# What do we mean when we say Bose condensation? And why can't fermions condense if they don't pair?

In common textbooks, we are told that bosons can condense in a single-particle state because of bose statistics and when the system undergoes a bose condensation, the bose field operator obtains a non-zero groundstate expectation value (GSEV). Apparently, this groundstate cannot be a particle conserved states otherwise the expectation value for the bose field operator must be zero. So is this non-zero field operator GSEV the main feature of Bose condensation?

For fermions, we all know that unpaired fermions cannot condense because of Pauli exclusion principle. But it seems that the fermi field operator can also obtain a non-zero GSEV if the groundstate of some "strange" system is the superposition of zero fermion state and one fermion state (i.e. $|GS\rangle =\prod_{\alpha}u_{\alpha} |0\rangle_{\alpha}+v_{\alpha}|1\rangle_{\alpha}$, where $\alpha$ is the index for single-particle state.). This is weird to me. I have never seen such a situation in literature. Can this strange groundstate exist? And if it exists, does this non-zero GSEV imply some kind of condensation?

• Maybe track down some articles on $He^3$ superfluidity? That's an example of Fermi-like particle condensation. – Carl Witthoft Jun 19 '14 at 15:40
• He3 superfluidity is an example of paired fermions that undergo condensation. In that case it is $c_{\alpha}c_{\beta}$ that obtains a non-zero GSEV but not $c_{\alpha}$. This paired case can be seen as some kind of bose condensation. – Waltergu Jun 19 '14 at 15:48

The defining feature of a Bose condensate is that the one-body density matrix $$\rho^{(1)}(\mathbf{r},\mathbf{r}^{\prime}) = \langle \Psi^{\dagger}(\mathbf{r})\Psi(\mathbf{r}^{\prime})\rangle,$$ has at least one eigenvalue that is macroscopically large, i.e. it is of order $N$, with $N$ the number of particles in the system. Here, $\Psi(\mathbf{r})$ is the field operator describing the particles in question. In fermionic systems, the corresponding condition is that the two-body density matrix $$\rho^{(2)}_{s_1s_2s_3s_4}(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3,\mathbf{r}_4) = \langle \Psi^{\dagger}_{s_1}(\mathbf{r}_1)\Psi^{\dagger}_{s_2}(\mathbf{r}_2)\Psi_{s_3}(\mathbf{r}_3)\Psi_{s_4}(\mathbf{r}_4)\rangle,$$ where $s$ indexes the spin variables, has at least one macroscopic eigenvalue.
In the vast majority of conceivable situations, including almost all experiments that have been ever performed$^{\ast}$, the number of particles is strictly conserved. It is therefore fundamentally incorrect to state that the field operator obtains a non-zero ground state expectation value (GSEV) in general. Assigning a non-zero GSEV to the field operator is just a useful calculational trick that leads rapidly to the observable physics. The drawback, of course, being that it also leads to all kinds of conceptual confusion. Therefore, although the presence of a non-zero GSEV implies some sort of condensation, the converse is definitely not true.
$^{\ast}$ Recently people have begun to experiment with more exotic systems such as photon condensates, where there is no superselection rule enforcing particle conservation.