# Chiral symmetry in the SSH model

According to "A short course on topological insulators", chapter 1, in the SSH model, the consequence of chiral symmetry for the states with $$E\ne 0$$ is the presence of another state with $$-E$$. The orthogonality of the wave functions corresponding to $$E>0$$ and $$E<0$$ i.e., $$\langle\psi_{E>0}|\psi_{E<0}\rangle=0$$ gives rise to the identical support of the two wave functions on the two sublattices. Namely, for both energies $$E>0$$ and $$E<0$$: $$\langle\psi_{E}|P_A|\psi_{E}\rangle=\langle\psi_{E}|P_B|\psi_{E}\rangle$$ where $$P_A$$ and $$P_B$$ are projectors on sublattices A and B.

Also, for $$E=0$$, we can choose the two states in such a way that one of them is supported by the sublattice A and the other with B. However, in this case, the orthogonality of the two states again results in equal support on the two sublattices. It seems a paradox!

Any help would be appreciated.

Suppose you have $$\psi_{A}$$ and $$\psi_{B}$$ at zero energy, with each supported on a different sublattice. Then $$P_{A}\psi_{A}=\psi_{A}$$ and $$P_{B}\psi_{B}=\psi_{B}$$ and $$P_{B}\psi_{A}=P_{A}\psi_{B}=0$$. These are orthogonal, and also at zero energy so $$H\psi_{A}=H\psi_{B}=0$$.
Now take a new pair, $$\psi_{1}=\frac{1}{\sqrt{2}}\psi_{A}+\frac{1}{\sqrt{2}}\psi_{B}$$ $$\psi_{2}=\frac{1}{\sqrt{2}}\psi_{A}-\frac{1}{\sqrt{2}}\psi_{B}$$ This is again an orthogonal pair of unit vectors, and $$H\psi_{1}=H\psi_{2}=0$$, but now \begin{align*} \left\langle \psi_{1},P_{A}\psi_{1}\right\rangle & =\frac{1}{2}\left\langle \psi_{A},P_{A}\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{B},P_{A}\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{A},P_{A}\psi_{B}\right\rangle +\frac{1}{2}\left\langle \psi_{B},P_{A}\psi_{B}\right\rangle \\ & =\frac{1}{2}\left\langle \psi_{A},\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{B},\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{A},0\right\rangle +\frac{1}{2}\left\langle \psi_{B},0\right\rangle \\ & =\frac{1}{2} \end{align*} and \begin{align*} \left\langle \psi_{1},P_{B}\psi_{1}\right\rangle & =\frac{1}{2}\left\langle \psi_{A},P_{B}\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{B},P_{B}\psi_{A}\right\rangle +\frac{1}{2}\left\langle \psi_{A},P_{B}\psi_{B}\right\rangle +\frac{1}{2}\left\langle \psi_{B},P_{B}\psi_{B}\right\rangle \\ & =\frac{1}{2}\left\langle \psi_{A},0\right\rangle +\frac{1}{2}\left\langle \psi_{B},0\right\rangle +\frac{1}{2}\left\langle \psi_{A},\psi_{B}\right\rangle +\frac{1}{2}\left\langle \psi_{B},\psi_{B}\right\rangle \\ & =\frac{1}{2} \end{align*} and thus $$\left\langle \psi_{1},P_{A}\psi_{1}\right\rangle =\left\langle \psi_{1},P_{B}\psi_{1}\right\rangle .$$ Just a few more minus signs needed for the same for $$\psi_{2}$$ .
• Thank you very much Terry Loring. Your answer is perfect and exact. I think you have mistakenly written $\psi_0$ instead of $\psi_1$. Jul 31 at 21:47