3
$\begingroup$

I've recently been watching this lecture series on Condensed Matter. The part I'm currently on covers band theory for the tight-binding model in a few different scenarios. We covered two different models in one-dimension. The first had a Hamiltonian given by $$H = -t \sum_i c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i,$$ where the sum runs over the $N$ sites in the one-dimensional chain with periodic boundary conditions and $c_i$ is a creation operator for an orbital at site $i$. We solved this model and got to the single-particle eigenenergies $$E(k) = - 2 t \cos(k a),$$ where $a$ is the lattice spacing and $k = \frac{2\pi n}{N a}$, with $n = 0, 1, \ldots, N-1$.

The second model had two sites per unit cell. We have $N$ unit cells and have the Hamiltonian $$H = - \sum_i t_1(c_{i,A}^\dagger c_{i,B} + c_{i,B}^\dagger c_{i,A}) + t_2(c_{i-1,B}^\dagger c_{i,A} + c_{i+1,A}^\dagger c_{i,B}),$$ where $A$ and $B$ label the two different sites per unit cell. The single-particle eigenenergies we found this time were $$E(k) = \pm \sqrt{t_1^2 + t_2^2 + 2 t_1 t_2 \cos(ka)},$$ where $a$ is the size of one unit cell and $k = \frac{2\pi n}{N a}$, $n = 0,1, \ldots, N_1$ (to the best of my knowledge).

What I find weird is that I would expect that the limit $t_1 \to t_2$ in the second model should reduce it to the first model with $t = t_1 = t_2$, albeit with $2N$ sites instead of $N$. However, that is not what happens, since the second model always has two energy bands, while the first has only one. My question is: why doesn't the second model reduce to the first one in the limit $t_1 \to t_2$?

$\endgroup$
6
  • 1
    $\begingroup$ Do the Hamiltonians converge to the same Hamiltonian? $\endgroup$
    – Mauricio
    Commented Jan 11 at 15:51
  • $\begingroup$ For me the second Hamiltonian has a completely different interpretation than the first, so I wouldn't expect both results to coincide even for $t_1=t_2$. I think this is along the lines what @Mauricio wants to point to (?). $\endgroup$ Commented Jan 11 at 15:56
  • $\begingroup$ How is your second Hamiltonian hermitian at all? The second term looks weird... $\endgroup$ Commented Jan 11 at 17:30
  • 1
    $\begingroup$ @TobiasFünke using the periodic boundary conditions, one can relabel $i\to i-1$ and get $\sum_i c_{i+1,A}^\dagger c_{i,B} = \sum_i c_{i,A}^\dagger c_{i-1,B}$ $\endgroup$ Commented Jan 11 at 17:34
  • 1
    $\begingroup$ Yes, you are right, I missed that, sorry. Just a minor (unrelated) additional comment: The energies $E(k)$ you give are the single-particle energies; the eigenvalues of $H$, an operator defined on the Fock space, are of the form $\sum_{k} E(k) n_k$, where the $n_k$ are the occupation numbers. $\endgroup$ Commented Jan 11 at 17:53

2 Answers 2

7
$\begingroup$

Here's a way in which we can make sense of the limits in a physical situation in which the limit $t_1\to t_2$ should yield the first Hamiltonian. We could think of the atomic species as being the same, but there being two different crystal structures. The first is one in which every atom is a distance $a$ apart. In the second, we shift every other atom some small distance $\alpha a$ to the right, with $\alpha < 1$. This would then introduce alternating hopping strengths due to the differing distances between adjacent atoms, but in the limit as we take $\alpha \to 0$, this becomes the original lattice. The lattices are illustrated below.

enter image description here

Crucially, we have doubled the size of the unit cell from $a$ to $2a$. This makes it so that in the second dispersion relation, we should take $a\to 2a$, yielding $$ E(k) = \pm \sqrt{t^2 + t_2^2 + 2 t t_2 \cos(2ka)}\,, $$ where I have set $t_1=t$. In the limit as $t_2\to t$, the above becomes $$ E(k) \to \pm 2t \cos(ka)\,. $$ It seems like this is a problem, because we get two branches (the plus and the minus). However, we haven't taken into account the fact that the first Brillouin zone must double in size in this degenerate limit, because the size of the unit cell gets cut in half ($2a\to a$) in this limit. Thus, we need to "unfold" the dispersion relation out of the reduced zone scheme picture we needed in going from a unit cell of size $a$ to a unit cell of size $2a$.

This is illustrated below. The black dashed curve is the dispersion relation for the simple case of the crystal structure with equal inter-atomic spacings $a$. The solid black curve is the dispersion relation for the diatomic lattice with $t_2/t = 1.3$. The dashed black curve is the dispersion relation for the monoatomic lattice, and we plot it over a larger region because the Brillouin zone is twice as large! Then, we can see that the positive branch of the limit dispersion relation (i.e., $+2t\cos(ka)$) seems as if it's a second branch of the dispersion relation, but really we need to shift those over by reciprocal lattice vectors (of the diatomic lattice) to where they should be in the first Brillouin zone of the monatomic lattice.

enter image description here

Thus, we can take the limit and get the same system, but you need to make sure that the unit cell sizes are consistent and that you are thinking carefully about the actual periodicity of your system and hence the correct size of the first Brillouin zone.

$\endgroup$
3
$\begingroup$

Taking your second equation an setting $t_1,t_2\to t$, we get $$E(k)=\pm t \sqrt2 \sqrt{1+\cos(ka)}=\pm 2 t \cos\left(\frac{ka}2\right)$$

where I used $1+\cos x = 2 \cos x/2$. The $k/2$ comes because $N$ is different by a factor of 2.

However notice that there is still a $\pm$. This happens because the equation used is not right, it was diagonalized as if I was handling a two-species unit cell, so we get a duplication of the bands.

$\endgroup$
15
  • $\begingroup$ Yes, but you still have the $\pm$-"problem" (i.e. you have two bands, which are non-degenerate, except at the BZ boundary, even if you have $t_1=t_2$). The point is, I guess, that the second model simply has two species (or two sites per unit cell), so it is fundamentally different from the first one to start with, no? $\endgroup$ Commented Jan 11 at 16:04
  • 1
    $\begingroup$ In that limit, you have chosen a non-primitive unit cell. The number of bands in your system is determined by the degrees of freedom in the unit cell, so if you chose a unit cell with four of the same orbitals, you would get four bands in a Brillouin zone which is four times smaller. The relevant keywords for this are band folding and extended zone schemes. $\endgroup$
    – G.Lang
    Commented Jan 11 at 16:13
  • $\begingroup$ Thanks @TobiasFünke for noticing, indeed it is dependent on how the diagonalization was done. $\endgroup$
    – Mauricio
    Commented Jan 11 at 16:26
  • $\begingroup$ I still don't understand the $\pm$ "issue". The eigenvalues of an operator do not depend on the procedure I use to diagonalize it, so why does it matter in this case? $\endgroup$ Commented Jan 11 at 17:02
  • $\begingroup$ @NíckolasAlves when you are diagonalizing you are still treating the Hamiltonian as a two species lattice, setting $t_i\to t$ does not solve this issue. You should first set $t_i\to t/2 $ and then diagonalize you will then recover the same result. $\endgroup$
    – Mauricio
    Commented Jan 11 at 17:08

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.