}$.

Now, in $d=3$ the field strenght Hodge-dual is a current $j^{\mu}:=\frac{1}{2}\epsilon^{\mu\nu\rho}F_{\nu\rho}$ and is conserved in virtue of the Bianchi Identity.

The questions are:

1) What is it meant when they say the current is conserved? Is it covariantly conserved (ie $D_{\mu}j^{\mu}$=0) or simply conserved (i.e $\partial_{\mu}j^{\mu}=0$)

2)Do I have just one vector current, or one for each generator of the gauge group? (i.e 4 in this case)

3) Can you explicitly carry out the computation of the conserved current and charge?

4) I am asked to state if the conserved charge arises because of the factor $U(1)$ of the gauge group (which has an algebra generated by $t_0$), because of the factor $U(1)$ which is the cartan subalgebra of $SU(2)$ (generated by $t_3$), or because both of them. [I really don't understand this question, what would you answer? Thanks.]

The part of the computation I did is the following.

$$F^0_{\mu\nu}=\partial_{\mu}A^0_{\nu}-\partial_{\nu}A^0_{\mu}$$ $$F^i_{\mu\nu}=\partial_{\mu}A^i_{\nu}-\partial_{\nu}A^i_{\mu}+g\epsilon^{ijk}A_{\mu}^jA^k_{\nu}$$

Therefore using Bianchi I have

$$0=D_\mu\epsilon^{\mu\nu\rho}F^0_{\nu\rho}=(\partial_{\mu}-igA_{\mu})\epsilon^{\mu\nu\rho}(\partial_{\nu}A^0_{\rho}-\partial_{\rho}A^0_{\nu})$$

while for the other side

$$0=D_\mu\epsilon^{\mu\nu\rho}F^i_{\nu\rho}=(\partial_{\mu}-igA_{\mu})\epsilon^{\mu\nu\rho}(\partial_{\nu}A^i_{\rho}-\partial_{\rho}A^i_{\nu}+g\epsilon^{ijk}A_{\mu}^jA^k_{\nu})$$

What can I do from here? It seems to me that the currents $$j^{\mu}_0=\epsilon^{\mu\nu\rho}(\partial_{\nu}A^0_{\rho}-\partial_{\rho}A^0_{\nu})$$ and $$j^{\mu}_i=\epsilon^{\mu\nu\rho}(\partial_{\nu}A^i_{\rho}-\partial_{\rho}A^i_{\nu}+g\epsilon^{ijk}A_{\nu}^jA^k_{\rho})$$

are both covariantly conserved...

Thanks a lot for answers and clarifications.

This post imported from StackExchange Physics at 2015-09-25 20:26 (UTC), posted by SE-user Federico Carta
asked Oct 26, 2013 in Theoretical Physics by Fedecart (90 points) [ no revision ]
retagged Sep 25, 2015
Most voted comments show all comments
This article explains in the first pages why the current is conserved in the case of d=3 QED (which is U(1) Yang-Mills) but still is quite different...

This post imported from StackExchange Physics at 2015-09-25 20:26 (UTC), posted by SE-user Federico Carta
You're asking us to do all the work without making an effort yourself. For 1), you write that the current is conserved because of the Bianchi identity. But in the next sentence you ask us to prove the conservation of the current. 2) is really textbook material. It's the same as asking whether there's only one "gluon" $A_\mu$ or several. 3.) is also a textbook question. Many of us know the answer but it's a waste of time to write down all these computations if they're done in every QFT textbook, e.g. Peskin-Schroeder (the chapter on non-Abelian gauge theories).

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Vibert
I have made an effort. I have done part of the computation (as far as I can get) but can't arrive at the correct answer. I can post it if you want to, or do not believe I first tried and then asked. I have also looked at Peskin-Schroeder but there is nothing similar to this in the whole book. They only treat ordinary Yang Mills in 4 dimension, and not in d=3. If it really is textbook material, could you suggest a book in which they treat Yang-Mills in d=3? I believe that there are 4 different currents, but then I don't understand point 4).

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Federico Carta
Well, the only difference between $SU(3)$ and your case really lies in the structure constants, correct? In turn, they show up in the expression for $F_{\mu \nu}.$ Try to write down (if you haven't already done so) the different $j_\mu^a$ in terms of the structure constants. Also in general, if you have done some work, you should always post it - this helps people to see what's going wrong and where you need some tips.

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Vibert
Correct. The only difference is in the structure costants. I will post it as soon as possible the computation I did. I have computed the four currents, and they are obviously different, but none of them is simply conserved. Is it possible? And how can one define a conserved charge without an ordinary continuity equation? Is it possible that a linear combination on the current associated to $t_0$ and the current associated to $t_3$ is simply conserved, thus answering to 4) that "the conserved charge" arises from both the U(1) factors?

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Federico Carta
Most recent comments show all comments
@FedericoCarta : Hints : For each matrix $X= X^aT_a$, the definition of $"D_\mu X"$ or $"[D_\mu,X]"$ (a notation for instance used in the Bianchi identities) is $D_\mu X = [D_\mu,X] = \partial_\mu X -ig [A_\mu, X]$ (see for instance formulae $4.29, 4.30$ in this paper). You don't need to have a detailed expression for the $j^\mu$, just use the definition of the $j^\mu$ in function of the $F_{\mu\nu}$

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Trimok
$\epsilon^{\mu\nu\rho}D_{\mu} F_{\nu\rho}=\epsilon^{\mu\nu\rho}\partial_{\mu}F_{\nu\rho}+\epsilon^{\mu\nu\rho}‌​[A^a_{\mu} t_a,F^b_{\nu\rho}t_b]=\epsilon^{\mu\nu\rho}\partial_{\mu}F_{\nu\rho}+\epsilon^{\‌​mu\nu\rho}f^{abc}A^a_{\mu}F^b_{\nu\rho}$ And therefore for $b=0$ (which is the current associated to $t_0$) I have that it is simply conserved. While for $b=i \quad i=1,2.3$ it is not simply but covariantly conserved. Therefore can I answer question 4) saying that the conserved topological charge arises from the factor $U(1)$ of the gauge group and not the $U(1)$ generated by $t_3$ ?!

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Federico Carta

1 Answer

+ 3 like - 0 dislike

With $X= X^at_a$, we have the following notation : $D_\mu X = [D_\mu,X] = \partial_\mu X -ig [A_\mu, X]$

The Bianchi identities are written :

$$D_\lambda F_{\mu\nu} + D_\nu F_{\lambda\mu} + D_\mu F_{\nu\lambda} = 0 \tag{1}$$ We may choose $\lambda, \mu, \nu = 0,1,2$, so we have :

$$D_0 F_{12} + D_2 F_{01} + D_1 F_{20} = 0 \tag{2}$$

From the definition of $j$, we have : $$j^0=F_{12}, j^1=F_{20}, j^2=F_{01}\tag{3}$$

From $(2)$ and $3$, we get :

$$D_\mu j^\mu = D_0j^0+D_1j^1 +D_2j^2 = 0\tag{4}$$

That is :

$$\partial_\mu j^\mu - ig[A_\mu, j^\mu]=0\tag{5}$$

Now, we may look at the $U(2)$ coordinates $(j^\mu)^a$ of $j^\mu$, we get :

$$\partial_\mu (j^\mu)^a +gf^{abc}(A_\mu)_b (j^\mu)_c=0\tag{6}$$ We know, that $f^{0bc}=0$ (because $[t_0,t_b]=0$ for $b=1,2,3$), so we get :

$$\partial_\mu (j^\mu)^0 =0\tag{7}$$

We see, that the current $(j^\mu)^0$ is conserved, and this corresponds to a conserved charge $Q^0 = \int d^2x (j^0)^0(x)$. The conserved $Q^0$ charge comes from the $U(1)$ generator $t_0$, which commutes with the $SU(2)$ generators $t_1,t_2,t_3$

The other currents $(j^\mu)^i$, $i=1,2,3$ are not conserved, because the $SU(2)$ generators $t_1,t_2,t_3$ do not commute with themselves, for instance, we have $\partial_\mu (j^\mu)^1 +g(A_\mu)_2 (j^\mu)_3=0$ (+ cyclic permutations).

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Trimok
answered Oct 28, 2013 by Trimok (950 points) [ no revision ]

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