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.


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