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I have the Kitaev chain tight-binding Hamiltonian into the BdG formalism (like the one in the "Bulk-edge correspondence in the Kitaev chain" section of Week 1 of this course http://www.topocondmat.org/) :

$H_{BdG}=-\sum_n \mu \tau_z\left|n\right\rangle\left\langle n\right|-\sum_n \left[(t\tau_z+i\Delta\tau_y)\,\left|n\right\rangle\left\langle n+1 \right| + \textrm{h.c.}\right].$

In the topological phase ($\mu < 2t$) we get two zero-energy solutions by diagonalizing the above matrix and if the go to the trivial phase ($\mu > 2t$) these two states will "split" into a pair with $\pm E$ due to the particle-hole symmetry of the system. Due to the redundancy of the formalism we look only at the excitations with positive energies, my question is: each zero-energy eigenstate is a single Majorana mode or a fermionic mode at zero energy? When two Majoranas fuse together we should take only the $+E$ eigenstate? The eigenvalues are ploted in the image below where I used $N=10$ sites with $\Delta=t=1.0$ and varied $\mu$. enter image description here

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  • $\begingroup$ There are always pairs of zero modes. If you want to just have a localized zero mode (say, at the end of the chain), then it will look like a Majorana bound state. But since they are degenerate, you can form linear superposition of them which is Dirac. $\endgroup$ – Meng Cheng Mar 12 '16 at 2:57
  • $\begingroup$ So both zero-energy eigenstates I obtained are two Majorana bound states (to obtain a Dirac at zero energy we form a linear combination from both). But when they hybridize and get a non-zero energy each state (+E and -E) corresponds to a Dirac fermion ? And we should consider only one of them?. I got confused because we can write this hybridization as $H=i2\varepsilon\gamma_{1}\gamma_{2}=\varepsilon(\hat{n}-1/2)$ so we get two fermionic states with two parities (empty and occupied states). The eigenstates we obtain by diagonalizing are not related to different parities? $\endgroup$ – Adonai Cruz Mar 12 '16 at 12:03
  • $\begingroup$ To make my point clearer: looking at the fig. we should only take states with $E>0$ (all other $E<0$ state is redundant), but at $\mu=0$ for instance there are two Majorana zero-modes. They hibridize after the phase transition (in my fig it's about $\mu\approx 1.5$ due to small number of sites) and each split branch is a Dirac fermion with finite energy. Again, is the $E<0$ state redundant isn't it? It's not related to occupations (parities) as I thought before. $\endgroup$ – Adonai Cruz Mar 23 '16 at 18:44
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The short answer is this: each line on your plot is denoting one fermionic mode (i.e. a fermionic creation/annihilation operator $c$), or equivalently each line corresponds to two Majorana modes (since a fermionic mode defines two Majorana operators $\gamma_1 = \frac{1}{2} \left(c+c^\dagger \right)$ and $\gamma_2 = \frac{1}{2i} \left(c-c^\dagger \right)$). Each line is represented twice though. It is not written in stone that you have to take the upper line, but just make sure you only count each set of two only once. Whether you take the upper or lower line comes down to choosing what you call the fermionic creation operator and the annihilation operator. I guess the clearest would be to just forget the bottom set of lines from the outset.

So take the point $\mu = 0$. On your plot we see two lines meeting there, but as you point out this is due to a redundancy in description, so in fact there is only one line there. In other words there is one fermionic mode $c$ with zero energy. This gives a twofold degeneracy: $|0\rangle$ and $c^\dagger |0\rangle$. Your plot shows that this mode does actually not stay at zero energy in the whole phase, but I guess this must be due to finite-size effects and as you take $L$ larger and larger, the line should approach zero until the phase transition point.

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