Slater determinant
This "Hartree product" gives us the wavefunction of a system (many-particle) as a combination of wavefunctions of the individual particles. It is inherently mean-field (assumes the particles are independent) and is the unsymmetrized version of the Slater determinant ansatz in the Hartree–Fock method. Although it has the advantage of simplicity, the Hartree product is not satisfactory for fermions, such as electrons, because the resulting wave function is not antisymmetric. An antisymmetric wave function can be mathematically described using the Slater determinant.