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Normally, we have the $z$-component Pauli matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$.

For the graphene with a magnetic field, the Hamiltonian is \begin{align} \hat{H} = v_{F} \left(\mathbf{P}-e \mathbf{A}\right) \cdot \boldsymbol{\sigma}, \end{align} it has the following commutation relations \begin{align} \left\{\hat{\sigma}_{z},\hat{H}\right\} = 0. \end{align} and satisfies the particle-hole symmetry, which is \begin{align} \hat{H}\hat{\sigma}_{z} {\psi} = -\hat{\sigma}_{z}\hat{H} {\psi} = -E\hat{\sigma}_{z} {\psi}. \end{align} But if we change the Hamiltonian into \begin{align} H = v_{F} \left(\boldsymbol{P} - e\boldsymbol{A}\right) \cdot \boldsymbol{\sigma} + \boldsymbol{v}^{\prime} \cdot \left(\boldsymbol{P} - e\boldsymbol{A}\right) I_{2 \times 2} . \end{align} It just like rotates the cone or rotates the z-axis and should also satisfies the particle-hole symmetry with a new operator.

So, what is the new $\hat{\sigma}_{z}$?

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  • $\begingroup$ I mean if you see the Fermi surface, the cone is tilted.You can also calculate the dispersion relationship numerically and there is a zero band which is the result of the particle-hole symmetry. $\endgroup$
    – Qian-Sheng
    Commented Mar 15 at 3:53

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