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Imaginary Partition function

  1. May 7, 2014 #1
    I have a partition function in euclidean quantum field theory. I have a parameter, let's say a charge, that I can change in the action that define the partition function.

    I found that for small charge the partition function is positive, but there is a critical charge, above the one the partition function becomes negative.

    Which is the meaning? Could this be interpreted as a phase transition?

    General question: the partition function must be positive? which is the meaning of having an imaginary or negative partition function?

    Thanks
     
  2. jcsd
  3. May 7, 2014 #2
    How can you get a minus sign, the partition function always has the action part exponentiated. Or are you asking something else?
     
  4. May 7, 2014 #3
    1) My action has an imaginary current as interaction (due to Wick rotation in from the theory in Minkowski space).
    2) From the symmetry if the action I know that the functional integral is positive so the partition function is well defined (it's like the integral of Exp(-x^2+ i x)
    3) I want to extimate the partition function through the saddle point method. I find a critical point (that in my case is imaginary) and I expand the theory around it, calculating the second order variation.
    4) For small charge, the second order variations are positive, but when I pass a critical charge I find two negative eigenvalues of the second order variation, that gives a factor minus 1 (the negativity of the partition function I refered).

    The question is: what does imply the fact that in this case the fact I have a negative contribution? Phase transition? The critical point around the one I was expanding the theory is no more good for large charge?

    Thanks
     
  5. May 7, 2014 #4
    You can evaluate the integral in closed form here, it will not be negative. Saddle point method is approximate, also which charge are you talking about? Are you using a scalar field theory with Wick rotation and coupling as charge, then it does not have a critical point.
     
  6. May 7, 2014 #5
    The action is Euclidean Yang Mills couled with an heavy static source (this is the imaginary current):
    $$F_{\mu\nu}F^{\mu\nu}+i\delta_{\mu0}A_{\mu}$$

    Solving the euclidean yang mills equation, I have an imaginary solution.
     
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