Degeneracy of two electrons on a ring The one-particle solution to the particle-on-a-ring problem is $\psi_m(\phi_j) = \frac{1}{\sqrt{2\pi}}\exp\left(-im \phi_j\right)$ for $m=0, \pm 1, \pm 2, \cdots$ corresponding to energies $E_m = \frac{m^2\hbar^2}{2I}$ where $I=MR^2$ is the moment of inertia.
I'm interested in the spatial wavefunction for two electrons on this ring. For the ground state, both can occupy the $m=0$ state:
$$
\Psi_0 = \psi_0(\phi_1)\psi_0(\phi_2) = \frac{1}{2\pi}.
$$
This state is, by my understanding, non-degenerate.
My question is: what is the degeneracy of the first excited energy level (ignoring spin)? My first thinking was that it should be 4, since each of the two electrons can have $m=\pm 1$ whilst the other has $m=0$.
$$
\Psi_1^{(a)} = \frac{1}{\sqrt{2}}\left[\psi_{0}(\phi_1)\psi_{+1}(\phi_2) + \psi_{+1}(\phi_1)\psi_{0}(\phi_2)\right]\\
\Psi_1^{(b)} = \frac{1}{\sqrt{2}}\left[\psi_{0}(\phi_1)\psi_{+1}(\phi_2) - \psi_{+1}(\phi_1)\psi_{0}(\phi_2)\right]\\
\Psi_1^{(c)} = \frac{1}{\sqrt{2}}\left[\psi_{0}(\phi_1)\psi_{-1}(\phi_2) + \psi_{-1}(\phi_1)\psi_{0}(\phi_2)\right]\\
\Psi_1^{(d)} = \frac{1}{\sqrt{2}}\left[\psi_{0}(\phi_1)\psi_{-1}(\phi_2) - \psi_{-1}(\phi_1)\psi_{0}(\phi_2)\right]\\
$$
But the answer is apparently not 4 and I have fallen into the trap of "failing to account for the indistinguishability of electrons". I thought that my symmetrized and antisymmetrized products did just that, though. Are these four states not distinct?
What is the correct way of thinking about this?
 A: The electrons can have different energies, so they don't both need to be in the one-particle excited state.
A: The states of the combined electron system must be antisymmetric, so your assertion that the ground state is 
$$ \psi(\phi_1,\phi_2) = \psi_0(\phi_1) \psi_0(\phi_2) $$
can not be the case. Further, the first excited state does not necessarily mean all combination of $\psi_{1,-1}$ (although it does in this case). It is the lowest energy level above (1,0), which is the lowest energy state possible whist obeying pauli exclusion. It has the wave function 
$$\psi(\phi_1,\phi_2) = \frac{1}{\sqrt{2}} \left( \psi_1(\phi_1) \psi_0(\phi_2) - \psi_0(\phi_1) \psi_1(\phi_2) \right) $$
up to an overall phase. Presumably, these particles are non-interacting, or else we would not be able to use the energy formula you listed to account for energy levels since the energy would depend on the interaction term... (and generally these wave functions would not diagonalize an arbitrary interaction Hamiltonian, so this exercise would be odd).
Since we are working with non-interacting particles, the total energy is just the sum of the single particle energies, which means the lowest energy is $E(0,\pm 1) = \frac{\hbar^2}{2I}$, there are two states here (0,1) and (0,-1); it's important to take note that the entire idea of anti-symmetrizing (or symmetrizing for bosons) the state is that (0,1) and (1,0) are not different states, they differ by only a phase which squares to 1. You can verify that the next lowest energy level is (1,-1).
The states that you listed above are almost correct but have two issues. First, they are not all anti-symmetric, and also you are working out ground state wave functions by using (m=0, m=1) for the reason I mentioned above. Try again with states (1,-1) and be sure that any state you write down differs by a minus sign when you interchange the values of $m$.
A: The miscount arises as follows.

*

*If the two electrons are in the antisymmetric singlet spin state, $S=0$ and
$\vert 00\rangle=\frac{1}{\sqrt{2}}\left(\vert \uparrow;\downarrow\rangle -\vert \downarrow;\uparrow\rangle\right)$, then the spatial part must be symmetric and you have correctly identified the two possible symmetric combinations.

*If the two electrons are in any one of the three symmetric triplet states, $S=1$ and
$$
\vert 11\rangle =\vert\uparrow;\uparrow\rangle\, ,\quad 
\vert 10\rangle=\frac{1}{\sqrt{2}}\left(\vert \uparrow;\downarrow\rangle + \vert\downarrow;\uparrow\rangle\right)\, ,\quad 
\vert 1,-1\rangle = \vert\downarrow;\downarrow\rangle
$$
then the spatial part must be antisymmetric and you have identified correctly these antisymmetric parts.

Thus, your error is that each antisymmetric spatial part has $3$ possible symmetric spin states.  This makes 6 possible states with an antisymmetric spatial part, plus 2 more with a symmetric spatial part.
