# How do I extract the groundstate of a quadratic field theory?

I have successfully solved a field theory quadratic in fermonic creation and annihilation operators via Bogolyubov transformation to a diagonal field theory. I now want to extract the groundstate of the theory, which is characterized by being in the kernel of each annihilation operator in the new diagonal theory. This translates, in my case, to the equations $$(c_j+c_{j+1}-c_j^\dagger+c_{j+1}^\dagger)\Psi_\text{GND}=0,~~~~j=1,2,\cdots N-1$$ $$(c_1+c_1^\dagger+c_N-c_N^\dagger)\Psi_\text{GND}=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ This constitutes a set of $N$ simultaneous linear equations. How do I now efficiently extract the unique solution $\Psi_\text{GND}$ to these equations (up to scaling by a complex number)? Is there a natural ansatz to take?

A straightforward way is to use the Bogolyubov transformation in unitary form. Since I don't know your exact transformation, consider a general one of the form $${\hat c}_j = (\cosh\theta_j) {\hat a}_j + (\sinh\theta_j) {\hat b}^\dagger_j\\ {\hat d}^\dagger_j = (\sinh\theta_j) {\hat a}_j + (\cosh\theta_j) {\hat b}^\dagger_j$$ Then $${\hat c}_j = {\hat U}^\dagger{\hat a}_j {\hat U},\;\;\; {\hat d}_j = {\hat U}^\dagger{\hat b}_j {\hat U}$$ for $${\hat U} = e^{\sum_j{\theta_j \left( {\hat a}^\dagger_j {\hat b}^\dagger_j - {\hat a}_j {\hat b}_j \right)}}$$ In this case the new vacuum for ${\hat c}_j$, ${\hat d}_j$ is the unitary transform of the non-interacting one for ${\hat a}_j$, ${\hat b}_j$: $${\hat a}_j |0\rangle = 0 \;\;\; \Rightarrow\;\;\; U^\dagger {\hat a}_j U U^\dagger |0\rangle = 0, \;\;\; {\hat c}_j U^\dagger|0\rangle = 0$$ and similarly $${\hat b}_j |0\rangle = 0 \;\;\; \Rightarrow\;\;\; {\hat d}_j U^\dagger|0\rangle = 0$$ See pgs.5,12,15, etc. in this review of "Canonical Transformations in Quantum Field Theory".