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?


1 Answer 1


There is nothing non-unitary going on here: the Hamiltonian matrix and its Majorana fermion representation are entirely equivalent. The confusion here is the mixing up of two distinct concepts: a zero-energy mode (relevant) and a zero-energy state (irrelevant).

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.

  • $\begingroup$ 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? $\endgroup$
    – Marco R.
    Aug 12, 2021 at 13:48
  • 1
    $\begingroup$ @MarcoR. You already have diagonalised the Hamiltonian right? The ground state is the energy eigenvector with the lowest eigenvalue. $\endgroup$ Aug 12, 2021 at 13:50

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