Analyzing 3 Electron States in Different Orbital Levels

[barefeet]

Homework Statement

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?

Homework 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:

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

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.

 

[DHIMAN]

According to classical mechanics the electrons are distinguishable.

But quantum mechanically you can't distinguish the electron and can't 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