Going to momentum space

The Majoranas are edge excitations that arise from the bulk-edge correspondence. You might wonder if there is a way to deduce the existence of Majorana modes by looking at the bulk? To answer this, eliminate the boundaries from the study of the Kitaev chain. You can imagine that the last site of the chain is reconnected to the first, so that the chain is closed in a ring (a “Kitaev ring”). In the absence of boundaries, the Bogoliubov-de Gennes Hamiltonian has a translational symmetry \(\left|n\right\rangle\,\to\,\left|n+1\right\rangle\), since all parameters \(t, \Delta\) and \(\mu\) do not depend on the chain site \(n\). In the presence of translational symmetries, it is always convenient to use Bloch’s theorem and write down the Hamiltonian in momentum space rather than in real space. In our case, a state with momentum \(k\) is given by

\[ \left|k\right\rangle =(N)^{-1/2} \sum_{n=1}^{N} e^{-ikn} \left|n\right\rangle.\]

We apply periodic boundary conditions, that is \(\left\langle k | n=0 \right\rangle=\left\langle k | n=N \right\rangle\). The momentum \(k\) is then a conserved quantum number with allowed values \(2\pi p /N\) where \(p=0, 1, 2, \dots, N-1\). Values of \(k\) which differ by \(2\pi\) are equivalent. One can also imagine that for very large \(N\), \(k\) is a continuous periodic variable with values in the interval \([-\pi,\pi]\), the Brillouin zone. Because \(k\) is a good quantum number, the Bogoliubov-de Gennes Hamiltonian in momentum space can be reduced to a \(2\times 2\) matrix:

\[ H(k) \equiv \left\langle k\right| H_\textrm{BdG} \left| k \right\rangle = (-2t\cos{k}-\mu)\,\tau_z + 2\Delta \sin{k}\,\,\tau_y.\]

The full Bogoliubov-de Gennes Hamiltonian is obtained by summing all these \(2\times 2\) blocks:

\[ H_\textrm{BdG} = \sum_k H(k) \left| k \right\rangle\left\langle k \right|.\]

In the limit of an infinite chain the sum becomes an integral over the Brillouin zone.

Particle-hole symmetry in momentum space

Going to momentum space does not affect the particle-hole symmetry of the Bogoliubov-de Gennes Hamiltonian. However, one must always be careful in dealing with anti-unitary operators when making a basis transformation. The reason is that the action of the complex conjugation operator might depend on the basis. In our case, we have

\[\mathcal{P}\left|k\right\rangle\!\left|\tau\right\rangle = \left(\sum_n\,e^{-ikn}\right)^*\,\left|n\right\rangle\,\tau_x\left|\tau\right\rangle^*=\left|-k\right\rangle\,\tau_x\left|\tau\right\rangle^*.\]

Note that the particle-hole symmetry operator changes \(k\) to \(-k\). Therefore, the action of \(\mathcal{P}\) on the Bogoliubov-de Gennes Hamiltonian written in momentum space is the following:

\[\mathcal{P}H_\textrm{BdG}\mathcal{P}^{-1} = \sum_k \tau_xH^*(k)\tau_x \left| -k \right\rangle\left\langle -k \right|=\sum_k \tau_xH^*(-k)\tau_x \left| k \right\rangle\left\langle k \right|\,.\]

In the last equality, we have used the fact that the allowed values of \(k\) always come in \((-k, k)\) pairs, plus the two symmetric points \(k=0\) and \(k=\pi\). Therefore, particle-hole symmetry \(\mathcal{P}H_\textrm{BdG}\mathcal{P}^{-1}= -H_\textrm{BdG}\) implies that

\[H(k)=-\tau_xH^*(-k)\tau_x.\]

You can verify that it is indeed the case:

\[\tau_x H^*(-k) \tau_x = (2t\cos{k}+\mu)\,\tau_z - 2\Delta \sin{k}\,\,\tau_y.\]

Given a solution with energy \(E\) and momentum \(k\), particle-hole symmetry dictates in general the presence of a solution with energy \(-E\) and momentum \(-k\).

Band structure

At this point, we only need to diagonalize this \(2\times 2\) matrix to obtain the band structure of the Kitaev chain model, that is the energy levels \(E(k)\). This can be done very easily, and results in two energy bands, one with positive energy and one with negative energy:

\[ E(k) = \pm\sqrt{(2t\cos{k}+\mu)^2 + 4\Delta^2\sin^2{k}}. \]

Let’s see what this band structure looks like (once again move the slider to change \(\mu\)):

You can see that the energy spectrum is gapped at all \(k\) for \(\mu=0\). This is natural: Since our chain does not have boundaries anymore, the energy spectrum does not contain the zero energy Majorana modes which we found in the previous subsection at \(\mu=0\). However, you can see very clearly the bulk gap closing which occurs at the points \(\mu=2t\) and \(\mu=-2t\). Indeed, for these values of \(\mu\) the two bands at negative and positive energy touch at \(E=0\), for \(k=\pi\) and \(k=0\) respectively.

At first sight, the band structure looks quite similar on both sides of the bulk gap closings. Just from the above plot, it is not clear that the bulk gap closings separate two distinct phases of the model. Nevertheless, we will see that it is possible to rigorously come to this conclusion by studying the properties of \(H(k)\) in more detail. But in the meantime, we will start by understanding better what happens close to the transition, by writing down a simple effective model from which we will be able to deduce a lot of information.

Study of the bulk transition with an effective Dirac model

Let’s look at the gapless points more in detail. We focus on the gap closing at \(\mu=-2t\), which happens at \(k=0\). Close to this point where the two bands touch, we can make a linear expansion of the Hamiltonian \(H(k)\),

\[H(k) \simeq m \tau_z + 2\Delta\,k\,\tau_y\,,\]

with \(m=-\mu-2t\). We can see that \(H(k)\) becomes a Dirac Hamiltonian - note that condensed matter physicists use this term quite loosely to refer to any Hamiltonian which is linear in the momentum operator. You can easily check that this Hamiltonian gives an energy spectrum \(E(k) = \pm\sqrt{m^2 + 4\Delta^2k^2}\), and by returning to the plot above you can indeed verify that this is a very good approximation of the exact band structure around \(\mu=-2t\).

The ‘’mass’’ \(m\) appearing in this Dirac Hamiltonian is a very important parameter to describe what is happening. Its magnitude is equal to the energy gap \(\left|\mu+2t\right|\) in the band structure close to the gap closing. Its sign reminds us of the two original phases which we encountered in the previous part of the lecture:

Previously, we had identified the point \(\mu=-2t\) as a phase transition between two phases with or without zero energy edge modes. By looking at the bulk Hamiltonian, the same point \(\mu=-2t\) appears as a point where the bulk gap closes and changes sign.

When \(m=0\), the Hamiltonian has two eigenstates with energy \(E=\pm 2\Delta k\). These states are the eigenstates of \(\tau_y\), hence they are equal weight superpositions of electron and holes. They are in fact Majorana modes, that are left-moving on the branch \(E = -2\Delta k\) and right-moving on \(E = 2\Delta k\). Now they are free to propagate in the chain since there is no bulk gap anymore. In our simple model, the speed of these modes is given by \(v=2\Delta\).

Majorana modes appearing at a domain wall between different phases

Let us consider the following question: What happens if the mass parameter \(m(x)\) varies continuously in space, and changes sign at a certain point?

To answer the question, let’s write down the Dirac Hamiltonian above in real space. Then we have

\[H = -v\,\tau_y\,i\partial_x + m(x)\, \tau_z.\]

As anticipated, the mass is now a function of position, \(m(x)\), with the property that \(m(x)\to \pm m\) for \(x\to\pm\infty\) and \(m(x=0)=0\). The point \(x=0\) is a domain wall, the border between two regions of space with opposite sign of the mass.

We already know that when \(m=0\), the Hamiltonian above has a zero energy Majorana mode as a solution. Let us look at it in more detail. We need to solve the equation \(H\Psi=0\), which can be rewritten as

\[\partial_x\Psi(x) = (1/v)\,m(x)\,\tau_x\,\Psi(x).\]

Only a single Pauli matrix \(\tau_x\) appears in this equation, which is therefore quite easy to solve. The solutions have the form

\[\Psi(x) = \exp\,\left(\tau_x\int_0^x \frac{m(x')}{v}\, dx'\right) \Psi(0).\]

Two linearly independent solutions are given by the eigenstates of \(\tau_x\):

\[\Psi(x) = \exp\,\left(\pm\int_0^x \frac{m(x')}{v}\, dx'\right) \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}\,.\]

Only one of the two is normalizable, thanks to the fact that \(m(x)\) changes sign at \(x=0\). In this way we obtain a wave function localized at \(x=0\), with two exponential tails on the sides. This solution is our Majorana mode which, in this case, is a bound state localized at the domain wall. Note that no zero-energy solution would exist if \(m(x)\) did not change sign. In this case \(\Psi(x)\) would not be normalizable at either side of the domain wall.

Physically, we are considering a situation where our system is in the topological phase for \(x<0\) and in the trivial phase for \(x>0\). Therefore we have just demonstrated that at the interface between these two regions there must be a zero energy mode.

To clarify the situation, we can represent the same domain wall with our domino tiles. When you join together two chains of dominoes paired up in a different way, one single unpaired Majorana must be left in the middle: