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I QFT vs Wave Function

  1. Jun 25, 2016 #1
    My level is not sufficient enough to easily understand QFT yet there is some basic question I need to understand in it - what in QFT corresponds to a wave function in QM, for a single particle case and, say, for a more general case of multiparticle nonseparable state (suppose the particles are identical). I do understand Fock space idea as well as creation-annihilation operators idea, yet I cannot grasp what is that operator field considered in QFT and how it may correspond to the general multiparticle case of QM. If this field defines an operator at each space point, then on which vector space does it act and how it may be related to the multiple particle wave function? Or maybe it is not the operator field itself which corresponds in QFT to the wave function/particles state, but something else?
    I tried to look through some discussions here and on other sites but did not succeed to see or grasp what is the solution.

    Thank you.
     
    Last edited: Jun 25, 2016
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  3. Jun 25, 2016 #2

    A. Neumaier

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    The quantum field ##\phi(x)## corresponds in meaning to the quantum mechanical ##q_j## (specifically, ##\phi\to q## and ##x\to j##, but with a continuum of indices). But its action on an ##N##-particle state gives (in the nonrelativistic case) a linear combination of ##(N-1)##-particle and ##(N+1)##-particle states.
    See https://en.wikipedia.org/wiki/Second_quantization.
     
  4. Jun 25, 2016 #3
    But how the quantum field and particle states are related? I was under an impression that quantum particles are an emergent property of a quantum field, that is particles are some discrete modes of oscillation of that quantum field. So I expected that a state of the quantum field itself may somehow be interpreted as particle state.

    From what I can see and understand in second quantization, operator field introduced there cannot contain any state-specific information.
     
  5. Jun 25, 2016 #4

    atyy

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    The creation operator acting on the vacuum state creates an excited state, which is a state with one particle.

    When you read about non-relativistic second quantization, first keep in mind non-relativistic quantum mechanics of many identical particles (ie. with the standard interpretation of the wave function etc.) Second quantization is simply non-relativistic quantum mechanics of many identical particles, but formulated in a different language.
     
  6. Jun 25, 2016 #5

    A. Neumaier

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    Just like in ordinary quantum mechanics, the position operator cannot contain any state-specific information. The latter is always in the state, not in the operators.
     
  7. Jun 26, 2016 #6

    Demystifier

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    In QFT there 3 different things which, in some sense, correspond to the wave function ##\psi({\bf x},t)## in QM.
    1. Field operator ##\hat{\phi}({\bf x},t)##
    2. Wave functional ##\Psi[\phi({\bf x}), t]##
    3. Wave function itself, which in QFT can be calculated as ##\psi({\bf x},t)=\langle 0|\hat{\phi}({\bf x},t)|\Psi\rangle##
     
  8. Jun 26, 2016 #7

    kith

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    Yes. The field operator ##\phi(\vec x)## is not the state but it acts on the state. One way to picture it is by its action on the vacuum. The vacuum state is the state with zero particles which is denoted by [itex]|0\rangle[/itex]. If I act on it with the field operator, I get a particle localized at location [itex]\vec x[/itex] (compare this with Demystifier's point 3 in post #6 above):
    [tex]\phi(\vec x) |0\rangle = |\vec x\rangle[/tex]
    In ordinary QM, the position wavefunction can be written as a superposition of momentum eigenstates. Analogously in QFT, the field operator can be written as a sum of operators which create or annihilate a particle with definite momentum. If I act with the field operator on the vacuum, it creates in effect a particle with definite location.

    (What I have written here is by no means rigorous and there are quite a few difficulties. For example, we cannot define a position operator with the usual properties for massless particles like photons. So the very concept of localizing particles isn't always well-defined.)
     
  9. Jun 26, 2016 #8

    kith

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    Your confusion about the "operator field" and the "state of a quantum field" may be related to the fact that the term "field" in these quotes refers to different things. I started a thread about this a couple of years ago.
     
  10. Jun 26, 2016 #9

    vanhees71

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    This has to be taken with very great care. E.g., in the case of photons, this "state" (it's a generalized state in the sense of a distribution in any case), this cannot be interpreted as a "generalized position eigenstate" since a photon doesn't permit the definition of a position operator in the strict sense. A photon has not a position as an observable at all (within standard QED)!
     
  11. Jun 26, 2016 #10

    bhobba

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    There are a number of equivalent formulations of QM:
    http://www.colorado.edu/physics/phys5260/phys5260_sp16/lectureNotes/NineFormulations.pdf

    QFT corresponds to formulation F - the second quantization formulation based on creation and annihilation operators. As operators they operate on states and that's how states come into it. The interesting thing about QFT is its based on operators with states being of secondary importance. They are there of course as they must be - but most of the time they take a back seat.

    Thanks
    Bill
     
  12. Jun 26, 2016 #11
    It'll take me some time to work on it. Some immediate questions to clarify:

    Are the ##|0\rangle## and ##|\Psi\rangle## appearing in the item 3 the wave functionals of the same kind as brought in item 2?

    Is the wave function brought in item 3 eqivalent to the single particle QM wave function or something else?

    Most importantly, how still can we get, say, a nonseparable wave function of two particles from the wave functional brought in item 2?
     
  13. Jun 26, 2016 #12

    bhobba

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    Read the link I gave on the second quantization formulation of QM.

    Thanks
    Bill
     
  14. Jun 26, 2016 #13

    DarMM

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    Ignoring distribution theoretic issues, the wave functions in QFT are:

    ##\Psi[\phi(\vec{x})]## in the Heisenberg picture. That is a function over the space of field configurations in ##\mathbb{R}^{3}##. Roughly speaking it is the amplitude to observe the field in that configuration.

    If the field is a free field then it turns out that this space of field wave functionals has a basis where it takes the form of a Fock space over a single particle space Hilbert space. Hence one can view the Hilbert space in either a particle or field manner.

    Technical details for others (I'll fill them in when I'm less tired!):
    If the space of classical fields is ##\mathcal{S}## and ##d\nu## is a measure on this space obeying the axioms of QFT and the specific identities of a free field, then one can prove that:

    ##L^{2}(\mathcal{S},d\nu) = \mathcal{F}(\mathcal{S}^{*})##

    (The single particle quantum Hilbert space is the dual of the classical field space)
     
    Last edited: Jun 28, 2016
  15. Jun 26, 2016 #14
    Thanks to everybody!
    I think I've got some directions and stuff to work on.
     
  16. Jun 26, 2016 #15

    gva

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    What is the relativistic quantum mechanics of many identical particles? Or spacetime background independence theory of many particles.. must we come to this latter for completeness?
     
  17. Jun 26, 2016 #16

    bhobba

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    Its called the Quantum Field Theory of Bosons or Femions which are all identical. We either have Bosons or Fermions in QFT - this is the famous Spin Statistics Theorem:
    http://www.worldscientific.com/worldscibooks/10.1142/3457

    Cant see how background independence has anything to do with this.

    Thanks
    Bill
     
  18. Jun 26, 2016 #17

    gva

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    Simple. General relativity is not yet united with quantum field theory. When you have general relativistic quantum field theory.. the particles will no longer occurred against a backdrop of fixed spacetime.. but dynamical where it is more than the particles curving spacetime... but particles and spacetime becoming united into a symmetry just like space and time where they become mere shadows. Is this not the final goal of QFT?
     
  19. Jun 26, 2016 #18

    bhobba

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    Thats a very common misconception, but totally false:
    http://arxiv.org/abs/1209.3511

    We have a perfectly valid theory up to about the Plank scale. QED is in the same boat - its only valid to a cutoff before the electroweak theory takes over and none would say QED is not EM united with QFT.

    Thanks
    Bill
     
  20. Jun 26, 2016 #19

    atyy

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    At the non-rigourous level, the relativistic quantum mechanics of many identical particles is conceptually the same as the non-relativistic quantum mechanics of many identical particles, except that particles can be created and destroyed. However, if you use second quantization in non-relativistic quantum mechanics, there you will already see that there are "particles" that can be created and destroyed, such as phonons in non-relativistic quantum mechanics, eg. http://yclept.ucdavis.edu/course/242/2Q_Fradkin.pdf.

    As DarMM points out above, the "particle" picture and Fock space are not rigourously correct when the field theory is interacting. So all the derivations in textbooks are only "physically" correct, although they are mathematically wrong.

    Background independence is not required for quantum field theory (or any theory with known physical relevance, including quantum gravity at low energies).
     
  21. Jun 27, 2016 #20

    Demystifier

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    Yes.

    Yes.

    Let me explain it for the simplest case of hermitian field operator. It can be written as
    $$\hat{\phi}(x)=\hat{\psi}(x)+\hat{\psi}^{\dagger}(x)$$
    where ##\hat{\psi}^{\dagger}(x)## contains the creation operators and ##\hat{\psi}(x)## the destruction operators. Then the non-separable 2-particle wave function is
    $$\psi(x_1,x_2)=\langle 0|\hat{\psi}(x_1)\hat{\psi}(x_2)|\Psi\rangle$$

    Even though ##\hat{\psi}(x_1)## and ##\hat{\psi}(x_2)## are local operators, the product ##\hat{\psi}(x_1)\hat{\psi}(x_2)## is not a local operator. This is why "local" QFT can have non-local (or non-separable) effects.
     
    Last edited: Jun 27, 2016
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