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Causality in QFT

  1. Jun 14, 2006 #1

    nrqed

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    I know that I have been asking that question before and that people are probably tired of it :redface:.

    I have done multiloop QFT calculations but when I get back to the fundamentals, something is really bugging me.


    There are several issues but let m start with a question that would seem very simple:

    Consider a scalar quntum field operator Phi(x). What is the meaning of the quantity [itex] c\equiv <0|\Phi(y)\Phi(x)|0>[/itex]?
    where x and y are spacetime points.

    Simple enough, no? It seems to me that this should be the amplitude corresponding to the creation of a particle at x and its annihilation at y. So the square of this (|c|^2) should be the probability that, given that a particle was created at x, it is annihilated at y.

    Peskin and Schroeder even describe this as the "amplitude for a particle to propagate from x to y" (page 27..their x and y are switched though).

    So we could imagine a reaction taking place at x, a particled described by x being created there, propagating to spacetime point y where there is a detector and being detected there.

    And yet, this quantity does not vanish for spacelike intervals!

    Then P&S sat, very mysteriously "to really discuss causality, we ask not whether particles can propagate over spacelike intervals, but whether a measurement performed at one point can affect a measurement at another point whose separation from the first is spacelike"

    I wonder why this is not bothering everyone!!
    If a particle can propagate faster than light, how would that not violate causality??

    (This is only one bothersome aspect of the above statement. There is also the vague use of "a measurement at one point "affecting" a measurement at another point". This *begs* for a discussion on EPR type of situation and why would the world "affecting" not apply here. But I don't want to get sidetrack on this in this thread, I want to focus on the QFT aspect)

    Another question concerns the "measurement" aspect. It is never clearly defined in the context of QFT,as far as I know (not in the way it is clearly defined in QM) but this is an issue that can wait for now. I would like first to understand the "faster than light propagation" issue and why it is deemed ok!
     
  2. jcsd
  3. Jun 14, 2006 #2
    Oh my! That is strange, indeed.

    To answer your last question, the definition of measurement in QFT comes straight from elementary quantum mechanics:

    The measurement of a physical system traslates roughly to an operator acting on a state of the system. So, if the state of the system is an eigenstate of the operator, then the correspnding eigenvalue will be the outcome of the measurement.
    Now, recall that two operators, A and B, are simultaneously diagonalizable if their commutator, [A, B], vanshes. The reason for this is that the vanishing commutator implies that the measurement of observable A does not interfere with (or in the context of QFT, affect) the measurement of observable B. So, in QFT we would hope that measurements separated by spacelike intervals do not affect one another. This saves us from having to dig out papers from the EPR paradox era. :)
     
    Last edited: Jun 14, 2006
  4. Jun 14, 2006 #3
    Hey, I'll give it a go again.

    Yeah.

    The way I see it is that propogation amplitudes correspond to the propogation of possibly virtual particles between two points and because they're virtual they may be off-shell and hence possibly acausal. As is shown in the Feynman functional approach, the amplitude consists of contributions from non-physical field configurations (the ones that don't obey the Euler-Lagrange equations of motion).

    Since this is a field theory I can construct my observables from the fields. These then correspond to measurements.

    I think "a measurement at one point 'affecting' a measurement at another point" translates into observables at some separation being (non-)commutative. In this instance, observables at space-like separation should commute. As was said above.
     
    Last edited: Jun 14, 2006
  5. Jun 14, 2006 #4

    CarlB

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    Maybe rewriting things in density matrix form will help explain:

    [itex] \langle0|\Phi(y)\Phi(x)|0\rangle[/itex]
    [itex]= tr( |\Phi(y)\rangle \langle \Phi(y)|\;\;
    |\Phi(x)\rangle \langle \Phi(x)|\;\; |0\rangle\langle 0|) \;\;/\;\;
    \langle \Phi(y)|\Phi(x)\rangle [/itex]

    The numerator is the trace of a product of three density matrices, the last of which is the density matrix of the vacuum, while the denominator is the complex number that solves the operator equation:

    [itex] |\Phi(y)\rangle\langle\Phi(y)|\;\; |\Phi(x)\rangle \langle\Phi(x)|
    = \kappa\;\;|\Phi(y)\rangle\langle\Phi(x)|[/itex]

    That the above equation can be solved with a complex number is due to the fact that the pure density matrices are "primitive idempotents". In other words, if they were matrices you could put them into diagonal form with all diagonal elements but one zero. Then the mixed matrix elements, such as [tex]|\Phi(y)\rangle\langle\Phi(x)|[/tex], are matrices with a single off diagonal element that is one, and the complex rule works accordingly. So from this point of view, the calculation is a coherency multiplied by the vacuum.

    The real mystery is what the vacuum is doing in a density matrix calculation. You know that you never had to include it when you were doing density matrix calculations as an undergraduate. The vacuum is something that QFT adds to QM without much explanation.

    The use of density matrices as a foundation for QM, when translated into QFT, means that one never writes a creation operator without some annihilation operator.

    This all gets back to the subject of does one truly require Hilbert space in the foundations of quantum mechanics. One can avoid it by using the density matrices as fundamental, but one of the results of this is that the vacuum works out to be an arbitrary state chosen in conjunction with the physical states. From there it's not surprising that one can make modifications of the vacuum that leave the physics unchanged (i.e. gauge transformations). But why would you?

    A recent paper that gets going in this direction is by Hiley (collaborator of D. Bohm):

    A Note on the Role of Idempotents in the Extended Heisenberg Algebra.
    B. J. Hiley.
    Motivated by a process explanation of quantum phenomena, we explore an algebraic representation-free approach to the quantum formalism. We show that the Schrödinger equation can be written in a representation-free way provided the appropriate algebra contains idempotents. To fit into the general scheme, the nilpotent Heisenberg algebra must be extended to include suitable idempotents. We show that the extended boson algebra, which includes the projector to the vacuum, can be used to generate these new idempotents using the Heisenberg and metaplectic groups. We briefly discuss some of the consequences of this approach.
    http://www.bbk.ac.uk/tpru/BasilHiley/IdpsinEEHA.pdf

    I've started a website with the subject of the density matrix as the foundation, but it hasn't got very far. Just two papers are linked in:
    http://www.DensityMatrix.com If anyone has a contribution, just type it into the wiki or leave a guest comment.

    On the question of the meaning of measurements in quantum mechanics, the best resource, I think, is Julian Schwinger's book "Quantum Kinematics and Dynamics". In it he makes measurements the basis of QM. It's very simple but beyond the scope of the discussion here. In it, Schwinger introduces the vacuum as a "fictitious" state that allows the conversion of measurements to a state vector. In other words, he explicitly adds in the vacuum to allow the conversion of his measurement algebra to work in a Hilbert space.

    In this sort of construction, there is some choice as to what field to use. There is an article where Schwinger's measurement algebra is converted to a quaternion based space instead of a Hilbert space:
    http://www.arxiv.org/abs/hep-th/9702080

    Carl
     
    Last edited: Jun 15, 2006
  6. Jun 14, 2006 #5

    nrqed

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    Thanks Perturbation! You have given me much insight before and I do appreciate you picking up the discussion (I was actually hoping you would still be around).

    Btw, I did think about the expression you had written (where Phi is written as an integral over the momentum of annihilation and creation operators corresponding to states of specific momenta) and you had said something about many particles being created. I think that actually it is still only one particle created and annihilated even though the momenta are summed over (so it allows for a single particle of arbitrary momentum).

    Going back to your comment here, I agree that this is the expression used to describe the propagation of virtual particles for internal lines in Feynman diagrams. But the obvious question is why can't it describe as well the propagation of a real particle? After all, a real particle can be created at x and be observed at y. Why can't this amplitude describe that? And the next obvious question is then, if this does not represent the ammplitude for the propagation of a real particle, what would? (my guess is that it is probably not defined at all since in order to create a particle at a specific point seems impossible in a relativistic field theory)

    Thanks for your input! I appreciate it!!

    Patrick

    Since
     
  7. Jun 14, 2006 #6

    nrqed

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    Thank for your reply. But then the obvious question is : is Phi(x) an observable? If yes, what are the eigenstates? What are the eigenvalues? I would think that Phi(x) is an observable, personally. But then it is very confusing to say that measurements must not affect one another for spacelike events and to use the fields as an example. What I *think* is happening is that they are saying :the real observables (like energy, momentum, angular momentum, etc) are built out of Phi's and derivatives of Phi's, so that if the Phi's commute for spacelike events, all observables will commute for space-like events. That is what I *think* is going on but this not at all clear from any book I have looked at!

    well, that's the problem. Because if we use "affect" in a broad sense, I would say that EPR type experiments *have* shown that measurements separated by spacelike intervals *do* "affect" one another in the sense that they are correlated. But I don't want to get sidetracked into that line of discussion in this thread.:wink:
     
  8. Jun 15, 2006 #7
    Yeah, it is only one particle, it was just a suggestion. By the time I realised what I'd said was wrong, it was too late to edit it.

    Well it can kind of, as in LSZ scattering theory, where one extracts expressions containing poles/branch-cuts near mass-shell. But in actual physics virtual/internal particles are immeasurable because they're involved in some manner of exchange, and one can never find a particle that is entirely on mass-shell, as there's always some uncertainty in its energy or momentum. Real particles are usually external lines, having definite four-momenta, described by the expression [itex]\langle 0|\phi (x)|\vec{p}\rangle[/itex], i.e. the field creates a state of definite momentum at x, which corresponds to the wave-function of a real quantum mechanical particle from non-relativistic quantum mechanics.

    The field operator isn't an observable, or at least any kind of useful physical observable. What would the field operator rerepsent as a physical quantity? All we're using it for is to create position and energy (or number) eigenstates with the creation/anihilation operators. I can perform any manner of scaling or in some cases gauge transformations without changing the physics. It's similar to the electromagnetic vector potential: it is somewhat arbitrary in that it is subject to gauge freedom/fixing and whatever but can represent actual physical fields, namely E and B.

    And yes, that's what is going on: one builds observables from the fields, as I've said. Remember that this is a field theory and we can therefore identify observables with the Noether currents of the theory. So for example one could taylor expand an observable in the field operator

    [tex]\mathcal{O}[\phi (x)]=\mathcal{O}(0)+\phi (x)\left.\frac{\delta\mathcal{O}}{\delta\phi(y)}\right|_{\phi =0}+\cdots[/tex]

    And this would obviously commute with another O at space-like separation if the fields do. [With fermion fields one has to be a bit more careful because of the anticommutation, but observables will be built from even numbers of fermion fields, so there's no problem there.]
     
    Last edited: Jun 15, 2006
  9. Jun 15, 2006 #8

    nrqed

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    Hi Perturbation. Thanks again for the interesting feedback

    Ok. Thanks for mentioning this. At least I know I am not losing my mind (yet).
    Yes, but then my question was why can <0|Phi(y)Phi(x)|0> NOT represent the propagation of a real particle? That's the key question. I have a guess but I have never seen this said clearly in print.

    But Phi(x) is supposed to create a particle at point x. So I am a bit confused about the meaning of that expression. It seems impossible to create a particle of definite momentum at a definite position! So I am not sure at all what this expression means. If it does mean the wavefunction at x of a particle of momentum p (which is not localized), then Phi(x) cannot represent the creation of a particle at x. :frown:

    Ok. Thank you, it is nice to see my view confirmed.
    I think that it is very misleading to see books saying "we will show that causality is preserved by showing that two measurements at spacelike intervals can't influence one another " and then proceed to work with the Phi with no further comment. I think it is really crucial to point out that they are not observables in themselves but can be used to build observables. I just find the presentation very confusing.
     
    Last edited: Jun 15, 2006
  10. Jun 15, 2006 #9

    vanesch

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    Just as a remark (even if you don't want to get sidetracked :smile: ): the locality of the *dynamics* of the fields (which is what you are describing here, namely the propagator from x to y) has nothing to do with any EPR correlations. Consider the propagator as the propagation of the different photons in an EPR experiment along their beams. It is not their propagation, which follows entirely a local dynamics, which is responsible for the EPR results, it is the initial entanglement of states.
    This is just as well valid in QFT as in ordinary QM ; only, in QFT one hardly uses entangled initial states (usually, one starts out with a product state - non-entangled - of different incoming particles, and one only calculates the transition amplitude (matrix element)) to the product state of different outgoing particles. One rarely if ever considers entangled initial states, and amplitudes which describe correlations between outgoing states.
    But it is entirely possible to do so because of the linearity of the calculation.
     
  11. Jun 16, 2006 #10

    Haelfix

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    Yea just to chime in with my opinion, similar to Vanesch.

    I think P&S are using weird terminology, specifically in their use of the word 'measurement' and 'particle'.

    As far as I can tell, no actual measurement has taken place, we haven't yet introduced a measuring apparatus. Thus the field operators are strictly obeying something like say the Klein Gordon equation, and are perfectly linear and obey special relativity in the sense that the fields at distant points outside the light cone cannot have causal impact. keep in mind I am not talking about particles, or virtual particles, but merely the actual *field* value at some point x and y.

    For instance, you compute the correlator of the two field operators and see that everything works out right.

    On the other hand, if indeed you make an actual *measurement* and entangle one of the states, then all of a sudden things are no longer in the KG poitn of view but are now subject to the collapse principle that is inherently nonlocal and nonlinear. This is what I call a quantum mechanics particle (it ticks in a detector) as opposed to a QFT particle (which is an excitation of a field and can be delocalized and wavy still)

    All bets are off once the measurement and measuring appartus come into play, EPR becomes important and I don't think there is a description with virtual particles at all.

    I hope that makes sense
     
  12. Jun 20, 2006 #11

    Hans de Vries

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    You often see this "leaking outside the lightcone" of the Klein Gordon
    propagator mentioned, not only in P&S but also in Zee for instance.
    Sometimes you also see Green functions which don't spread outside
    the light cone.

    I do wonder if this SR violating leaking isn't the result from mixing up
    past and future lightcones in the propagator.This is always a bit tricky.

    One might try for instance to understand the Klein Gordon propagator
    like this:

    [tex]
    \frac{1}{p^2-m^2}\ =\ \frac{1}{p^2} + \frac{m^2}{p^4} +
    \frac{m^4}{p^6} + \frac{m^6}{p^8} + ....
    [/tex]

    The first term at the right hand side is a massless propagator which
    is purely on the lichtcone only, like Lienard Wiechert. The propagation
    It is a convolution with the future lightcone.

    The second term is a convolution of a convolution with the light cone,
    that is, a fraction of what is propagated by the first term becomes a
    source itself for propagation, and so on. The third term takes the
    secondary propagation as a source, et-cetera.

    Now if one restricts this to future lightcones only there won't be any
    leaking outside the lightcone at all. SR violating leaking only occurs
    if the past light cones of the indirect propagators aren't excluded.


    Regards, Hans
     
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