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Causality in QFT

  1. Oct 9, 2009 #1
    The Feynman propagator in QFT is not zero for space-like separation, but we say this does not mean that causality is violated, we should check the commutator of field operators instead, and the commutators vanish for space-like separation.

    My question is: why do we use commutators to check causality instead of propagators? Is there any examples that commutators are related to observation? And what is the 'causality criteria' for causality in classical relativistic field theory?
     
  2. jcsd
  3. Oct 9, 2009 #2
    In quantum theories, the relevant criterion here is that a commuting set of operators form a set of "observable" properties. E.g. since the equal-time commutator [tex][\phi(x,t_{0}),\phi(y,t_{0})]=0[/tex] fields at different spatial points, but the same time, can be used simultaneously as physical properties of the field. More generally, if the fields commute for any 2 space-like separated points of space-time commute, then fields at all causally-separated places can be used in a single framework for computational purposes. That is, the fields at space-like separations don't fundamentally care about each other; the measurement of one is not intrinsically dependent upon the other. Of course, there can be *correlations* between such measurements (encoded in propagators).

    If two operators A and B do not commute, we can still consider cases in which a system has property (corresponding to) A at one time, and property B at another time...this is consistent. But for a relativistic theory in which space-time is a single space we generally want to use operators that commute on the whole space-time to be able to work with a single framework of physical properties on the whole space-time (rather than different sets of commuting operators for each time slice).
     
  4. Oct 9, 2009 #3
    The case of non-commuting operators raises a curious paradox. In classical QM, if operators A and B don't commute, you can't find a set of shared eigenstates for them. So, if you measure A and get a certain value, and then you measure B, the system is no longer in the eigenstate of A.

    Fast forward to QFT. We have two non-commuting operators, one in the causal future of the other. If you measure A and then you measure B, by the same logic, the second measurement kicks the system out of A's eigenstate, making the result of the first measurement invalid. In other words, your second measurement changes your past!
     
  5. Oct 10, 2009 #4
    There is no paradox at all. If two operators as you say are localized at timelike separated points (A is to the past of B lets say) then A*B term in the commutator cannot be interpreted "measuring B and THEN measuring A" for obvious reasons that once can never "measure" B (which is located to the future of A) before A. The state on which the measurement is being done also evolves and moves out of the whatever operators eigenstate it was anyway (Schroedinger picture).
     
  6. Oct 10, 2009 #5

    Hans de Vries

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    Science Advisor



    Lots more can be said about the propagators:


    1)======================================================

    The part of the Greens function outside the light-cone diminishes over time.
    For an electron it's say 10-13m at t=0 but less then 10-20m after 1 microsecond

    2)======================================================

    There's confusion between the propagator from source and the self-propagator.
    Books discuss the propagator from source as if it is the self-propagator, like:
    An electron at x has and amplitude of D(y-x) to propagate to y. This should be:
    A source at x has an amplitude of D(y-x) to produce an electron at y.

    3)======================================================

    The Klein Gordon self propagating Greens function is:

    [tex]
    {\cal D}^{\cal S}(x^\mu) = \left \{ \qquad \begin{matrix}
    \frac{1}{2 \pi^2}~\frac{rt}{s}~ \frac{\partial}{\partial s}\delta(s^2) - \frac{m^2}{4 \pi}~\frac{rt}{s^2}~ H_2^{(1)}(ms) & \textrm{ if }\, s^2 \geq 0 \\ & \\
    -\frac{m^2}{ 2 \pi^2 }~\frac{rt}{s^2} K_2(ims) & \textrm{if }\, s^2 < 0
    \end{matrix} \right.
    [/tex]

    Which is obtained by taking the 3d Fourier transform of a point at the origin
    in a single time slice and then using the on the shell energy for the time -
    evolution of all the individual 3-momentum components.

    This Greens function is explicitly zero outside the lightcone at t=0. (!!)

    There is still a part of the Greens function which is non-zero outside the
    light cone for t>0 (which is diminishing over time as well). Still, there is
    no faster then light propagation since the wave-function which is a point
    at t=0 zero is non-zero at t<0 and the part ouside the lightcone can be
    considered as being a time-like propagation from the wave-function
    immediately prior to t=0.


    Regards, Hans
     
  7. Oct 12, 2009 #6
    JavierR: I think you are right. That explains why causality requires commutator=0 for space-like separations. But I am still trying to figure out a rigid proof of the inverse, ie., commutator=0 for space-like separattions => causality is respected.

    Hans: Although I did not understand how you derived the self propagating Green's function, you reminded me that the "source" of a Feynman propagator is in fact not a delta function. In non-relativistic theory, Green's functions are correlation functions between point sources (delta functions in spacetime), yet in relativistic theory, we renormalize the source so that they contain spacially separated ingredients, so no wonder relativistic propagator correlates space-like separations.
     
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