# How to write lattice $\phi^4$ hamiltonian in terms of Pauli matrices?

I want to decompose lattice~$$\phi^4$$ hamiltonian in terms of Pauli matrices. Particularly, how can I decompose

$$H_\text{Lattice}=a^d\sum_{{n}\in{Z}}\left[\frac{1}{2}\Pi_{n}^2+\frac{1}{2}\left(\nabla_a\Phi_{n}\right)^2+\frac{m^2}{2}\Phi_{n}^2+\frac{\lambda}{4!}\Phi_{n}^4\right]$$ $$\left(\nabla_a\Phi_{n}\right)^2=\sum_{{e}\in\mathcal{N}}\left(\frac{\Phi_{{n}+{e}}-\Phi_{n}}{2} \right)^2$$

in terms of in terms of Pauli matrices?

Honestly - I don't know the context or the meaning of your variables so I have no clue. But maybe the following will be helpful (originally a comment but too long):

Just like a vector can be decomposed into its components $$\langle i | \psi \rangle$$ in an orthonormal basis $$\{|i\rangle \}$$ as

$$|\psi \rangle = \sum_i \langle i |\psi \rangle |i\rangle$$

we can also decompose a matrix $$A$$ into its components $$Tr(\sigma_i^\dagger A)$$ in an orthonormal basis of matrices $$\{ \sigma_i\}$$:

$$A = \sum_i Tr(\sigma_i^\dagger A)\sigma_i$$

but we would have to normalize the Pauli matrices by multiplying each one by a factor of $$\frac{1}{2}$$ compared to the usual normalization. Then we'd have the needed $$Tr(\sigma_i \sigma_j) = \delta_{ij}$$.

The $$Tr()$$ operation here is called the Frobenius Inner Product.