Factorizing Slater determinant into product of spin-up and spin-down slater determinant I am reading the book on MONTE CARLO METHODS IN AB INITIO QUANTUM CHEMISTRY by Hammond and Lester. (Chapter 5-Variational Trial Functions, Section 5.3-Hartree Fock and Beyond, Sub-Section 5.3.3 Hartree Fock and Correlation Energies)In section 5.3.3, the book says the following,

"In Monte Carlo, we may factor the determinant into electron spin components, i.e., $\psi_{D}=\psi_{\alpha}\psi_{\beta}$. This factorization is justified because all possible permutations of this term that are present in the full antisymmetric $\psi$  are operationally equivalent: each term simply corresponds to a relabeling of the electrons. Expectation values, and even local values of any observable, are unchanged."

I couldn't understand how these two are equal. Is it really possible to factorize this into two slater determinants? How does one justify that this is true?
When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function.
Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product. It is like up- and down-spin wave functions occupy different subspaces and interact via the electron–electron correlation term in the Hamiltonian.