I think I have a confusion on some basics of quantum mechanics. To explain my problem I constructed this following simple example. Let's consider an infinite 1D system made by two sub lattices $A$ and $B$, with $a_{1}$,$a_{2}$ the two lattice lengths. In this system an electron can jump from one site to the other with probability amplitude given by the hopping parameters $t$ and $t^{'}$ as in the figure.

enter image description here

The corresponding tight binding Hamiltonian for an electron in the system (only first nearest neighbor hopping) is:

$\hat{H}=\sum_{<i,j>}t_{i,j}\cdot(|i\rangle \langle j|+|j\rangle \langle i|)=\sum_{i\in A}t^{'} (|i\rangle \langle i-a_{2}|)+t(|i\rangle \langle i+a_{1}|)+h.c.$

where by $|i\rangle $ I mean the i-th site position state.

Now, I want to build the matrix for H. To do that I choose to represent my wave function as a two component spinor where the two components correspond to the two sub lattices $A,B$:

$\psi(x)=\begin{pmatrix}\psi_{a}(x)\\\psi_{b}(x) \end{pmatrix}$.

So my basis would be a collection of spinors containing delta functions centered on each lattice site:

$\phi_{i,a}=\begin{pmatrix}\delta(x_{i})\\0 \end{pmatrix}$ and $\phi_{j,b}=\begin{pmatrix}0\\\delta(x_{j}) \end{pmatrix}$

where $i$ goes over all sites of the $A$ sub lattice and $j$ same for $B$.

Now comes my problem: let's say I want to calculate the matrix for $\hat{H}$ in momentum space. The system is translational invariant so I can just focus on the 2x2 block (corresponding to a given momentum $k$):

$h(k)=\begin{pmatrix} \langle k_{A}|\hat{H}|k_{A}\rangle & \langle k_{A}|\hat{H}|k_{B}\rangle \\ \langle k_{B}|\hat{H}|k_{A}\rangle & \langle k_{B}|\hat{H}|k_{B}\rangle \end{pmatrix} $

where by $|k_{A}\rangle$ I indicate a state in which the electron has momentum k and is completely delocalized on sub lattice $A$ (hence no amplitude on sub lattice B) and viceversa for $|k_{B}\rangle$ .

Now, I know that the non zero elements of the previous matrix should be the off-diagonal ones. But lets look at $\langle k_{A}|\hat{H}|k_{B}\rangle$:

$\langle k_{A}|\hat{H}|k_{B}\rangle=\sum_{i \in A}t^{'}\langle k_{A}|i\rangle \langle i-a_{2}|k_{B}\rangle+t\langle k_{A}|i\rangle \langle i+a_{1}|k_{B}\rangle$

Here is the crucial part:the term $\langle k_{A}|i\rangle $ is just (by definition) the conjugate wave function (written in real space) of a state with momentum k delocalized on sub lattice A.

So using the "spinor" basis I described above: $\langle k_{A}|i\rangle= \begin{pmatrix} e^{-ik r_{i}} \\0 \end{pmatrix}^{T}$.

By the same argument $\langle i-a_{2}|k_{B}\rangle=\begin{pmatrix} 0 \\e^{ik (r_{i}-a_{2})} \end{pmatrix}$.

But this would give me that the product $\langle k_{A}|i\rangle \langle i-a_{2}|k_{B}\rangle$ is zero (I just did the scalar product between the two previous spinors just like for vectors).

The same goes for the second term $\langle k_{A}|i\rangle \langle i+a_{1}|k_{B}\rangle$ hence that matrix element seems to be zero. If I argue similarly I get $0$ also for the other matrix elements, hence the entire Hamiltonian matrix is zero!

Can someone tell me what I'm doing wrong here?

  • $\begingroup$ I think you don't want to build the matrix for $H$ rather you actually want to diagonalize it, i.e. finding eigenvalues and eigenvectors. On a finite lattice the actual procedure depends if you have even or odd number of sites and on the boundary conditions. In any case your model is equivalent to the SSH model for which you should be able to find a lot of material. $\endgroup$
    – lcv
    Oct 19, 2022 at 20:34
  • $\begingroup$ I'm not interested in eigenvalues or eigenvectors. I purposely built this example because I'm interested in the calculation of those matrix elements of H. But my problem is that if I use the "spinor" basis I described above the H matrix elements are 0. Whereas if I use a basis of normal functions (so my wave function is not a spinor with the two components corresponding to the two sub lattices, but just a function ) I get a non zero result (which is the correct one, reported in many resources). $\endgroup$
    – Mathew
    Oct 20, 2022 at 6:54
  • $\begingroup$ What I want to understand is why if I separate the two sub lattices in two different components of a spinor I get zero? Is it because I'm doing the calculation above wrong or is it wrong to choose that spinor basis? $\endgroup$
    – Mathew
    Oct 20, 2022 at 6:55

1 Answer 1


The answer to last question in the comments is yes (there is an error in the calculation).

Note that in order to write the matrix elements of the Hamiltonian there is no need to make any reference to the two sublattices.

For example, ordering the states as in your figure, the matrix of the Hamiltonian in position basis is

$$ H=\left(\begin{array}{ccccc} 0 & t\\ t & 0 & t'\\ & t' & 0 & t\\ & & t & 0 & \ddots\\ & & & \ddots & \ddots \end{array}\right). $$

The reason why we define the two sublaticces is that the Hamiltonian (with appropriate boundary conditions)is invariant if we translate the system by two, or rather by $a_1 +a_2$.

Hence using Bloch theorem if you want, we can write the Hamiltonian as

$$ H = \sum_{k\in BZ} h(k), $$

where BZ is the Brillouin zone $h(k)$ is a $2\times2$ matrix and the sum is in fact a direct sum.

The way to do that is first write the lattice as consisting of cells of two sites each and then write all the various hopping between neighboring cells and between neighboring sites in the same cell. This is just a reindexing of the original position states.

However in the new form it is apparent that translation by one cell leaves the Hamiltonian invariant.

At this point we can go to momentum space for the $x$ variable that was indexing the cell.

The two steps can of course be combined but keeping them separate is helpful for checking against errors.

After the first step above the Hamiltonian looks:

$$ H=\sum_{x} \left ( t|x,1\rangle\langle x,2|+t'|x,2\rangle\langle x+\ell,1|+\mathrm{h.c.} \right ) $$

where $\ell$ is the distance between one cell and the next ($=a_1+a_2$ in your example but you can set it to one of course as it is a simple label here).

Note that I have not been careful about boundary conditions since they have not been specified and I assume the lattice is as you depicted it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.