2
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.