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.


1 Answer 1


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
    Commented Jul 31, 2021 at 21:47
  • $\begingroup$ Glad that helped. I think the subscripts are fixed. $\endgroup$ Commented Jul 31, 2021 at 21:54

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.