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I am currently studying Short Course On Topological Insulator by J. K. Asb´oth, L. Oroszl´any, A. P´alyi. In Chapter 3.2, I have a few questions to ask regarding it:

  1. Is the phase $e^{i\phi}$ in (3.31) arbitrarily introduced?

  2. I do not understand why $p_{0}$ and $p_\pi$ in (3.33) can take the values of -1 if they follow from (3.32).

  3. If $k\in \{ \delta_k,2\delta_k,3\delta_k,..., N\delta_k \}$ with $\delta_k=\frac{2\pi}{N}$ for $N$ unit cells, what is the importance of values of $k=\pi$ and $k=0$ as mentioned before (3.32)?

Help will be much appreciated.

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I'm also studying this course recently,so I will attempt to give my understanding. 1.The phase $e^{i\phi(k)}$ is arbitrary.In (3.30) $\hat{H}(-k)\hat{\pi}\left|u(k) \right>=E(k)\hat{\pi}\left|u(k) \right>$ is a eigen function,so we say the related eigenvector $\left|u(-k) \right>$ is supposed to be the formula $\hat{\pi}\left|u(k) \right>$,but there's still a phase uncertainty between these two states,that's $e^{i\phi(k)}$.
2.In (3.32) $p_0$ can take value of -1 because both -1 and +1 are eigenvalues of $\hat{\pi}$.You can get this from (3.31) for k =0,that is$$\left|u(-0) \right>= \left|u(0) \right>=e^{i\phi(0)}\hat{\pi}\left|u(0) \right>$$as you can see $\left|u(0) \right>$ is the eigen state of $\hat{\pi}$ with eigenvalue $e^{-i\phi(0)}$.Then we can take any values of $\phi(0)$ because it is arbitrary,but we simply take $0$ related to even parity state or $\pi$ related odd parity state for convenience.In other word,taking $p_0=1$ or -1 depending on state $\left|u(0) \right>$ is even or odd.Case $k=\pi$ is the same,because of the periodic condition,$\left|u(-\pi) \right>=\left|u(\pi) \right>,H(-\pi)=H(\pi)$.
3. I don't know the physical importance of these two points neither.The latter chapters in this book might explain that,I guess.

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  • $\begingroup$ Thanks for the reply. So for (3.32), there should be a $\pi$ operator acting on the left hand side of each equation? @Quantum life $\endgroup$ – C.C. Oct 31 '20 at 5:56
  • $\begingroup$ Oh,I made a mistake(my apology for my negligence).I didn't see that the operator $\hat{\pi}$ should be Hermitian from (3.27),so it's eigenvalues must be real,that's $\pm 1$ due to it's hermitian and projector operator(instead of choice of phase).That's to say,for (3.31),we first take the arbitrary phase $e^{i\phi(k)}$ value of 1 for convenience,and it turns to $\left|u(-k) \right>=\hat{\pi}\left|u(k) \right>$,this formula is eigen function for state $k=0,k=\pi$,related to eigenvalues of +1(even state) and -1(odd state). $\endgroup$ – Guoqing Oct 31 '20 at 15:07
  • $\begingroup$ If $|u(-k) \rangle = \hat{\pi} |u(k) \rangle$, then for the state $k=0$, shouldn't the eigenvalue be 1 since its $|u(0) \rangle$ for both the left and right side of the equation ? $\endgroup$ – C.C. Oct 31 '20 at 16:21
  • $\begingroup$ It would make sense that there is a $\hat{\pi}$ missing on the left hand side of (3.32) so that $\hat{\pi} |u(0)\rangle=p_{0}|u(0)\rangle$ and $\hat{\pi} |u(\pi)\rangle=p_{\pi}|u(\pi)\rangle$. Then the result of (3.36) would make sense when $\hat{\pi}$ acts to the left ($\langle u_{0}|$ to give $p_{0}$) or to the right ($|u_{M}\rangle$ to give $p_{\pi}$). $\endgroup$ – C.C. Oct 31 '20 at 16:37
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    $\begingroup$ Yes,you're right.The complete expression for (3.32) should be $\left|u(-0) \right>=\hat{\pi}\left|u(0) \right>=p_0\left|u(0) \right>$. $\endgroup$ – Guoqing Nov 1 '20 at 1:00

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