I am trying to understand the SSH model. I am studying the book A Short Course on Topological Insulators by J. Asbóth. I have a problem because he claims that eigenstates may be decomposed into external and internal degrees, i.e.,

$$|m, \alpha\rangle = |m\rangle \otimes |\alpha\rangle,$$

where $|m\rangle$ denotes a unit cell and $\alpha$ means sublattice $A$ or $B$.

Then he introduces a basis in momentum space, i.e.,

$$|k\rangle = \frac{1}{\sqrt{N}}\sum_{m=1}^N e^{imk}|m\rangle \quad k\in \mathrm{First~Brilloune~Zone}.$$

And now he defines eigenstate as follows

$$ |\psi_n(k)\rangle = |k\rangle \otimes|u_n(k)\rangle, $$

$$|u_n(k)\rangle = a_n(k)|A\rangle + b_n(k) |B\rangle,$$ where $|u_n(k)\rangle$ are the eigenstates of the bulk momentum Hamiltonian which is defined as $$ H(k) = \sum_{\alpha,\beta\in\{A,B\}} \langle k | H_{bulk} | k \rangle = \sum_{\alpha,\beta\in\{A,B\}} \langle k, \alpha | H_{bulk} | k, \beta \rangle \cdot |\alpha\rangle\langle\beta|. $$

And I am baffled by all these signs. What are $H(k)$ and how he gets this form after the second equal to? Moreover, he defines $|\psi_n(k)\rangle$ and doesn't use it. Instead, he projects Hamiltonian od $k$ states, but $|u_n(k)\rangle$ are eigenstates of this Hamiltonian. Can somebody please explain it to me?


1 Answer 1


The entire Hilbert space of the system may be decomposed into external (inter-cell physics) and internal (intra-cell physics) degrees. This means you can express eigenstates in the form of a tensor product exactly as you said, i.e.,

$$|m, \alpha\rangle = |m\rangle \otimes |\alpha\rangle,$$, where $|m\rangle\in \cal{H_{ext}}$ and $|\alpha\rangle\in \cal{H_{int}}$ whose basis is {$|A\rangle$,$|B\rangle$}.

So far so good. Now, in the thermodynamic limit (when the number of unit cells $N\rightarrow\infty$), we should focus on the bulk, since it is much larger than the boundaries. By setting periodic (Born-von Karman) boundary conditions, we restore system's translational invariance. This plays an essential role in solving this problem since if there is translation invariance, Bloch's theorem applies and then we can look for eigenstates in a plane wave form:

$$|k\rangle = \frac{1}{\sqrt{N}}\sum_{m=1}^N e^{imk}|m\rangle \quad k\in \mathrm{First~Brillouin~Zone}.$$

The thing is those $|k\rangle$ eigenstates do not belong to $\cal{H}$, but to $\cal{H_{ext}}$. In order to describe the whole system physics, we must combine them with the eigenstates of the internal part $|u_n(k)\rangle=a_n(k)|A\rangle + b_n(k) |B\rangle$:

$$ |\psi_n(k)\rangle = |k\rangle \otimes|u_n(k)\rangle $$

Finally, it is said that the set of {$|u_n(k)\rangle$} are eigenstates of $H(k)$. That's why it does not depend on the external degrees of freedom. From this, you can infer that $H(k)$ represents the bulk Hamiltonian in k-space. In the second equal, he just takes the $H_{bulk}$ elements $\langle k, \alpha | H_{bulk} | k, \beta \rangle$ in the whole Hilbert space basis and project them onto $\cal{H_{int}}$ by means of the corresponding projector $\sum_{\alpha,\beta\in\{A,B\}}|\alpha\rangle\langle\beta|$.


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