Total energy vs. energy in a finite region

[giova7_89]

I was thinking about the following thing: we know that if the Lagrangian in field theory doesn't depend on the spacetime position, the Noether's theorem says that the stress-energy tensor is conserved, and that T^00 is the energy density at spacetime point x.

Then if one integrates this h(x) on the hypersurface x^0 = t, one gets the total energy at time t (and this total energy (operator) called H(t) doesn't depend on t, etc.). Then, I also think that it must be true that (because of relativistic causality):

[h(x),h(y)] = 0 if (x-y)^2 < 0

With that said, I thought that if I integrate h(x) on a finite region (which I will call R_t) which is contained in the hypersurface x^0 = t, the observable I will get will be "the energy in the region R_t": I will call this observable H(R_t).

Now I calculate the commutator of H(t) with H(R_t). This is equal to:

∫∫d^4xd^4y [h(x),h(y)]

where x belongs to the hypersurface x^0 = t and y belongs to R_t. Now this commutator is equal to 0 because for each couple of x and y one has (x-y)^2 < 0 because x^0 = y^0 = t(except when one has also x^i = y^i, but then one has [h(x),h(x)] which is 0, too).

Then I concluded that these two observables commute, and then the eigenvectors of the total energy must be eigenvectors of the energy contained in R_t.

I trusted this result being correct, and so I wanted to verify it in the free case, where one knows explicitly the eigenvectors of the total energy.

I "charged head on", and tried to apply directly the operator H(R_t) to the vacuum (the one which satisfies H(t)|0> = 0|0> = 0)...), to see which was its eigenvalue, but from my calculations it didn't even seem that the vacuum was an eigenvector of H(R_t).

So I'm beginning to doubt my reasoning about those two commuting observables... Can anyone give me some advice?

 

[Bill_K]

Then I concluded that these two observables commute, and then the eigenvectors of the total energy must be eigenvectors of the energy contained in R_t.

Let's see, how do you know this? That's not what the theorem says - it only says they have a set of eigenvectors in common.

Counterexample: J² and J_z commute. But an eigenvector of J² is not necessarily an eigenvector of J_z. And an eigenvector of J_z is not necessarily an eigenvector of J².

 

[giova7_89]

Yep, you're right...