How can I rewrite the coupling between a system and a bath in terms of Hermitian operators (Deriving Lindblad eqn)?

Question

I'm trying to derive a Lindblad equation for a system where there is tunneling between a bath and a reservoir. This means that my interaction Hamiltonian is $H_I =\Sigma_s A^†B_s + B_s^†A$, where $A^†$ creates a particle in the system and $B_s^†$ creates one in state s of the bath.

Many of the derivations I find in texts/online use Hermitian operators in the interaction Hamiltonian, and may even say that is the most general possible interaction Hamiltonian. This fact then gives the bath correlation functions certain properties (Kramers-Kronig).

Exercise 14 of https://www1.itp.tu-berlin.de/schaller/download/NEQME2.pdf says:

"Show that it is always possible to choose hermitian coupling operators $A_α = A^†_α$ and $B_α = B^†_α$ using [the fact] that $H_I = H_I^†$."

But I don't see how to show it in the general case, or in the specific case I have. Can anybody show me how to rewrite my interaction Hamiltonian in terms of Hermitian operators?

Answer 1

For the exercise 14, assume you have an interacting term $AB$ in $H$, because it is hermitian, you’ll necessarily also have the term $A^\dagger B^\dagger$ as well. You therefore need to reduce $AB+ A^\dagger B^\dagger$ in the desired form. This can easily be done by analogy with the real part of the product of complex numbers since $A,B$ commute. You get: $$ AB+ A^\dagger B^\dagger=\\ \frac{1}{2}( (A+A^\dagger)(B+B^\dagger)+(A-A^\dagger) (B-B^\dagger)) $$

Applying the general result to your example, writing (by analogy with the harmonic oscillator): $$ x= \frac{1}{\sqrt 2}(A+A^\dagger) \\ x_s= \frac{1}{\sqrt 2}(A_s+A_s^\dagger) \\ p= \frac{-i}{\sqrt 2}(A-A^\dagger) \\ p_s= \frac{-i}{\sqrt 2}(A_s-A_s^\dagger) $$

your new hamiltonian is written as: $$ H=\sum_s xx_s+pp_s $$ So a sum of product of hermitian operators.

Hope this helps and tell me if something’s not clear.