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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?

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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).

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