How do I find the matrix for the Su-Schrieffer-Heeger model's Hamiltonian?

Question

I will not bother to write down the tensor product in the joint basis of A and B here in this post, where A and B are atoms/electrons and m denotes a unit cell.

The Hamiltonian for this model is given by:

$H = v \sum_{m=1}^{N} \left( |m,B\rangle \langle m,A| + \text{h.c.} \right) + w \sum_{m=1}^{N-1}\left( |m+1,B\rangle \langle m,A| + \text{h.c.} \right)$

The matrix is given by:

\begin{bmatrix} 0 & v & 0 & 0 & 0 & 0 & 0 & 0\\ v & 0 & w & 0 & 0 & 0 & 0 & 0\\ 0 & w & 0 & v & 0 & 0 & 0 & 0\\ 0 & 0 & v & 0 & w & 0 & 0 & 0\\ 0 & 0 & 0 & w & 0 & v & 0 & 0\\ 0 & 0 & 0 & 0 & v & 0 & w & 0\\ 0 & 0 & 0 & 0 & 0 & w & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & v & 0\\ \end{bmatrix}

I think the sum in the Hamiltonian is being done over the unit cells which are denoted by m and A,B are the atoms/electrons respectively. I assume no spin or Couloumb interaction. I tried to write the Hamiltonian with the completeness relation on either side to figure out the matrix like this:

IHI, where I is the completeness relation written using the basis $|m, A\rangle|m, B\rangle$

However I think I did it wrong as when I supplied the completeness relation, I ended up with kronecker delta's in A,A' and B,B' with a leftover basis in the ket, but I do not know how to proceed or what to do next. My QM knowledge is at the level of Griffiths/Zettili. I can't seem to derive the matrix representation of this Hamiltonian for N=4.

Answer 1

$A$ and $B$ aren't atoms and electrons. The Su-Schrieffer-Heeger chain is a bipartite chain of atoms of the form

$$A\overset{v}{-}B\overset{w}=A-B=A-B=A-B=A-B$$

The standard unit cell of the chain is one $[A-B]$ unit. The amplitude for hopping between $A$ and $B$ sites within a given unit cell is $v$, and the amplitude for hopping between $A$ and $B$ sites on adjacent unit cells is $w$.

The real-space basis is given by $|m,\sigma\rangle$ where $m\in\{1,2,3,4\}$ and $\sigma\in\{A,B\}$. The matrix elements in the $8\times 8$ matrix are

$$\pmatrix{\langle 1,A|H|1,A\rangle & \langle1,A|H|1,B\rangle & \cdots & \langle 1,A|H|4,B\rangle \\ \langle 1,B|H|1,A\rangle & \langle 1,B|H|1,B\rangle & \cdots & \langle 1 ,B|H|4,B\rangle \\ \langle 2,A|H|1,A\rangle & \langle 2,A|H|1,B\rangle & \cdots & \langle 2,A|H|4,B\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle 4,A|H|1,A\rangle & \langle 4,A|H|1,B\rangle & \cdots & \langle 4 ,A|H|4,B\rangle\\ \langle 4,B|H|1,A\rangle & \langle 4,B|H|1,B\rangle & \cdots & \langle 4 ,B|H|4,B\rangle}$$

Computing several of them by hand should be sufficient to give you a sense of the pattern.