Bloch theorem, Energy, Free electron I'm trying to learn on my own a bit of solid physics to tackle semiconductors afterwards. I'm struggling with the Energy versus $k$ diagrams for a free electron which shows that for a single value of $k$ we can have many energies value although that there's only one band (in this example).
Bloch's theorem tells us that we can label the energies the system can take with a vector $k$ and an integer $n$ (band index). And the theorem tells us also that for any vector $G$ in the reciprocal lattice we have:
$E(n,k)=E(n,k+G)$
Let's apply this to the free electron. This tells us that (there's only one band index)  
$$\frac{\hbar^2}{2m}\lvert k\rvert^2 = \frac{\hbar^2}{2m}\lvert k +G\rvert^2 $$ 
for all $G$ in the reciprocal lattice$ \lvert k\rvert=\lvert k+G\rvert$ which is absurd.
We find another contradiction by making the following reasoning:
Since the potential is constant, we can consider that the potential is periodic with respect to a lattice with arbitrary vectors $a_1,a_2,a_3$. (In 1D, all that I'm saying is that a constant function can be seen as 1-periodic or 100-periodic or 0.0001 periodic..). So by using this fact we can prove that there is only one possible value for $E$ which is also absurd. 
Can someone please explain in detail what am I doing wrong?
 A: It is more difficult to think about the free particle case, because it is unnatural to put it on a lattice and not gap the Brillioun Zone edges. However, if you do want to do such a thing, the dispersion would look like (b) in the figure below for an arbitrary choice of lattice spacing $a$:
nearly free elecrton model http://users-phys.au.dk/philip/pictures/solid_metalquantum/nearfree.gif
Why is this so? It is because you have made something in real space periodic. When you do that, reciprocal space also becomes periodic. (This is precisely the reason that a reciprocal lattice exists.) Now, look at the energy of the bottom band labeled next to (d) in the figure. (The gap here is irrelevant, but it makes it clearer what I'm referring to as the bottom band). This is indeed periodic just like your equation and you can see that for a specific band:
\begin{equation}
\frac{\hbar^2k^2}{2m}=\frac{\hbar^2(k+G)^2}{2m}
\end{equation}
Also, for a specific value of $k$ you can have more than one band. I think you are mistaken in thinking that there is only one band for the free electron model. In fact, if you break real space up into $N$ different units of spacing $a$, you will get $N$ bands, though most of them will be unoccupied. You can see that in the limit that $a\rightarrow\infty$, we recover the usual free particle dispersion.
A: The formula $$ E = \frac{\hbar^{2}}{2m} |k|^{2}$$ is valid in the vicinity of $k = 0$, this is called parabolic band approximation, which is - as the name suggests - only an approximated formula. Regarding the actual calculation of energy, you need to calculate the eigenvalues of the Hamiltionian of the periodic system, then the theorem is trying to say that $$\mathcal{H}_{k} \psi(k) = \mathcal{H}_{k+K} \psi(k+K)$$ from which the equivalence of the values of energy occurs.
Regarding the second question, it's really important to realize, that the actual force on a particle in a potential does not come from a constant value of a potential, but rather from it's gradient. Again, if you were to calculate the eigenvalues of Hamiltonian in a potential where $V = 0$ everywhere, you would get the same result as if you were calculating the same equation with, lets say, $V = 42$.
