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Position operator in QFT

  1. Oct 29, 2006 #1
    I'm reading some QFT and have been puzzled by the following question:
    What's the physical meaning of the position OPERATOR X_\mu in QFT? whose position does it measure?:confused: Thanks for any help.
  2. jcsd
  3. Oct 29, 2006 #2
    I believe, in QFT, that in order to obtain sensible results, you end up demoting the position operator into a parameter of the problem, like time in "classica" quantum mechanics, while you promote the wave function to the status of an operator.
  4. Oct 30, 2006 #3
    Thanks, StatMechGuy, I agree on the point you talked about, there is a 'position' that simply is the parameter for the position dependent operator function \phi(x). But what I asked about is the position OPERATOR whose eigenstate is |x>, that's the operator in the expression: <x'|\phi|x''>=\delta(x'-x'')\phi(x')
  5. Oct 30, 2006 #4
    In qft the fields really don't mean much, it is correlation function we are looking to calculate. We ask the question, if a particle is created at x, what is the probably it will be destroyed (detected) at x'. This will be proportional to

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