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In Hartree Fock equations, The concept of Pauli exclusion principle is included in the exchange term (writing the wave function in a Slater determinant form, causes the exchange term). I have written a code in which I am using the SCF method to solve Hartree-Fock equations for electrons inside a 1 dimensional infinite quantum Well, and I am using Sin functions as the initial wave functions. My energies are converging and It can find the ground state energies.

My question: To test my code, I was thinking to set the potential equal to zero and only keep h1, which is the hamiltonian of a single particle in a 1D infinite square Well. So that I can compare the numerical results (computed wave functions) from my code with the simple 1D infinite square Well wave functions for which I know the analytical solution. The problem is by doing that I am loosing the exclusion term (which has the fermionic behavior of particles in its heart). Does any one have any suggestion how can I turn off the interction potential and still maintain the fermionic behavior of particles?

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The Slater determinant guarantees antisymmetry and this ansatz wavefunction is valid irrespective of the potential.

The difficulty you will have is that the boundary condition of your problem change completely when you go from an infinite well (a confining potential with hard boundaries) to no potential (obviously not confining).

From what I understand you should be able to just keep the single particle part (which would be $h_1$ for the infinite well) and find that the antisymmetry conditions forces you to distribute your fermions one (or two if you keep track of spin) at the time in each of the single particle levels, i.e. the ground state for $3$ particles would be (up to normalization and ignoring spin) $$ \psi(x_1,x_2,x_3)= \left\vert\begin{array}{ccc} \phi_1(x_1)&\phi_1(x_2)&\phi_1(x_3)\\ \phi_2(x_1)&\phi_2(x_2)&\phi_2(x_3)\\ \phi_3(x_1)&\phi_3(x_2)&\phi_3(x_3)\end{array}\right\vert \tag{1} $$ with energy $E_{tot}=E_1+E_2+E_3$ where $E_n$ is the $n$'th eigenvalue of the single particle in an infinite well, and $\phi_n(x)$ the corresponding eigenfunction. After all, (1) is the ground state of 3 non-interacting fermions, and you can verify that Pauli exclusion is still in effect.

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  • $\begingroup$ Thank you for your answer. So does it mean that my initial basis functions should be also slater determinant? for example the antisymmetric form of the Sin functions? The thing is that it would very difficult to substitute it in interaction and exchange integrals. $\endgroup$
    – Delaram Nematollahi
    Commented Jul 12, 2017 at 14:18
  • $\begingroup$ @DelaramNematollahi if I understand what you have in mind yes the initial basis should be in the form of Slater determinants. The interaction and exchange integrals are indeed the difficult part. $\endgroup$
    – ZeroTheHero
    Commented Jul 12, 2017 at 15:33

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