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Scalar field as quantum operator.

  1. Nov 30, 2011 #1
    Hallo,

    I was wondering what is the physical significance of scalar field [itex]\Phi (x)[/itex] as an quantum operator. [itex]\Phi (x)[/itex] have canonical commutation relation such as [ [itex]\Phi (x) , \pi (x)[/itex] ] so it must be an opertor, thus what are his eigenstates?

    Thanks,
    Omri
     
  2. jcsd
  3. Nov 30, 2011 #2

    A. Neumaier

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    For each x, Phi(x) is an operator, typically with the real line as spectrum, and without eigenstates. (Many other operators of physical significance including position and momentum, have no eigenstates either.)
     
  4. Nov 30, 2011 #3

    dextercioby

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    The field [itex] \Phi[/itex] is mathematically speaking a double distribution and its generalized eigenstates have no physical relevance. Only the true eigenstates of N (number operator, Fock space operator-valued distribution) have physical relevance; they are the free states with definite number of particles which in turn have well-defined energies/momenta and spin projections.
     
  5. Nov 30, 2011 #4

    samalkhaiat

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    See posts #26,27,28 and 31 in the following
    www.physicsforums.com/showthread.php?t=388556

    Sam
     
  6. Nov 30, 2011 #5

    dextercioby

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    Sam, now that you brought it up, do you have a reference for point 1) from post 26 of the thread you gave a reference to ?

    1) the field ϕ in Ψ[ϕ] is an ordinary function with the following transformation law

    ϕˉ(xˉ)=D(a,A)ϕ(x), A∈SL(2,C)


    where D is a finite-dimentional, irreducible, non-unitary matrix representation of the Poicare' group.
    The wave functional transforms as

    Ψˉ[ϕ]=U(a,A)Ψ[D−1ϕ]


    The representation U(a,A) is faithful, UNITARY and INFINITE-DIMENTIONAL but NOT IRRIDUCIBLE. So, we do have Poincare'-invariant norms (probabilities) in the Hilbert space of wave functionals.
     
  7. Dec 1, 2011 #6
    Thank you all!
    It was very helpful.
     
  8. Dec 1, 2011 #7

    samalkhaiat

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    Jan topuszanski, "An Introduction to Symmetry and Supersymmetry in Quantum Field Theory", World Scientific, 1991.

    I learnt the stuff from Roman Jackiw, "Analysis on Infinite Dimensional Manifolds; Schrodinger Representation for Quantized Field", In the V Jorge Andre' Swieca Summer School, Sao Paulo, 1989; Particles and Fields.
     
  9. Dec 1, 2011 #8

    dextercioby

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    The reason it popped up to me is that I've never seen finite dimensional representations of the Poincare group...I'll check Lopuszanski. No mention of what you wrote in Lopuszanksi's text and the proceedings you mentioned are not within my reach.
     
    Last edited: Dec 1, 2011
  10. Dec 1, 2011 #9

    tom.stoer

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    There are no finite-dimensional, unitary representations of non-compact groups.
     
  11. Dec 1, 2011 #10

    dextercioby

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    Yes, but he specifically said non-unitary (see above), so this means finite matrices, just like for spinors for SU(2)/SL(2,C).
     
  12. Dec 2, 2011 #11

    A. Neumaier

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    Every finite-dimensional Lie algebra has a finte-diemsnional faithful representation, by the theorem of ADO.
    For the Poincare algebra, one doesn't need this (fairly hard to prove0 theorem, but can see directly that it has a 5-dimensional representation, obtained by augmenting the Lorentz representation in momentum space by one coordinate.
     
  13. Dec 2, 2011 #12

    Physics Monkey

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    Perhaps this is what A. Neumaier meant just above, but one can obtain such a representation of Poincare using a representation of the conformal group. In that case a finite dimensional represenation is provided by the appropriate higher dimensional Lorentz group for a metric (-1 -1 1 ... 1).
     
  14. Dec 2, 2011 #13

    A. Neumaier

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    The conformal group SO(2,4) has a 6-dimensional representation, and any Poincare group inside the conformal group is represented on a 5D subspace. This gives something equivalent to what I had in mind, which is a bit more elementary:

    The Poincare group is the semidirect product of the translation group T=R^4 and the Lorentz group acting on T, and there is a canonical construction of a representation of the semidirect product of R^n and GL(n) in n+1 dimensions that restricts to one of the semidirect product of R^n and any subgroup of GL(n)
     
  15. Dec 2, 2011 #14

    Hans de Vries

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    It's also in Roman Jackiw's book: Diverse Topics in Theoretical and Mathematical Physics
    Paper IV.4 page 383: "Analysis on Infinite Dimensional Manifolds; Schrodinger Representation for Quantized Field"

    Hans
     
  16. Dec 2, 2011 #15

    dextercioby

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    Thanks, Hans. I'll check it there, if I can get it. Nice to see you and Arnold back.
     
    Last edited: Dec 2, 2011
  17. Dec 2, 2011 #16

    DrDu

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    I wouldn't sign this! Think of the operator of the electromagnetic field. It!s eigenstates are coherent states formed from superpositions of different number eigenstates. While these are easy to prepare experimentally, Eigenstates of photon number n are hard to create.
     
  18. Dec 2, 2011 #17

    dextercioby

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    OK, checked Jackiw. No sign of finite dim. representations of the Poincare group.

    DrDu, point taken.
     
  19. Dec 2, 2011 #18

    DrDu

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    You have a similar construction already for the rotation translation group and its subgroups, e.g. in crystallography. Suppose an operation of your group consists of a rotation matrix R and a translation vector a, the composition (R',a')(R,a)=(R'R,a'+R'a) cannot be implemented on a three vector x, but if you augment X=(x^T,1)^T,
    you can operate on it with the matrix
    ( R a )
    ( 0 1 )

    (Sorry, I am too lazy to tex this).
    Generalization to 4 dimensional space time (either Galilean or Poincare) should now be obvious.
     
  20. Dec 2, 2011 #19

    samalkhaiat

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    Don’t be silly; spinor, vector and irreducible tensor fields all transform by finite dimensional matrix representation [itex]D(A)[/itex], where [itex]A \in SL(2,C)[/itex]. The representation matrix [itex]D[/itex] (which is a product of finite number of [itex]SL(2,C)[/itex] matrices [itex]A[/itex] and [itex]\bar{A}[/itex]) act not on states in Hilbert space but on finite-component classical fields and therefore they need not be unitary. This is the reason why fields are the necessary ingredients in relativistic QFT’s. Below is the 4-dimensional representation of the Poincare’ group [itex]\{T, SL(2,C)\}[/itex];
    [tex]
    \tilde{\phi}_{\alpha \dot{\alpha}}(x) = A_{\alpha}{}^{\beta}\bar{A}_{\dot{\alpha}}{}^{\dot{\beta}}\phi_{\beta \dot{\beta}} (\{-a , A^{-1}\}x)
    [/tex]

    Chapter 4 is about the representations of Poincare' group. However, he treats the fields as operators from the start.

    The notes are about the Schrodinger representation of QFT, they do not teach you about the representation theory of the Poincare group. I mentioned them to you because I thought you wanted to learn about doing QFT in the Schrodinger representation.

    Sam
     
  21. Dec 4, 2011 #20

    A. Neumaier

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    But this representation is not faithful, as translations are represented trivially. Thus it is usually referred to just as a representation of the Lorentz group. But in momentum space, one has a faithful representation in 5D, which is the one given by Dr. Du's recipe, with the Lorentz group in place of rotations.
     
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