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Total energy vs. energy in a finite region

  1. Jul 27, 2012 #1
    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?
  2. jcsd
  3. Jul 27, 2012 #2


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    Science Advisor

    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 Jz commute. But an eigenvector of J2 is not necessarily an eigenvector of Jz. And an eigenvector of Jz is not necessarily an eigenvector of J2.
  4. Jul 27, 2012 #3
    Yep, you're right...
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