# From Kitaev model to an experiment¶

We have a special guest to begin this week's lecture, Yuval Oreg from the Weizmann Institute in Rehovot.

# Small parameters¶

We are now all set to make Majoranas in a real system. Or at least to invent a way to make Majoranas in a real system.

The way we approach this problem is by considering the Kitaev chain a 'skeleton', and 'dressing' it with real physics phenomena until it becomes real.

Interestingly, this is not at all how the condensed matter community came to this model. Instead, the path to it was from complex to simple. The whole story started from what we'll consider in the very end of the course, fractional particles.

Then it was simplified to topological superconductors (that still do not exist in nature, as far as we know). Majoranas were then predicted to exist (week 7) in a combination of a 3D topological insulator (week 6), which was then simplified to a two-dimensional topological insulator (week 5), and only after a few more simplification steps, the nanowire model was developed.

So once again, here is our 'skeleton', the Kitaev model Hamiltonian written in momentum space:

$$H_{Kitaev} = (-2 t \cos k -\mu) \tau_z + 2 \Delta \tau_y \sin k.$$

The model seems OK for a start, because it has some superconducting pairing $\Delta$ and some normal dispersion given by terms proportional to $\mu$ and $t$.

Before we proceed further, let's understand the relation between these parameters.

First of all, we want to make a controllable system, so that we can tweak its parameters. That means that we need a **semiconductor**. In semiconductors the electron density is very low, so that the chemical potential is near the bottom of the band. This makes it easier to define $\mu$ with respect to the bottom of the band:

$$\mu \rightarrow \mu - 2t.$$

Now the transition between trivial and non-trivial states happens when $\mu = 0$.

Of course semiconductors are never additionally superconducting. Luckily this is easy for us to resolve. We just paste a superconductor and semiconductor together into a hybrid structure, and let the superconductor induce superconductivity in the semiconductor. Making such a hybrid is extremely challenging from the material science point of view, but it's definitely not our problem for now.

The next thing we should consider is that $\mu$ will always stay small compared to the bandwidth, so $\mu \ll 2t$. The same holds for superconducting pairing: $\Delta \ll t$. This is because superconductivity is a very weak effect compared to the kinetic energy of electrons. These two inequalities combined mean that we can expand the $\cos k$ term and only work with the continuum limit of the Kitaev model:

$$H = (k^2/2m - \mu) \tau_z + 2 \Delta \tau_y k.$$

The effective electron mass $m$ is just the coefficient of the expansion. Let's take a look at the band structure in this regime, both in the topological regime and in the trivial regime:

# The need for spin¶

Still, there is one obvious thing missing from the model, namely electron spin. This model works with some hypothetical spinless fermions, that do not really exist. So to make the model physical, we need to remember that every single particle has spin, and the Hamiltonian has some action in spin space, described by the Pauli matrices $\sigma$.

The simplest thing which we can do is to just add the spin as an extra degeneracy, that is to multiply every term in the Hamiltonian by $\sigma_0$. Obviously this doesn't change the spectrum, and a zero energy solution stays a zero energy solution.

Just kidding, this would be very bad! The problem about adding spin is that the whole point of a Kitaev chain is to create *unpaired* Majorana modes. If we add an extra spin degeneracy to these Majoranas, the edge of our chain will host two Majoranas, or in other words one regular fermion fine-tuned to zero energy.

What's the correct way of introducing spin then? We still need to add it. Let's add spin such that the Kitaev chain corresponding to one spin species is topologically trivial, and the Kitaev chain corresponding to the other spin species non-trivial. We know that the chemical potential $\mu$ controls whether a Kitaev chain is topological or trivial, so if say spin up has $\mu > 0$ and spin down $\mu < 0$, we're back in business.

We achieve this by adding Zeeman coupling of the spin to an external magnetic field:

$$H = (k^2/2m - \mu - B \sigma_z) \tau_z + 2 \Delta \tau_y k.$$

Whenever the Zeeman energy $|B|$ is larger than $\mu$ we have one Majorana fermion at the end of the chain.

Let's look at what happens with the dispersion as we increase the magnetic field from zero to a value larger than $\mu$.

We now see that we resolved the first problem:

A high enough

Zeeman splittingallows to separate the different spins. Then we can make one spin species trivial, while the other one is topological and hosts Majoranas.

# Realistic superconducting pairing¶

The next part for us to worry about is the superconductor.

Something that you probably saw in the Kitaev chain Hamiltonian is that the superconducting pairing $\Delta$ has a peculiar form. It pairs electrons from *neighboring* sites, and not those from the same site. In momentum space this means that the superconducting pairing is proportional to $\Delta k$.

Of course, in a Kitaev chain the superconducting pairing cannot couple two electrons from the same site since there is just one particle per site!

Real world superconductors are different. Most of them, and specifically all the common superconductors like $Al$, $Nb$, $Pb$, $Sn$ have $s$-wave pairing. This means that the pairing has no momentum dependence, and is local in real space. The Kitaev chain pairing is proportional to the first power of momentum and so it is a $p$-wave pairing.

High temperature superconductors like cuprates or pnictides do have a momentum-dependent pairing, but it's yet another type ($d$-wave, or a more exotic $s\pm$-wave).

So if we want to invent a way to make Majoranas, we will need to use $s$-wave pairing. And then, as you should remember from the previous week, due to the fermionic statistics the pairing function should be antisymmetric. In a Kitaev chain the antisymmetry is due to the real space structure of the pairing, but in an $s$-wave superconductor, the antisymmetry of the pairing should arise due to its spin structure.

This leaves only one option. All the $s$-wave superconductors are spin-singlet:

$$H_{pair} = \Delta(c_\uparrow c_\downarrow - c_\downarrow c_\uparrow) + \text{h.c.}$$

This means that now we need to modify the pairing, but before that we'll need to do one other important thing.

### Important and useful basis change.¶

When you see Bogoliubov-de-Gennes Hamiltonians in the literature, you will find them written in two different bases. One variant is the one which we introduced last week:

$$ H_\textrm{BdG} = \begin{pmatrix} H & \Delta \\ -\Delta^* & -H^* \end{pmatrix}. $$

It has the particle-hole symmetry $H_\textrm{BdG} = - \tau_x H^*_\textrm{BdG} \tau_x$. In this basis, the $s$-wave pairing is proportional to $\sigma_y$.

However for systems with complicated spin and orbital structure, there is a different basis which makes the bookkeeping much easier.

If we have a time-reversal symmetry operator $\mathcal{T} = U \mathcal{K}$, we can apply the unitary transformation $U$ to the holes, so that in the new basis we get the Bogoliubov-de-Gennes Hamiltonian

$$ H_\textrm{BdG} = \begin{pmatrix} H & \Delta' \\ \Delta'^\dagger & -\mathcal{T} H \mathcal{T}^{-1}\end{pmatrix}, $$

with $\Delta' = \Delta U^\dagger$.

Why is this basis useful?

- First of all, because in this new basis the $s$-wave pairing is a unit matrix regardless of system we consider.
- Second, because it's easy to get the Hamiltonian of holes. We take the Hamiltonian for electrons, and change the signs of all terms that respect time-reversal symmetry, but not for those that break it, such as the term proportional to the magnetic field $B$. So if the electrons have a Hamiltonian $H(B)$, the Hamiltonian of the holes just becomes $-H(-B)$.

There is one disadvantage. The particle-hole symmetry now becomes more complicated. For our system with only one orbital and spin it is $\mathcal{P} = \sigma_y \tau_y \mathcal{K}$. But, let us tell you, the advantages are worth it.

# s-wave superconductor with magnetic field¶

Let's look at how our chain looks once we change the superconducting coupling to be $s$-wave. The Zeeman field (or anything of magnetic origin) changes sign under time-reversal symmetry.

This means that the Zeeman field has the same form for electrons and for holes in the new basis, and the full Hamiltonian is now:

$$ H_\textrm{BdG} = (k^2/2m - \mu)\tau_z + B \sigma_z + \Delta \tau_x. $$

This Hamiltonian is easy to diagonalize since every term only has either a $\tau$ matrix or a $\sigma$ matrix. At $k=0$ it has 4 levels with energies $E = \pm B \pm \sqrt{\mu^2 + \Delta^2}$.

We can use this expression to track the crossings. We also know that when $B=0$ the system is trivial due to spin degeneracy. Together this means that we expect the system to be non-trivial (and will have a negative Pfaffian invariant) when

$$ B^2 > \Delta^2 + \mu^2.$$

Are we now done? Not quite.

### Problem with singlets¶

A singlet superconductor has an important property: Since electrons are created in singlets, the total spin of every excitation is conserved. Zeeman field conserves the spin in $z$-direction, so together every single state of our system has to have a definite spin, *including the Majoranas*.

And that is a big problem. Majoranas are their own particle-hole partners, and that means that they cannot have any spin (energy, charge, or any other observable property at all).

So does this now mean that we "broke" the bulk-edge correspondence? Let's look at the band structure (tweak the Zeeman energy):

Of course we didn't break bulk-edge correspondence. Majoranas in our system would have to have a spin, which isn't possible. That in turn means that they cannot appear, and that means that the system cannot be gapped.

We can also approach this differently. From all the spin Pauli matrices, only $\sigma_z$ appears in the Hamiltonian, so there's a conservation law. The two bands that cross at zero energy in the band structure above belong to opposite spin bands, and so cannot be coupled.

Now we need to solve this final problem before we are done.

# How to open the gap?¶

The final stretch is straightforward.

We know that there is no gap because of conservation of one of the spin projections, so we need to break the spin conservation.

If we don't want to create an inhomogeneous magnetic field, we have to use a different term that couples to spin. That term is spin-orbit interaction. In it's simplest form this interaction appears in our wire as

$$H_{SO} = \alpha \sigma_y k,$$

so it is like a Zeeman field pointing in $y$-direction with a strength proportional to the particle momentum. Note that this term is invariant under time reversal symmetry (both $\sigma_y$ and $k$ change sign). So now we have our final Hamiltonian:

$$ H_\textrm{wire} = (k^2/2m + \alpha \sigma_y k - \mu)\tau_z + B \sigma_z + \Delta \tau_x. $$

At $k = 0$, spin-orbit coupling vanishes, so it has no effect on the system being topologically trivial or non-trivial.

Let's now check that it does what we want, namely open the gap at a finite momentum: