Hartree-Fock wave function for a mixture of two oppositely charged gases

[iibewegung]

Hi,

Suppose there is half-half a mixture of an electron gas and a gas of some hypothetical particle with the same mass, spin but an opposite charge (like positrons that don't decay).

Would anyone be able to tell me how the Slater determinants combine in this case?
The HF wave function for just the electron gas would be a determinant of single-electron wave functions...
so does this mean I can construct a similar looking determinant for the positrons and add (or multiply) it to the electron gas determinant?

Any insights greatly appreciated!

 

[BigJDubs]

Yes, you can combine the Slater determinants for the electron gas and the positron gas by taking the product of the two determinants. This will result in a wavefunction that is a product of the Slater determinants for the electron and positron gases. The resulting wavefunction will take into account the effects of Coulomb repulsion between the electrons and positrons.

 

[charng]

Hello,

Thank you for your question regarding the Hartree-Fock wave function for a mixture of two oppositely charged gases. In this case, the wave function would be a combination of two Slater determinants, one for the electron gas and one for the positron gas.

To construct the wave function, you would need to consider the single-electron wave functions for both gases and combine them in a similar manner as you would for a single gas. However, since the positron has an opposite charge, its wave function would have a negative sign compared to the electron's wave function.

In terms of adding or multiplying the two determinants, it would depend on the specific properties of the gases and the interactions between them. Generally, the wave function would take the form of a linear combination of the two determinants, with coefficients determined by the Hamiltonian of the system.

I hope this helps clarify the construction of the Hartree-Fock wave function for a mixture of two oppositely charged gases. Feel free to reach out if you have any further questions. Thank you.