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I have the Hamiltonian of pristine graphene \begin{equation} H=v_{F}.\boldsymbol{\gamma}.\boldsymbol{p} \end{equation} with $\boldsymbol{p}=(p_{x},p_{y})$ is the momentum operator, $v_{F}$ is the Fermi velocity and $\boldsymbol{\gamma}=(\gamma_{x},\gamma_{y})$ given by \begin{equation} \gamma_{x}=\sigma_{z}\otimes\tau_{x}\otimes s_{0}, \qquad \gamma_{y}=\sigma_{z}\otimes\tau_{y}\otimes s_{0} \end{equation} The unit $2\times2$ matrices $\sigma_{0}$, $\tau_{0}$ and $s_{0}$ together with the pauli matrices $\boldsymbol{\sigma}$, $\boldsymbol{\tau}$ and $\boldsymbol{s}$ act on the valley-$1/2$, sublattice-$1/2$ and spin-$1/2$ two-dimensional subspaces of graphene, respectively. We can be written $H$ in the matrix form \begin{align} &H= \begin{pmatrix} 0 & v_{F}\left(p_{x}-ip_{y}\right) &0 &0 \\ v_{F}\left(p_{x}+ip_{y}\right) & 0 &0 &0 \\ 0 &0&0&-v_{F}\left(p_{x}-ip_{y}\right) \\ 0 &0 &-v_{F}\left(p_{x}+ip_{y}\right) &0 \end{pmatrix} \end{align} My question is that: why we have a $4\times4$ matrix ? Normally the tensor product of the three matrix given matrix $8\times8$.

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$s_0$ is just the unit matrix. It is not explicitly written in the Hamiltonian - every entry in the $4\times4$ matrix is a $2\times2$ unit matrix itself, but since it's just a unit matrix, a simplified notation is used.

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  • $\begingroup$ is there the possibility of giving some explanations concerning that because I understood well and thank you so much for your answer $\endgroup$ – Elmanara Apr 9 at 21:59

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