# Free parameter in Bose Einstein Condensate

In Kapusta and Gale's Finite-Temperature Field Theory book, BEC is derived for a complex scalar by Fourier expanding $$\phi _1 = \sqrt2 \zeta \cos \theta + \sqrt{\frac{\beta}{V}}\sum_{n,\bar p}e^{i(\bar p \cdot\bar x + \omega _n \tau)}\phi_{1,n}(\bar p)$$ $$\phi _2 = \sqrt2 \zeta \sin \theta + \sqrt{\frac{\beta}{V}}\sum_{n,\bar p}e^{i(\bar p \cdot\bar x + \omega _n \tau)}\phi_{2,n}(\bar p)$$ then calculating the partition function to get $$\ln Z = \beta V (\mu ^2 - m^2) \zeta^2 - V \int \frac{d^3p}{(2 \pi)^3}\left( \beta \omega + \ln(1-e^{-\beta(\omega - \mu )}) + \ln(1-e^{-\beta(\omega + \mu )}) \right) \,.$$ We note that $$\theta$$ is eliminated as expected by symmetry considerations, while for $$\zeta$$ we require for a fixed temperature and chemical that $$\partial_\zeta \ln Z= 2\beta V (\mu^2 -m^2)\zeta = 0 \,,$$ and this is "where the magic happens", for when $$|\mu| = m$$ we will get some $$\zeta>0$$, i.e. a condensate.

My question is, why do we need to set $$\partial _\zeta \ln Z = 0$$? What is the physical meaning and justification for this condition?

But your equation $$\partial_\zeta \ln Z=0$$ does not fix $$\zeta$$! Either $$\zeta=0$$, or, when $$\mu= m$$, any value is allowed. This latter case corresponds to any condensate density $$\langle \phi\rangle^2= |\zeta|^2$$ being possible. Thus when we have a non-interacting bose gas with a condensate you will have $$\mu$$ exactly equal to $$m$$. It is only when you have an interaction that there will be a non-trvial relation between $$\mu$$ and the condensate fraction $$\zeta$$ given by minimizing something like $$V(\phi)= \frac 12 (m^2-\mu^2) |\phi|^2 +\frac 1{4!} (|\phi|^2)^2$$ to find $$\langle \phi\rangle=\zeta$$.
• Thanks, I wasn't clear in the question, will edit post. I didn't mean to talk about what happens when $|\mu| = m$, only the condition itself - why do we need $\partial_\zeta \ln Z = 0$ in the first place?
• @Yoni becuase you are replacing the annihilation operator $\phi_0$ for, say, the $n=0$ by c-number (you are bit vague in your question but I assume that you are doing quantum thermodynamics so the $\phi_n$ are annihilation operators) and you need to determine it by some equation of motion. Sep 2 '19 at 17:50