Is degeneracy split a particular case of Hamiltonian diagonalization in 1D nearly free electron model?

Question

In the nearly free electron model, the diagonalization of the Hamiltonian leads to the equations: $$E_0(k,h_n)A_n + \sum v_mA_{n-m}= EA_n$$ where $E_0$ is the energy of the free electron model, $h_n = \frac{2\pi n}{a}$ ($a$ is the lattice distance), $A$'s are the coefficients of the Fourier expansion of the eigenfunction, and $v$'s are the coefficients of the Fourier expansion of the potential. Taking the potential as $$V(\frac{1}{2} + cos(\frac{2\pi x}{a})) = v_{-1}\exp (\frac{-2i\pi x}{a}) + v_0 + v_1\exp (\frac{2i\pi x}{a})$$ with all $v$'$= \frac{V}{2}$, and writing only the equations for $n = 0$ and $n = 1$ we get the system: $$(E_0(k,h_0) - E + v_0)A_0 + v_{-1}A_{1} + v_1A_{-1} = 0$$ $$(E_0(k,h_1) - E + v_0)A_1 + v_{-1}A_{2} + v_1A_{0} = 0$$ If we disregard the terms with $A_{-1}$ and $A_{2}$, the result is exactly the linear system to estimate the shift of energy due to perturbation for degenerated case. It seems ok to exclude $A_{-1}$ because $n>0$ by definition of $h_n$, but I don't see a good reason to exclude $A_{2}$. All the references that I read mention the degenerate case after the formula for the diagonalization of the Hamiltonian, but without making an explicit connection between them.

Answer 1

When you write "only the equations for $n=0$ and $n=1$", you are already making a mistake, because you've gotten rid of the equations for $A_{-1}$ and $A_2$. Crucially, the equations for $A_{-1}$ and $A_2$ explain why you can get rid of them.

Let's write out the matrix Schrodinger equation in $k$-space explicitly, not getting rid of any terms. Note that I am setting $v_0=0$ because it is an irrelevant offset: $$ \begin{bmatrix} \ddots & \vdots & & \vdots & & \vdots & \\ \cdots & \epsilon_{k-2g} & v_{-1} & v_{-2} & v_{-3} & v_{-4} & \cdots \\ & v_1 & \epsilon_{k-g} & v_{-1} & v_{-2} & v_{-3} & \\ \cdots & v_2 & v_1 & \epsilon_{k} & v_{-1} & v_{-2} & \cdots \\ & v_3 & v_2 & v_1 & \epsilon_{k+g} & v_{-1} & \\ \cdots & v_4 & v_3 & v_2 & v_1 & \epsilon_{k+2g} & \cdots \\ & \vdots & & \vdots & & \vdots & \ddots \end{bmatrix} \begin{bmatrix} \vdots \\ A_{k-2g} \\ A_{k-g} \\ A_k \\ A_{k+g} \\ A_{k+2g} \\ \vdots \end{bmatrix} = E \begin{bmatrix} \vdots \\ A_{k-2g} \\ A_{k-g} \\ A_k \\ A_{k+g} \\ A_{k+2g} \\ \vdots \end{bmatrix}\,. $$ (I've written the $\epsilon_{k-ng} = E(k,h_n)$, where $g=2\pi/a$.) Now, this is an infinite matrix, and we want to take only a portion of it. To understand how we can only take a portion of it, we appeal to second-order perturbation theory and the following diagram.

This diagram is the free-particle dispersion relation $E=\frac{\hbar^2k_x^2}{2m}$ plotted in reduced-zone scheme, i.e., for $\pi/a\leq k_x\leq \pi/a$. Each point in the diagram represents one free-particle state, but we can see that since we are in reduced-zone scheme, there many states that share a $k$ in the first Brillouin zone: these are all the points in the dispersion relation that lie on the same vertical line. It is only these states that are connected to each other by the Schrodinger equation, because the periodicity of the Hamiltonian enforces conservation of quasi-momentum: that is, the only free-particle states that are connected to each other by the Schrodinger equation are those whose wave-vectors $k_x$ differ by a reciprocal lattice vector $n2\pi/a$. These are the red dots shown, and they are labeled by their free-particle wave vectors $k_x$.

Now, let's suppose we want to find the lowest-energy Bloch states near the Brillouin zone boundary. We choose $k$ in the first Brillouin zone as shown and write the matrix equation above. The free-particle energies of the red-dotted states are, from bottom to top, $\epsilon_k$, $\epsilon_{k-g}$, $\epsilon_{k+g}$, $\epsilon_{k-2g}$, etc., where $g=2\pi/a$, and the corresponding amplitudes are $A_k$, $A_{k-g}$, $A_{k+g}$, $A_{k-2g}$, etc.

At this point, second-order perturbation theory comes to the rescue. We know that the correction to the energy of the lowest state is given by $$ E^{(2)} = \sum_{n\neq 0} \frac{ \left\lvert v_{-n} \right\rvert^2 } {\epsilon_k-\epsilon_{k+ng}}\,, $$ which means that the contribution decreases with the energy difference between the two states. Well, if we're computing the energy of the ground state, this means that the higher-lying excited states contribute only a small amount to changes in the energy and the wave function. As a consequence, it is a good approximation to drop the higher-energy plane wave states in the expansion and keep only the low-energy ones.

If $k$ is pretty near the Brillouin zone boundary, then there are two free-particle states---in this case with $k_x=k$ and $k_x=k-g$---that are very close in energy, with the other free-particle states having energies very far away from the energies of these two states. Thus, it is a reasonable approximation to keep just these two amplitudes. The result is to take the block of our matrix equation in the ($k$,$k-g$) subspace, which yields the equation $$ \begin{bmatrix} \epsilon_{k-g} & v_{-1} \\ v_1 & \epsilon_{k} \end{bmatrix} \begin{bmatrix} A_{k-g} \\ A_k \end{bmatrix} = E \begin{bmatrix} A_{k-g} \\ A_k \end{bmatrix}\,. $$

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Postscript: there is a labeling mismatch between the OP and here that I don't really have the energy to track down, but the idea in the post is sound. It could be that in the OP, they're writing the equations down for $k$ in the left half of the first Brillouin zone, in which case $n=0$ would couple to $n=1$ instead of $n=-1$. Alternatively, it could be that what I've labeled $n=-1$---i.e., the free-particle state with $k_x=n-2\pi/a$---the OP has labeled the $n=1$ state.