Symmetrisation of wave function for fermions

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SUMMARY

The wave function for fermions must be anti-symmetric with respect to the exchange of electron positions, as established by the Pauli exclusion principle. In a two-electron system, the correct formulation is Ψ(r1,k1,r2,k2) = - Ψ(r2,k1,r1,k2), indicating that the wave vector and position cannot be exchanged simultaneously. The overall state must remain anti-symmetric, and if an additional degree of freedom, such as spin, is involved, the full state must also be anti-symmetric. For systems described by Coulomb wave functions, the spatial wave function can be symmetric if the spin part is anti-symmetric, particularly in singlet states.

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djelovin
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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)
 
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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.
 
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.
 
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