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Field Theory Partition Function

  1. May 22, 2013 #1
    The 'partition function' in QFT is written as [itex]Z=\langle 0 | e^{-i\hat H T} |0\rangle[/itex], but I'm having a difficult time really understanding this. I'm assuming that [itex]|0\rangle[/itex] represents the vacuum state with no particles present. If that's the case, and the Hamiltonian acting on such a state would just produce the ground state energy, which can be defined to be zero. How does this give you anything interesting then? If [itex] e^{-i\hat H T} |0\rangle=e^{-i (0)T}|0\rangle[/itex], then I fail to see how this quantity is of any use. (I know that by adding a source to the Hamiltonian, you can calculate the interaction energy of two particles, but for a source free H I'm not sure what it does). Also, if you have a Hamiltonian of the form [itex]:\hat H:=\sum_{k}\hat a^{\dagger}_{k}\hat a_{k}[/itex], where [itex]\hat a_{k}|0\rangle=0[/itex], then I really don't get what's going on here when you write out Z. Does it only make sense to calculate this when you have a source term added?
    Thanks in advance for any help
     
  2. jcsd
  3. May 22, 2013 #2

    fzero

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    The vacuum state is a very complicated issue in QFT. Since we usually can't solve the interacting theory exactly, we resort to perturbation theory, where

    $$ H = H_0 + H_I,$$

    with ##H_0## the free Hamiltonian and ##H_I## the interacting part. There are then really two vacuum states that we can define. For example, in Peskin & Schroder, ##|0\rangle## is the vacuum for the free Hamiltonian, while they use ##|\Omega\rangle## to stand for the vacuum of the full theory.

    The Hamiltonian really isn't [itex]:\hat H:=\sum_{k}\hat a^{\dagger}_{k}\hat a_{k}[/itex], rather the normal-ordered free Hamiltonian has this form. Along with the normal-ordering, we're typically dropping the zero-point energy of the oscillators, so we can define the zero of energy by choosing ##H_0 |0\rangle =0##. However, once we do this, the interaction terms generally are not normal ordered, so ##H_I|0\rangle \neq 0##. The interactions can create virtual particles from the vacuum (in such a way that all charges are conserved). Therefore, the partition function is generally non-trivial and corresponds to a sum over so-called bubble diagrams, which have no external lines.
     
  4. May 22, 2013 #3
    Thank you, I didn't think too much about what the Hamiltonian really is. Including the interaction terms gives you a non-trivial calculation, and I think I overlooked the addition of that to the total Hamiltonian.
     
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