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5.11: Appendix III- Example Bose Condensation Problem

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    18768
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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 \(n\).


    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.
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