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How can we see unpaired Majoranas for a Kitaev chain in topological non-trivial phase? By looking at the equation below (obtained by making $c_{j}=\frac{e^{-i\phi/2}}{2}\left(\gamma_{B,j}+i\gamma_{A,j}\right)$ in the original Kitaev chain):

$$H=-\frac{i\mu}{2}\sum_{j=1}^{N}\gamma_{B,j}\gamma_{A,j}+\frac{i}{2}\sum_{j=1}^{N-1}\left[\left(|\Delta|-t\right)\gamma_{B,j}\gamma_{A,j+1}+\left(|\Delta|+t\right)\gamma_{A,j}\gamma_{B,j+1}\right]$$

And assuming $|\mu| < 2t$ and $|\Delta| \neq 0$, there should appear uncoupled Majoranas at the ends of the chain, but how is it possible if all Majoranas operators are coupled? See figure below ...

It's clear for $|\Delta| \pm t$ and $\mu = 0$, but I'd like to see It for more general parameters.

enter image description here

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  • $\begingroup$ The uncoupled Majorana operators will be linear combinations of several Majorana operators localized around the edge (in fact, all of them with an exponentially decaying amplitude). $\endgroup$ – Norbert Schuch Aug 10 '15 at 20:52
  • $\begingroup$ Kitaev's paper arxiv.org/abs/cond-mat/0010440 is explicitly clear about that. Did you try to read it ? Actually, the coupling via p-wave gap is between A and B sub-index, and so you let two unpaired operators. In your counting, the topological non-trivial situation corresponds to the red OR the black coupling scheme (not both at the same time). You explicitly see the two remaining $\gamma$'s (for red pairing: A,1 and B,n). In fact, these modes did not appear in the Hamiltonian since they have no energy, but counting the mode (or pairing the modes) explicitly let two unpaired modes. $\endgroup$ – FraSchelle Aug 11 '15 at 9:28
  • $\begingroup$ @FraSchelle, as I said before, it's clear to me that the zero-modes appear at the ends for $|\Delta| = \pm t$, my point is that for general conditions ($|\mu| < 2t$ and $\Delta \neq 0$) all Majoranas operators are, in principle, coupled together and the zero-modes can't be seen as isolated operators anymore. As NorbertSchuch noticed, these zero-modes are now linear combinations of several Majoranas operators. It appears in that Kitaev's paper [link]( arxiv.org/abs/cond-mat/0010440) but I couldn't reproduce his results to understand this better. Have you ever gone through this paper? Thank you. $\endgroup$ – Adonai Cruz Aug 11 '15 at 19:02
  • $\begingroup$ I don't think this is any more involved than computing/understanding the normal modes of a chain of masses coupled by springs. All masses are coupled, but in the end you obtain a basis of decoupled "masses" (which are not individual masses any more). $\endgroup$ – Norbert Schuch Aug 11 '15 at 22:44
  • $\begingroup$ As far as I remember, Kitaev shows that there are modes the wave function of which decay when going inside the bulk. It's a standard argument for edge modes (also called surface modes). I guess Norbert gives you a generic argument about modes: add coupling between them and you need to change the basis to find the eigenmodes with coupling ... Kitaev gives you the method to find those modes which are exponentially decaying. I'm not sure you can apply Norbert's argument up to a constructive point. It's unclear how many $\gamma$ from the original basis will contribute to an exponential decay... $\endgroup$ – FraSchelle Aug 12 '15 at 3:43
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Perhaps a way to understand all this in a slightly different way: when $\mu=0$ and $\Delta=t$, a half-second look at your Hamiltonian shows that $\gamma_{j}^{A}\gamma_{j+1}^{B}$ is an eigenmode of the problem. Similar with $\gamma_{j+1}^{A}\gamma_{j}^{B}$ when $\mu=0$ and $\Delta=-t$. In both cases two Majorana are left unpaired, because the sum runs only to $N−1$. When $\Delta=t=0$ then the usual Majorana paired $\gamma_{j}^{A}\gamma_{j}^{B}$ are eigenmodes. Outside these limiting cases, none of the $\gamma$'s written in your Hamiltonian is an eigenmode, and you have to identify the correct basis which renders the Hamiltonian diagonal. That's just a rephrasing of Norbert's argument (see comment).

So in general argument, to obtained the unpaired Majorana modes, you need either:

  • 1) find the new eigenmodes which read $\alpha_{ij}^{CD}\gamma_{i}^{C}\gamma_{j}^{D}$ (sum implied, the task is to find all the $\alpha$'s) and

  • 2) find those eigenmodes which are localised around the edge (Kitaev's argument discussed shortly in comments, see also this answer somewhere else on SE for similar arguments) OR

  • 2-bis) find the eigenmodes which are inside the gap, i.e. the eigenmodes with associated energies below $\Delta$.

The points 2 and 2-bis are equivalent (in this simple system at least). The argument to be sure the edge-localised modes are unpaired is given in Kitaev's paper: cut a wire in two parts, or glue two wires: edge modes should appear, or disappear, respectively.

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  • $\begingroup$ FraSchelle,first of all thank you for you answer, but I'm not sure I understand when you said for instance that $\gamma_{j}^{A}\gamma_{j+1}^{B}$ is an eigenmode for $\mu=0$ and $\Delta = t$. Because in this regime we have $H=it\sum_{j=1}^{N-1}\gamma_{j}^{A}\gamma_{j+1}^{B}$, so in order to obtain the eigenmodes we should first diagonalize the Hamiltonian, isn't it? At least I couldn't see it in half-second. Anyway, outside this limiting case, the first step would be to write the Hamiltonian matrix and find it's eigenvectors for zero-energy.But I couldn't find Kitaev's condition for $x_{\pm}$. $\endgroup$ – Adonai Cruz Aug 14 '15 at 0:59
  • $\begingroup$ @AdonaiCruz Have you tried to go through the calculation to diagonalize the Hamiltonian? Then, it would be best if you describe in detail what you did and where you got stuck. $\endgroup$ – Norbert Schuch Aug 14 '15 at 6:08
  • $\begingroup$ I wrote the hamiltonian matrix $A$ using Majorana operators as basis, and wrote the equation for eigenvalue equals zero. To satisfy this equation each row of my eigenvectors should be zero.I obtained rows like $\mu \gamma_{1}^{A} + (t+|\Delta|)\gamma_{2}^{A} + (-t+|\Delta|)\gamma_{3}^{A}$ and similar ones with $\gamma_{j}^{B}$. I'm stucked in obtaining the $|x_{\pm}|$ condition and see that both zero-modes can be written by $b^{'}=\sum_{j}(\alpha^{'}_{+}x_{+}^{j} +\alpha^{'}_{-}x_{-}^{j})\gamma_{j}^{A}$ and $b^{''}=\sum_{j}(\alpha^{''}_{+}x_{+}^{-j} +\alpha^{''}_{-}x_{-}^{-j})\gamma_{j}^{B}$. $\endgroup$ – Adonai Cruz Aug 14 '15 at 13:51
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Good question. There is in fact a really simple way of seeing there are Majorana modes on the edges, using the idea of `symmetry fractionalization'. In fact this principle works for all symmetry protected topological phases in 1D and it classifies all of them! The below explanation might seem a bit long, but I assure you it is all quite intuitive, and moreover you can then immediately apply the same logic to all SPTs you might know or encounter in the future :)

Generally, suppose your Hamiltonian has an on-site symmetry $U$ (meaning it acts on every site $U = U_1 \otimes U_2 \otimes \cdots$). If we have periodic boundary conditions, and the symmetry is not spontaneously broken in the ground state, then we expect that acting $U$ on the ground state should leave it invariant1. If we now cut the system to create (open) boundaries, then we expect our ground state to only change near the edges with a length scale given by the correlation length $\xi$. So since the ground state in the bulk is still the same, and we knew that ground state had to be invariant, we know that acting $U$ can only change our state near the left and right edge (with a scale given by $\xi$). So effectively the symmetry $U$ acts as $U = U_L U_R$ with these operators localized on the edges (it is said the symmetry `fractionalizes' onto the edges). Let's now see how this can classify and detect topological phases.

Note that the Kitaev chain only has terms which have an even number of fermionic operators. Physically this means that although the Hamiltonian doesn't preserve particle number (indeed its physical origin is related to superconductivity, where pairs of particles can be created), it does preserve whether you have an even or odd number of fermions. This is measured by the fermionic particle number parity $P = \prod (1-2n_i)$, which is a symmetry of the Hamiltonian. Rewriting this in terms of the Majorana operators, this is $\boxed{P = \prod i \gamma_{A,i} \gamma_{B,i}}$. Let's look at the case where it is clear there are Majorana modes: $\boxed{H = -\sum i \gamma_{A,i+1} \gamma_{B,i}}$. (Then $\gamma_{A,1}$ and $\gamma_{B,N}$ are free.) Note that these terms in the Hamiltonian commute and so for the ground state state subspace $i \gamma_{A,i+1} \gamma_{B,i} = 1$. Plugging this into $P$ one gets that in the ground state subspace $P = i\gamma_{A,1} \gamma_{B,N}$. Note that this only acts on the edge, consistent with the above general argument! In other words we can write $\boxed{P = P_L P_R}$ with $P_L = i\gamma_{A,1}$ and $P_R = \gamma_{B,N}$.

Now here comes the punchline: note that $P_L$ and $P_R$ are fermionic, so $P_L P_R = - P_R P_L$. Even if we now add extra couplings as in the original post, of course the expressions for $P_L$ and $P_R$ will change, however the anti-commutation cannot suddenly/discretely change to a commutation (and it has to be one of the two: since $P$ is bosonic, $P_L$ and $P_R$ are either both fermionic or bosonic). The only way it can change is if $P_L$ and $P_R$ start overlapping, which means $\xi$ has to diverge. This means it can only change if there is a phase transition! So as long as the system remains gapped, $P_L P_R = -P_R P_L$. But note that we then2 have two anticommuting symmetries. But by general linear algebra arguments3, the presence of two anticommuting symmetries implies a ground state degeneracy. This is exactly detecting the non-local fermionic mode formed by our two Majorana zero modes!


Footnotes:

1. The intuition is the following: suppose $U$ does not leave your ground state invariant and suppose our ground state is translationally invariant--consistent with using periodic boundary conditions--then by translation symmetry it changes the state everywhere, but that means $|\psi_\textrm{gs}\rangle$ and $U|\psi_\textrm{gs}\rangle$ are globally different, in which case we expect there to be spontaneous symmetry breaking.

2. Clearly $P$ is a symmetry, but you might wonder why $P=P_L P_R$ would imply that $P_L$ and $P_R$ are separately symmetries. Well the way to see this: decompose $H=H_L + H_R$ such that $H_R$ has no support near the left edge, same for $H_L$ and the right edge. Then $[P_L,H_R] = 0$. But since $[P,H_R] = 0$, that means that $[P_R,H_R] = 0$. Combined with $[P_R,H_L] = 0$, this shows that $P_R$ is a symmetry of $H$.

3. Suppose $|\psi_\textrm{gs}\rangle$ is a ground state. Then either $P_L|\psi_\textrm{gs}\rangle$ or $P_R |\psi_\textrm{gs}\rangle$ has to be a different state, otherwise $P_L P_R |\psi_\textrm{gs}\rangle = P_R P_L |\psi_\textrm{gs}\rangle$, yet we know they anticommute.

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