# Some questions in Asboth, Oroszlany, Palyi 's lecture notes on Topological insulators

I am reading Asboth, Oroszlany, Palyi 's lecture notes on Topological insulators. I am having some diffculty with the mathematics on page 4, Section 1.2.1.

We have the SSH hamiltonian: $$\hat H_{bulk} = \sum_{m=1}^N(v|m, B\rangle\langle m, A|+w|(m\text{ mod }N)+1, A\rangle\langle m, B|)+\text{h.c.}$$

and we want to find its eigenstates and eigenvalues $$\hat H_{bulk}|\Psi_n(k)\rangle=\hat E_n(k)|\Psi_n(k)\rangle$$

So, first we Fourirer transform $$|m\rangle$$ to go in momentum space: $$|k\rangle = \frac{1}{N}\sum_{m=1}^Ne^{imk}|m\rangle$$

Now, the authors define the following Bloch states $$|\Psi_n(k)\rangle = |k\rangle\otimes|u_n(k)\rangle$$ where $$|u_n(k)\rangle = a_n(k)|A\rangle+b_n(k)|B\rangle$$. My first question is: What kind of Bloch states are these? As far as I know, Bloch states are of the form $$\psi_k(x) = e^{ikx}u(x)$$ with $$u(x)$$ having same periodicity as that of the lattice.

Now according to authors, $$|u_n(k)\rangle$$ are eigenstates of the bulk momentum-space Hamiltonian $$\hat H(k)$$ defined as: $$\hat H(k)=\langle k|\hat H_{bulk}|k\rangle = \sum_{\alpha, \beta \,\in\,\{A, B\}}\langle k, \alpha|\hat H_{bulk}|k, \beta\rangle\cdot|\alpha\rangle\langle\beta|$$

My second question is how did they get the second equality? My guess is they used the completeness relation. I am giving my calculations below. Since, I have almost no knowledge of tensor product, I want to know what would be the correct way of doing this: \begin{align} \langle k|\hat H_{bulk}|k\rangle &= \langle k|I\hat H_{bulk}I|k\rangle\\ &= \langle k|\left(\sum_{k';\alpha\in\{A, B\}}|k', \alpha\rangle\langle k', \alpha|\right)\hat H_{bulk}\left(\sum_{k'';\beta\in\{A, B\}}|k'', \beta\rangle\langle k'', \beta|\right)|k\rangle\\ &= \left(\sum_{k'}\langle k|k'\rangle\langle k'|\otimes\sum_{\alpha\in\{A, B\}}| \alpha\rangle\langle\alpha|\right)\hat H_{bulk}\left(\sum_{k''}|k''\rangle\langle k''|k\rangle\otimes\sum_{\beta\in\{A, B\}}|\beta\rangle\langle\beta|\right)\\ &= \sum_{\alpha, \beta\in\{A, B\}}\left(\langle k|\otimes|\alpha\rangle\langle\alpha|\right)\hat H_{bulk}\left(|k\rangle\otimes|\beta\rangle\langle\beta|\right)\\ &= \sum_{\alpha, \beta\in\{A, B\}}\langle k, \alpha| H_{bulk}|k, \beta\rangle\cdot|\alpha\rangle\langle\beta|\\ \end{align} To be more precise, I want to know what would be correct way for going from step 2 to step 3 and then from step 4 to step 5. Or is my approach totally wrong?

Lastly, the authors say that $$|u_n(k)\rangle$$ are eigenstates of $$\hat H(k)$$: $$\hat H(k)|u_n(k)\rangle=\hat E_n(k)|u_n(k)\rangle$$ Is this true for all values of $$a_n(k)$$ and $$b_n(k)$$? I don't think this is true. Let $$H_{\alpha\beta} = \langle k, \alpha| H_{bulk}|k, \beta\rangle$$.Then $$$$\hat H(k) = H_{AA}|A\rangle\langle A|+H_{AB}|A\rangle\langle B|+H_{BA}|B\rangle\langle A|+H_{BB}|B\rangle\langle B|$$$$ \begin{align} \hat H(k)|u_n(k)\rangle &= (a_n(k)H_{AA}+b_n(k)H_{AB})|A\rangle+(a_n(k)H_{BA}+b_n(k)H_{BB})|B\rangle \end{align} Wouldn't this be true only for some values of $$a_n(k)$$ and $$b_n(k)$$?

It is more convenient to work with the following notation instead of the one you are using:

Let $$u,w$$ be two complex numbers.

Your Hilbert space is $$\mathcal{H}= \ell^2(\mathbb{Z})\otimes\mathbb{C}^2$$ (note I am using an infinite system in the bulk instead of a finite system to avoid worrying about periodic boundary conditions--you could just as well replace $$\mathbb{Z}$$ with $$\{1,\dots,N\}$$ with periodic boundary conditions if you prefer and then work with the finite Fourier series instead of the Fourier series).

On $$\mathcal{H}$$, the Hamiltonian $$H$$ as you defined it in your first formula is (not the usual notation for the SSH model but indeed equivalent to it) $$H = v \mathbb{1}\otimes\sigma_{\downarrow}+w R\otimes\sigma_{\uparrow} + \text{h.c.}\,.$$ Here, $$R$$ is the right-shift operator on $$\ell^2(\mathbb{Z})$$ ($$(R\psi)_n \equiv \psi_{n-1}$$ for $$n\in\mathbb{Z}$$ and $$\psi\in\ell^2(\mathbb{Z})$$) and $$\sigma_{\uparrow}\equiv\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$ takes a spin down state and converts it into a spin up state. It is zero on spin up states. (We encode the $$A/B$$ sub-lattice degree of freedom as spin-1/2).

Now the Bloch decomposition is implemented via the Fourier series $$\mathcal{F}:\ell^2(\mathbb{Z})\to L^2(\mathbb{S}^1)$$ which is defined as $$\hat{\psi}(k):=\sum_{n\in\mathbb{Z}}\exp(-\mathrm{i}kn)\psi_n\qquad(k\in\mathbb{S}^1)\,.$$

This transformation factorizes through tensor products. Furthermore, it is clear unitarity that $$\mathcal{F}\mathbb{1}_{\ell^2(\mathbb{Z})}\mathcal{F}^{\ast} = \mathbb{1}_{L^2(\mathbb{S}^1)}$$ and by translation invariance of $$R$$ that $$\mathcal{F}R\mathcal{F}^{\ast} = M_r$$ where $$M_r$$ is the multiplication operator on $$L^2(\mathbb{S}^1)$$ by the function $$r(k) \equiv \exp(-\mathrm{i}k)\qquad(k\in\mathbb{S}^1)\,,$$ i.e., $$(\mathcal{F}R\mathcal{F}^{\ast}\hat{\psi})(k) = \exp(-\mathrm{i}k)\hat{\psi}(k)\qquad(k\in\mathbb{S}^1)\,.$$

Hence, thanks to Bloch decomposition we may work pointwise in $$k$$, that is, after this Bloch decomposition, $$H$$ is the following matrix-valued multiplication operator $$h(k) = \begin{bmatrix} 0& w\exp(-\mathrm{i}k) + \bar{v}\\ \bar{w}\exp(\mathrm{i}k) + v&0 \end{bmatrix}$$ and we need only solve a $$2\times 2$$ eigensystem.

It's worth mentioning that the Zak phase is precisely the winding number of $$\mathbb{S}^1\ni k \mapsto w\exp(-\mathrm{i}k) + \bar{v}\in\mathbb{C}\setminus\{0\}\,.$$