5.11: Appendix III- Example Bose Condensation Problem

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\)
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A three-dimensional gas of noninteracting bosonic particles obeys the dispersion relation \(\ve(\Bk)\,=A\,\big|\Bk\big|^{1/2}\).

• Obtain an expression for the density \(n(T,z)\) where \(z=\exp(\mu/\kT)\) is the fugacity. Simplify your expression as best you can, adimensionalizing any integral or infinite sum which may appear. You may find it convenient to define \[{Li}\ns_\nu(z)\equiv{1\over\RGamma(\nu)}\int\limits_0^\infty\!\! dt\>{t^{\nu-1}\over z^{-1}\,e^t -1} =\sum_{k=1}^\infty {z^k\over k^\nu}\ . \label{zetadef}\] Note \({Li}\nd_\nu(1)=\zeta(\nu)\), the Riemann zeta function.
• Find the critical temperature for Bose condensation, \(T_\Rc(n)\). Your expression should only include the density \(n\), the constant \(A\), physical constants, and numerical factors (which may be expressed in terms of integrals or infinite sums).
• What is the condensate density \(n\nd_0\) when \(T=\half\,T_\Rc\)?
• Do you expect the second virial coefficient to be positive or negative? Explain your reasoning. (You don’t have to do any calculation.)

We work in the grand canonical ensemble, using Bose-Einstein statistics.

• The density for Bose-Einstein particles are given by \[\begin{split} n(T,z)&=\int\!\!{d^3\!k\over (2\pi)^3}\,{1\over z^{-1}\,\exp(Ak^{1/2}/\kT)-1}\\ &={1\over\pi^2}\bigg({\kT\over A}\bigg)^{\!\!6}\int\limits_0^\infty\!\!ds\,{s^5\over z^{-1}\,e^s-1}\\ &={120\over\pi^2}\bigg({\kT\over A}\bigg)^{\!\!6}\> {Li}\nd_6(z)\ , \end{split}\] where we have changed integration variables from \(k\) to \(s= Ak^{1/2}/\kT\), and we have defined the functions \({Li}\ns_\nu(z)\) as above, in Equation [zetadef]. Note \({Li}\nd_\nu(1)=\zeta(\nu)\), the Riemann zeta function.
• Bose condensation sets in for \(z=1\), \(\mu=0\). Thus, the critical temperature \(T_\Rc\) and the density \(n\) are related by \[n={120\,\zeta(6)\over \pi^2}\bigg({\kT_\Rc\over A}\bigg)^{\!\!6},\] or \[T_\Rc(n)={A\over \kB}\bigg({\pi^2\,n\over 120\,\zeta(6)}\bigg)^{\!\!1/6}\ .\]
• For \(T<T_\Rc\), we have \[\begin{split} n&=n\nd_0+{120\,\zeta(6)\over\pi^2}\bigg({\kT\over A}\bigg)^{\!\!6}\\ &=n\nd_0+\bigg({T\over T_\Rc}\bigg)^{\!\!6}\,n\ , \end{split}\] where \(n\nd_0\) is the condensate density. Thus, at \(T=\half\,T_\Rc\), \[n\nd_0\big(T=\half T_\Rc\big)=\frac{63}{64}\,n.\]
• The virial expansion of the equation of state is \[p=n\kT\Big(1 + B_2(T)\,n + B_3(T)\,n^2 + \ldots\Big)\ .\] We expect \(B_2(T)<0\) for noninteracting bosons, reflecting the tendency of the bosons to condense. (Correspondingly, for noninteracting fermions we expect \(B_2(T)>0\).)

  For the curious, we compute \(B_2(T)\) by eliminating the fugacity \(z\) from the equations for \(n(T,z)\) and \(p(T,z)\). First, we find \(p(T,z)\): \[\begin{split} p(T,z)&=-\kT\!\int\!\!{d^3\!k\over (2\pi)^3}\,\ln\Big(1- z\,\exp(-Ak^{1/2}/\kT)\Big)\\ &=-{\kT\over\pi^2}\bigg({\kT\over A}\bigg)^{\!\!6}\int\limits_0^\infty\!\!ds\,s^5\,\ln\big(1-z\,e^{-s}\big)\\ &={120\,\kT\over \pi^2}\bigg({\kT\over A}\bigg)^{\!\!6} {Li}\nd_7(z). \end{split}\] Expanding in powers of the fugacity, we have \[\begin{split} n&={120\over \pi^2}\,\bigg({\kT\over A}\bigg)^{\!\!6}\,\Big\{z+{z^2\over 2^6} + {z^3\over 3^6} + \ldots \Big\}\\ {p\over\kT}&={120\over \pi^2}\,\bigg({\kT\over A}\bigg)^{\!\!6}\,\Big\{z+{z^2\over 2^7} + {z^3\over 3^7} + \ldots \Big\}\ . \end{split}\] Solving for \(z(n)\) using the first equation, we obtain, to order \(n^2\), \[z=\bigg({\pi^2 A^6\,n\over 120\, (\kT)^6}\bigg)-{1\over 2^6}\, \bigg({\pi^2 A^6\,n\over 120\, (\kT)^6}\bigg)^{\!\!2} + \CO(n^3)\ .\] Plugging this into the equation for \(p(T,z)\), we obtain the first nontrivial term in the virial expansion, with \[B_2(T)=-{\pi^2\over 15360}\,\bigg({A\over \kT}\bigg)^{\!\!6}\ ,\] which is negative, as expected. Note that the ideal gas law is recovered for \(T\to\infty\), for fixed \(n\).

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1. For a review of the formalism of second quantization, see the appendix in §9.↩
2. Several texts, such as Pathria and Reichl, write \(g\ns_q(z)\) for \({Li}_q(z)\). I adopt the latter notation since we are already using the symbol \(g\) for the density of states function \(g(\ve)\) and for the internal degeneracy \(\Sg\).↩
3. Note the dimensions of \(g(\omega)\) are \(({frequency})^{-1}\). By contrast, the dimensions of \(g(\ve)\) in Equation [BDOS] are \(({energy})^{-1}\cdot({volume})^{-1}\). The difference lies in the a factor of \(\CV\ns_0\cdot\hbar\), where \(\CV\ns_0\) is the unit cell volume.↩
4. If \(\omega(\Bk)=Ak^\sigma\), then \(\CC=2^{1-d}\>\pi\nsub^{-{d\over 2}}\,\sigma^{-1}\,A\nsub^{-{d\over\sigma}}\,\Sg\,\big/\,\RGamma(d/2)\,\).↩
5. OK, that isn’t quite true. For example, if \(g(\ve)\sim 1/\ln\ve\), then the integral has a very weak \(\ln\ln(1/\eta)\) divergence, where \(\eta\) is the lower cutoff. But for any power law density of states \(g(\ve)\propto \ve^r\) with \(r>0\), the integral converges.↩
6. It is easy to see that the chemical potential for noninteracting bosons can never exceed the minimum value \(\ve\ns_0\) of the single particle dispersion.↩
7. Note that in the thermodynamics chapter we used \(v\) to denote the molar volume, \(\NA\,V/N\).↩
8. The \(\Bk\ne 0\) particles are sometimes called the overcondensate.↩
9. IBG condensation is in the universality class of the spherical model. The \(\lambda\)-transition is in the universality class of the \(XY\) model.↩
10. Recall that two bodies in thermal equilibrium will have identical temperatures if they are free to exchange energy.↩
11. The phonon velocity \(c\) is slightly temperature dependent.↩
12. Many reliable descriptions may be found on the web. Check Wikipedia, for example.↩
13. Explicitly, one replaces \(\zeta(3)\) with \(\zeta(2)=\frac{\pi^2}{6}\), \({Li}\ns_3(y)\) with \({Li}\ns_2(y)\), and \(\big(\kT/\hbar{\bar\omega}\big)^3\) with \(\big(\kT/\hbar{\bar\omega}\big)^2\).↩
14. Note that writing \(v=(2n+1)\,i\pi + \eps\) we have \(e^{\pm v}=-1\mp\eps -\half\eps^2 + \ldots\\), so \((e^v+1)(e^{-v}+1)=-\eps^2 + \ldots\) We then expand \(e^{vD}=e^{(2n+1)i\pi D}\big(1+\eps D + \ldots)\) to find the residue: \({Res}=-D\,e^{(2n+1)i\pi D}\).↩
15. I thank my colleague Tarun Grover for this observation.↩
16. We’ve used \(-\frac{2}{V}Q'(\mu)=-{1\over V}{\pz^2\!\Omega\over\pz\mu^2}= n^2\kappa\ns_T\).↩
17. Note that we have written \(\mu n = {\bar\mu}n + {1\over 2} U n^2\), which explains the sign of the coefficient of \(n^2\).↩
18. The Gibbs-Duhem relation guarantees that such an equation of state exists, relating any three intensive thermodynamic quantities.↩
19. A theorem due to Nagaoka establishes that the ground state is ferromagnetic for the case of a single hole in the \(U=\infty\) system on bipartite lattices.↩
20. See J. P. F. LeBlanc , Phys. Rev. X 5, 041041 (2015) and B. Zheng , Science 358, 1155 (2017).↩
21. The best case for stripe order has been made at \(T=0\), \(U/t=8\), and hold doping \(x=\frac{1}{8}\) ( \(n=\frac{7}{8}\)).↩
22. In the normalization integrals, each \(\int\!d^d\!x\) implicitly includes a sum \(\sum_\zeta\) over any internal indices that may be present.↩