From the definition of Slater determinant it seems that the set of single-particle wavefunctions is chosen to be a orthonormal one. Is orthogonality required for the Slater determinant to describe an antisymmetric N-electron wavefunction?
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$\begingroup$ Wikipedia says yes. en.wikipedia.org/wiki/Slater_determinant $\endgroup$– probably_someoneCommented May 31, 2018 at 23:07
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3 Answers
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No, it's not necessary but if the spin-orbitals aren't orthogonal it's much more complicated to compute matrix elements of operators between such Slater determinants (Slater-Condon rules). Valence-bond theory famously uses non-orthogonal orbitals and as a result has to face a host of computational problems because of it, so much so that most approaches prefer to rely on orthogonal orbitals.
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As Lorents answer correctly points out, it is not necessary (to use orthonormal single-particle wavefunctions), but just much more convenient. In addition, it is no real restriction, since $$\left|\begin{matrix} \chi_1(\mathbf{x}_1) & \chi_2(\mathbf{x}_1) & \cdots & \chi_N(\mathbf{x}_1) \\ \chi_1(\mathbf{x}_2) & \chi_2(\mathbf{x}_2) & \cdots & \chi_N(\mathbf{x}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \chi_1(\mathbf{x}_N) & \chi_2(\mathbf{x}_N) & \cdots & \chi_N(\mathbf{x}_N) \end{matrix}\right|=\left|\begin{matrix} \chi_1(\mathbf{x}_1) & \chi_2(\mathbf{x}_1)+c_{12}\chi_1(\mathbf{x}_1) & \cdots & \chi_N(\mathbf{x}_1)+c_{1N}\chi_1(\mathbf{x}_1) \\ \chi_1(\mathbf{x}_2) & \chi_2(\mathbf{x}_2)+c_{12}\chi_1(\mathbf{x}_2) & \cdots & \chi_N(\mathbf{x}_2)+c_{1N}\chi_1(\mathbf{x}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \chi_1(\mathbf{x}_N) & \chi_2(\mathbf{x}_N)+c_{12}\chi_1(\mathbf{x}_N) & \cdots & \chi_N(\mathbf{x}_N)+c_{1N}\chi_1(\mathbf{x}_N) \end{matrix}\right|$$ and more generally for $C=\begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1N} \\ c_{21} & c_{22} & \cdots & c_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ c_{N1} & c_{N2} & \cdots & c_{NN} \end{bmatrix}$ we have $$\left|\begin{bmatrix} \chi_1(\mathbf{x}_1) & \chi_2(\mathbf{x}_1) & \cdots & \chi_N(\mathbf{x}_1) \\ \chi_1(\mathbf{x}_2) & \chi_2(\mathbf{x}_2) & \cdots & \chi_N(\mathbf{x}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \chi_1(\mathbf{x}_N) & \chi_2(\mathbf{x}_N) & \cdots & \chi_N(\mathbf{x}_N) \end{bmatrix}C\right|=\left|\begin{matrix} \chi_1(\mathbf{x}_1) & \chi_2(\mathbf{x}_1) & \cdots & \chi_N(\mathbf{x}_1) \\ \chi_1(\mathbf{x}_2) & \chi_2(\mathbf{x}_2) & \cdots & \chi_N(\mathbf{x}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \chi_1(\mathbf{x}_N) & \chi_2(\mathbf{x}_N) & \cdots & \chi_N(\mathbf{x}_N) \end{matrix}\right|\operatorname{det}(C)$$ Hence we can orthogonalize the wavefunctions without changing the resulting many-particle state. (This also shows that the specific basis of the subspace is not important, as long as one takes care of the phase factor.)
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This should be a comment, but I don't have enough rep.
If I'm not mistaken, what @Thomas Klimpel showed is really orthogonalizing a set of basis vectors, not the underlying AO's from which they are constructed.
Gerrat's book on VB discusses orthogonalizing basis vectors at length, and mentions in passing that that's distinct from the question of whether or not the AO's are orthogonal.
I'm not clear on whether this is always possible or not.
