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I Why no position operator for photon?

  1. Mar 8, 2017 #1

    referframe

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    Apparently, in QM, the photon does not have a position operator. Why is this so?

    As usual, thanks in advance.
     
  2. jcsd
  3. Mar 8, 2017 #2
    It can be shown that in the context of relativistic quantum theory the position operator leads to violations of causality. The demonstration can be found, for example, in the first chapter of Ticciati's "Quantum Field Theory for Mathematicians". This is why QFT dispenses with the position operator altogether, relegating position to the role of a parameter.
     
  4. Mar 9, 2017 #3

    vanhees71

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  5. Mar 9, 2017 #4

    referframe

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    So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?
     
  6. Mar 9, 2017 #5
    QFT is what's needed to avoid the causality violations that result from applying QM and SR in a simple-minded way to particles. But yes, they're present for massive particles as well. I had forgotten that Markus Luty, a physicist at UC Davis, had provided a demonstration of causality violation in a video lecture of his. It start at 1:40 and ends around 16:18
     
    Last edited: Mar 9, 2017
  7. Mar 9, 2017 #6

    vanhees71

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    There are no violations of causality in QFT, because QFT is precisely constructed such that it is causal (see e.g., Weinberg Quantum Theory of Fields, vol. 1).
     
  8. Mar 9, 2017 #7
    I once complained to some physicists that isn't it amazing that there is no spatial representation for the relativistic quantum mechanical states. They responded to me that if you define the excitations of the electromagnetic field the usual way with lots of operators, then it will be equivalent to the spatial wave functions of photons obeying the Maxwell's equations. To me it sounds like that if there exists a spatial wave function for photon, then there also exists a position operator, so is this answer that I got back then contradicting the answer you guys are putting forward in this thread?
     
  9. Mar 9, 2017 #8
    If you try to interpret the ##A_{\mu }## of Maxwell's equations as the wave function of a single photon, then you'll run into problems (negative energies, causality violation, etc.). But the equation can be salvaged by instead viewing ##A_{\mu }## as a quantum field that creates/anninilates photons at different points of space and time. That's the basic trick of QFT: the relativistic wave equations don't describe the time evolution of wave functions, but of quantum fields.
     
    Last edited: Mar 9, 2017
  10. Mar 9, 2017 #9

    kith

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    For the phrasing of your question see vanhees71's reply. If your question is whether a position operator can be defined for massive relativistic particles, the answer is yes, see https://en.wikipedia.org/wiki/Newton–Wigner_localization.
     
  11. Mar 11, 2017 #10
    On the first page of the first chapter the author says

    That almost sounds like that humans' decision about the meaning of [itex]|x\rangle[/itex] would affect the behavior of Schrödinger's equation. Like if humans decide to give [itex]|x\rangle[/itex] a meaning, then the solutions of [itex]i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle[/itex] will become such that they violate causality. If humans instead decide to leave the notation [itex]|x\rangle[/itex] without any meaning, then the solutions of [itex]i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle[/itex] will remain such that causality doesn't get violated. We of course don't wont the solutions of [itex]i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle[/itex] to become such that they would start violating causality, so it is very important that we don't assign any meaning to the notation [itex]|x\rangle[/itex].
     
  12. Mar 11, 2017 #11

    PeterDonis

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    All it's saying is that including position eigenstates in the model means the model predicts violation of causality. You can't change reality by changing your model; all you can do is make the model match or not match reality.
     
  13. Mar 11, 2017 #12
    Even the model's predictions should not depend on a choice of assigning a meaning to [itex]|x\rangle[/itex], because the model's prediction should come from the time evolution giving Schrödinger's equation.

    What the author is really saying is that the behavior of relativistic quantum systems is so strange that it looks paradoxical (which is typical with special relativity, of course), and then the author is trying to convince the reader that the paradox can be nicely dealt with by refusing to think about it (which is a very typical way of "solving" paradoxes, when you don't know the real solutions).

    "Never use [itex]|x\rangle[/itex], and then you are not going to see the difficult paradox (related to special relativity)".
     
  14. Mar 11, 2017 #13
    To expand on PeterDonis’ point, when developing a model of a system one has to decide which Hermitian operators will be elevated to the status of observables. The usual way of approaching this task is to identify the symmetries that the system obeys, which gives one the system’s symmetry group(s). The generators of the symmetry group(s) are then identified as observables of the system. This approach leads to many of the usual suspects: energy, momentum, angular momentum, charge, etc. In non-relativistic quantum mechanics, in which the relevant symmetries are encapsulated by the Galilean group, if one elevates the position operator to the status of an observable and then calculates its commutation relations with the generators of the Galilean group, one finds that the generators can be expressed completely in terms of the position operator and a naturally defined velocity operator (this is in fact how one derives the position representation of the Schrödinger equation, as is shown, for example, in the third chapter of Ballentine’s “Quantum Mechanics: A Modern Development”). However, if one attempts something similar with the relativistic Poincaré group, one ends up with some funky commutation relations that can’t be easily made sense of, as is explained in the Wikipedia article linked to by kith.
     
  15. Mar 11, 2017 #14

    PeterDonis

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    First, Schrodinger's equation is non-relativistic, so if we're talking about relativistic quantum theory, it's not the right equation.

    Second, "time evolution" is problematic in any relativistic (more precisely, Lorentz invariant) theory, because there is no preferred frame and therefore no preferred "time". So that's not really a good way of thinking about the model's predictions.

    That said, the model's predictions certainly do depend on what states you include in the state space, because the states in the state space are the possible states of the system (in the model). So including position eigenstates in the state space is saying they are possible states of the system, and what the author appears to be saying (at least based on what you quoted--I do not have the book itself) is that if position eigenstates are possible states of the system, you can get violations of causality.
     
  16. Mar 11, 2017 #15
    When I wrote down Schrödinger's equation, by [itex]|\psi(t)\rangle[/itex] I meant something very general, like a wave functional, that contains the information about all the excitation states of the quantum fields, and so on. QFT texts usually speak about things like Heisenberg picture or interaction picture, but you should keep in mind that they are defined in such way that ultimately they should be equivalent with some Schrödinger picture. Perhaps if we replace the differential equation by

    [tex]
    |\psi(t)\rangle = e^{-\frac{it}{\hbar} H}|\psi(0)\rangle
    [/tex]

    it will look more familiar for the QFT contex?

    I am confident that if a theory is really good, it will only look like that you could try to generate many time evolutions by using different frames of references, but eventually a closer look will reveal that the different time evolutions are actually equivalent. That is how Lorentz invariance often works out.

    There are two possibilities concerning the time evolution giving operator [itex]H[/itex]. Either it gives a time evolution that is consistent with the special relativity, or then it does not give that. I am not convinced that a decision to use or to avoid using [itex]|x\rangle[/itex] would have such effect on the behavior of [itex]H[/itex], because it does not look like that [itex]H[/itex] could depend on such thing. The decision to use or to avoid using [itex]|x\rangle[/itex] will obviously affect how much we can learn about the behavior of [itex]H[/itex] though, so the decision to avoid [itex]|x\rangle[/itex] looks like a decision to deliberately not study the behavior of [itex]H[/itex].
     
  17. Mar 12, 2017 #16

    PeterDonis

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    That doesn't change the fact that Schrodinger's Equation is not relativistically correct. You can't make it relativistically correct by redefining what you mean by ##\psi##.
    No. Have you read any textbooks on QFT? I strongly suspect that you haven't, which means you don't have the requisite background for this discussion. In QFT, there are no wave functions ##\psi(t)##. There are quantum fields, which are operators defined at particular points in spacetime.
     
  18. Mar 13, 2017 #17

    referframe

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    I've always been a little confused regarding this line of reasoning. It's true, especially in SR and GR, that we require the laws of physics to be independent of reference frame. But, at some point, especially when we do an experiment, we do choose a specific reference frame - the reference frame of the lab. In that experiment, we can measure energy and momentum separately, we are not forced to measure just four-momentum. So, can we not also consider the time evolution (within Minkowski spacetime), of the system within that frame?
     
  19. Mar 13, 2017 #18

    PeterDonis

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    We can do this, yes, and we usually do, for reasons of convenience. But that doesn't mean we are forced to by the laws of physics. We're not.

    Yes. But you will find that anything you want to express as a physical law or an actual observable, can be expressed in any frame, and cannot depend on the "time evolution" in your particular chosen frame.

    Another way of putting this is: "time evolution in a particular frame" must be taken to mean looking at the coordinate times of various events, as assigned in that frame. But the only kind of "time" that has any invariant physical meaning is proper time along a particular worldline. And for a general physical system, there will be no coordinate chart in which coordinate time is the same as proper time along all worldlines belonging to the system: the best you can do, in general, is to construct a chart in which coordinate time equals proper time along one worldline. (The only exception I'm aware of is a system in which all particles move inertially and are always at rest relative to each other, in flat spacetime--in that case you can just use inertial coordinates in which the particles are all at rest.)
     
  20. Mar 13, 2017 #19
    Er, you make it relativistically correct by redefining what you mean by the Hamiltonian Evolution operator - the Schrodinger Equation is a postulate of quantum mechanics for a reason.

    Just because people haven't worked out the correct evolution operator doesn't mean it's not possible to get relativistically consistent evolution operators in the Schrodinger Equation.

    The actual, correct answer to the op, is that we haven't worked out a Hamiltonian operator for QED yet, so we can't have a totally correct position operator for a photon, there are suggested approximations in the literature.
     
  21. Mar 13, 2017 #20

    PeterDonis

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    Not of quantum field theory. We are talking about quantum field theory, right?

    In other words, you are not talking about actual mainstream physics but about your personal theory. Please review the PF rules on personal theories.

    The same comment applies to the rest of your post.
     
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