Understanding Liouville Space Super-Operators: A Simple Computation?

[Einstein Mcfly]

"straightforward computation..."

Hello. I am reading a paper that discusses moving from a regular [tex] NxN [/tex] dimensional Hilbert space into and [tex]N^2 X N^2[/tex] dimensional Liouville space. The density matrix can be re-written as a [tex]1 X N^2 [/tex]vector by stacking the rows and the Liouvile equation can be re-written as [tex]N^2 X N^2 [/tex]"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
[tex] H_o=\left(\begin{array}{cc}
-hw & 0 \\
0 & hw
\end{array}\right) [/tex]
and
[tex] H_1=\left(\begin{array}{cc}
0 & d1 \\
d1 & 0
\end{array}\right) [/tex]

The Liouville space super operators describing these in the expanded space are:
[tex] 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) [/tex]
and

[tex] 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) [/tex]

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.