Is the confinement mechanism understood in 1+1 QCD ('t Hooft's model)?

Question

I was following the coleman lectures on "Aspects of symmetry" particularly the chapter about the 't Hooft's model. Then I have wandered into older papers like the 't Hooft's papers and many others. And concerning the reason why there are no free quarks in this model it seems to me that their reasoning is the following:

Calculate the dressed propagator and you will get something like :

$$\frac{ip_-}{2p_+p_- -M^2-\frac{g^2|p_-|}{\lambda\pi}+i\varepsilon}$$

This is a propagator depending on the cut-off $\lambda$.Then, because of the infra-red divergence, we have to restore gauge invariance taking the $\lambda\to 0$ limit. The pole of this propagator is shifted towards $p_-\to \infty$. We conclude that there is no physical single quark state.

But, Einhorn claims (in PhysRevD.14.3451) that the dependence on $\lambda$ has nothing whatever to do with the confinement mechanism. In order to prove that the 't Hooft's argument is wrong, Einhorn switch off the coulomb potential but retain a constant gauge-dependent term. He finds that the interaction between $\bar{q}q$ pairs cancels the term in the self energy, so that free quarks are produced. So the confinement must be obtained by other means.

By the way, It seems that Coleman is using the principal value method and not the original 't Hooft's regularization. So it is not obvious for me to realize whether Coleman agrees or not.

Is the underlying reason of confinememnt (in 't Hooft's model) clear currently? Are Einhorn's arguments wrong?

Answer 1

I think that in the same lectures you mention, an elegant answer to your question is contained. Namely, in lightcone gauge, the only non-null component of the gauge field $A_0$ is non-dynamical as, on shell, $\partial^2_1 A_0 = - g \overline{\Psi}^A \gamma^0 \Psi^A =- g (\overline{\Psi}^\dagger)^A\Psi^A = -g j^0 =0$. Now, the Green function of this problem is: \begin{equation} \partial^2_1 G(x) = -g \delta^{(2)}(x) \Longrightarrow \tilde{G}(p) = -\frac{g}{p^2_1} \Longrightarrow G_0(x) = -\frac g 2 |x_1| \delta(x_0) \end{equation} The solutions of the homogeneous are obvious. This means that the potential is: \begin{equation} A_0 = -\frac g 2 \int dx_1 |x_1-y_1|\,j^0(x_0,y_1) + B x^1 + C \end{equation} $B$ and $C$ may be removed by a linear shift once we assume that there is no background field. By putting this in the Lagrangian, the term $A_0\overline {\Psi}^A\gamma_0\Psi^A$ gives : \begin{equation} \mathcal{L} \supset \mathcal \int dx_1\,dx_2\,j^0(x_0,x_1)|x_1-y_1|\,j^0(x_0,y_1) \end{equation} which is exactly a linear confining potential between two quarks.

I like this derivation, as it makes the linear potential evident.