Consider a real scalar field theory in finite temeperature. According to the book by Kapusta and Gale, Finite-Temperature Field Theory, its retarded Green's function is given by $$iD^R(x,x') = Tr\{\hat{\rho}[\hat{\phi}(x),\hat{\phi}(x')]\}\theta(x^0 - x'^0)\,, $$ where $\hat{\rho} = \exp[-\beta(H-\mu N)]/Z$, and the square brackets represents a commutator. My question: what would be the path-integral representation for it? Intuitivelly, I would write $$ iD^R(x,x') = \frac{1}{Z} \int \mathcal{D}[\phi(x)]\; \{\phi(x),\phi(x')\} e^{-\beta(H-\mu N)}\theta(x^0 - x'^0)\,,$$ where the curly brackets represents the Poisson brackets. But I am neither not sure if this is correct and if so, how to derive it.

Edit: I went to look on Le Bellac book Thermal Field Theory, where he introduces the propagators $$ D^<_C(x,x') = D^>_C(x',x) = \langle \hat{\phi}(x')\hat{\phi}(x)\rangle\,.$$ Thus, I'm thinking about computing $\langle \hat{\phi}(x')\hat{\phi}(x)\rangle$ without regard of time ordering and then manually split the result of $x^0 > x'^0$ and $x'^0 > x^0$. However, It is not clear for me that this inversion will yield different results when using a path-integral formulation.

Edit 2: I implemented what should be almost the correct procedure. I used the standard trick of introducing a classical current coupled to the field to be able to compute $\langle \phi(y) \phi(x) \rangle$. What I found, after taking continuum and zero temperature limit was $$\langle \phi(y) \phi(x) \rangle = -\frac{1}{2}\int \frac{d^4 p}{(2 \pi)^4} \frac{e^{-i p^\mu(x-y)_\mu }}{p_\mu p^\mu - m^2}\,.$$

Then I evaluated the integral on $p_0$ using a path that selects the positive pole under our assumption that $x^0 > y^0$ $$ \langle \phi(y) \phi(x) \rangle = - \frac{i}{2} \int \frac{d^3\vec{p}}{(2\pi)^3}\frac{e^{-i \omega_p (y_0-x_0)+i\vec{p}\cdot (\vec{y}-\vec{x})}}{2 \omega_p}\,.$$

Now, If I say $\langle[\phi(x),\phi(y)]\rangle = \langle \phi(x) \phi(y)\rangle - \langle \phi(y) \phi(x)\rangle$ I obtain the right answer. But if I try to follow the suggestion on the comments to time order the operators and say that $\langle [\phi(x),\phi(y)]\rangle = \langle \phi(y^0,\vec{x}) \phi(x^0,\vec{y})\rangle - \langle \phi(y^0,\vec{y}) \phi(x^0,\vec{x})\rangle$, It will yield 0 since the integrand will be odd in $\vec{p}$. However, I cannot justify the first method, because in $\langle \phi(x) \phi(y)\rangle$ the time of $\phi(x)$ is bigger than in $\phi(y)$. Thus I should not use my previous result. If I use the right result guess what happens? I get 0 again! So what is going on here?

  • $\begingroup$ Commutators are tricky inside of path integrals because you get ordering ambiguities. This blog post might be of help. The idea is that your Hamiltonian contains derivatives, which are rather formal because most paths (or in the case of QFT, field configurations) that contribute are not smooth, and the way that you discretize the path integral changes the results. $\endgroup$ Commented May 18, 2017 at 0:14
  • $\begingroup$ Path integral always computes time-ordered correlators. Commutators are encoded in the singularities of Green's functions when the arguments coincide. In another words, you would want to calculate $\left< a(t + \varepsilon) b(t) \right> - \left< a(t) b(t + \varepsilon) \right>$ which would give you the expectation of the quantum commutator $[a, b]$. $\endgroup$ Commented May 18, 2017 at 4:16
  • $\begingroup$ @FelipeLopes The blog post helped me understand how one would treat equal time commutators. Very much thank you. However, on the case presented, $x^0 > x'^0$. If I try to time order the operators in each term of the commutator, I will get 0 and the issue persists. $\endgroup$
    – WilhelmM
    Commented May 19, 2017 at 13:50
  • $\begingroup$ Can you point to where in the book he makes this claim about the retarded greens function coming with a commutator? Its not entirely obvious to me that this should be the expression for the retarded green's function in the first place. $\endgroup$
    – Anonjohn
    Commented May 19, 2017 at 14:13
  • $\begingroup$ @Anonjohn It is in the begining of the chapter about linear response theory (Chapter 6, pg. 86 on the second edition). Mandl-Shawn book also shows that you can get the propagator in a complex scalar theory from the communtator $[\phi(x),\phi(y)]$, so for me it does sound beliavable. $\endgroup$
    – WilhelmM
    Commented May 19, 2017 at 14:28


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.