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
\begin{equation}
\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|
\end{equation}
\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 $b_n(k)$?
 A: 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 $$ \mathbb{S}^1\ni k \mapsto w\exp(-\mathrm{i}k) + \bar{v}\in\mathbb{C}\setminus\{0\}\,. $$
