Splitting Scalar into Holomorphic and Anti-Holomorphic Parts

Question

I am reading Tong’s string theory lecture notes. On page 78, he splits the 2d free scalar into left- and right-moving parts, seemingly using the classical equation of motion as justification.

Why is he able to do this? I thought that the classical equations of motion only hold on the mass shell. Certainly it’s true that not every free scalar satisfies the Klein-Gordon equation. Is there an implicit assumption of being on-shell in his argument? If this is the case, are there any resources for off-shell CFT that can derive the same results of the spectrum, stress tensor, etc. without this assumption?

If this ability to split into left- and right-movers is manifest, why would that be the case?

Answer 1

In the Heisenberg picture the field operators satisfy the equation of motion, and that is how the usual decomposition is made.

In a Euclidean path integral formulation, one can follow Mandelstam and define an off shell decomposition $$ \phi_R(x,y)= \frac 12 \phi(x,y )+ \frac i 2 \int_{-\infty}^x \partial_y \phi(x,y) dx, $$ which makes the no-locality of the decomposotion manifest. (In real-time there would be no $i$). This follows from writing $$ \phi(z,\bar z)= \int_{-\infty}^z \partial_z \phi(z,\bar z) dz+ \int_{-\infty}^\bar z \partial_{\bar z} \phi(z,\bar z) d\bar z $$ and recalling that $$ \partial_z = \frac 12(\partial_x-i\partial_y), \quad etc. $$

Answer 2

1. Tong is here using the operator formalism, which is manifestly on-shell. In the Heisenberg picture, the Heisenberg EOMs are operator identities, cf. answer by mike stone.

2. In contrast, the path integral formalism is off-shell.

See also e.g. this Phys.SE post.