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 $b_n(k)$?


1 Answer 1


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 $$ \mathbb{S}^1\ni k \mapsto w\exp(-\mathrm{i}k) + \bar{v}\in\mathbb{C}\setminus\{0\}\,. $$


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.