For weyl semimetal whose Hamiltonian is $\hat H= v_f \ \vec k \cdot \vec\sigma $ [1,2], its eigen value is $\epsilon_ {\pm}=\pm v_f k $, thus the hamiltonian can be written as $\hat H= \frac{1}{2}\epsilon_+(I+\hat k\cdot \sigma)+\frac{1}{2}\epsilon_-(I-\hat k\cdot \sigma)$. Here I is the unit matrix. Then the projection operator can be written as $P^{\pm}=\frac{1}{2}(I \pm \hat k\cdot \sigma)$.

Whereas for the topological insulator with three band crossings [3], its Hamiltonian is $\hat H= v_f \ \vec k \cdot \vec S $, here $\vec S$ denote the angular momentum operator with quantum number $1$. Here its eigen values are $\epsilon_ {\pm}=\pm v_f k $ and $\epsilon_ {0}=0$. Here it seems not so accurate to diagonalize the Hamiltonian as the way with weyl semimetal due to the eigen state with eigen value $0$. So how can the projection operator be obtained here?


[1] Raghu, Srinivas, et al. "Collective modes of a helical liquid." Physical review letters 104.11 (2010): 116401.

[2] Chen, Yu, and Hui Zhai. "Hall conductance of a non-Hermitian Chern insulator." Physical Review B 98.24 (2018): 245130.

[3] Lin, Yu-Ping. "Chiral flat band superconductivity from symmetry-protected three-band crossings." arXiv preprint arXiv:1911.01436 (2019).


If you know the eigenvalues, you know equation obeyed by $H$ is $$ 0=(H-E_1)(H-E_2)(H-E_3). $$ Then the algebraic identity $$ 1= \frac{(H-E_2)(H-E_3)}{(E_1-E_2)(E_1-E_3)}+ \frac{(H-E_1)(H-E_3)}{(E_2-E_3)(E_1-E_3)}+ \frac{(H-E_1)(H-E_2)}{(E_3-E_1)(E_3-E_2)} $$ becomes the orthogonal projector resolution of the identity $$ {\mathbb I}= P_1+P_2+P_3 $$ where $$ P_1= \frac{(H-E_2)(H-E_3)}{(E_1-E_2)(E_1-E_3)} $$ is the projector onto the eigenvector with eigenvalue $E_1$, and so on.

  • $\begingroup$ Thanks a lot! Just wondering whether the second term in the RHS of algebraic identity should be $\frac {(H-E_1)(H-E_3)}{(E_2-E_1)(E_2-E_3)}$, thus the full equation should be $$1=\frac {(H-E_2)(H-E_3)}{(E_1-E_2)(E_1-E_3)}+\frac {(H-E_1)(H-E_3)}{(E_2-E_1)(E_2-E_3)}+\frac {(H-E_1)(H-E_2)}{(E_3-E_1)(E_3-E_2)}$$. And also would you mind giving the reference about this derivation? Best regards, Xinxin $\endgroup$ – Xinxin Peng Dec 16 '19 at 0:49
  • $\begingroup$ Yes! My error cutting and pasting... I learned this equation from Dirac's book on QM. It's in the appendix in my book with Paul Goldbart "Mathematics for Physics". The proof is that the LHS munis the RHS is a polynomila of degree two, but it is obviously zero for $H$ being any of $E_1,E_2 ,E_3$ but a cubic poly with more than two roots is zero. $\endgroup$ – mike stone Dec 16 '19 at 0:50
  • $\begingroup$ I meant quadratic poly with more that 3 roots must be zero (this is the remainder theorem from high-school algebra) . Obviously the formula generalizes to $M$-by-$N$ matrices, but the degree $N-1$ polynomials in $H$ become harder to use as projectors for explicit eigenvectors. $\endgroup$ – mike stone Dec 16 '19 at 0:59
  • $\begingroup$ It's more than clear now:) Thanks again! $\endgroup$ – Xinxin Peng Dec 16 '19 at 1:27

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