How we can get the "Fermion Parity" and "Ground states" for Majorana fermions in Bernevig's talk PiTP 2015? I have two questions regarding the talk, Topological Superconductors, Majorana...and Interactions, by Bernevig in PiTP 2015.

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*How he gets the "Fermion Parity" for the ground states in the talk? for example when he talked about 2 Majorana wires and he gets the ground states $|0>$ and $f^{\dagger}|0>$, and the 6 wires system which has $f_1^{\dagger}|0>$ and $f_2^{\dagger}|0>$. You can watch it 56:24

*I do not understand how he got the ground states for the 6 wires. Even when he considered the first term why he did not write down other combinations like $f_3^{\dagger}f_2^{\dagger}|0>$ or $f_3^{\dagger}f_1^{\dagger}|0>$( I guess the changing of fermionic operators just give a minus that is why he didn't consider them?). Also, why he did not write down $f_3^{\dagger}|0>$ as a ground state? For this part, please watch the video 58:20 to 1:01.

Bernevig's talk: https://www.youtube.com/watch?v=nywYAwKp2ac
 A: Fermion parity means the parity of the number of fermions in a given state. The vacuum $\vert 0\rangle$ has zero fermions, an even number, while $\hat{f}^{\dagger}\vert 0\rangle$ has one (odd).
To obtain fermion parity you can either find the eigenvalue of the number operator $\hat{n} = \hat{f}^{\dagger}\hat{f}$, or, equivalently, of the "parity operator"
\begin{align}
\hat{P} &= -i\hat{\gamma}_1\hat{\gamma}_2 \\
&= -i(\hat{f}^{\dagger} + \hat{f})(i\hat{f}^{\dagger} - i\hat{f}) \\
&= \hat{f}^{\dagger} \hat{f}^{\dagger} -  \hat{f}^{\dagger} \hat{f} +   \hat{f} \hat{f}^{\dagger} -  \hat{f} \hat{f} \\
&= 1 - 2  \hat{f}^{\dagger} \hat{f} \\
&= 1 - 2 \hat{n}.
\end{align}
Notice how the two operators commute, so they share a common set of eigenvectors.
As for your second question, notice the structure of the energy term he was considering:
\begin{align}
\hat{E}_{12} &= \alpha(\hat{f}_1^{\dagger}\hat{f}_1 - \frac{1}{2})(\hat{f}_2^{\dagger}\hat{f}_2 - \frac{1}{2}) \\
&= \alpha (\hat{n}_1 - \frac{1}{2})(\hat{n}_2 - \frac{1}{2})
\end{align}
The eigenvalues of the terms in parenthesis are $\pm 1/2$. If only one of the terms is negative ($-1/2$), then the whole thing is $-\alpha/4$. If none or both of the terms are positive ($+1/2$), then the energy is $+\alpha/4$. In the lecture, Bernevig takes $\alpha$ to be positive, as in these systems this is a Coulomb (repulsive) interaction.
This means the ground state for $\hat{E}_{12}$ must have either of the modes $1$ or $2$ occupied, but not both. The state $\hat{f}^{\dagger}_3\vert 0\rangle$, for example, has no particles in modes $1$ and $2$, such that both terms in parenthesis will be negative, yielding a resulting energy of $+\alpha /4$, so that it is not in the groundstate manifold.
Also, notice how $\hat{E}_{12}$ does not depend on $\hat{n}_3$ and that it comutes with $\hat{f}^{\dagger}_3$. This means that if $\vert \psi \rangle$ is in the ground state manifold, then $\hat{f}^{\dagger} \vert \psi \rangle$ is also in that manifold:
\begin{align}
\hat{E}_{12} \hat{f}^{\dagger}_3 \vert \psi \rangle
&= \hat{f}^{\dagger}_3 \hat{E}_{12} \vert \psi \rangle \\
&= (-\alpha/4) \hat{f}^{\dagger}_3 \vert \psi \rangle.
\end{align}
This means that since $\hat{f}^{\dagger}_1\vert 0 \rangle$ and $\hat{f}^{\dagger}_2\vert 0 \rangle$ are ground states, then $\hat{f}^{\dagger}_1\hat{f}^{\dagger}_3\vert 0 \rangle$ and $\hat{f}^{\dagger}_2\hat{f}^{\dagger}_3\vert 0 \rangle$ are ground states as well (remember $\hat{f}^{\dagger}_i$ and $\hat{f}^{\dagger}_j$ ant commute for $i\neq j$).
Finally, when we consider both terms $E_{12}$ and $E_{23}$, we can apply the same reasoning and find the smallest contributions from both terms. The states $\hat{f}^{\dagger}_2\hat{f}^{\dagger}_3\vert 0 \rangle$ and $\hat{f}^{\dagger}_1\vert 0 \rangle$ have an energy of $-\alpha/4$ for the first term, but an energy of $+\alpha/4$ for the second term, such that their total energy is zero. Likewise, the terms $\hat{f}^{\dagger}_1\hat{f}^{\dagger}_2\vert 0 \rangle$ and  $\hat{f}^{\dagger}_3\vert 0 \rangle$ have energies $+\alpha/4$ and $-\alpha/4$ for the first and second terms, repsectivelly, for a total of zero energy too. It is only the states $\hat{f}^{\dagger}_1\hat{f}^{\dagger}_3\vert 0 \rangle$ and $\hat{f}^{\dagger}_2\vert 0 \rangle$ that have contributions $-\alpha/4$ for  both terms, such that total energy is $E=-\alpha/2<0$.
So these last states are the least energetic states and, by definition, the groundstates.
