To quantize a scalar field, we impose the equal time commutation relations $$ [\Phi(t,\mathbf{x}),\partial_t\Phi(t,\mathbf{x}')] = i\hbar\delta^{(3)}(\mathbf{x-x'}). $$ This can also be generalized to spacetimes which are not flat by choosing a spacelike slicing. Now as per equation 2.13 in this paper, the canonical commutation relations for a flat background in null coordinates $U = t-x,V=t+x$ read: $$ [\Phi(U_1,V,\zeta_1),\partial_U\Phi(U_2,V,\zeta_2)] = \frac{i\hbar}{2}\delta(U_1 - U_2)\delta^{(d-1)}(\mathbf{\zeta_1-\zeta_2})\tag{ 2.13} $$ Where $\zeta$ represents transverse coordinates. I understand this as follows: I write the lagrangian in null coordinates, $$ \mathcal{L} = 2\partial_U\Phi\partial_V\Phi + ... $$ If I take the $V$ coordinate as time, then the conjugate momentum should be $$ \Pi = \frac{\partial \mathcal{L}}{\partial(\partial_V\Phi)} = 2\partial_U\Phi $$ Which gives me the canonical commutation relation for equal $V$ as written above. However there are two things that I do not understand about this:
How is it justified to take $V$ as the time coordinate-- since it is null and $V = constt.$ is not a spacelike but null hypersurface.
Since it is the same theory we are quantizing in two different coordinate systems, the two commutation relations must be equivalent. Is there some way to explicitly see that?
