# Three fermions

1. Feb 20, 2015

### barefeet

1. The problem statement, all variables and given/known data
Consider three electrons in three different orbital levels n, m and p. We
assume that there is one electron in each orbital level. How many states
are then possible?

2. Relevant equations
Equations for constructing symmetric and asymmetric wavefunctions:

Symmetric under pair permutation:
$$\frac{1}{\sqrt{6}} \{\left |nmp\right \rangle + \left |mpn\right \rangle + \left |pnm\right \rangle +\left |npm\right \rangle + \left |pmn\right \rangle + \left |mnp\right \rangle \}$$

Anti-symmetric under pair permutation:
$$\frac{1}{\sqrt{6}}\{\left |nmp\right \rangle + \left |mpn\right \rangle + \left |pnm\right \rangle - \left |npm\right \rangle - \left |pmn\right \rangle - \left |mnp\right \rangle \}$$

Symmetric under cyclic permutation:

$$\frac{1}{\sqrt{3}} \{ \left |nmp\right \rangle + \varepsilon \left |mpn\right \rangle + \varepsilon^* \left |pnm\right \rangle \}$$
$$\frac{1}{\sqrt{3}} \{ \left |npm\right \rangle + \varepsilon \left |pmn\right \rangle + \varepsilon^* \left |mnp\right \rangle \}$$

With:

$$\varepsilon = e^{i\frac{2 \pi}{3}}$$
$$\varepsilon^* = \varepsilon^2$$

Anti-symmetric under cyclic permutation:

$$\frac{1}{\sqrt{3}} \{ \left |npm\right \rangle + \varepsilon^* \left |pmn\right \rangle + \varepsilon \left |mnp\right \rangle \} \}$$
$$\frac{1}{\sqrt{3}} \{ \left |nmp\right \rangle + \varepsilon^* \left |mpn\right \rangle + \varepsilon \left |pnm\right \rangle \}$$

3. The attempt at a solution
I could get the symmetric spin part of the wave function and use that with the anti-symmetric orbital part:
$$\left | \uparrow \uparrow \uparrow \right \rangle$$
$$\frac{1}{\sqrt{3}} \{ \left | \downarrow \uparrow \uparrow \right \rangle + \left | \uparrow \downarrow \uparrow \right \rangle + \left | \uparrow \uparrow \downarrow \right \rangle \}$$
$$\frac{1}{\sqrt{3}} \{ \left | \uparrow \downarrow \downarrow \right \rangle + \left | \downarrow \uparrow \downarrow \right \rangle + \left | \downarrow \downarrow \uparrow \right \rangle \}$$
$$\left | \downarrow \downarrow \downarrow \right \rangle$$

There should be 8 states, but this only gives me 4. I don't know how to get the asymmetric spin part.

2. Feb 20, 2015

### DHIMAN

According to classical mechanics the electrons are distinguishable.

But quantum mechanically you cant distinguish the electron and cant say i-th electron is in j-th state.What can you do? You can just distinguish the angular momentum component along the z direction of the electron system.And so the solution above given is right.As the electron is spin half particle so the given four solutions represents just the 3/2,1/2,-1/2,-3/2 angular momentum states