# What happens to a system of three free fermions when you turn on a two states potential?

In textbooks, we usually study the symmetry and antisymmetry of a system of $N$ particles in a $m=N$ states system. The most basic example is the symmetry and antisymmetry of two bosons and two fermions in a two states system:

$$\psi_{_b} = \frac{1}{\sqrt{2}}\left[\phi_{1}(r_{1})\phi_{2}(r_{2})+\phi_{1}(r_{2})\phi_{2}(r_{1})\right]$$ $$\psi_{_f} = \frac{1}{\sqrt{2}}\left[\phi_{1}(r_{1})\phi_{2}(r_{2})-\phi_{1}(r_{2})\phi_{2}(r_{1})\right]$$

where $r_{i}$ is the coordinate of the particle-$i$ and $\phi_{j}$ is one of the possible states. Notice that this is only one of all possible wavefunctions for bosons. Since the bosonic wavefunction must be symmetric we could also have all bosons in the same state.

Now, if we have a three particles/three states system we would have 10 possible wavefunctions for the bosons and just one for the fermions:

Bosons:

(1) All bosons in the same state (3 possibilities)

(2) One boson in each state (1 possibility)

(3) Two bosons in the same state and one in a different one (6 possibilities)

Fermions:

One possibility due to the necessity of having an antisymmetric wavefunction

What happens if we have three particles in a two states system? For bosons, (2) is not available anymore, (1) has now two possibilities and (3) has two. These wavefunctions are

$$\psi_{_{b,I}} =\phi_{1}(r_{1})\phi_{1}(r_{2})\phi_{1}(r_{3})$$ $$\psi_{_{b,II}} =\phi_{2}(r_{1})\phi_{2}(r_{2})\phi_{2}(r_{3})$$ $$\psi_{_{b,III}} = \frac{1}{\sqrt{3}}\left[\phi_{1}(r_{1})\phi_{2}(r_{2})\phi_{2}(r_{3})+\phi_{2}(r_{1})\phi_{1}(r_{1})\phi_{2}(r_{3})+\phi_{2}(r_{1})\phi_{2}(r_{2})\phi_{1}(r_{3})\right]$$ $$\psi_{_{b,IV}} = \frac{1}{\sqrt{3}}\left[\phi_{2}(r_{1})\phi_{1}(r_{2})\phi_{1}(r_{3})+\phi_{1}(r_{1})\phi_{2}(r_{1})\phi_{1}(r_{3})+\phi_{1}(r_{1})\phi_{1}(r_{2})\phi_{2}(r_{3})\right]$$

Here begin my problems. For fermions, my first guess was that you cannot have three particles in a two-state system due to the exclusion principle, but that was not very satisfying. So I used the Slater determinant to find the wave function for three fermions in a three states system and then I choose $\phi_{3}=\phi_{2}$ to reduce the number of states, building then the wavefunction I wanted from one I was sure was correct. This procedure gave me

$$\psi_{_{f,I}} = \phi_{2}(r_{1})\bigg[\phi_{1}(r_{2})\phi_{2}(r_{3})-\phi_{1}(r_{3})\phi_{2}(r_{2})\bigg] + \phi_{2}(r_{2})\bigg[\phi_{1}(r_{3})\phi_{2}(r_{1})-\phi_{1}(r_{1})\phi_{2}(r_{3})\bigg] + \phi_{2}(r_{3})\bigg[\phi_{1}(r_{1})\phi_{2}(r_{2})-\phi_{1}(r_{2})\phi_{2}(r_{1})\bigg]$$

and a similar wavefunction for the choice $\phi_{3}=\phi_{2}$. This wavefunction is antisymmetric and for $r_{i}=r_{j}$, $\psi_{_{f,I}}=0$ as expected, but a an issue is troubling me:

i) If you rearrange the terms in $\psi_{_{f,I}}$ you'll see that it is always zero, no matter what. So my initial line of thought was right: you cannot have three fermions in a two states system. This, however, poses another question: What if you have three free fermions and then turn on a two-state potential. What happen to the particles? The wavefunction becomes null but they cannot be annihilated in this formalism since it is non-relativistic.

Thank you.

Any physical potential that you could "turn on" in a laboratory would also admit a continuum of scattering states. These will exist whenever the total energy $E$ may be greater than the potential energy $V$. If your potential initially admits three fermions and you modify it so that it can only contain two, you are essentially forcing one of them out, providing $E$ as you increase $V$, making such scattering states accessible.
As for your errors farther up: you will not have enough freedom to make three orthogonal states if $\phi_3 = \phi_2$; this is reflected in the fact that by doing so, you find your solution is zero.
• Thank you very much. So, could you say that it is impossible for three fermions to inhabit a two-state system and that $\psi_{f}=0$ is the mathematical description of this fact? So, what you pointed out as an error... is it really? Doesn't it patch itself up by returning zero? Once again thank you very much for your thoughts on the matter. – Gabu Jun 19 '17 at 20:11