# Triangular Tight Binding Model

I have the following Hamiltonian for a triangular tight binding model:

$$H = e^{i\phi} \Big(|2\rangle \langle1| + |3\rangle \langle2| + |1\rangle \langle3| \Big) + e^{-i\phi} \Big(|1\rangle \langle2| + |2\rangle \langle3| + |3\rangle \langle1| \Big)$$

where the states $$|1\rangle, |2\rangle, |3\rangle$$ are the basis states and $$\phi \in \mathbb{R}$$.

I wanted to find the eigenvalues and eigenstates of this Hamiltonian. The matrix turns out to be: $$H = \begin{bmatrix} 0 & e^{-i\phi} & e^{i\phi}\\ e^{i\phi} & 0 & e^{-i\phi}\\ e^{-i\phi} & e^{i\phi} & 0 \end{bmatrix}$$

I got $$\lambda_1 = \lambda_2 =\lambda_3 = 2cos(\phi)$$. Now, one of the eigenvectors is $$v_1 = \frac{1}{\sqrt{3}}\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}$$

How do I find the other eigenvectors?

Also, how do I find the time-dependent solution $$|\psi (t)\rangle$$ of the Schrödinger equation with the given Hamiltonian?

• Did you mean $\phi\in\mathbb{R}$? Oct 1, 2023 at 2:44

First of all, if all three eigenvalues were really equal, then $$\textbf{any}$$ vector would be an eigenvector with that same eigenvalue. But that would also mean that the Hamiltonian is the identity matrix (up to an overall constant factor).

For the given Hamiltonian only one of the eigenvalues is $$\lambda_1 = 2\cos\phi$$. The other two are $$\lambda_{2,3} = -\cos\phi \pm \sqrt{3}\sin\phi$$. All of them can be determined from the usual equation defining the eigenvalues:

$$\det|H-\lambda I| = 0,$$

where $$I$$ is the $$3\times 3$$ identity matrix.

Knowing the eigenvalue, one can find the corresponding eigenvector by solving the linear system of 3 equations ($$i$$ could be 1, 2 or 3)

$$H v_i = \lambda_i v_i,$$

or more explicitly

$$\begin{pmatrix} 0 && e^{i\phi} && e^{-i\phi} \\ e^{-i\phi} && 0 && e^{i\phi} \\ e^{i\phi} && e^{-i\phi} && 0 \end{pmatrix} \begin{pmatrix} v_{i1} \\ v_{i2} \\ v_{i3} \end{pmatrix} = \lambda_i \begin{pmatrix} v_{i1} \\ v_{i2} \\ v_{i3} \end{pmatrix}.$$

To find the time-dependent solution one needs first to decompose the initial state into the eigenstates of the Hamiltonian, since each of them evolves in a known simple way. So we start start by finding a decomposition

$$v = \sum_{i=1}^3 c_i v_i$$

where $$v$$ is the initial state vector, $$v_i$$ are the Hamiltonian eigenvectors defined above, $$c_i$$ are complex numbers - coefficients of the decomposition. Eigenvectors of the Hamiltonian change with time by being multiplied by a time-dependent phase:

$$v_i(t) = e^{-\frac{i}{\hbar}\lambda_i t} v_i,$$

which follows from the time-dependent Schrodinger equation for the eigenvector:

$$i\hbar \frac{\partial}{\partial t} v_i(t) = \lambda_i v_i(t).$$

Finally, the time-evolved state $$v(t)$$ is given by the same linear combination of the time-evolved eigenstates as the initial state $$v$$ in terms of initial eigenstates $$v_i$$ which results in

$$v(t) = \sum_{i=1}^3 c_i e^{-\frac{i}{\hbar}\lambda_i t} v_i.$$

• Oh, yes! My bad. I did check that $\lambda_2, \lambda_3$ that you wrote were in fact the eigenvalues but I am struggling to find the eigenvectors for those two eigenvalues. Now, I am doubting if I even calculated the Hamiltonian correctly. Could you have a look? Oct 1, 2023 at 18:53
• The Hamiltonian that you wrote in matrix form matches the one in the basis decomposition that you wrote in the first equation of your question, if that's what you are asking. Oct 2, 2023 at 3:08
• Yes, that is what I was asking. The eigenvector corresponding to $\lambda_1 = 2cos(\phi)$ turns out to be $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ Now, for $\lambda_2, \lambda_3$, the calculation is becoming dirty with some weird result. Is there any trick involved here for these two eigenvalues? Oct 2, 2023 at 11:18
• Sometimes the results are weird and there is nothing you can do about it :) You can check (for example) with Mathematica that you have found the eigenvectors correctly. It can also help you decompose the initial state into these weird eigenvectors to figure out the time evolution. Oct 2, 2023 at 14:14
• I did check it with Mathematica. Here is the link. The first eigenvector is the same as I found it. But look at the second and the third. Do you think that is normal? Oct 2, 2023 at 15:16