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In the Kane-Mele model, the spin-orbit coupling is defined in real space as

$$\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}$$

where the sum is over next-nearest-neighbor sites on a honeycomb lattice, and $\nu_{ij} = - \nu_{ij} = \pm 1$ depends on the orientation of the next-nearest-neighbor bonds (I don't believe the details of how $\nu_{ij}$ is calculated is relevant for Hermiticity). I am having difficulty understanding how this term is Hermitian. Taking the Hermitian conjugate seemingly gives

$$\left(\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}\right)^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} (-i) t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha} c_{j \beta}^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{j \beta}^\dagger c_{i \alpha} \\ = -\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta} $$ where in the final line we have relabeled indices and used that $\nu_{ij} = - \nu_{ji}$, $s^z_{\alpha \beta} = s^z_{\beta \alpha}$. I must be missing something obvious, but this seems to show that the term is anti-Hermitian, instead of Hermitian. What am I missing here?

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2 Answers 2

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In the first expression after the equals sign on the first line you have $c_{i\alpha}c^\dagger_{j\beta}$. It should be $c^\dagger_{j\beta}c_{i\alpha}$ since $(AB)^\dagger = B^\dagger A^\dagger$.

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  • $\begingroup$ Oh shoot, you're totally right. Thanks a lot for pointing it out. $\endgroup$ Commented Nov 6, 2019 at 22:45
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From the first to second expression on the first line, the Conjugation operation should transpose the matrix $v_{ij}$ to $v_{ji}$, giving the missing minus sign.

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