Is the non-perturbative approach to QFT a path integral approach? If so then how, given we don't have simple path integral formula for Dirac equation?

Here is my understanding of the scenario. Please correct me if i go wrong somewhere.

Initially, the perturbative approach to QED (Feynman diagrams​) was very successful. But the same approach to QCD proved cumbersome, and hence some sort of 'path integral' approach on vanishing lattice spacetime was developed. This approach was actually non-perturbative.

So, my understanding is that there are two approaches that everyone learns to QFT. The canonical (or perturbative or Feynman diagram approach) and the path integral (or lattice spacetime approach).

Now, having read Feynman and hibb's path integral book a little, i am under impression that the path integral approach works only for Schrodinger's equation, and becomes really difficult when we try to incorporate the relativistic Dirac equation. For the same reason, the same book treats QED as an approximation, using Schrodinger's equation and avoiding Dirac's.

So, i am surprised that in QCD, the path integral approach returns. How did they manage this? How did they overcome the problem of providing the path integral equations for Dirac's equation? Did they?

The Dirac equation is the classical equation of motion for the free Dirac Lagrangian $i \bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi} \psi$, so in a sense it's not quantum-mechanical at all (although of course the complex numbers and the spin degrees of freedom carried by the spinors that it describes are most naturally interpreted in a quantum context). The QED Feynman diagrams come in when you consider the spinor's coupling to an EM gauge field.
• > The Dirac equation is the classical equation of motion for the free Dirac Lagrangian ... Do you mean that $\psi$ is complex valued tuple and not an operator? That is not enough to call the equation classical, using the word classical in that way is quite confusing. – Ján Lalinský May 27 '17 at 12:48
• @JánLalinský No, I mean that the Dirac equation is the Euler-Lagrange equation for the free Dirac Lagrangian, so it describes the on-shell field configurations $\psi(x)$ that extremize the action. – tparker May 27 '17 at 17:38