0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $\langle \phi\rangle=\zeta$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Yoni
    Sep 2 '19 at 12:51
  • $\begingroup$ @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. $\endgroup$
    – mike stone
    Sep 2 '19 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.