Graduate Covariance of equal time commutation relations

[Betty]

I have a question regarding the covariance of the equal time commutation relations in relativistic quantum field theory. In the case of a scalar field one has that the commutator is (see Peskin, pag. 28 eq. (2.53) )
$ [\phi(0), \phi(y)] = D(-y) - D(y) $
is an invariant function, which is zero outside the light cone.

The commutator between the field and the conjugate momentum is

$[\phi(0), \pi(y)] = [\phi(0), \dot \phi(y)] = \partial_{y^0} [\phi(0), \phi(y)] , $

which also implies that $ [\phi(0), \pi(y)] = 0 $ outside the light cone. The equal time commutation rules which lead to second quantization read

$[\phi(0), \pi(y)] = \partial_{y^0} [\phi(0), \phi(y)] = i \delta^3(\vec{y})$ for $ y^0 = 0 $.

However this point seems to me a little bit odd, since this relation, being expressed as the derivative of an invariant quantity, is not invariant. This is also confirmed by the fact that the $\delta^3$ is also not invariant. Therefore, I wonder how we get a covariant theory starting from second quantization rules which are not invariant, or why this fact does not lead to any contradiction.

 

[PeroK]

You might want to fix your latex. Use two hashes or two dollar signs.

 

[Demystifier]

See http://de.arxiv.org/abs/hep-th/0202204 Eq. (13).

 

[Betty]

Thank you for your comments. I had a look to Eq. (13) in http://de.arxiv.org/abs/hep-th/0202204 and this (the result of (13) shouldn't be $$ i \chi(x)/\sqrt(g_{00})$$ ?) just confirms that the formulation of the equal time commutation relations depends on the choice of $$ \Sigma $$. Even if one expresses the relation in an invariant form as $$n^\mu \partial_\mu [\phi(0),\phi(x)] = i \frac{\delta(x)}{\sqrt{g_{00} g^3}}$$ the quantity on the r.h.s. is not an invariant unless $$ g_{0i}=0$$.

There is some other reference where this topic is treated more in detail?