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Three fermions

  1. Feb 20, 2015 #1
    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. jcsd
  3. Feb 20, 2015 #2
    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
     
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