Abstract
Recent studies have considered the possibility that the exact ground-state wavefunction from any Hamiltonian with two-particle interactions may be generated from a single finite two-body exponential transformation acting on an arbitrary Slater determinant [Piecuch et al., Phys. Rev. Lett. 90, 113001 (2003)]. Using the Campbell-Baker-Hausdorff relation, we show that it is difficult for the variational minimum of this trial wave function to satisfy the contracted Schrödinger equation which is a necessary and sufficient condition for the wave function to satisfy the Schrödinger equation. A counterexample is presented through the Lipkin quasispin model with 4–50 fermions. When the number of fermions exceeds four, the wave function from a finite two-body exponential transformation is shown to be inexact. If the trial wave function ansatz is extended to include products of finite two-body exponential transformations acting on an arbitrary Slater-determinant reference, then we show that the ansatz includes the exact ground-state wave function from any Hamiltonian with only two-particle interactions. Connections between the two-body exponential transformation of the wave function and recent research on two-body exponential similarity transformations of the Hamiltonian [S.R. White, J. Chem. Phys. 117, 7472 (2002)] are discussed.
- Received 13 May 2003
DOI:https://doi.org/10.1103/PhysRevA.69.012507
©2004 American Physical Society