I might be totally wrong, but my understanding is that the condensate by definition cannot be written as a state created by a creation operator. Afaik, the condensate isn't even a state at all but an expectation value of the field operator in the ground/vacuum state (vacuum expectation value or "vev"). Given a field operator $\Phi$ you get:
$vev=\langle\Phi\rangle(x) =\langle0 |\Phi(x) |0\rangle=\langle0 |\Phi_0(x)+\delta\Phi(x) |0\rangle = \langle0 |\Phi_0(x)|0\rangle+\langle0 |\delta\Phi(x) |0\rangle$
where the splitting of $\Phi$ in $\Phi_0$ and $\delta\Phi$ is such that
$\langle0 |\Phi(x) |0\rangle=\langle0 |\Phi_0(x)|0\rangle$ and $\langle0 |\delta\Phi(x) |0\rangle=0$
Physically, $\Phi_0(x)$ describes the condensate and $\delta\Phi(x)$ the fluctuations around the condensate. Now, if we want to quantize the field, we write the commutator relations not for the whole field $\Phi(x)$ but for the fluctuating part $\delta\Phi(x)$ only (although many QFT books assume $\Phi_0=0$ such that the field is just its fluctuation). Also the creation and annihilation operators enter in the mode expansion of $\delta\Phi(x)$, not $\Phi(x)$. But since $\langle0 |\delta\Phi(x) |0\rangle=0$ the creation operators inside $\delta\Phi(x)$ will never generate a condensate.
This is a somewhat floppy answer and I'm not even completely sure whether it is correct, so I'd be glad to get corrections and comments!