# The tensor product in the Hamiltonian of graphene

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 $$8\times8$$.

$$s_0$$ 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.