Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook