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 $\psi_{2}$ .

Sorry for the math notation. It is how I think.

  • $\begingroup$ Thank you very much Terry Loring. Your answer is perfect and exact. I think you have mistakenly written $\psi_0$ instead of $\psi_1$. $\endgroup$
    – H. Khani
    Jul 31 at 21:47
  • $\begingroup$ Glad that helped. I think the subscripts are fixed. $\endgroup$ Jul 31 at 21:54

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