I am learning about Majorana fermions in topological quantum computation, and more particularly about the Kitaev chain, described by $$ H = -\mu \sum_{i=1}^N c_i^\dagger c_i - \sum_{i=1}^{N-1} \left(t c_i^\dagger c_{i+1} + \Delta c_i c_{i+1} + h.c.\right) $$ where $c_i = (\gamma_{2i-1} + i\gamma_{2i})/2$ is the annihilation operator written as a sum of two Majorana fermions $\gamma_{2i-1}$ and $\gamma_{2i}$. In the special case where $t=|\Delta|$ and $\mu = 0$, this Hamiltonian simplifies to $$ H = 2t \sum_{i=1}^{N-1} \left[ d_i^\dagger d_i - \frac{1}{2} \right] $$ where $d_i = (\gamma_{2i+1} + i\gamma_{2i})/2$ is a new annihilation operator defined "in between" two fermions. This leaves a Majorana zero-mode that can defined with both ends of the chain through $d_0 = (\gamma_{1} + i\gamma_{2N})/2$. Using this operator, we can now define a computational basis $\{|0\rangle, |1\rangle \}$ through $d_0 |0\rangle = 0$ and $|1\rangle = d_0^\dagger |0\rangle$, which are the two degenerate ground states of the Hamiltonian above.

My question is the following. How are these two states $\{|0\rangle, |1\rangle \}$ protected against errors that can physically happen in the system? For instance, if the first element of the Kitaev chain loses (or gains) a fermion, the state $|0\rangle$ will collapse into $c_1^{(\dagger)}|0\rangle \neq |0\rangle$. Wouldn't we then lose information?


The OP is right: if the system is open to an environment allowing for single-particle tunneling into the system, then the Majorana edge mode is, in fact, not stable. This was studied by Budich, Walter and Trauzettel in Phys. Rev. B 85, 121405(R) (2012) (or check out the freely-accessible preprint: arXiv:1111.1734).

Kitaev's claim about the absolute stability---as quoted by AccidentalFourierTransform in his/her answer---is presuming a closed system. In that case, one can indeed argue that fermion parity symmetry cannot be broken in a local system, such that the edge mode becomes absolutely stable (for energy scales below the bulk gap).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.