# Inversion Symmetry on Internal Hilbert Space of Bulk Lattice

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.

## 1 Answer

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.

• 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 – C.C. Oct 31 '20 at 5:56
• 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). – Guoqing Oct 31 '20 at 15:07
• 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 ? – C.C. Oct 31 '20 at 16:21
• 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 $p_{\pi}$). – C.C. Oct 31 '20 at 16:37
• Yes,you're right.The complete expression for (3.32) should be $\left|u(-0) \right>=\hat{\pi}\left|u(0) \right>=p_0\left|u(0) \right>$. – Guoqing Nov 1 '20 at 1:00