# How are Majorana zero-modes protected against fermionic operators?

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?