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Let us take the famous Kane-Mele(KM) model as our starting point.

Due to the time-reversal(TR), 2-fold rotational(or 2D space inversion), 3-fold rotational and mirror symmetries of the honeycomb lattice system, we can derive the intrinsic spin-orbit(SO) term. Further more, if we apply a spatially uniform electric field perpendicular to the 2D lattice(now the mirror symmetry is broken), a (extra) Rashba-type SO term will emerge.

To present my question more clearly, I will first give a more detailed description of the above symmetry operations in both first- and second- quantization formalism. In the follows, a 3D Cartesian coordinate has been set up where the 2D lattice lies in the $xoy$ plane.

First-quantization language:

(1) TR symmetry operator $\Theta:$ $\Theta\phi(x,y,z)\equiv \phi^*(x,y,z)$, hereafter, $\phi(x,y,z)$ represents an arbitrary wave function for single electron.

(2) 2-fold rotational operator $R_2:$ $R_2\phi(x,y,z)\equiv \phi(-x,-y,z)$, where we choose the middle point of the nearest-neighbour bond as the origin point $o$ of the coordinate.

(3) 3-fold rotational operator $R_3:$ $R_3\phi(\vec{r} )\equiv \phi(A\vec{r})$, where $A=\begin{pmatrix} \cos\frac{2\pi}{3}& -\sin\frac{2\pi}{3}& 0\\ \sin\frac{2\pi}{3}& \cos\frac{2\pi}{3} & 0\\ 0& 0 & 1 \end{pmatrix}$ $\vec{r}=(x,y,z)$ and we choose the lattice site as the origin point $o$ of the coordinate.

(4) Mirror symmetry operator $\Pi:$ $\Pi\phi(x,y,z)\equiv \phi(x,y,-z)$.

Second-quantization language:

(1) TR symmetry operator $T:$ $TC_{i\uparrow}T^{-1}=C_{i\downarrow}, TC_{i\downarrow}T^{-1}=-C_{i\uparrow}$, where $C=a,b$ are the annihilation operators referred to the two sublattices of graphene.

(2) 2-fold rotational operator $P_2:$ $P_2a(x,y)P_2^{-1}\equiv b(-x,-y), P_2b(x,y)P_2^{-1}\equiv a(-x,-y)$, $P_2$ is unitary and we choose the middle point of the nearest-neighbour bond as the origin point $o$ of the coordinate.

(3) 3-fold rotational operator $P_3:$ $P_3C(\vec{x})P_3^{-1}\equiv C(A\vec{x}), \vec{x}=(x,y),C=a,b$, where $A=\begin{pmatrix} \cos\frac{2\pi}{3}& -\sin\frac{2\pi}{3}\\ \sin\frac{2\pi}{3}& \cos\frac{2\pi}{3} \\ \end{pmatrix}$ and $P_3$ is unitary, we choose the lattice site as the origin point $o$ of the coordinate.

(4) Mirror symmetry operator $M:$ ????

as you see, that's what I want to ask: how to define the mirror symmetry operator $M$ in terms of second-quantization language for this 2D lattice system? Or maybe there is no well defined $M$ for this model? Thanks in advance.

Remarks:

(1) A direct way to verify your definition of $M$ being correct or not is as follows: The intrinsic SO term $i\lambda\sum_{\ll ij \gg }v_{ij}C_i^\dagger\sigma_zC_j$ should be invariant under $M$ while the Rashba term $i\lambda_R\sum_{<ij>}C_i^\dagger \left ( \mathbf{\sigma}\times\mathbf{p}_{ij}\right )_zC_j$ will not be.

(2)Here mirror operation is just reflection in one of the three spatial axes (i.e. $(x,y,z)\rightarrow (x,y,-z)$), not the “parity” operation in the context of "CPT symmetry" in field theory.

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up vote 2 down vote accepted

Firstly, You do not in fact seem to be asking about "second quantization" - the mirror operation works in second quantization the same way other symmetries work, which I deduce that you understand from your question.

What is seems you are actually asking is: "how does the mirror symmetry work in the tight binding model?". As you probably realized the lattice sites are invariant and Bloch wavefunctions are invariant under the mirror symmetry. However the electron spin is not invariant. Keeping in mind that the Pauli matrices form a "pseudo-vector" you see that the correct transformation is: $C\rightarrow \sigma_z C\sigma_z$.

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@ BebopButUnsteady, thanks for your good comment. I have one more question, mirror symmetry is a symmetry about spatial degrees of freedom, how is it related to the spin space? Furthermore, if the electron is spinless in our tight binding model, is your definition $C\rightarrow \sigma_zC\sigma_z$ still work? –  K-boy Jun 29 '13 at 9:13
    
@ BebopButUnsteady, to confirm one thing: Does the symbol $\sigma_z$ in your answer mean $\sigma_z=C_\uparrow^\dagger C_\uparrow-C_\downarrow^\dagger C_\downarrow$? If so, the operator $\sigma_z $ represents a symmetry transformation(unitary operator) **only if** the single occupation condition $C_\uparrow^\dagger C_\uparrow+C_\downarrow^\dagger C_\downarrow=1$ satisfies. –  K-boy Jun 29 '13 at 9:21
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