Undergrad Doubt about partition functions in QFT and in stat Mechanics

[Iliody]

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)

 

[Iliody]

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:
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}})
Z_{F-D}=det(1+e^{-\beta\sqrt{p^2+m^2}})
2-Dirac in QFT
Z_{Dirac-QFT}=det(\beta(m+i \displaystyle{\not} \partial))

 

[Iliody]

Explaining a little more:
-The trouble I hace is un the difference between \sum_n \rangle n | e^{-\beta \hat{H}} | n \langle [\tex] and \int \prod_{i\in fieldcomponents} e^{-\beta H} [\tex]<br /> <br /> 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?<br /> Something similar appears to happen in the bosonic case (difference of partition function "per mode")

 

[Iliody]

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.

 

[protonsarecool]

In my opinion, posting the solution would be more useful.

 

[Iliody]

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 = \hbar \beta), we have as log of partition function <br /> 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<br />. The former is equal to the fermi-dirac partition function plus an energy shift and with a different normalization.