# Basic questions about the Kitaev chain

I am trying to understand the first 5 pages of Introduction to topological superconductivity and Majorana fermions http://arxiv.org/abs/1206.1736

I read it 2-3 times and thought about it but a few things still confuse me about the Kitaev chain.

1) We want Majorana-Fermions (MF) from this model. Fermions are 3D particles, but MF is a Non-Abelian Anyonen which is a 2D only particle. This already seems to make no sense. We construct with the Kitaev Chain a 1D system (a chain) and want to find there a 2D only particle? Technically, I guess it should even be 0D since the MF only shows up at the end of the chain. How is this supposed to make sense?

2) We rewrite the chain from the c fermion (electrons?) into a chain of MF and than again in form of another type of fermion $\tilde c$. How does this change the physical stuff the chain is made of? How do we translate this math-trick into a physical thing?

3) It is stated that a MF can be thought of as a electron and a hole. It is also stated that two MF can make any fermion. So an electron can be thought of as two MF which in turn can be thought of as an electron + hole. So one electron is two electrons and two holes?

• I would also appreciate any links to more simpler explanations of the kitaev chain. (I have not found any so far) – Thomas Elliot Jul 20 '15 at 15:08
• Here is a link explaining the Kitaev chain: topocondmat.org/w1_topointro/… – zed111 Dec 26 '17 at 8:15

You should forget about the name "Majorana fermions", although people have been using it (unfortunately) a lot in the literature on topological superconductivity. A much better terminology is "Majorana zero modes". "Majorana fermions" is more appropriate when you have propagating modes like chiral edge modes of a 2D topological superconductor, but not for these localized zero-energy modes.

There are no anyons in 1D. However, Majorana zero modes share many properties of the true 2D anyonic excitations (i.e. Majorana zero modes in vortices of 2D chiral topological superconductors). But maybe one should first put aside these analogies (since they confused you) and just understand what Majorana zero modes actually are. As the name suggests, it is a zero-energy mode of the BCS Hamiltonian. We know that BCS Hamiltonian mixes electrons and holes, and one can diagonalize the Hamiltonian by forming Bogoliubov quasiparticle operators which are linear superposition of electrons and holes. Majorana zero mode is a special one of those Bogoliubov quasiparticle operators, which is also self-Hermitian and square to 1 (thus "Majorana"). In the Kitaev chain, such zero-energy operators are easily found if you rewrite $c$ as Majorana operators, then you see that the two Majorana operators decouple from the Hamiltonian and thus have exactly zero energy. The whole point of doing this transformation is to expose the zero modes clearly.

For your last question, if you take these zero modes and write them in terms of the electron operators, they look like $c+c^\dagger$. So they are linear superpositions of $c^\dagger$ (which creates an electron) and $c$ (which annihilates an electron, or creates a hole). Of course you can invert the linear transformations and write $c$ as the linear combination of two Majorana operators. In a word, your counting makes little sense, since we are really talking about linear superposition (in the quantum mechanical sense), not some bound state of an electron and a hole.

• Does this mean that the MF has zero energy? If it does I would think it does not exist because "everything" that "is" has energy of some sort. – Thomas Elliot Jul 20 '15 at 17:50
• Majorana zero mode creates an excitation of the system on top of the ground state, and the correct question to ask is what is the excitation energy. So yes, it has zero excitation energy(namely, the operator has to commute with the Hamiltonian as you can check for the Kitaev model). It basically means the ground state is 2-fold degenerate. – Meng Cheng Jul 20 '15 at 18:07
• Why do you call it an excitation if it takes zero energy to get there? Shouldn't it be part of the groundstate? – Thomas Elliot Jul 20 '15 at 18:19
• You can of course call it part of a degenerate ground state. But in any generic finite-size system, the Majorana zero modes do not have exactly zero energy, instead there is a small splitting which scales like $e^{-L/\xi}$ where $L$ is the size of the system and $\xi$ is the coherence length of the superconductor. This happens, for example if the chemical potential of the Kitaev chain is not exactly $0$, or if the magnitudes of the hopping and pairing are not equal. But the point is the excitation energy approaches zero exponentially fast if you increase the size of the system. – Meng Cheng Jul 20 '15 at 19:19

A few less technical details than the ones you can find in Meng Cheng answer on this page are perhaps welcome.

For practical applications, we are not interested in Majorana modes, since they are a simple mathematical rewriting of the fermionic creation and annihilation operators. Say differently, to any creation $c^{\dagger}$ and annihilation $c$ operators you can associate two Majorana operators according to the definition

$$\gamma_{1}=\dfrac{c+c^{\dagger}}{\sqrt{2}}\;;\;\gamma_{2}=\mathbf{i}\dfrac{c-c^{\dagger}}{\sqrt{2}}$$

and obviously $\gamma_{1,2}^{\dagger}=\gamma_{1,2}$, the property of the Majorana operator. In fact, talking about conventional fermion representation or Majorana fermions representation is perfectly equivalent, since there is a unitary transformation between them. Remark that the transformation is exactly the same one usually uses extensively for bosons, passing from position $X$ and momentum $P$ operators to normal modes operators $a$ and $a^{\dagger}$. So the Majorana representation is a kind of phase-space representation of the fermions, so to say. Note that for more degrees of freedom (spin, momentum, ...) one needs more indices, but the construction is still perfectly allowed.

What we are looking for applications in quantum informations are uncoupled Majorana modes. The best introduction is definitely the original one by Kitaev. We want a Majorana mode which in fact is perfectly delocalised along the wire. In the Kitaev construction, you pair the Majorana in such a way that two extra degrees of freedom remain at the end of the wire (in the topological phase). These are the so-called edge modes, and usually coined zero-energy Majorana mode. As extensively explained by Meng-Cheng somewhere else, they appear in superconductors when spin symmetry is broken, while particle-hole symmetry remains (well, otherwise it's no more a superconductor ;-).

Unfortunately, the difference between Majorana mode (or fermion, but in condensed matter there is no fundamental particle, and one prefers to use the word mode) and unpaired Majorana modes is never clearly discussed, except in the paper by Kitaev. Every time you read Majorana mode, please think of them in term of unpaired Majorana mode. More certainly it's a kind of self-evidence that people in the field forget to precise when they introduce the topic.