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Particles and Fields – a neverending story

  1. Jun 29, 2008 #1
    It is often said that the field operator [tex]\hat{\phi}(x)[/tex] of some free field theory, e.g. Klein-Gordon or Dirac, acting on the vacuum state, creates "one particle localized at x" or "in a position eigenstate at x".

    For example, from the Fourier expansion of the real Klein-Gordon field operator, we get

    [tex]\hat{\phi}(x) |0\rangle = \int \mathrm{d}^3 \tilde{p} \: f_p^*(x) a^\dagger(p) |0\rangle \ , [/tex]

    where [tex]\mathrm{d}^3 \tilde{p} \equiv \mathrm{d}^3 p/2p^0[/tex] is the Lorentz covariant measure and [tex]f_p(x)[/tex] are the plane wave solutions to the Klein-Gordon equation.

    Now, this state could be regarded as a localized one-particle state if [tex]f_p^*(x)[/tex] is the ordinary QM position space wave-function of the particle in momentum eigenstate [tex]|\mathbf{p}\rangle[/tex], namely [tex]f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle[/tex]. Writing [tex]a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle[/tex], the above equation would turn into

    [tex]\hat{\phi}(x) |0\rangle = \int \mathrm{d}^3 \tilde{p} \: |\mathbf{p} \rangle \langle \mathbf{p} |\mathbf{x} \rangle \ ,[/tex]

    which would somehow be related to [tex]|\mathbf{x}\rangle[/tex] since the first part resembles an identity operator. But two problems occur:

    1. What is [tex]|\mathbf{x}\rangle[/tex]? While I can define the state of "one particle with momentum p" to be [tex]a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle[/tex], I have no idea how a state like [tex]|\mathbf{x}\rangle[/tex] can canonically arise in a QFT. My only idea is to promote the equation

    [tex]f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle[/tex]

    to a defining axiom for [tex]|\mathbf{x}\rangle[/tex], that is, postulating that the free field solutions to the real Klein Gordon equation can be interpreted as one-particle wave functions in the ordinary QM sense. But this would mean you have to add a new axiom relating QM and QFT?!

    2. I have some doubts that I can regard [tex]\int \mathrm{d}^3 \tilde{p} \: |\mathbf{p} \rangle \langle \mathbf{p} |[/tex] as an identity operator because of the covariant measure.

    Can anyone help me? Thanks!!
    Last edited: Jun 29, 2008
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  3. Jun 30, 2008 #2


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    The problem with the states
    [tex]|\mathbf{x}\rangle = \hat{\phi}(\mathbf{x},0)|0\rangle[/tex]
    is that they are not orthogonal.
    [tex]\langle\mathbf{x}'|\mathbf{x}\rangle \neq \delta^3(\mathbf{x}-\mathbf{x}')[/tex]
    Note, however, that they become orthogonal in the non-relativistic limit.
  4. Jun 30, 2008 #3
    Thanks for the reply! I a book about group theory, the basic relation

    [tex]\psi(x) = \langle 0 | \hat{\psi}(x)| \mathbf{p}\rangle[/tex]

    is used to obtain the transformation behavior of the field operator from the transformation behavior of the single-particle wave function. In a way, I use the representation theory of the Poincaré group to construct QM single-particle basis states and relate their transformation behavior to field operators of QFT by the above equation. Is this right?

    But I don't get the connection; I think this is a subtle point: from the viewpoint of second quantization, everything is clear because one explicitly constructs the possibility to create and annihilate particles, so the above equation just follows mathematically.

    But from the viewpoint of field quantization, I never introduced single-particle wave functions, I just applied the QM axioms to fields so that there is no obvious interpretation of the function which is represented by the above matrix element.
    Does this make any sense?
    Last edited: Jun 30, 2008
  5. Jun 30, 2008 #4


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    There is no generally accepted answer to your question.
    Nevertheless, a discussion that might be helpfull is provided in Secs. 8 and 9 of
    http://xxx.lanl.gov/abs/quant-ph/0609163 [Found. Phys. 37 (2007) 1563]
  6. Jun 30, 2008 #5
    Wow, this is odd! I have found this nowhere in standard textbooks on that topic...instead I spent hours blaming myself for not getting this allegedly obvious relation...

    Thanks very much!
  7. Jun 30, 2008 #6
    There is no position operator and no states in position space in QFT!

    Read the first 7 pages of these https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1784740" [Broken].
    Last edited by a moderator: May 3, 2017
  8. Jul 1, 2008 #7


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    It seems that you gave a wrong link. Can you correct it?
    Last edited by a moderator: May 3, 2017
  9. Jul 1, 2008 #8
    http://www2.physics.utoronto.ca/~luke/PHY2403/References_files/lecturenotes.pdf [Broken]

    Otherwise, Srednicki writes in his book on page 10, that keeping X and promoting t to an operator can also be done to get a relativistic particle theory.

    Which leaves me puzzled.
    Last edited by a moderator: May 3, 2017
  10. Jul 1, 2008 #9


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    It contains some frequent wrong claims.

    For example, it says that a particle cannot be localized in a region smaller than its Compton length. However, if it was true, it would imply that a massless photon could not be localized in any region smaller than its Compton length. On the other hand, the position of the photon can be measured by certain accuracy.

    Next, it says that in relativistic QM there is no such thing as a 2, 1, or 0 particle state. But for free relativistic fields, such things certainly do exist. Moreover, a 1 and 0 particle states are stable even for interacting particles.
    Last edited by a moderator: May 3, 2017
  11. Jul 1, 2008 #10
    Position of a photon? But you know that a photon has no rest frame...

    Right. I think every QFT text starts with this observation. What is wrong with it?
  12. Jul 1, 2008 #11


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    1. So what? Take for example a classical photon. It has no rest frame too. Yet, at any time it has a definite position. The position, of course, changes with time, but it does not mean that the position does not exist.

    2. For example, what is the state |0>, if not the state with exactly zero number of particles? And no, not every QFT text starts with this observation.
    Last edited: Jul 1, 2008
  13. Jul 1, 2008 #12
    This is a point related to my question. What does particle localization mean in QFT? I am able to superpose some one-particle momentum eigenstates. But to determine the position of the "quantum object" thus obtained I need a position operator which does not exist. I cannot see more than a formal analogy to second quantization where it is just a mathematical implication that the fourier transform of the superposition weight function is the one-particle wave function where an ordinary QM position operator can act on.
  14. Jul 2, 2008 #13


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    It depends first on how you define "particle". I.e., which group you start with for which
    you find the unitary irreducible representations. The Poincare algebra doesn't come
    with a position generator, so some people advocate constructing one (Newton-Wigner)
    from the enveloping ring. Other people are unimpressed with such "schtick". :-)

    The x parameters in orthodox QFT (and Minkowski space in general) just serve as
    a representation space for the Poincare group. The vacuum state has deterministic
    energy-momentum of 0, and therefore cannot be a localised state in the usual sense.
    Some people have suggested that this fact is at the core of the Reeh-Schlieder
    paradox (which "proves" that fields over "there" can be recovered entirely from
    fields "here" -- where "here" and "there" mean spacelike-separated regions).

    Here's some fairly recent papers that debate these still-controversial points.

    1) Fleming, "Reeh-Schlieder Meets Newton-Wigner", available via
    http://philsci-archive.pitt.edu/archive/00000649/00/RS_meets_NW,_PDF.pdf [Broken]

    The Reeh-Schlieder theorem asserts the vacuum and certain other states to be spacelike
    superentangled relative to local fields. This motivates an inquiry into the physical status
    of various concepts of localization. It is argued that a covariant generalization of Newton-
    Wigner localization is a physically illuminating concept. When analyzed in terms of
    nonlocally covariant quantum fields, creating and annihilating quanta in Newton-Wigner
    localized states, the vacuum is seen to not possess the spacelike superentanglement that
    the Reeh-Schlieder theorem displays relative to local fields, and to be locally empty as
    well as globally empty. Newton-Wigner localization is then shown to be physically
    interpretable in terms of a covariant generalization of the center of energy, the two
    localizations being identical if the system has no internal angular momentum. Finally,
    some of the counterintuitive features of Newton-Wigner localization are shown to have
    close analogues in classical special relativity.

    2) Halvorson, "Reeh-Schlieder defeats Newton-Wigner: On alternative localization
    schemes in relativistic quantum field theory".
    Available as quant-ph/0007060.

    Many of the “counterintuitive” features of relativistic quantum
    field theory have their formal root in the Reeh-Schlieder theorem,
    which in particular entails that local operations applied to the vacuum
    state can produce any state of the entire field. It is of great interest
    then that I.E. Segal and, more recently, G. Fleming (in a paper entitled
    “Reeh-Schlieder meets Newton-Wigner”) have proposed an alternative
    “Newton-Wigner” localization scheme that avoids the Reeh-Schlieder
    theorem. In this paper, I reconstruct the Newton-Wigner localization
    scheme and clarify the limited extent to which it avoids the counterin-
    tuitive consequences of the Reeh-Schlieder theorem. I also argue that
    there is no coherent interpretation of the Newton-Wigner localization
    scheme that renders it free from act-outcome correlations at spacelike

    3) DeBievre, "Where's that quantum"? Available as math-ph/0607044

    The nature and properties of the vacuum as well as the meaning
    and localization properties of one or many particle states have at-
    tracted a fair amount of attention and stirred up sometimes heated
    debate in relativistic quantum field theory over the years. I will review
    some of the literature on the subject and will then show that these is-
    sues arise just as well in non-relativistic theories of extended systems,
    such as free bose fields. I will argue they should as such not have given
    rise either to surprise or to controversy. They are in fact the result
    of the misinterpretation of the vacuum as “empty space” and of a too
    stringent interpretation of field quanta as point particles. I will in par-
    ticular present a generalization of an apparently little known theorem
    of Knight on the non-localizability of field quanta, Licht’s character-
    ization of localized excitations of the vacuum, and explain how the
    physical consequences of the Reeh-Schlieder theorem on the cyclic-
    ity and separability of the vacuum for local observables are already
    perfectly familiar from non-relativistic systems of coupled oscillators.
    Last edited by a moderator: May 3, 2017
  15. Jul 2, 2008 #14
    Thanks very much for the detailed reply! I will have a look at some of these papers. But I'm still confused why in standard textbooks on QFT, say, a Gaussian superposition of momentum eigenstates is called localized (blurred around a certain point x) in space. Which reasoning is behind this?
  16. Jul 2, 2008 #15


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    Did you have specific textbook quote in mind?

    IMHO, I think that's just what people would like to think. I.e., we'd like to
    think that the x's we use when constructing orthodox QFT do indeed correspond
    meaningfully to points in physical space. But one must delve into the subtleties
    of position operators, understand the Reeh-Schlieder theorem, etc, (and maybe
    also Haag's theorem), before it becomes apparent that something is deeply
    puzzling about this whole subject.

    Of course, such delving is clearly not necessary to calculate the stunningly
    accurate predictions of QFT, hence most people don't worry very much about it.
  17. Jul 3, 2008 #16
    Srednicki for example, Chapter 5 (LSZ formula), Eqs. (5.6) and (5.7). He speaks of creating a particle localized in momentum and position space. (A draft of the book is available at http://www.physics.ucsb.edu/~mark/qft.html" [Broken].)

    Okay, thanks to you all for your answers! I think I know now what to read to get on...
    Last edited by a moderator: May 3, 2017
  18. Jul 3, 2008 #17


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    Strangerep, thanks for posting those references. I'll definitely take a closer look at the last one and maybe the other two as well some time soon.

    You seem to know this stuff pretty well already, so I'd like to ask you specifically about something that I have believed to be true until now. (Right now I'm really confused. I have e.g. never heard of the Reeh-Schlieder theorem before). I've been saying things like this:
    Is this completely false? I guess I always thought that since the f above is pretty much the same in relativistic QM as in non-relativistic QM, its Fourier transform should also be pretty much the same as in non-relativistic QM.
  19. Jul 3, 2008 #18


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    The right questions are: Can this Fourier transform be interpreted as the probability amplitude? In other words, is such a probability conserved? And is it defined in a relativistic covariant way?
    See e.g. the Appendix in
    http://xxx.lanl.gov/abs/0804.4564 [not yet published]
  20. Jul 3, 2008 #19
    If I get this right, this is the same as asking whether you can interpret a first-quantized field as a particle system obtained by second quantization which obviously is done without ever asking for reasons.
  21. Jul 3, 2008 #20


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    (Urgle!?) No, I'm not an expert. I'm aware that the various puzzles exist, but I
    don't know how to resolve them satisfactorily. (Actually, I don't think
    anyone does - hence the ongoing debate.)
  22. Jul 4, 2008 #21


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    No, that's not the question here. The [itex]a^\dagger(\vec p)|0\rangle[/itex] states are eigenstates of the momentum operator that you can construct from the quantum field. (Noether's theorem tells us how). The question is about the meaning of the Fourier transform of the coefficients when an arbitrary state is expressed as a "linear combination" of momentum eigenstates.

    In non-relativistic QM, that Fourier transform is the wave function, and as you know the square of its absolute value is a probability density describing where in space the particle is likely to be found. I've been assuming that the interpretation as a probability density holds in relativistic QM too, but I'm still not sure how correct that is. (I still haven't had time to really think about Demystifier's answer. It looks like a very good one but I need to think some more about what it means).
  23. Jul 4, 2008 #22
    But this is exactly what I meant, isn't it? You treat a first-quantized field like a second quantized theory by interpreting

    [tex]\psi(x_1, \hdots, x_n) := \langle 0 | \hat{\psi}(x_1) \cdots \hat{\psi}(x_n) | \alpha \rangle[/tex]

    as a many-particle wave function, where [tex]|\alpha\rangle[/tex] is some superposition of momentum eigenstates. But the point is that the matrix element [tex]\langle 0 | \hat{\psi}(x_1) \cdots \hat{\psi}(x_n) | \alpha \rangle[/tex] itself gives no key to this interpretation since the field operator is obtained by first quantization. And in fact, as has been pointed out here, there are reasonable doubts to do so.
  24. Jul 4, 2008 #23


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    In such areas, it helps to go back and think about how the multi-particle Fock space
    is constructed. One starts with a (single-particle) Hilbert space [itex]H_1[/itex] carrying
    a +ve energy representation of the Poincare group. By various abstract nonsense
    (the spectral theorem from Functional Analysis) we can then assert that [itex]H_1[/itex]
    is spanned by momentum eigenstates, subject to the mass-shell condition
    [itex]\delta^{(4)}(p^2 - m^2)[/itex].

    If we try to Fourier-transform to position space (i.e., to obtain an alternate basis
    of position eigenstates), it ceases to be physically meaningful if we don't include
    the [itex]\delta^{(4)}(p^2 - m^2)[/itex] factor in the Fourier transform.
    Remember now that a product in momentum space Fourier-transforms into a
    convolution in position space, so you don't get very nice position eigenstates.
    (IMHO, this is another reason why we hit paradoxes like Reeh-Schlieder.)

    Now, what is a 2-particle Hilbert space? We take the tensor product of two
    copies of [itex]H_1[/itex], i.e., [itex]H_2 := H_1 \otimes H_1[/itex].
    Remember: I'm still thinking of these in momentum basis, or rather, two tensored
    copies of momentum space, each with their separate [itex]\delta^{(4)}(p^2 - m^2)[/itex]
    constraints. Now think about the horrors when you try to Fourier-transform
    this tensor product to a 2-position basis. Convolutions everywhere. (And even
    that's only if you can somehow make the Fourier transform itself make sense.)

    And yet, we try to kid ourselves that the tensor product of 2 separately-transformed
    [itex]H_1[/itex] spaces can be handled in the simple way shown at the start of
    this post. Really, it becomes mathematical nonsense very quickly. You can't
    work with infinite-dimensional spaces and their duals sensibly without
    paying careful attention to topological matters, unboundedness of operators,
    and the like.

    (That probably didn't help, I know. :-)
  25. Jul 4, 2008 #24
    If you assume that the position and momentum space wave functions [tex]\psi(x)[/tex] and [tex]\hat{\psi}(p)[/tex] are related by the Fourier transforms, define the single particle state as

    |\psi\rangle = \int d^3p\; \hat{\psi}(p) a_p^{\dagger}|0\rangle,

    derive the time evolution

    |\psi(t)\rangle = \int d^3p\; \hat{\psi}(p) e^{-iE_p t} a_p^{\dagger}|0\rangle

    from the Shrödinger's equation (after fixing the Hamilton's operator from whatever QFT principles you want to postulate initially...) (here [tex]E_p=\sqrt{|p|^2+m^2}[/tex]) and then recognize the evolving momentum space wave function

    \hat{\psi}(t,p) = \hat{\psi}(p) e^{-iE_p t},

    it follows that in the position space representation the time evolution is given by the old-fashioned relativistic Schrödinger's equation

    \psi(t,x) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(x),

    which satisfies the normalization conserving condition

    \int d^3x\; |\psi(t,x)|^2 = \int d^3x\; |\psi(x)|^2 = 1

    at all times. To me this looks all good, and I must say I'm slightly confused about why this (assuming I've understood the mainstream view correctly) is widely considered incorrect. The QFT books always start doing some tricks with the [tex]1/(2E_p)[/tex] factor, and that's where I usually get lost.
  26. Jul 5, 2008 #25


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    Aren't you in trouble already, even at this early stage? You don't
    have the [itex]E_p[/itex] factor in the denominator, so maybe
    this expression is not Lorentz-covariant?
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