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.