- #1

giant_bog

- 13

- 0

I'm aware of a vector calculus identity that makes [itex](\nabla\phi)^2 = 1/2 (\nabla^2[\phi^2]) - \phi \nabla^2 \phi[/itex].

That's almost what we have here, but the [itex]\frac{1}{2}(\nabla^2[\phi^2])[/itex] term is missing in the [itex]\pi(x)[/itex] commutator.

Did anybody see where it went?