i'm trying to understand antiscreening in QCD. I'm following Peskin & Schroeder's Introduction to QFT and i'm having trouble to follow their argument. In section 16.17, p542-543, we get an equation that says $$ D_i E^{ai} = g \rho^{a},$$ which I'm fine with. then they say that, using $ (D_\mu \phi)^a = \partial _\mu \phi^a + g f^{abc} A_\mu^b \phi^c$ and $f^{abc} = \epsilon^{abc}$ in SU(2), we get $$ \partial _i E^{ai} = g \delta^{(3)}(\vec{x})\delta^{a1} + g \epsilon^{abc} A^{bi}E^{ci}.$$ My first question (a) is, I don't understand why there's no minus sign in front of the second right hand term ? Afterwards, they say that the first RHS term gives a $1/r^2$ field of type $a=1$ radiating from $\vec{x} = 0$.

Then, somewhere in space, this field will cross paths with a fluctuation of the vacuum in the form of a bit of vector potential $A^{ai}$. "For definiteness, let's assume that this fluctuation has $a=2$.

Why can they assume this ? Shouldn't they look at the case where a=3 too ? Then they say the second term on the RHS, with a = 2 for the vector potential and c = 1 from the electric field, will make a field sink where they cross. The new electric field from this sink will be parallel or antiparallel to A depending on which side we are ; there is a source with a=1 closer to the origin and a sink with a=1 farther away.

So my other questions are :

b) how do they choose the values of $a,b,c$ ?

c) how do they know there will be a souce on the side of the origin and a sink on the other side ? figure 16.11 of P&S.

  • $\begingroup$ It is analyse of equations to understand physical meaning.. $\endgroup$
    – Nikita
    Jan 1, 2020 at 17:09
  • $\begingroup$ thanks i gathered as much. what i don't understand is how they do it ... $\endgroup$ Jan 1, 2020 at 17:16

1 Answer 1


a) In my notes on P&S I have a minus sign as well. Maybe this is just a typo?

b) and c) P&S just take these values of the indices as an example. They assume a point charge at the origin of the physical space in the group direction $a=1$. They then consider the $A^{2i}$ gauge field in the "bubbling'' vacuum. That gauge field then interacts with $E^{1i}$ according to the second term in (16.139) and the antisymmetric symbol $\epsilon^{c12}$ implies a source field in the $E^{3i}$ direction. That electric field will also interact with the $A^{2i}$ field, via $\epsilon^{d23}$ in the $E^{1i}$ direction.

Let us consider the signs - I assume there is a minus sign in (16.139) - and work iteratively. The vacuum disturbance $A^{2i}$ implies a contribution to Gauss law in the group direction 3: \begin{equation} \partial_i E^{3i} = +\cdots - g \epsilon^{321} A^{2i}E^{1i} = +\cdots - g (-1) A^{2i}E^{1i} = +\cdots + g A^{2i}E^{1i} \end{equation} That that disturbance has an opposite sign of the Gauss Law. This $E^{3i}$ then gives a contribution to Gauss Law in direction 1: \begin{equation} \partial_i E^{1i} = +\cdots - g \epsilon^{123} A^{2i}(-E^{3i}) = +\cdots + g A^{2i}E^{3i} \end{equation} And so the non-Abelian character of the theory results in an additional electric field contribution on top of the one that comes from the abelian part of the vacuum fluctuations. P&S say that if you work out the details of this - referring to the original literature - then you find that there this vacuum fluctuation results in a strengthening of the field, rather than a weakening. One thus need to work out the dampening effect of the abelian screening and the strengthening effect of the non-Abelian screening.

I must admit that I don't find P&S very clear about the details although the general gist is. Maybe you should revert to the original article?

  • $\begingroup$ Thanks for taking the time to answer. I understand how ti works now. Just a remar k actually it works too if it's really a + sign since we use the equation twice... We just get a sink for $E^3$ in place of a a source. I'll try to check the original article to find out $\endgroup$ Jan 2, 2020 at 16:09
  • $\begingroup$ yes - it works with both signs indeed. $\endgroup$ Jan 2, 2020 at 16:12

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