The definition of Kitaev chain and the solution is in this article:

The Hamiltonian has the form:
$H = -\mu\sum_n c^\dagger_n c_n -t\sum_n c^\dagger_{n+1} c_n +\Delta\sum_n (c_n c_{n+1}+\text{h.c.})$

I am interested in the case of periodic boundary condition where the famous Majorana edge modes are not presented. To deal with the pairing terms $\Delta c_nc_{n+1}$ and $\Delta^* c_n^\dagger c_{n+1}^\dagger$, this article uses a Bogoliubov transformation and then diagonalize the Hamiltonian. Here is my understanding of the solution: I view the Bogoliubov transformation as a unitary transformation whose basis are electron states with different number of electrons. So, although the eigenstates are single-particle states of Bogoliubov particles, each eigenstate is a superposition of electron states with different number of electrons.

My questions are:
(1) Is my understanding correct? Is each eigenstate of Kitaev chain a superposition of states with different number of electrons?
(2) If so, when we measure the electron number operator on an eigenstate, do we get different number of electron every time?
(3) Are the eigenstates many-body? (ie. there is no way to write the eigenstates as a tensor product of singe-electron states).


1 Answer 1


1) This is correct, each energy eigenstate is not an eigenstate of the number operator, although they are eigenstate of the parity operator $(-1)^N$, meaning that each eigenstate is either a superposition of states with an even number of particles or states with an odd number of particles.

2) Yes, this directly follows from (1).

3) This isn’t possible, because a tensor product state always has a definite number of electrons.


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.