# What is the simplest possible Hamiltonian that yields an Antisymmetric Wavefunction?

I am using a Split-Operator Fourier Transform (SOFT) technique to solve the time-dependent electronic Schrödinger Equation (TDSE) for a Hydrogen molecule under the Born-Oppenheimer approximation. So I have two electrons (1 and 2) which are moving and two protons ($$a$$ and $$b$$) which are stationary. The wavefunction $$\psi$$ is thus a function of 6 variables, the three position coordinates of each electron.

Because I am using SOFT, there are no orbitals or Slater determinants, everything is a scalar grid of either real or complex values in position or momentum space. That includes the wavefunction and the operators.

In this setup I am $$not$$ assuming that a many-electron wavefunction can be broken down into a Slater determinant of one-electron spin-orbitals, in fact I am not assuming anything about the wave function. $$\psi$$ is initialized as a random grid and then ground state is reached by iterativelly solving the TDSE, propagating it in imaginary time.

In this regime, I was thinking that the mechanism by which the electronic wavefunction is guaranteed to be antisymmetric is gone. After all, the SE does not take into account the antisymmetry, automatically, it has to be hand-pushed into it by using an appropriate ansatz along with abstract spin "eigenstates".

At this point I realized that using SOFT to solve the many-body TDSE may be fundamentally flawed due to the lack of anti-symmetry. Not wanting to give it up just yet I wondered if maybe I could move the anti-symmetrization from the ansatz of $$\psi$$ to the Hamiltonian.

So I have to ask: Is there a way to construct a many-body Hamiltonian for the SE that will guarantee a ground-state anti-symmetric wavefunction?

• Why can't you just say the antisymmetrized version of $\psi$ is your trial wavefunction? – Jahan Claes Mar 15 '19 at 19:14

Is there a way to construct a many-body Hamiltonian for the SE that will guarantee a ground-state anti-symmetric wavefunction?

Yes. To do this, first note that configuration space of multiple identical fermions is not simply a set of all combinations of positions and spins of these particles. It's a smaller space. As I explain in this answer, two e.g. 1D particles in a box of width $$L$$, when distinguishable, have a square configuration space $$L\times L$$. But if they are indistinguishable, the configuration space is a triangle, with one side being the line of collision of same-spin particles.

For two 3D particles like your case of two electrons, naive configuration space also has a redundant half. When you are looking for an antisymmetric spin-orbital, you need to introduce homogeneous Dirichlet boundary condition on the locus of collision of the electrons with the same spins.

If you are working initially in momentum space, you'll actually want to tweak your basis so that it has a node at the locus where electrons have the same spin-momentum. The basis will become twice as smaller, since you'll throw away all the symmetric basis functions. Then this will automatically ensure that you only get antisymmetric solutions.

• Thanks a lot. Your other answer actually also helped me think about spin as well, which is a problem if you want to compute the wavefunction of a two-fermion system. – urquiza Mar 17 '19 at 21:16
• Do I write the operators (kinetic and potential) also in triangular form? Does that make sense? – urquiza Mar 17 '19 at 21:34
• @urquiza I'm not sure what you mean. The operators should be written in the form where they "assume" that the function vanishes at the locus of same-spin collision. If you do tweak the basis you expand them in, excluding the symmetric basis states, your expansion of the operators will automatically satisfy the antisymmetry requirements. – Ruslan Mar 18 '19 at 5:25

You can't do this with a Hamiltonian alone.

If $$\hat{H}$$ is the Hamiltonian, then let $$\hat{P}$$ be the operator that switches the positions of particle one and particle two: $$\hat{P}|r_1 r_2\rangle=|r_2 r_1\rangle$$. Then $$[\hat{P},\hat{H}]=0$$, so it's possible to simultaneously diagonalize both $$\hat{H}$$ and $$\hat{P}$$. In other words, a generic $$\hat{H}$$ that treats both particles identically will have eigenstates in both the $$\hat{P}=+1$$ and $$\hat{P}=-1$$ subspace, and if you iteratively solve Schrodinger's equation you will simply get whichever eigenstate is nearest your target energy, and you will have to accept whatever symmetry/antisymmetry you get.

This is why people generally use an explicitly antisymmetrized ansatz.

• I understand now that your comment on the question was in reference to what Rusian explained, but since he was a whole lot more detailed and really helped me understand something that I was missing I figured I should give him the win =P – urquiza Mar 17 '19 at 21:20