# A Symmetrisation of wave function for fermions

1. Apr 27, 2017

### djelovin

The wave function for fermions has to be anti-symmetric with respect to exchange of positions of electrons, but what if it depends on wave vector as well. Does they have to be exchanged as well, in other words, for two-electron system what is correct

Ψ(r1,k1,r2,k2) = - Ψ(r2,k1,r1,k2)

or

Ψ(r1,k1,r2,k2) = - Ψ(r2,k2,r1,k1)

2. Apr 27, 2017

### Orodruin

Staff Emeritus
The wave vector and the position are just different representations of the same Hilbert space. You can use either to represent your state but not both at the same time.

That being said, it is the overall state that should be antisymmetric under the exchange of the electrons. If you have an additional degree of freedom that the state depends on (spin comes to mind) then you need to make the full state antisymmetric. If the state in the additional degree of freedom is already antisymmetric your spatial wavefunction will be symmetric. Positronium in the spin-0 state comes to mind.

3. Apr 27, 2017

### djelovin

Thanx for quick replay,
That somewhat clarifies my problem.
However I have particles in continuum, in the presence of some potential, that are described by Coulomb (Coulomb-like to be more precise) wave function that does depend on both, wave vector and position at the same time.
https://en.wikipedia.org/wiki/Coulomb_wave_function
The system is in singlet state, so spin related part can be taken out.

Last edited: Apr 27, 2017