# Eliminating an eigenvalue from the Hamiltonian

I have a momentum space Hamiltonian $$H(\vec k)$$ for a Kagome lattice and I want to find its eigenvalues which may be dependent on $$\vec k$$. Now, I'm told that one of the eigenvalues for such Hamiltonians is "flat" (it has no $$\vec k$$-dependence). Obtaining the analytical form of the eigenvalues is difficult since the Hamiltonian is quite complex. I was told that one can do this easily by eliminating the flat eigenvalue and the resultant Hamiltonian is easy to work with. I want to know how this is done.

I think this paper does the same thing (see right-column on page 2), however I don't fully understand this. Following the paper, I was able to get the non-flat energy bands, but didn't quite understand what I was doing. Is there a name for this technique? Can anybody provide some links on it?

Edit 1: Here's the arXiv link of the paper: https://arxiv.org/abs/1009.3792 . The relevant portion is on the second page.

Edit 2: I thought this was a common thing to do while calculating for energy bands, but following @ZeroTheHero's advice, here's what I am doing:

I have an electrical circuit in the form of Kagome lattice. This circuit shows topological behaviour. By using KCL and applying Fourier transform we can get the circuit's Laplacian $$J$$ defined as $$I=JV$$ From this, we can get the Hamiltonian for this lattice-circuit (in momentum space): \begin{align*} H &= -iJ\\ &= \begin{bmatrix} -\frac{1}{\omega L}+2\omega(C_A+C_B) && -\omega C_A -\omega C_B e^{-i (\frac{k_xd}{2}+\frac{\sqrt{3}k_yd}{2})} && -\omega C_A -\omega C_B e^{-i k_xd}\\ -\omega C_A -\omega C_B e^{i (\frac{k_xd}{2}+\frac{\sqrt{3}k_yd}{2})} && -\frac{1}{\omega L}+2\omega(C_A+C_B) && -\omega C_A -\omega C_B e^{-i (\frac{k_xd}{2}-\frac{\sqrt{3}k_yd}{2})}\\ -\omega C_A -\omega C_B e^{i k_xd} && -\omega C_A -\omega C_B e^{i (\frac{k_xd}{2}-\frac{\sqrt{3}k_yd}{2})} && -\frac{1}{\omega L}+2\omega(C_A+C_B)\\ \end{bmatrix} \end{align*} Here, $$\omega$$ is the frequency of AC current, $$C_A$$ and $$C_B$$ are capacitances of the capacitors used in the circuit, $$L$$ is the inductance and $$d$$ is the lattice parameter.

Now let's define $$\psi_A = \frac{1}{\sqrt{3}}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\text{,} \quad \psi_B = \frac{1}{\sqrt{3}}\begin{bmatrix}e^{iP} \\ e^{iQ} \\ e^{iR}\end{bmatrix}\text{,} \quad \psi = \frac{1}{\sqrt{3}}\begin{bmatrix}\psi_A && \psi_B\end{bmatrix} = \frac{1}{\sqrt{3}}\begin{bmatrix}1 && e^{iP} \\ 1 && e^{iQ} \\ 1 && e^{iR}\end{bmatrix}\text{,} \quad D = 3\begin{bmatrix}-\omega C_A && 0\\ 0 && -\omega C_B\end{bmatrix}\text{and}\quad h = \psi D \psi^\dagger$$

We can equate $$H$$ to $$h$$ (upto a constant times identity matrix), and get the appropriate values of $$P$$, $$Q$$ and $$R$$. Here, $$P=0$$, $$Q = \frac{k_xd}{2}+\frac{\sqrt{3}k_yd}{2}$$ and $$R = k_xd$$

We'll see that $$H = h+\left(3\omega(C_A+C_B)-\frac{1}{\omega L}\right)I_{3\times 3}$$

We can ignore the 2nd term (containing the identity matrix) as that changes the eigenvalues only by a constant and does not change the eigenvectors and work only with $$h$$. Let $$O = \psi^\dagger\psi$$.

The wave function for non-ﬂat energy bands is given as linear combination of $$\psi_A$$ and $$\psi_B$$ ($$\phi_A$$ and $$\phi_B$$ are scalars) $$\gamma = \phi_A\psi_A+\phi_B\psi_B=\psi\phi \qquad \text{where } \phi=\begin{bmatrix}\phi_A\\ \phi_B\end{bmatrix}$$ Writing the Schrodinger equation $$h\gamma=E\gamma \Rightarrow \psi D \psi^\dagger \, \psi \phi = E\psi\phi \Rightarrow ODO\phi=EO\phi$$ Multiplying by ($$O^{1/2})^{-1}$$, we get $$h_\psi \phi^\prime = E\phi^\prime$$ where $$h_\psi = O^{1/2}DO^{1/2}$$ and $$\phi^\prime = O^{1/2}\phi$$

$$h_\psi$$ is a $$2\times 2$$ matrix and its eigenvalues can be given by $$\lambda_\pm = \frac{1}{2}\left(\text{tr}(h_\psi)\pm\sqrt{\text{tr}^2(h_\psi)-4\text{ det}(h_\psi)}\right)$$ We can easily calculate det($$h_\psi$$) and tr($$h_\psi$$) and get the non-flat energy bands. Also, $$h_\psi \phi^\prime = E\phi^\prime$$ can be easily solved for eigenvectors.

So, my questions specific to this calculations are:

1. When we write $$h = \psi D \psi^\dagger$$, how does this eliminates the flat band?
2. Is there something special about $$\psi_A$$ and $$\psi_B$$ that's enabling us to eliminate the flat band?
3. Why is the eigenvector of $$h$$ (i.e., $$\gamma$$) a linear combination of $$\psi_A$$ and $$\psi_B$$?
• The paper is behind a paywall. Is there an ArXiv link? Commented Jun 22, 2020 at 13:38
• @mikestone please check the edit Commented Jun 22, 2020 at 14:59
• This is an interesting question but it requires additional details on your part as people are unlikely to plot through an arXiv submission to know what you’re talking about. Commented Jun 27, 2020 at 23:46
• @ZeroTheHero I've added my calculations in the edit, hope this helps. Commented Jun 29, 2020 at 19:15