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I Doubt about partition functions in QFT and in stat Mechanics

  1. Dec 9, 2017 #1
    Hi, I was studying for my final exam on statistical physics and a doubt raised on my head that was truly strong and disturbing (at least, for me), and that I couldn't answer to myself by now.

    The doubt is: Given that we have in d dimensions a fermion non interacting gas, the statistical mechanics partition function will be

    Z=∏p∈R^{d+1}, spin∈Z2 (1+e^{-β√(p2+m2)}))
    "=det(1+e^{-β√(p2+m2)}))",
    and in (d-1)+1 dimensions for a Dirac action will have as wick-rotated partition function

    Z(β)=det(β(iγ⋅∂+m))

    Are both related in some way?

    I mean, they are systems with the same Hamiltonians, and given that there is a lot of analogies between the formalisms of QFT and statistical physics I though that both needed to be a little more similar...

    Is there a relation between the two that I am not seeing? What's wrong with my intuition? I will be very grateful for any answer (if it isn't on the line of "Quantum Field Theory is a myth, the real thing is Aliens").

    Sorry if this question was against the rules (I don't know for now if it could be).

    (After posting this question here, I posted in phys*** st**kexc***ge, but I deleted from that place... Is that wrong? Sorry)
     
  2. jcsd
  3. Dec 9, 2017 #2
    Because I can't edit the post (I know that there are good reasons for that), I will retype the mathematical expressions using LaTex, for making them more understandable:
    1-Fermi-Dirac in Statistical mechanics:
    [tex] Z_{F-D}=\prod_{p\in \mathbb{R}^{d+1}} \prod_{s\in\{-\frac{1}{2},\frac{1}{2}\}} (1+e^{-\beta\sqrt{p^2+m^2}}) [/tex]
    [tex] Z_{F-D}=det(1+e^{-\beta\sqrt{p^2+m^2}}) [/tex]
    2-Dirac in QFT
    [tex] Z_{Dirac-QFT}=det(\beta(m+i \displaystyle{\not} \partial)) [/tex]
     
  4. Dec 10, 2017 #3
    Explaining a little more:
    -The trouble I hace is un the difference between [tex] \sum_n \rangle n | e^{-\beta \hat{H}} | n \langle [\tex] and [tex] \int \prod_{i\in fieldcomponents} e^{-\beta H} [\tex]

    If a field is grassman-like, in the first case we sum between "mide is turn on"+" mode is turn off", but in the other case we only take into account the case un which it's turn con only?
    Something similar appears to happen in the bosonic case (difference of partition function "per mode")
     
  5. Dec 13, 2017 #4
    Homer simpson doh:
    I remembered today, talking with a friend, that this thing was there in the advanced qft course that we had taken. It's useful that I post here the answer in a few days, if someone has the same question? Or it's better for this post to be deleted and be burned in the oblivion ice's?

    Cheers to the community.
     
  6. Dec 14, 2017 #5
    In my opinion, posting the solution would be more useful.
     
  7. Dec 14, 2017 #6
    OK, well... this is a pretty standard thermal QFT calculation:

    *First: When we do the Wick rotation and make the time periodic (With time period = [itex] \hbar \beta [/itex]), we have as log of partition function [itex]
    log Z_{Dirac-QFT}=Tr[ln(\prod_{n+1\in 2\mathbb{Z}} \beta(m+i \gamma \cdot \nabla + \gamma_0 \frac{2\pi n}{\beta}))]=\frac{1}{2}Tr[ln(\prod_{n+1\in 2\mathbb{Z}} (\beta^2(m^2+p^2)+4\pi^2 n^2))]=Tr[ln(\prod_{n-1\in 2\mathbb{N}} (\frac{\beta^2(m^2+p^2)}{4\pi^2 n^2}+1))]+const=Tr[ln(cosh(\frac{\beta\sqrt{m^2+p^2}}{2}))]+const=Tr[ln(1+exp(-\beta\sqrt{m^2+p^2}))]+Tr[ln(exp(\frac{\beta\sqrt{m^2+p^2}}{2}))]+const
    [/itex]. The former is equal to the fermi-dirac partition function plus an energy shift and with a different normalization.
     
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