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# 5.11: Appendix III- Example Bose Condensation Problem

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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.