Posted by Steve Simon on May 28, 2013, 6:54 am, in reply to "Reciprocal lattice vectors in nearly free electron models"
This is the essence of Bloch's theorem! Yes it is true that if you multiply a free (or nearly free) electron wavefunction by e^{2 i pi x/a} you do get a different free electron wavefunction. However, let us think in the modified plane wave form that Bloch guarantees. We must have eigenfunctions of the form psi_k^a(x) = e^{i k x} u_k^a(x) where for each k in the BZ, there are multiple possible eigenfunctions (corresponding to the different bands) which we label with different a's. Here u is a function periodic in the unit cell. Note that the particular form of the eigenfunctions, u, depends on the details of the Hamiltonian (as do the corresponding eigenenergies). Now if you take such a modified plane wave and you decided to multiply it by e^{2 i pi x/a} you could completely absorb this factor into the period function u, so that the form of the periodic wavefunction still looks exactly the same. While this sounds like a trivial comment, think about doing the following. Take the original Schroedinger equation (including periodic potential) and substitute in this modified planewave form. For each value of k you choose, you will get an eigenvalue/eigenfunction problem for the function u and its energy E as the eigenvalue (with boundary conditions that u must be periodic in the unit cell). If you now shift k by 2pi/a and try plugging that into the Schroedinger equation, you get exactly the same energy eigenvalues (as well as the same functions psi_k^a ) because you are simply redefining u to include this extra periodic factor of e^{2 i pi x/a}. While this does not mean that multiplying by e^{2 i pi x/a} is the same as doing nothing, it does mean that the space of eigenfunctions and eigenvalue is unchanged by shifting k by 2 pi/a. Does this help? 
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