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^2vector 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}<br /> -hw &amp; 0 \\<br /> 0 &amp; hw<br /> \end{array}\right)
and
H_1=\left(\begin{array}{cc}<br /> 0 &amp; d1 \\<br /> d1 &amp; 0<br /> \end{array}\right)

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

L_1=<br /> \left(\begin{array}{cccc}<br /> 0 &amp; -d1 &amp; d1 &amp; 0 \\<br /> -d1 &amp; 0 &amp; 0 &amp; d1\\<br /> d1 &amp; 0 &amp; 0 &amp; -d1\\<br /> 0 &amp; d1 &amp; -d1 &amp; 0<br /> \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.