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What you said is true. But what I'm trying to say is there exist eigenstates of H that are not eigenstates of D.You still emphasize the wrong point. D(a) is a pedagogical tool to get the intuition right. For a completely periodic potential, you will get a whole family of operators corresponding to D(na), where n is an integer. Any of them represents a symmetry of the system. Any complex [itex]\lambda[/itex] for D(a) will be a real [itex]\lambda[/itex] for D(na) for some n. The reason why you get wavefunctions that do not share the periodicity of the lattice is that all Bloch wave functions in a fully periodic potential need to be simultaneous eigenstates of the Hamiltonian and D(na) for one n, not necessarily of the Hamiltonian and D(a).