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Position Representation in QFT?

  1. Sep 4, 2011 #1

    referframe

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    I have read that, in QFT, unlike QM, there is no position probability density function because position is not considered an observable.

    Then how is a position measurement represented/modeled in QFT?

    As always, thanks in advance.
     
  2. jcsd
  3. Sep 4, 2011 #2
    Position operator and position representation can be defined in relativistic QFT just as well as in ordinary quantum mechanics. You may want to check

    E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?", Found. Phys., 32 (2002), 673.

    before forming your own opinion.

    Eugene.
     
  4. Sep 5, 2011 #3

    tom.stoer

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    Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
     
  5. Sep 5, 2011 #4
    I agree completely, the field argument 'x' in QFT has no relationship to the position observable and its eigenvalues. The true position operator should be built by the Newton-Wigner recipe.

    Eugene.
     
  6. Sep 5, 2011 #5

    Demystifier

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    Even though it is true in the standard formulation of QFT, I think there is a way to reformulate QFT such that x becomes an operator just as in many-particle QM. See Sec. 3 of
    http://xxx.lanl.gov/abs/0904.2287 [Int. J. Mod. Phys. A25:1477-1505, 2010]
     
  7. Oct 8, 2011 #6
    As you well say, position is not an observable in QFT and cannot be measured/modeled. As a consequence, any initial position dependence must be eliminated from the equations before going to the lab, and this is the reason for which the only physically relevant QFT object is the S-matrix, which does not depend of position (spatial coordinates are integrated out and eliminated from the matrix).

    As Mandl and Shaw note in his classic QFT textbook, this is a fundamental difference between QFT and QM.
     
    Last edited: Oct 8, 2011
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