Massless Thirring Model in 1+1 Dimensions

Question

In Coleman's paper, "Quantum sine-Gordon equation as the massive Thirring Model" (link to Phys Rev D article), he pointed out that the massless Thirring Model is exactly scale invariant. More over, the massless part scales as $$H_0(x) \rightarrow \lambda^2 H_0(\lambda x)$$ What does it mean here by scale invariance, and why does it imply that the Hamiltonian scales as this?

Note the Thirring Model has Lagrangian $$\cal{L} = i\bar{\psi}\partial_{\mu}\gamma^{\mu}\psi - \frac{g}{2} j_{\mu} j^{\mu}$$

Answer 1

Scale invariance refers to invariance under scaling the coordinates i.e., $x^\mu \rightarrow \lambda\ x^\mu$ ($\mu=0,1$ in this case). One needs to associate a (naive) scaling dimension to the fields -- this is done as follows. Suppose that $$ \psi(\lambda\ x) = \lambda^\Delta\ \psi(x)\ . $$ Plug this into the action and use the kinetic term to figure out a value for $\Delta$. For your Lagrangian, one obtains $\Delta=-1/2$. Since $j_\mu$ is a fermion bilinear, it transforms as $j_\mu(\lambda\ x)= \lambda^{-1} j_\mu(x)$. Thus, the term $j_\mu j^\mu$ goes as $\lambda^{-2}$ -- this exactly cancels the $\lambda^2$ coming from the integration measure $d^2x$ in the action. This implies that the action is scale-invariant classically. Equivalently, one says that the coupling constant $g$ has zero scaling dimension. Quantum corrections can change this conclusion and Coleman says that this is true quantum mechanically as well -- that is what is meant by the adjective "exact". Suppose you add a mass term -- that will break scale-invariance as you can/must check.