Few particle fermion system wavefuction

Question

Suppose I have 3 fermions($\left|\psi_1\right\rangle$, $\left|\psi_2\right\rangle$, $\left|\psi_3\right\rangle$) and a system with 3 states ( $\left|1\right\rangle$, $\left|2\right\rangle$, $\left|3\right\rangle$)

By exclusion principle we can say that the total wave function of the system, $\left|\Psi\right\rangle$ , can only have terms like $\left|ijk\right\rangle$, where ijk is some permutation of $123$

Suppose the wave function of the particle $1$ is prepared as $\left|\psi_1\right\rangle= a \left|2\right\rangle+ b\left|3\right\rangle$ Then will it be true that the total wave function, $\left|\Psi\right\rangle$ will not have terms $\left|123\right\rangle$ and $\left|132\right\rangle$?

In general is it possible to prepare particle wave functions of a system of fermions to make sure that the total wave function of the system admits only certain configurations of the states?

Answer 1

Once you consider all three fermions, their wave function must be completely antisymmetric which, in your case, means that it must be a multiple of $$ \frac{ |123\rangle+ |231\rangle+ |312\rangle- |132\rangle- |321\rangle- |213\rangle }{\sqrt{6}} $$ You may imagine that at the very beginning when you only know about the single fermion, it has the wave function you wrote down. But once you consider all three, the wave function of all of them has to be the antisymmetric part of the tensor product, and not just the tensor product itself, which guarantees that the terms starting with "1" are represented just like all others.

Wave functions that are not fully antisymmetric are simply not allowed for fermions.