# What is the wavefunction of a Bose-Einstein condensate made of composite particles?

The textbook answer is that a problem involving bosons is described by a wavefunction $\Psi(R_1,...,R_n)$ which is completely symmetric, and if non-interacting particles Bose condense, the wavefunction is $\phi(R_1)\phi(R_2)...\phi(R_n)$, where $\phi$ is the ground state wavefunction.

But in reality, all particles that bose condense are composite particles, made out of fermions, so they should have a completely antisymmetric wavefunction. For example, we could say each boson is made out of a pair of fermions, and $R_i=\frac{r_{i1}+r_{i2}}{2}$ is really the center of mass of the two fermions. We should be able to write down a totally antisymmetric wavefunction in terms of the $r_{ij}$s, $\Psi(r_{11},r_{12},...,r_{n1},r_{n2})$ that represents the condensed state. What is this wavefunction, and how can we see that it represents condensation?

• As I understand it, Asher Peres’ Quantum Theory, Concepts and Methods has a good discussion of how composite bosons factor out the fermion-ness of its component particles to a global bosonic one. – Emilio Pisanty Jun 26 '17 at 18:44
• Similar issue arises when BCS-BEC crossover is considered and the Legget's pairing ansatz is used (see page 7, capri-school.eu/Capri15/ac1.pdf). – Alexey Sokolik Jun 26 '17 at 19:46
• Addition: in the pairing ansatz, the wave function is $\Psi(\vec{r}_1\ldots\vec{r}_N)=A\phi(\vec{r}_1-\vec{r}_2)\ldots\phi(\vec{r}_{N-1}-\vec{r}_N)$ and it is equivalent to the BCS wave function $|\Psi\rangle=\prod_{\vec{k}}(u_\vec{k}+v_\vec{k}c^+_\vec{k}c^+_{-\vec{k}})|0\rangle$ where the "Cooper pair wave function" $u_\vec{k}v_\vec{k}$ is the Fourier transform of $\phi(\vec{r})$. – Alexey Sokolik Jun 26 '17 at 19:53
• @EmilioPisanty That gave a treatment for two bosons made of two fermions each, which is fine. But my confusion comes in when I try to generalize this to writing down the wavefunction of three or more bosons in the same state. – Jahan Claes Jun 26 '17 at 22:40