"left" specification for rho_0

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andrea_simion
Posts: 11
Joined: Sun Aug 15, 2021 9:42 pm

"left" specification for rho_0

Post by andrea_simion »

I have a short question. In the following line, "left" means that the hamiltonian used to define the initial density operator was built by left product between operators? Am I right? Or what is the meaning of "left"?

parameters.rho0=equilibrium(spin_system,hamiltonian(assume(spin_system,'labframe'),'left'));
kuprov
Posts: 201
Joined: Mon Mar 29, 2021 4:26 pm

Re: "left" specification for rho_0

Post by kuprov »

That requests a left side product superoperator. That is because the equilibrium state is computed in Liouville space as exp(-H_left/kT)*unit_vector, and the commutation superoperator (default) would not be correct here. You can see this directly from the Taylor expansion of that definition, it's basically a column stretch of exp(-H/kT).
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