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I know that in general, interacting fermions cannot necessarily be described by a single Slater determinant. Can anyone provide a simple example of a state that has no such representation?

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2 Answers 2

up vote 6 down vote accepted

How about $\Psi(x_1,x_2) \equiv (x_1 -x_2)e^{-(x_1-x_2)^2}$?

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BebopButUnsteady answer is obviously correct. Actually most wave functions cannot be written as single Slater determinants and certainly no energy eigenstates of N-fermion systems (atom, molecules or solids) interacting by two-body potentials (like the Coulomb potential). The Slater determinant is just a nice simple approximation to ground state (close-shells) wave functions of these systems, but it can be easily shown that the actual wave function is not really a single Slater determinant. The Slater determinants minimize the energy of a quantum system where the fermions interact via an effective average Coulomb potential. They really cannot minimize the energy of the true Coulomb potential. This problem is usually called the lack of dynamic correlation in Quantum Chemistry.

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