Free electron in weak periodic potential I am new to solid state physics and I want to ask a question regarding the free electron with weak periodic potential in 1D case. Suppose the Hamiltonian of the system can be written as follow:
\begin{equation}
H = \frac{p^{2}}{2m} + V(x) ~~,~~ V(x + a) = V(x) 
\end{equation}
Suppose we neglect the perturbation $V(x)$ and focus on the free theory. We know that for free electron, the energy spectrum is $E(k) = \frac{ \hbar^{2} k^{2}}{2m}$. Therefore, for the states $|k \rangle$ and $|-k\rangle$ they share the same energy $E(k)$. Therefore, all states are degenerate. However, in David Tong's lecture note on Solis State physics(p.34), it says that two states $|k\rangle$ and $|-k\rangle$ have the same energy does not necessarily mean that we have to use degenerate perturbation theory. This is only true if the perturbation causes the two states to mix. And then, he computed the $\langle k |V| k'\rangle$ to see the criteria of mixing.
There are two statements that I do not understand: "two states share same energy does not necessarily mean that we need to use degenerate perturbation theory" and "Perturbation causes the two states mixing". When I learnt perturbation theory, I always use the degenerate perturbation theory for degenerate states and therefore I do not understand why David Tong can use non-degernate perturbation theory when the unperturbed states are degenerate. Besides, I do not understand the concept of mixing of two states. Could someone explain why he can do such operation and mixing of states ? Thank you.
 A: When you apply degenerate perturbation theory, the standard approach is:

*

*Identify the subspace of degenerate states, in which you are interested.

*Write the perturbation in this subspace.

*Diagonalize the resulting matrix.

With this out of the way, lets focus on the concept of mixing states. Saying that an operator mixes states is just a way of saying that the matrix element between the two states are non-zero. This is a "fair" term because, when these elements are non-zero, we expect the eigenbasis of the full Hamiltonian to contain states that are superposition of the original states, even at the perturbation theory level.
In this case, the matrix for the potential of the subspace generated by {$|k\rangle$,$|-k\rangle$}, is the 2$\times$2 matrix
$$
\left( 
\begin{array}{cc}
\left\langle k|V|k\right\rangle  & \left\langle k|V|-k\right\rangle \\
\left\langle -k|V|k\right\rangle  &\left\langle -k|V|-k\right\rangle \\
\end{array} 
\right).
$$
If the off-diagonal elements are zero, then the results of degenerate and non-degenerate perturbation theory are the same (to first order at least), as this matrix is already diagonal.
If the off-diagonal elements are non-zero, then when you diagonalize this matrix, its eigenbasis will be a superposition of both states.
This becomes more complex when the size of the degenerate subspace increases, but in essence it is the statement that, if the perturbation matrix can be put in a block-diagonal form, just by swapping columns or lines, then you only need to do perturbation theory with the elements on each block of the diagonal. How can see this by noticing that diagonalizing the matrix can be done by diagonalizing one block at a time.
