# The overlap of two Slater determinant states

Suppose I have two fermionic number states in different bases, with the same particle number $$N$$ - call them $$|\Psi\rangle$$ and $$|\Phi\rangle$$. In the position basis, I can write the many-body wavefunctions as the Slater determinants: $$\Psi(x_1,x_2,\dots)=\frac{1}{\sqrt{N!}}\det(\overline{\psi})$$ $$\Phi(x_1,x_2,\dots)=\frac{1}{\sqrt{N!}}\det(\overline{\phi})$$ where: $$\overline{\psi}=\begin{bmatrix}\psi_1(x_1)&\psi_2(x_1)&\dots&\psi_N(x_1)\\\psi_1(x_2)&\psi_2(x_2)&\dots&\psi_N(x_2)\\\vdots&\vdots&\ddots&\vdots\\\psi_1(x_N)&\psi_2(x_N)&\dots&\psi_N(x_N)\end{bmatrix}$$ $$\overline{\phi}=\begin{bmatrix}\phi_1(x_1)&\phi_2(x_1)&\dots&\phi_N(x_1)\\\phi_1(x_2)&\phi_2(x_2)&\dots&\phi_N(x_2)\\\vdots&\vdots&\ddots&\vdots\\\phi_1(x_N)&\phi_2(x_N)&\dots&\phi_N(x_N)\end{bmatrix}$$

Suppose now that I know the $$\psi_n,\phi_n$$, and I wish to calculate $$\langle\Psi|\Phi\rangle=\int d\mathbf{x} \Psi^*(\mathbf{x})\Phi(\mathbf{x})$$. Is there a tractable way to compute this? I guess I could use the fact that $$\det(\overline{\Psi}^\dagger\overline{\Phi})=\det(\overline{\Psi}^\dagger)\det(\overline{\Phi})$$ to construct the integrand, then integrate over all spatial dimensions. But this seems truly horrendous: an $$N\times N$$ matrix multiplication and determinant calculation, then $$N$$ spatial integrations. I don't believe it could actually be that complicated. Is there a simplification I'm missing that will let me tractably compute this? I am hoping it would reduce to something like this: $$\sum_n\int dx \phi_n^*(x)\psi_n(x)$$ but properly antisymmetrized so that it is invariant under exchange of particle identities. I'm totally stumped on how to simplify to a form like that, though.