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I Particle energy in QFT..,

  1. Apr 12, 2017 #1
    Hi all,

    Question for which .I feel silly asking - but since I'm still learning:

    A particle state in QFT is considered to be an asymptotic state with a well defined energy. Now, if I take an ensemble of particles after a very large number of interactions (say, e.g., a macroscopic object like a person's body, and perhaps since the birth of the universe), can one say that the ensemble has a bounded energy?

    It would seem that the energy would have to be bounded at any time, even in an interacting picture (otherwise there would be no bound on spacetime curvature, etc., and yes, I realize I'm mixing in general relativity here). I think I'm basically asking about the spectrum of the Hamiltonian for the system - but since I'm still struggling to get through several texts, I'm unsure.

    Also, given the above, there would be a maximum energy that a body could emit over a given time, correct? (If the above is correct)

    Thanks!
     
  2. jcsd
  3. Apr 13, 2017 #2
    Ok, scratch the above as a very poorly formed question - I'll ask this instead:

    Still talking about particles in QFT (subtleties aside) - although in an interacting picture a particle may not have a well-defined energy, this does not mean it may have any energy. What is the correct method to compute the distribution of possible energies (after a larger number of interactions)? And what would the support of this function be?

    Thanks.
     
  4. Apr 14, 2017 #3
    Ok, one more time - hopefully this will be more clear, would really appreciate anyone's help to set me straight:

    A particle can be considered as a wave packet. If we consider the momentum wave function, for any real particle is it not the case that the wave function should exactly vanish outside of a specified, finite, interval? Otherwise, if the momentum wave function is non-zero over ±∞, then the particle could, with some non-zero probability, have any momentum and hence any energy (arbitrarily close to infinite). Take, e..g., the Gaussian wave packet, for example, which has non-zero tails.

    Note here that I'm not concerned with the expectation value of the energy, rather the probability that the energy of a particle is enormous (which depends on the particle of course - but should never go off to infinity). This is what I was trying to get at above.

    Apologies in advance for not being more precise and for missing the obvious, I'm sure - lots to learn...
     
  5. Apr 14, 2017 #4

    PeterDonis

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    No. But see below.

    There's a more general problem in QFT, which is that, if we just take the simple math at face value, even processes in which the actually measured energy is finite can contain virtual particle states with arbitrarily large energies. This is called "ultraviolet divergence". It is typically handled either by imposing an arbitrary cutoff or by renormalization.
     
  6. Apr 14, 2017 #5

    vanhees71

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    A cutoff is just a (not too clever) way to regularize undefined infinite expressions called "proper vertex functions" as one step to get to the physically well defined expressions and measurable quantities, which are the S-matrix elements of renormalized perturbation theory.
     
  7. Apr 14, 2017 #6
    Thanks PeterDonis and vanhees71.

    Ok, so let's consider the regularized and renormalized case - if particles going in are asymptotic states with well-defined energies, then after scattering, the particles coming out may not have well-defined energies immediately, but their total energy cannot exceed the input energy (conservation of energy)?

    More generally - consider, say, the wave packet for, e.g., an electron that happens to be part of a desk in my office - and now the momentum wave function for this particle... if I understand PeterDonis, then there is some finite, non-zero probability (because the wave function is non-zero everywhere), that the particle in my desk has more momentum, and hence more energy, that the most energetic cosmic ray ever to hit the Earth....

    The above is an extreme example, but illustrates the point. Again, speaking in the renormalized case, how can the above be possible?
     
  8. Apr 14, 2017 #7

    PeterDonis

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    Which is only an approximate model anyway, and you are simply pushing it into a regime where the approximation breaks down. Similar remarks apply to non-renormalized QFT; you can use it if you put in a cutoff, but that just amounts to admitting that you know you're just using an approximation and you need to avoid the regime where the approximation breaks down.

    The fact that total energy is conserved in particle physics processes, however, is not an approximation, so we don't expect it to break down in the regime where the approximations you are using are no longer valid.
     
  9. Apr 14, 2017 #8
    Interesting - I didn't realize the wave packet was just an approximate model. Specifically in the renormalized case (i.e., using renormalized QFT), does one confront the same issue?

    So, then, can I say that the momentum (and hence the energy) of the electron in my desk is bounded to a 'reasonable' value (or range, I should say, since the momentum is not well defined)? How does one determine what is 'reasonable', i.e., where the wave packet model breaks down?
     
    Last edited: Apr 14, 2017
  10. Apr 14, 2017 #9

    PeterDonis

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    It's approximate because it's non-relativistic. So it can't really be applied to trying to calculate the probability of an electron (for example) having an arbitrarily large momentum, because that would not be non-relativistic. One of the key things about any approximation is learning what not to use it for. :wink:

    Obviously, because electrons don't shoot out of your desk, they stay in it.

    There is no exact boundary of what is "reasonable". You have to look at each individual case and try to apply common sense.
     
  11. Apr 14, 2017 #10
    Ah, ok, thanks PeterDonis. Moving to the relativistic case (which would be relativistic QFT), then, it should be possible to assign an exact bound to the range of energies of the electron, no? Or the computation is done in such a way that this is not an issue...

    Let's just consider the 'electron in my desk' case - in relativistic QFT, how would the energy of the particle be determined?

    Thanks for the help by the way!
     
  12. Apr 14, 2017 #11

    PeterDonis

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    You don't need relativistic QFT to do that. You just need, as I said before, to know what questions not to ask your non-relativistic model. For most atoms the non-relativistic approximation gives pretty good predictions for the ionization energy of the atom (how much energy you have to add to an electron in the atom to take it out of the atom and make it a free electron); that energy is a good upper bound for the energies of the electrons in the atom. The fact that the non-relativistic model, mathematically, assigns non-zero values for the electron's wave function to points in momentum space that, classically speaking, are not consistent with that energy upper bound is one of the questions you should not ask the model, because that is getting into the range where the model's approximation breaks down.
     
  13. Apr 14, 2017 #12
    Ah, ok got it! Out of curiosity, if you did use relativistic QFT, could you get an exact bound (instead of the approximation)? (presumably yes? - and I should have thought about the relativistic case earlier and not conflated this with the QM solution)

    Last question, promise :smile:
     
  14. Apr 15, 2017 #13

    PeterDonis

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    What do you mean by "an exact bound"? Do you mean a more accurate value for the ionization energy of the atom? Yes, in principle including relativistic corrections should give you that.
     
  15. Apr 15, 2017 #14

    vanhees71

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    I think, here's a lot of confusion around.

    First of all it is important to realize that in relativistic QT a proper particle interpretation is only possible for asymptotic free states, and of course you have wave-packets for single-particle states as the only proper, i.e., normalizable, single-particle states (plane waves are not representing states but generalized eigenfunctions of the momentum operator).

    To understand this, read the section, where the S-matrix is introduced in the textbook by Peskin and Schroeder.

    "Virtual particles" is slang for "propagators". It's represented by internal line of Feynman diagrams, and strictly speaking Feynman diagrams do not depict real processes (although on a heuristic level they indeed do) in nature but are a clever notation of formulae to evaluate the S-matrix in perturbation theory.
     
  16. Apr 15, 2017 #15
    A wave packet is just a specific linear combination of plane wave states. As vanhees71 stated, plane wave states are useful because we can construct physical states out of them (by way of Fourier transform), not because they are themselves physically relevant. These (normalizable) physical states, obtained as a linear combination of basis plane wave states, is what we call a wave packet. There's nothing non-relativistic about the concept. You can build wave packets out of solutions of the Schrödinger equation, the Klein-Gordon equation, Dirac, Rarita-Schwinger, whatever. It's just a Fourier transform.

    What the OP was talking about is the fact that in, for example, a Gaussian wave packet, there is a nonzero amplitude for a measurement of momentum to give arbitrarily large values. Formal infinities in loop integrals are something altogether different.

    Renormalization is the art of teasing out measurable parameters from a theory written in a theorist's notebook in terms of some abstract object such as a Lagrangian. In principle it has nothing to do with the appearance of infinities: even wholly finite theories require renormalization. The procedure that parameterizes infinities so they are expressed in a form suitable for renormalization is called regularization. Cutoffs are one example of regularization, although one typically reserved for textbook toy models rather than realistic theories.
     
  17. Apr 15, 2017 #16

    vanhees71

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    BTW, there is a way to formulate perturbative QFT such as to give finite results from the very beginning, the Epstein-Glaswer approach of "causal perturbation theory", but it's more complicated then the standard way. The infinities are due to the naive multiplication of operator valued distributions like field operators; strictly speaking already the Lagrangian or Hamiltonian are not well-defined as we write them naively down.

    It leads to the same result as the hand-waving physicists' way but it's good to read about it to understand the troublesome infinities a bit better. A standard source is

    G. Scharff, Finite Quantum Electrodynamics, Springer
     
  18. Apr 15, 2017 #17
    Hi all, thanks for your responses... I'm afraid now I'm a bit more confused.
    This is in fact the question I'm wondering about - if you take into account relativistic effect, using e.g., relativistic QFT (although I realize there's no clear definition for a particle except as an asymptotic free state) do you still end up with wave packets where there is a nonzero amplitude for a measurement of momentum to give arbitrarily large values. This seems impossible to me.
    vanhees71, I'm not sure - above are you talking about finite results for momentum, so you get, e.g. proper bounded values for the momentum of particles (or field excitations ...) .

    Basically, my original post was concerning the possibility of arbitrarily large measured momentums - my example was an electron in the desk in my office having some non-zero probability of having more momentum than an extraordinarily energetic cosmic ray...
     
  19. Apr 15, 2017 #18

    vanhees71

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    Momenta are always finite. How could they be infinite in any experiment?
     
  20. Apr 15, 2017 #19
    Yes, being finite in any experiment makes sense, of course, but this still doesn't answer the question about the values of measured momenta being bounded in the relativistic case - or maybe I'm just not understanding your answer.

    E.g. (sorry for beating a dead horse here): I have a relativistic wave packet that describes an 'electron-like' field excitation (that is part of an everyday macroscopic object, so nothing exotic) - is the momentum wave function for the wave packet exactly zero outside of an envelope?

    If not, I may measure a huge momentum, granted with very small probability, but still ... and this is what doesn't make sense to me. PeterDonis answered this above and I got it, but there seems to be some disagreement.

    Thanks.
     
  21. Apr 15, 2017 #20

    PeterDonis

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    I think it is important here to carefully distinguish the model from reality. In the model, the wave packet is not strictly zero anywhere. But that is because it's a lot easier to work with such wave packets (e.g., Gaussians) than to try to work with the messy things you get when you insist on the amplitude being strictly zero outside of some finite interval.

    In reality, we don't measure energies to be unbounded in ordinary objects, so whatever thing in reality we are attempting to model with our wave packets presumably is zero outside of some finite interval (heuristically speaking). So the model is an approximation, as I said; we simply refrain from asking it what the probability is of measuring some arbitrarily large energy for an electron in an ordinary object, because we know the answer it gives us ("small but nonzero") doesn't match what we actually observe.
     
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