# 5.10: Appendix II- Ideal Bose Gas Condensation


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We begin with the grand canonical Hamiltonian $$K=H-\mu N$$ for the ideal Bose gas, $K=\sum_\Bk (\ve\ns_\Bk-\mu)\,b\yd_\Bk b\nd_\Bk - \sqrt{N}\sum_\Bk\big(\nu\ns_\Bk \,b\yd_\Bk + {\bar\nu}\nd_\Bk\, b\nd_\Bk\big)\quad.$ Here $$b\yd_\Bk$$ is the creation operator for a boson in a state of wavevector $$\Bk$$, hence $$\big[b\nd_\Bk\,,\,b\yd_{\Bk'}\big]=\delta\ns_{\Bk\Bk'}$$. The dispersion relation is given by the function $$\ve\ns_\Bk$$, which is the energy of a particle with wavevector $$\Bk$$. We must have $$\ve\ns_\Bk-\mu\ge 0$$ for all $$\Bk$$, lest the spectrum of $$K$$ be unbounded from below. The fields $$\{\nu\ns_\Bk,{\bar\nu}\ns_\Bk\}$$ break a global $$\SO(2)$$ symmetry.

Students who have not taken a course in solid state physics can skip the following paragraph, and be aware that $$N=V/v\ns_0$$ is the total volume of the system in units of a fundamental "unit cell" volume. The thermodynamic limit is then $$N\to\infty$$. Note that $$N$$ is not the boson particle number, which we’ll call $$N\ns_b$$.

Solid state physics boilerplate : We presume a setting in which the real space Hamiltonian is defined by some boson hopping model on a Bravais lattice. The wavevectors $$\Bk$$ are then restricted to the first Brillouin zone, $${\hat\ROmega}$$, and assuming periodic boundary conditions are quantized according to the condition $$\exp\!\big(i N_l\, \Bk\,\cdot\Ba\ns_l\big) = 1$$ for all $$l\in\{1,\ldots,d\}$$, where $$\Ba\ns_l$$ is the $$l^{th}$$ fundamental direct lattice vector and $$N\ns_l$$ is the size of the system in the $$\Ba\ns_l$$ direction; $$d$$ is the dimension of space. The total number of unit cells is $$N\equiv \prod_l N\ns_l$$ . Thus, quantization entails $$\Bk=\sum_l (2\pi n\ns_l/N\ns_l)\,\Bb\ns_l$$ , where $$\Bb\ns_l$$ is the $$l^{th}$$ elementary reciprocal lattice vector ($$\Ba\ns_l\cdot\Bb\ns_{l'}=2\pi\delta\ns_{ll'}$$) and $$n\ns_l$$ ranges over $$N\ns_l$$ distinct integers such that the allowed $$\Bk$$ points form a discrete approximation to $${\hat\ROmega}$$ .

To solve, we first shift the boson creation and annihilation operators, writing $K=\sum_\Bk (\ve\ns_\Bk-\mu)\,\beta\yd_\Bk\beta\nd_\Bk - N\sum_\Bk {|\nu\nd_\Bk|^2\over \ve\ns_\Bk-\mu}\quad,$ where $\beta\nd_\Bk= b\nd_\Bk - {\sqrt{N}\,\nu\nd_\Bk\over \ve\ns_\Bk-\mu}\qquad,\qquad \beta\yd_\Bk= b\yd_\Bk - {\sqrt{N}\,{\bar\nu}_\Bk\over \ve\ns_\Bk-\mu}\quad.$ Note that $$\big[\beta\nd_\Bk\,,\,\beta\yd_{\Bk'}\big]=\delta\nd_{\Bk\Bk'}$$ so the above transformation is canonical. The Landau free energy $$\Omega=-\kT\ln\Xi$$ , where $$\Xi=\Tra\,e^{-K/k\ns_\RB T}$$, is given by $\Omega=N\kT\!\!\impi d\ve\>g(\ve)\,\ln\big(1-e^{(\mu-\ve)/k\ns_\Rb T}\big) - N\sum_\Bk { |\nu\nd_\Bk|^2\over \ve\nd_\Bk - \mu} \quad,$ where $$g(\ve)$$ is the density of energy states per unit cell, $g(\ve)={1\over N}\sum_\Bk\delta\big(\ve-\ve\ns_\Bk\big)\bmapright{N\to\infty}\int\limits_{\hat\ROmega}\!\!{ d^d\!k\over (2\pi)^d}\>\delta\big(\ve-\ve\ns_\Bk\big) \quad.$ Note that $\psi\ns_\Bk\equiv {1\over\sqrt{N}}\,\blangle b\nd_\Bk \brangle = -{1\over N}\,{\pz\Omega\over\pz{\bar\nu}\nd_\Bk} = {\nu\nd_\Bk\over\ve\ns_\Bk-\mu}\quad.$ In the condensed phase, $$\psi\ns_\Bk$$ is nonzero.

The Landau free energy (grand potential) is a function $$\Omega(T,N,\mu,\nu,{\bar\nu})$$. We now make a Legendre transformation, $Y(T,N,\mu,\psi,{\bar\psi})=\Omega(T,N,\mu,\nu,{\bar\nu}) + N\sum_\Bk\big(\nu\ns_\Bk{\bar\psi}\ns_\Bk + {\bar\nu}\ns_\Bk\psi\ns_\Bk\big)\quad.$ Note that ${\pz Y\over\pz{\bar\nu}\ns_\Bk}={\pz\Omega\over\pz{\bar\nu}\ns_\Bk} + N\psi\ns_\Bk=0\quad,$ by the definition of $$\psi\ns_\Bk$$. Similarly, $$\pz Y/\pz\nu\ns_\Bk=0$$. We now have $Y(T,N,\mu,\psi,{\bar\psi})=N\kT\!\!\impi d\ve\>g(\ve)\,\ln\big(1-e^{(\mu-\ve)/k\ns_\Rb T}\big) + N\sum_\Bk(\ve\ns_\Bk-\mu)\,|\psi\nd_\Bk|^2\quad.$ Therefore, the boson particle number per unit cell is given by the dimensionless density, $n={N\ns_b\over N}=-{1\over N}{\pz Y\over\pz\mu} = \sum_\Bk |\psi\nd_\Bk|^2 +\! \impi d\ve\>{g(\ve)\over e^{(\ve-\mu)/k\nd_\RB T} - 1 } \quad,$ and the condensate amplitude at wavevector $$\Bk$$ is $\nu\ns_\Bk = {1\over N}{\pz Y\over\pz{\bar\psi}\ns_\Bk\vphantom{A^B}} = (\ve\ns_\Bk-\mu)\,\psi\ns_\Bk\quad.$

Recall that $$\nu\ns_\Bk$$ acts as an external field. Let the dispersion $$\ve\ns_\Bk$$ be minimized at $$\Bk=\BK$$ . Without loss of generality, we may assume this minimum value is $$\ve\ns_\BK=0$$ . We see that if $$\nu\ns_\Bk=0$$ then one of two must be true:

• $$\psi\ns_\Bk=0$$ for all $$\Bk$$
• $$\mu=\ve\ns_\BK$$ , in which case $$\psi\ns_\BK$$ can be nonzero.

Thus, for $$\nu={\bar\nu}=0$$ and $$\mu>0$$, we have the usual equation of state, $n(T,\mu)=\! \impi d\ve\>{g(\ve)\over e^{(\ve-\mu)/k\nd_\RB T} - 1 } \quad, \label{GDE}$ which relates the intensive variables $$n$$, $$T$$, and $$\mu$$. When $$\mu=0$$, the equation of state becomes $n(T,\mu=0) = \stackrel{n\ns_0}{\overbrace{\sum_\BK |\psi\nd_\BK|^2}} + \! \stackrel{n\ns_>(T)}{\overbrace{\impi d\ve\>{g(\ve)\over e^{\ve/k\nd_\RB T} - 1 }}} \quad, \label{condeqn}$ where now the sum is over only those $$\BK$$ for which $$\ve\ns_\BK=0$$ . Typically this set has only one member, $$\BK=0$$, but it is quite possible, due to symmetry reasons, that there are more such $$\BK$$ values. This last equation of state is one which relates the intensive variables $$n$$, $$T$$, and $$n\ns_0$$ , where $n\ns_0=\sum_\BK |\psi\nd_\BK|^2$ is the dimensionless condensate density. If the integral $$n\ns_>(T)$$ in Equation [condeqn] is finite, then for $$n>n\ns_0(T)$$ we must have $$n\ns_0>0$$. Note that, for any $$T$$, $$n\ns_>(T)$$ diverges logarithmically whenever $$g(0)$$ is finite. This means that Equation [GDE] can always be inverted to yield a finite $$\mu(n,T)$$, no matter how large the value of $$n$$, in which case there is no condensation and $$n\ns_0=0$$. If $$g(\ve)\propto\ve^\alpha$$ with $$\alpha>0$$, the integral converges and $$n\ns_>(T)$$ is finite and monotonically increasing for all $$T$$. Thus, for fixed dimensionless number $$n$$, there will be a critical temperature $$T\ns_\Rc$$ for which $$n=n\ns_>(T\ns_\Rc)$$. For $$T<T\ns_\Rc$$ , Equation [GDE] has no solution for any $$\mu$$ and we must appeal to Equation [condeqn]. The condensate density, given by $$n\ns_0(n,T)=n-n\ns_>(T)$$ , is then finite for $$T<T\ns_\Rc$$ , and vanishes for $$T\ge T\ns_\Rc$$ .

In the condensed phase, the phase of the order parameter $$\psi$$ inherits its phase from the external field $$\nu$$, which is taken to zero, in the same way the magnetization in the symmetry-broken phase of an Ising ferromagnet inherits its direction from an applied field $$h$$ which is taken to zero. The important feature is that in both cases the applied field is taken to zero after the approach to the thermodynamic limit.

This page titled 5.10: Appendix II- Ideal Bose Gas Condensation is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.