Understanding Liouville Space Super-Operators: A Simple Computation?

[Einstein Mcfly]

"straightforward computation..."

Hello. I am reading a paper that discusses moving from a regular $$NxN$$ dimensional Hilbert space into and $$N^2 X N^2$$ dimensional Liouville space. The density matrix can be re-written as a $$1 X N^2$$vector by stacking the rows and the Liouvile equation can be re-written as $$N^2 X N^2$$"super-operators" acting on the new density-matrix "vector". The actual form of this super-operator is given by "a straightforward computation" where the field free and interaction Hamiltonians are given by
$$H_o=\left(\begin{array}{cc} -hw & 0 \\ 0 & hw \end{array}\right)$$
and
$$H_1=\left(\begin{array}{cc} 0 & d1 \\ d1 & 0 \end{array}\right)$$

The Liouville space super operators describing these in the expanded space are:
$$L_o= \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & -hw &0 & 0\\ 0 & 0 & homework & 0\\ 0 & 0 & 0 & 0 \end{array}\right)$$
and

$$L_1= \left(\begin{array}{cccc} 0 & -d1 & d1 & 0 \\ -d1 & 0 & 0 & d1\\ d1 & 0 & 0 & -d1\\ 0 & d1 & -d1 & 0 \end{array}\right)$$

It looks like they're takind a direct product with something, but I can't tell with what or why. They make is sound like it's all just a pretty standard way to write a matrix in a larger space, so if anyone recognizes what's going on here, I'd really happy for your help. Thanks in advance.

 

[fresh_42]

$L_1$ looks more like a tenso product than a direct product, since the secondary diagonal isn't empty. Tensor products are also a standard method to extent e.g. the scalar domain.

The dyads over the Hilbert space build a basis for the Liouville space.