How to tell if a state written in second quantization is a Slater determinant?

Question

How do I tell if a state written in second quantization is a Slater determinant? I was solving some basic quantum-many body systems, and, for numerical purposes, I would like to determine if the ground state that I found is a Slater determinant.

Answer 1

Let $\mathfrak h$ denote a single-particle Hilbert space, and $H_N:=\wedge^N \mathfrak h$ the Hilbert space of $N$ identical fermions. We call a normalized vector $\psi \in H_N$ a Slater determinant if there exists $\varphi_1,\varphi_2,\ldots,\varphi_N \in\mathfrak h$ with $\langle \varphi_i,\varphi_j\rangle_\mathfrak h=\delta_{ij}$ for $i,j\in \{1,2,\ldots, N\}$ such that

$$\psi = \varphi_1\wedge \varphi_2\wedge \ldots \wedge \varphi_N = \frac{1}{\sqrt N} a^\dagger (\varphi_1)a^\dagger(\varphi_2)\cdots a^\dagger(\varphi_N)\, |0\rangle \quad . \tag 1 $$

Given any $\psi\in H_N$, one can define an operator $\gamma_\psi:\mathfrak h\rightarrow \mathfrak h$, the so-called one-body reduced density matrix, via $$\langle \varphi, \gamma_\psi \phi\rangle_\mathfrak h := \langle \psi, a^\dagger(\phi)a(\varphi)\psi\rangle_{H_N} \quad . \tag 2 $$

Theorem: A normalized vector $\psi \in H_N$ is a Slater determinant if and only if $\gamma_\psi^2=\gamma_\psi$.

The proof is by direct computation and can be found in e.g. Many-Particle Theory. E. Gross and E. Runge. Chapter 6, p. 49.

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Thus, given your (normalized) ground state $\psi$, you can choose a suitable orthonormal single-particle basis $\{e_j\}_{j=1,2,\ldots, \dim \mathfrak h}$, construct the matrix $(\gamma_\psi)_{ij}:=\langle \psi, a^\dagger(e_j)a(e_i)\psi\rangle_{H_N}$ and simply check whether it is idempotent.

That being said, most interacting systems do not admit Slater determinant ground states. A notable exception is e.g. the Hubbard model with $t=0$, where at least one ground state is a Slater determinant.

Answer 2

A many-fermion state that can be written as a single Slater determinant if and only if it satisfies the Plucker relations.