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Could someone explain to me what this snippet of text means?

Although it is possible for DMC to be used as a benchmark for quantum-chemistry methods and vice versa, DMC does not operate in a Slater determinant space, but rather a real space representation of the wavefunction. As such, it would require quantum-chemical calculations to be converged to high accuracy with respect to the basis set size before any meaningful comparisons could be drawn.

DMC stands for Diffusion Monte Carlo. I know what a Slater determinant is, but what is a Slater determinant space? And real space is just $\Psi(\mathbf{r}, \mathbf{t})$, right?

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It seems that they refer to the difference between the "individual" fermion wave functions $\psi(x_i)$ instead of the "total" wave function $\Psi(\{x_i\})$. –  Vibert Apr 7 '13 at 9:31

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Let us consider (for a moment) a finite dimensional one particle Hilbert space $\mathcal{H}$of dimension $N$. Consider a set of $n \le N$ identical fermions, each of which living in $\mathcal{H}$. The state of the system of the $n$ identical fermions must be of the form of a linear combination of Slater determinants each built from a basis of $\mathcal{H}$.

The Hartree-Fock approximation idea is to allow only states comprised of a single Slater determinant. These states can all be obtained from a single state via a unitary transformation of the Hilbert space $\mathcal{H}$ basis, these transformations form the unitary group $U(N)$. However, not all of these transformations are independent; because a transformation which do not mix the occupied and non-occupied one particle states will only result a global phase in the Slater determinant. Thus the space of all possible Slater determinants modulo a global phase is the complex Grassmannian coset space

$ Gr(n,N) = \frac{U(N)}{U(n) \times U(N-n)}$

This is the Slater determinant space of dimension $2n(N-n)$ . The solution in the Hartree-Fock approximation amounts the determination of a single point on this Grassmannian minimizing the Hartree-Fock functional.

In the case of an infinite dimensional one particle Hilbert space, the Slater determinant space becomes an infinite dimensional Grassmannian. To apply Monte-carlo methods, it is important to know how to sample on these types of manifolds, please see for example the following article by: Raczkowski,Fong, Schultz, Lippert and Stechel.

For a review of the application in the infinite dimensional case, please see the article "Stiefel and Grassmann manifolds in Quantum Chemistry"

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