Let’s now imagine that experimentalists are not only able to build such a network, but also to move the position of the domain walls and swap the positions of two Majoranas, for instance by performing the following trajectory:

Let’s suppose that the trajectory takes a time \(T\). During the trajectory, the system is described by a time-dependent Hamiltonian \(H(t)\), \(0\leq t \leq T\). This Hamiltonian contains all the details of the system, such as the positions of the domain walls where the Majoranas are located. Because the final configuration of the system is identical to the initial one, for instance all the domain walls are in the same positions as in the beginning, we have that \(H(0)=H(T)\). In other words, we are considering a closed trajectory which brings the Hamiltonian back into itself. To ensure that the wave-function for the system does not leave the ground state manifold of states \(|\Psi\rangle\), we need to change the Hamiltonian \(H(t)\) slowly enough to obey the adiabatic theorem.

So let’s imagine that we are in the adiabatic limit and that we exchange two Majoranas \(\gamma_n\) and \(\gamma_m\). As usual in quantum mechanics, the initial and final quantum states are connected by a unitary operator \(U\) (\(U^{-1}=U^\dagger\)),

\[\left|\Psi\right\rangle \,\to\, U \left|\Psi\right\rangle\,.\]

Because the quantum state \(\left|\Psi\right\rangle\) never leaves the ground state manifold, which has \(2^N\) states, the operator \(U\) can be written a \(2^N\times 2^N\) unitary matrix.

We can derive the exact form of \(U\) without a direct calculation, which would require knowing \(H(t)\), but only based on the following, general considerations. First, the adiabatic exchange of two Majoranas does not change the parity of the number of electrons in the system, so \(U\) commutes with the total fermion parity, \([U, P_\textrm{tot}]=0\). Second, it is reasonable to assume that \(U\) only depends on the Majoranas involved in the exchange, or in other words that it is a function of \(\gamma_n\) and \(\gamma_m\), and of no other operator. And because it has to preserve fermion parity, it can only depend on their product, that is on the parity operator \(-i\gamma_n\gamma_m\), which is Hermitian. Finally, the exponential of \(i\) times a Hermitian operator is a unitary operator. So, in general \(U\) must take the form

\[U\equiv\exp(\beta \gamma_n \gamma_m) = \cos(\beta) + \gamma_n\gamma_m \sin(\beta)\,,\]

up to an overall phase. Here, \(\beta\) is a real coefficient to be determined, and in the last equality we have used the fact that \((\gamma_n\gamma_m)^2=-1\). To determine \(\beta\), it is convenient to go to the Heisenberg picture and look at the evolution of the Majorana operators in time. We have that

\[ \gamma_n\,\to\, U\,\gamma_n\,U^\dagger\,,\\ \gamma_m\,\to\, U\,\gamma_m\,U^\dagger\,. \]

Inserting our guess for \(U\) we obtain:

\[ \gamma_n\,\to\, \cos (2\beta)\,\gamma_n - \sin(2\beta)\,\gamma_m\,,\\ \gamma_m\,\to\, \cos (2\beta)\,\gamma_m + \sin(2\beta)\,\gamma_n\,. \]

Now we have to remember that at time \(T\) we have completed a closed trajectory, so that the Majorana \(\gamma_n\) is now in the place initially occupied by \(\gamma_m\), and vice versa. This condition leads to the choice \(\beta = \pm \pi/4\). It is not strange that we find that both signs are possible - this distinguishes the clockwise and the counterclockwise exchange of the Majoranas.

Thus, we can write the unitary operator that exchanges the Majorana modes \(\gamma_n\) and \(\gamma_m\) in an explicit (and somewhat non-trivial looking!) form as: \[U = \exp \left(\pm\frac{\pi}{4}\gamma_n \gamma_m\right) = \tfrac{1}{\sqrt{2}}\left(1\pm\gamma_n\gamma_m\right)\]

To fix our ideas and study the consequences of \(U\) more closely, it is convenient to just focus on four Majoranas \(\gamma_1\,\gamma_2,\gamma_3\) and \(\gamma_4\). For this discussion we will assume that counter-clockwise exchanges pick the \(+\) sign in \(U\). Their ground state manifold has four states, which in the notation introduced before we write down as

\[\left|00\right\rangle, \left|11\right\rangle, \left|01\right\rangle, \left|10\right\rangle\,,\]

where the first digit is the occupation number of the fermionic mode \(c^\dagger_1=\tfrac{1}{2}(\gamma_1+i\gamma_2)\) and the second digit the occupation number of \(c^\dagger_2=\tfrac{1}{2}(\gamma_3+i\gamma_4)\). The most generic possible wave function is a superposition

\[\left|\Psi\right\rangle = s_{00}\left|00\right\rangle + s_{11} \left|11\right\rangle + s_{01} \left|01\right\rangle + s_{10} \left|10\right\rangle\,,\]

which we can also represent as a vector with four entries, \(\left|\Psi\right\rangle = (s_{00}, s_{11}, s_{01}, s_{10})^T\). The operator \(U\) at this point can be written as a \(4\times 4\) matrix. In order to do so, you just have to compute the action of a product of Majoranas on the basis states. This a simple but tedious operation, which we skip here. It results in the following matrices for the operators \(U_{12}, U_{23}\) and \(U_{34}\) exchanging neighboring Majoranas:

\[ U_{12} = \exp\left(\frac{\pi}{4}\gamma_1 \gamma_2\right) \equiv\begin{pmatrix} e^{-i\pi/4} & 0 & 0 & 0 \\0 & e^{i\pi/4} & 0 &0 \\0 & 0& e^{-i\pi/4} &0 \\ 0&0& 0& e^{i\pi/4} \end{pmatrix}\,, \]

\[ U_{23} = \exp\left(\frac{\pi}{4}\gamma_2 \gamma_3\right) \equiv\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i & 0 & 0\\ -i & 1 & 0& 0\\ 0& 0& 1 & -i\\ 0& 0& -i & 1 \end{pmatrix}\,, \]

\[ U_{34} = \exp\left(\frac{\pi}{4}\gamma_3 \gamma_4\right) \equiv\begin{pmatrix} e^{-i\pi/4} & 0 & 0 & 0\\ 0& e^{i\pi/4} & 0& 0\\ 0& 0& e^{i\pi/4} & 0\\ 0& 0& 0& e^{-i\pi/4} \end{pmatrix}\,. \]

These matrices indeed act in a very non-trivial way on the wave function. For instance, if we start from the state \(\left|00\right\rangle\) and we exchange \(\gamma_2\) and \(\gamma_3\), we obtain

\[\left|00\right\rangle\,\to\,U_{23}\left|00\right\rangle=\tfrac{1}{\sqrt{2}}\left(\left|00\right\rangle-i\left|11\right\rangle\right)\,,\]

which is a superposition of states! Hence we have seen explicitly that the effect of the exchange two Majoranas on the wavefunction amounts to much more than just an overall phase, as it happens for bosons and fermions.

Let’s now try a sequence of two exchanges. In this case, we have to multiply the corresponding \(U\)s, ordering them from right to the left according to the order of the exchanges. Given that the matrices above are not diagonal, it is not surprising that the order in the product matters a lot. For instance you can check that

\[U_{23}U_{12}\neq U_{12}U_{23}\,\]

We have just shown that exchanging two Majorana modes leads to a non trivial rotation in the ground state manifold, and that changing the order of the exchanges changes the final result. These properties make Majorana modes non-Abelian anyons. The exchange of two non-Abelian anyons is usually called braiding, a name which is suggestive of the fact that, when thinking of the trajectories of the different particles, a sequence of exchanges looks like a braid made out of different strands.

Finally, you might object to the fact that the network of nanowires drawn in the figures only allows to exchange neighbouring Majoranas, even though our derivation of \(U=\exp(\pi\gamma_n\gamma_m/4)\) seems to hold for any pair of Majoranas. This geometric constraint is not a big problem: by carefully composing many exchanges between neighbours, we can exchange any pair of Majoranas. As an example, you have that \(U_{13}\equiv\exp\left(\pi\gamma_1 \gamma_3/4\right) = U_{12}^\dagger\,U^\dagger_{23}\,U_{12}\).