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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted