Zero energy eigenstates of the Kitaev chain Hamiltonian

I am thinking about a two-site Kitaev Hamiltonian. Namely, $$H = -\mu c^\dagger_1 c_1 -\mu c^\dagger_2 c_2-t c^\dagger_{2} c_1-t c^\dagger_{1} c_2 +\Delta c_1 c_{2}+\Delta c_2^\dagger c_{1}^\dagger.$$

I focus on the topological regime $$\mu=0$$ and $$\Delta=t=1$$.

When I switch to a Majorana representation and solve this I end up with two fermionic states which vanish from the Hamiltonian and thus reside at zero energy.

(see 5.3.1 in these notes)

However, if I directly diagonalize the Kitev Hamiltonian (in the basis of $$|00\rangle$$,$$|10\rangle$$,$$|01\rangle$$,$$|11\rangle$$) I end up having no zero energy eigenstates (see 2.4.2 here).

Any unitary transformation should preserve the eigenspectrum which made me conclude that via changing to a "Majorana representation" I must have done something non-unitary to my Hamiltonian?

Is the change from $$|00\rangle$$,$$|10\rangle$$,$$|01\rangle$$,$$|11\rangle$$ to a Majorana representation non unitary?

A zero-energy mode is a mode into which you can add an excitation without changing the energy of the system. So if the mode creation operator is $$\hat{a}^\dagger$$ and $$|0\rangle$$ is the ground state with energy $$E_0$$, then $$\hat{a}^\dagger|0\rangle$$ is also an energy eigenstate with energy $$E_0$$, i.e. the ground state is degenerate. The actual value of $$E_0$$ is completely irrelevant and can be chosen at will by redefining the zero of energy. In particular, there is no requirement that $$E_0 = 0$$: it does not matter if the energy eigenvalues are zero or not. Instead, the presence of a zero mode is signalled by degeneracies in the spectrum.
• Hi, if I understand your answer well you say that Majorana fermions are actually excitations of "ground states" which cost no additional energy. Then the question is how I determine what is a "ground state" or $E_0$ of the Kitaev chain? Aug 12, 2021 at 13:48