# Temperature of quantum fields and periodicity

I have read this PSE post Finite Temperature Quantum Field Theory, saying that

In a QFT at finite temperature, we consider the Euclidean time to be periodic, i.e. we consider a theory on the manifold $$\mathbb{R}^{d−1}×S^1$$, where the spatial coordinates are in $$\mathbb{R}^{d−1}$$ and the time coordinate in $$S^1$$. The temperature of the system is then the inverse of the circumstance $$T=1/\beta,\ \tau\in[0,\beta)$$.

Why are we allowed to identify the imaginary time as the inverse of the temperature and why do we enforce periodicity on the circle of circumference $$\beta$$, i.e. why does the condition $$\phi(0,\vec{x})=\pm\phi(\beta,\vec{x})$$ (for $$\vec{x}$$ being a vector in $$\mathbb{R}^{d-1}$$), where $$\phi$$ is some scalar field, has to hold?

I have also read another PSE post Why is Euclidean Time Periodic?, saying that

it turns out that thermal averages of operators are periodic with respect to the Wick rotated time $$\tau$$

Is this motivation enough for us to impose this periodicity in thermal field theories?

I realize that this may be a textbook question, so if it is, please point me to the right direction. If not, please answer this post.

• I'm not totally sure that I understand your question. But if you agree that the partition function in thermal FT is given by path integral of periodic path, then you can prove that the fields are periodic. This is because, when you do the calculation, you get some $\delta (\phi(t_0,\vec{x}) - \phi(t_f,\vec{x}))$ term that tells you that you have to consider periodic path.
– Jean
Commented Mar 31, 2023 at 13:06
• Hi @jean159753. I think this is actually what I am asking... Why would the partition function of a thermal field theory be given by path integral whose path is periodic?? Commented Mar 31, 2023 at 13:29
• I just replied with an answer, hope it helps.
– Jean
Commented Mar 31, 2023 at 13:50

I will develop my comment. Consider the general amplitude term $$\tag{1}\label{1}\langle\phi(t_f)| e^{-it_f H} |\phi(t_f)\rangle.$$ The technique to get a path integral form of this expression is to consider $$N$$ time slices. We rewrote the previous expression $$\langle\phi(t_f)| e^{-it_f H} |\phi(t_f)\rangle = \langle\phi(t_f)| e^{-i\Delta t H} \cdots e^{-i\Delta t H} |\phi(t_f)\rangle$$ with $$t_f = N \Delta t$$. Then you insert $$N$$ identity relations of the form $$\int d\phi_i |\phi_i\rangle \langle\phi_i|$$ inside the amplitude term (you also add identities for the conjugate field $$\pi = \dot{\phi}$$). In the end you get

$$\langle\phi(t_f)| e^{-i\Delta t H} \cdots e^{-i\Delta t H} |\phi(t_f)\rangle = \int \Pi_i d\phi_i d\pi_i\langle\phi(t_f)|\pi_N\rangle \langle \pi_N|e^{-i\Delta t H} \cdots e^{-i\Delta t H} \langle\phi_1|\phi(t_f)\rangle .$$ Taking $$N \rightarrow \infty$$ you get the path integral for the amplitude term. But because $$\langle \phi_1|\phi(t_f)\rangle = \delta(\phi_1 - \phi_{t_f})$$, the fields in the path integral will be periodic, and so you will have to consider periodic path in the path integral of the partition function. When you take the trace over $$e^{-\beta H}$$, you will simply redo the calculus but with imaginary time.

EDIT :

In thermal field theory, $$Z = Tr [e^{-\beta H}] = \int \phi \langle \phi| e^{-\beta H} |\phi\rangle ,$$ where the integrand is equivalent to ($$\ref{1}$$), where $$\beta$$ plays the role of the complex time.

• Your starting point is an amplitude of the form $\langle\phi(t_f)|e^{-iHt_f}|\phi(t_f)\rangle$. Why is that? It makes more sense to consider integrals of the form $\langle\phi(t_f)|e^{-iH(t_f-t_i)}|\phi(t_i)\rangle$, where $t_i$ is some initial time. What I mean, I guess is "why time-evolve a state and take the overlap with the same state?" Commented Mar 31, 2023 at 13:59
• This is because in TQFT we take the trace operator to compute the partition function, I added an edit in the answer.
– Jean
Commented Mar 31, 2023 at 14:13
• Please consider using angle brackets $\langle$ and $\rangle$ (\langle and \rangle) instead of greater/lesser-than signs Commented Mar 31, 2023 at 14:29
• @Mauricio Much better indeed thx.
– Jean
Commented Mar 31, 2023 at 14:33
• @schris38 For $T=0K$ QFT, the partition function is indeed calculated for arbitrary path (see e.g. here where $Z(0)$ in the first equation is the partition function at $T=0K$). In TQFT we want to calculate objects such as $\langle A(t) B(t') \rangle$ where $A$, $B$ are some operator. $\langle ... \rangle$ is the thermal mean value, i.e. the trace of this operators times the Boltzmann distribution. So the calculations are very similar to $0K$ QFT.
– Jean
Commented Mar 31, 2023 at 15:54