# Effective field theory approach to Bose-Einstein condensation

I am currently reading this paper, about an effective field theory approach to Bose-Einstein condensation. I quote

We start by considering a simple non-relativistic many-body problem of spinless bosonic particles at finite temperature. In units of $$\hbar=1$$ and $$k_B = 1$$, its Euclidean action functional is given by $$I[\psi, \psi^\dagger] = \int_0^\beta\text d \tau\int_{\mathbb{R}}\text d^3x \mathcal L$$ with the Lagrangian density $$\mathcal L = \frac 1 2 [\psi^\dagger(x) \partial_\tau\psi(x)-\psi(x)\partial_\tau\psi^\dagger(x)] - \frac{1}{2m}\nabla\psi^\dagger(x)\cdot\nabla\psi(x) - \lambda[\psi(x)^\dagger\psi(x)]^2$$ where $$\beta = 1/T$$ denotes the inverse of temperature. Here, we write $$x = (\text x,\tau)$$ and $$\tau$$ is the imaginary "time". All fields fulfill the boundary conditions as of $$\psi(\text x, \tau + \beta) = \psi(\text x, \tau)$$.

Can somebody explain the thoughts/intuition which lead to this Lagrangian? I do not understand what this "imaginary time" is supposed to be since it is here related to the temperature of the system.

I recognize the similarities to the Schrödinger action with dirac-delta interaction but there is something odd about the time derivatives in this theory which I do not understand.

I am also confused about the boundary condition $$\psi(\text x, \tau + \beta) = \psi(\text x, \tau)$$ which is probably related to my confusion about $$\tau$$. Any insight is highly appreciated.

My Knowledge: I am familiar with the mean-field description of a Bose-Einstein condensate using the Hartree method and the resulting Gross-Pitaevskii equations. And I am familiar with the Thomas-Fermi approximation to the Gross-Pitaevskii equations. I have also some knowledge of scalar quantum field theory and the Higgs-Mechanism. But I am really struggling to make sense of the above effective field Lagrangian.

Maybe I can say something about the relationship with Wick Rotation and temperature which might help. Ultimately, given a quantum system described by a Hamiltonian $$\hat{H}$$, suppose that you want to describe the system at a temperature $$T$$. Then, the partition function should be given by: $$Z = \text{tr} \left[ e^{-T\hat{H}} \right] = \sum_n \langle n | e^{-T\hat{H}} |n \rangle = \sum_n e^{-T E_n}$$
where the $$|n\rangle$$ states are the eigenstates of $$\hat{H}$$ with energies $$E_n$$; we recognize the right hand side as the classical partition function of a system at temperature $$T$$ and energy levels $$E_n$$. Now, the key equation in some sense that links the partition function to imaginary time is this one: $$Z \simeq \int D \Phi \ e^{-S_E [\Phi]}, \quad S_E[\Phi] = \int_0^\beta d \tau \int d^3 x \ \mathcal{L}_E$$
Note that this is just some mathematical result, (proof of which follows lines of usual path integral proofs), for a reference, there is one in Gatringer and Lang Chapter 1 (this wick rotation is important to lattice qcd because it allows us to use monte carlo techniques). The path integral is over configurations $$\Phi$$ that satisfy the boundary conditions $$\Phi(0) = \Phi(\beta)$$, this is due to the fact that in our definition of $$Z$$ we have to perform a trace.
As for the intuition, what happens when you wick rotate to imaginary time is that the usual time propagator $$e^{-it\hat{H}}$$ becomes the euclidean operator $$e^{-\tau \hat{H}}$$. In Minkowski time, higher energy states essentially just pick up phases very quickly, whereas in euclidean time, higher energy states actually get suppressed exponentially. So, for instance if you took the operator $$\lim_{\tau \to \infty} e^{-\tau \hat{H}}$$, this should map states to some vacuum, because all the positive energy states will get suppressed. Now, when you instead compactify time to $$[0,\beta]$$, you 'aren't letting the states decay to zero', instead they are somehow weighted by their energies, and maybe you can intuitively see why $$\beta$$ should correspond to inverse temperature (as $$T \to 0$$, you expect the thermal system to freeze into the lowest energy state, which corresponds to $$\beta \to \infty$$)