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Showing that commutator is invariant under orthchronous LTs

  1. Feb 21, 2010 #1
    I'm having difficulty deciphering my notes which 'proove' that the commutor of two real free fields φ(x) and φ(y) (lets call it i∆) ie. i∆=[φ(x),φ(y)] are Lorentz invariant under an orthocronous Lorentz transformation. Not sure if it helps but φ(x)=∫d3k[α(k)e-ikx+(k)eikx].

    Now, apparently I have to 'observe' that:

    i∆=∫d4k.δ(k2-m2)ε(k0)e-ikx

    where ε(k0) = 1 if k0>0 and -1 if k0<0.

    Firstly, I can't see at all, how this expression comes about: why there is a delta function δ(k2-m2) (this looks like the mass shell condition, but why is it relavent here?)

    The "ε(k0) keeps positive k0 positive and negative k0 negative". I guess this makes sense because an orthocronous LT maps future directed vectors to future directed vector and past to past.

    But I still can't see why that expression proves that ∆(Lx)=∆(x).

    Any thought?
     
  2. jcsd
  3. Feb 21, 2010 #2
    i/delta is Feynman propagator.
    since field theory is relativistic hence all the expression must be Lorenz Invariant.
     
  4. Feb 21, 2010 #3
    Hi vertices,
    I'm only studying this stuff myself atm, but think I might be able to make a few helpful observations (pending corrections by my superiors :wink:)

    Firstly, the definition of the field you've written down doesn't explicitly use a lorentz invariant measure, as in general 3D subsets of 4D minkowski space aren't preserved under arbitrary lorentz transformations. (I'm guessing this will manifest itself in the transformation properties of your creation and annihilation operators- are they defined containing the reciprocal of the energy?) However, the 3d subspace of minkowski momentum space defined by the on-shell condition, coupled with the step function, is lorentz-invariant; it's not co-ordinate dependent, but is defined wrt the lorentz invariant inner product.

    Once you've observed that the measure is lorentz invariant, you just have to observe that the exponent is defined wrt a Minkowski inner product, so it's obviously lorentz invariant.
     
  5. Feb 21, 2010 #4
    Write the delta function as a sum of two delta functions (to get rid of (k0)2), integrate over k0, and the expression should reduce to the "familiar" propagator for scalar fields.
     
    Last edited: Feb 21, 2010
  6. Feb 21, 2010 #5
    Thanks for the replies - they've certainly been helpful.

    I have found a neat proof that shows that the integration over k-space is indeed equal to the expression d4k.δ(k2-m2)ε(k0), along the lines saaskis recommended..

    Can I ask a rather stupid question:

    What makes that expression a Lorentz Invariant measure?
     
  7. Feb 21, 2010 #6
    If you want to calculate [tex]\Delta (\Lambda x) [/tex] and show that it is equal to [tex]\Delta (x) [/tex], I think you'll have to make a change of variables in the integration over k-space. This change of variables is itself a Lorentz transformation, and since expressions like [tex]k^2=k_{\mu}k^{\mu}[/tex] are Lorentz-invariant, the claim follows. The orthochronocity (is this English?) is required as well, perhaps you can see where.
     
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