Weakly Interacting Homogeneous Bose Gas

Typical introductory statistical mechanics courses examine BECs by assuming that they consist of many non-interacting atoms. That simple model does an excellent job of predicting the condensation temperature and fraction of atoms in the condensate, however it quantitatively and even qualitatively misses some of the properties of real BECs. The issue is that real atoms interact with each other and these interactions can alter many properties of a BEC. Fortunately, a simple mean-field treatment of the interactions can create an excellent model that captures much of the behavior seen in real BECs as will be shown in the following sections.

Weakly Interacting Bose Gas near $T=0$

We can start to account for atom-atom interactions by adding a collisional term to the hamiltonian. We can consider a collision as a process that annihilates a particle with momentum $p$ and a particle with momentum $k$, then creates two particles with momenta $p^{\prime }$ and $k^{\prime }$. By momentum conservation we may write $p^{\prime }=p+q$ and $k^{\prime }=k-q$. We let $U_{q}$ be the matrix element for this process, and so we can write collisional hamiltonian as the sum of all possible collisions (with a factor of 2 to avoid double-counting input states)

$H_{I}={\frac {1}{2V}}\sum _{k,p,q}U_{q}a_{p+q}^{\dagger }a_{k-q}^{\dagger }a_{k}a_{p}\,.$

This hamiltonian is far too complicated to solve in the general case, so we must make some approximations. First, for typical BEC parameters, the spacing between atoms is much larger than the collisional scattering length of the atoms. Therefore the complicated atomic interaction potential can be well approximated by replacing it with a delta function potential. In particular, if the atomic separation is $r$, then we may write the potential as $U(r)=U_{0}\delta (r)$ where $U_{0}=4\pi \hbar ^{2}a/m$ and $a$ is the $s$-wave scattering length. Now $U_{q}$ is the Fourier transform of $U(r)$, and since the Fourier transform of a delta function is a constant function, we have that $U_{q}=U_{0}$. So we may write

$H_{I}\approx {\frac {U_{0}}{2V}}\sum _{k,p,q}a_{p+q}^{\dagger }a_{k-q}^{\dagger }a_{k}a_{p}\,.$

Unfortunately this Hamiltonian is still too complicated to solve. The reason is that it is extremely difficult to diagonalize hamiltonians that are a product of four operators. Therefore we need some way to simplify things down to two operators. We do this with the Bogoliubov approximation, which says that when there are a large number $N_{0}$ of atoms in the condenstate, we may approximate $N_{0}+1\approx N_{0}$. And since $a_{0}^{\dagger }|N_{0}\rangle ={\sqrt {N_{0}+1}}|N_{0}+1\rangle$ and $a_{0}|N_{0}\rangle ={\sqrt {N_{0}}}|N_{0}-1\rangle$, we may then approximate those operators with c-numbers $a_{0}\approx a_{0}^{\dagger }\approx {\sqrt {N_{0}}}$. Furthermore, since $N_{0}$ is large, we see that the terms in the hamiltonian that will dominate are the ones in which there are two or more occurrences of $a_{0}$ and/or $a_{0}^{\dagger }$. We may therefore approximate the hamiltonian as

$H_{I}\approx {\frac {U_{0}N_{0}^{2}}{2V}}+{\frac {U_{0}N_{0}}{2V}}\sum _{k\neq 0}a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}+2a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

The full hamiltonian is simply this plus the kinetic energy term

$H\approx {\frac {U_{0}N_{0}^{2}}{2V}}+\sum _{k}\epsilon _{k}a_{k}^{\dagger }a_{k}+{\frac {U_{0}N_{0}}{2V}}\sum _{k\neq 0}a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}+2a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

At this point we have made enough approximations to arrive at a solvable hamiltonian, since it only involves quadratic products of operators.

To show that it is possible to diagonalize this hamiltonian, we will outline the method. We start by replacing $N_{0}$ with $N$ using the relation

$N=N_{0}+{\frac {1}{2}}\sum _{k\neq 9}a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

to give

$H\approx {\frac {U_{0}N^{2}}{2V}}+{\frac {1}{2}}\sum _{k\neq 0}\left(\epsilon _{k}+{\frac {NU_{0}}{V}}\right)\left(a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\right)+{\frac {NU_{0}}{V}}\left(a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}\right)\,.$

Each term in the sum has the form

$H_{k}=E_{0,k}\left(a^{\dagger }a+b^{\dagger }b\right)+E_{1}\left(a^{\dagger }b^{\dagger }+ab\right)\,.$

where we've define $E_{0,k}=\left(\epsilon _{k}+{\frac {NU_{0}}{V}}\right)$, $E_{1}={\frac {NU_{0}}{V}}$, $a=a_{k}$ and $b=a_{-k}$. Now the first term in $H_{k}$ looks nice, it's just the hamiltonian for harmonic oscillators. The second term however is inconvenient, so we'd like to get rid of it. This can be done by performing the Bogoliubov transformations, which rewrites the hamiltonian in terms of different operators $\alpha$ and $\beta$, which are superpositions of $a$ and $b$. In particular we write

Failed to parse (syntax error): {\displaystyle a=u\alpha -v\beta^\dagger \\ b=u\beta -v\alpha^\dagger }

where we have the freedom to choose $u$ and $v$. The first requirement we impose is to make sure $[\alpha ,\alpha ^{\dagger }]=[\beta ,\beta ^{\dagger }]=1$ so that $\alpha$ and $\beta$ obey the bosonic commutation relations. This forces $u^{2}-v^{2}=1$, but still leaves one degree of freedom. When plugging $\alpha$ and $\beta$ into $H_{k}$, we see we'll get something of the form

$H_{k}=(\ldots )\left(\alpha ^{\dagger }\alpha +\beta ^{\dagger }\beta \right)+(\ldots )\left(\alpha ^{\dagger }\beta ^{\dagger }+\alpha \beta \right)$

Therefore, we use our other degree of freedom to choose $u$ and $v$ such that the second prefactor is zero, thereby getting rid of the troublesome terms. We are just left with

$H_{k}=(\ldots )\left(\alpha ^{\dagger }\alpha +\beta ^{\dagger }\beta \right)$

for which we know the solutions are just the harmonic oscillator eigenstates.

Sound propagation in Bose-Einstein condensates

From the previous section, we see that we may write the hamiltonian under the Bogoliubov approximation as

$H=\sum _{k}E_{k}a_{k}^{\dagger }a_{k}+{\text{const.}}$

where we've changed notation a bit so that $a$ now represents what we had called $\alpha$, the annihilation operator for a quasiparticle (not an atom). We've also used

$E_{k}={\sqrt {\left({\frac {\hbar ^{2}k^{2}}{2m}}\right)^{2}+(\hbar ck)^{2}}}\,.$

and $c={\sqrt {NU_{0}/Vm}}$ represents the speed of sound in the BEC. This dispersion relation shows us that the low lying excitaitons are phonons. At $k\rightarrow 0$, $E_{k}=\hbar ck$ that of sound, while at $k\rightarrow \infty$, $E_{k}=\hbar ^{2}k^{2}/2m$, a free particle. Free particles start with a quadratic dispersion relation, while phonons and other Bose systems start with a linear dispersion relation.

File:Superfluid to Mott insulator transition-bec3-sound1

File:Superfluid to Mott insulator transition-bec3-sound2

The Bogolubov solution has a great deal of physics in it. It gives the elementary excitation, and the ground state energy. In the simple model that we have a mean field, the ground state energy is

$E_{0}={\frac {U_{0}h}{2}}\left(1+{\frac {128}{15}}{\sqrt {na^{3}/\pi }}\right)\,.$

The extra correction term on the right is a small term, recently observed by the Innsbruck group, due to collective effects. The Bogolubov solution also gives the ground state wavefunction,

$|\psi _{0}\rangle =|0{\rangle }^{\otimes N}+\epsilon \,,$

where $\epsilon$ is the quantum depletion term, which makes the wavefunction satisfy

$\langle n_{k}\rangle ={\frac {v_{k}^{2}}{1-v_{k}^{2}}}$

and

$\langle n_{0}\rangle =N-\sum \langle n_{k}\rangle =N\left[{1-{\frac {8}{3}}{\sqrt {na^{3}/\pi }}}\right]\,.$

The quantum depletion term, which arises from the fact that the gas is weakly interacting, has now been experimentally observed. Recall that in the Bogolubov approximation, the original interaction

$H'=U_{0}\sum a_{p}^{\dagger }a_{q}^{\dagger }a_{r}a_{s}$

is approximated by

$H'=U_{0}a_{0}a_{0}\sum a_{p}^{\dagger }a_{-p}^{\dagger }$

The quantum depletion this leads to is very small. The effect can be more readily experimentally observed by increasing the mass of the particle, and this can be done by placing the particles in a lattice. Plotting the quantum depletion which can be obtained as a function of lattice depth, in such an experiment, one gets:

File:Superfluid to Mott insulator transition-bec3-qdepletion

Beyond a quantum depletion fraction of $0.6$, the Bogolubov approximation breaks down, as the condensate goes through a superfluid to Mott-insulator transition.

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