I'm reading A simple derivation of the Lindblad equation.

It introduces a Hamiltonian for a system consisting of a principal system $S$, a heat bath $B$ and an interaction term: $\hat{H}=\hat{H}_S+\hat{H}_B+\alpha\hat{H}_{SB}$.

It then switches to using operators $\hat{H}(t)$ and $\hat{\rho}(t)$ in the interaction picture.

On page 3 it says "$\bar{H}_{SB}$ can always be defined in a manner in which the first term on the right hand size of Eq. (10) is zero."

Here is Eq. (10):

$\frac{d}{dt}\hat{\rho}_S(t)=-\frac{i}{\hbar}\alpha \text{Tr}_B\left\{[\hat{H}(t),\hat{\rho}(0)]\right\}+\ldots$

As $H_{SB}$ is a given, I'm not sure how we can choose anything to make that term zero. What does the statement mean?

  • 1
    $\begingroup$ As far as I can tell, those notes contain at least two statements that are at best imprecise and at worst misleading/downright wrong. A good reference for this derivation, which is equally simple, is Breuer and Petruccione Section 3.3.1. $\endgroup$ Commented Jul 25, 2016 at 0:58

1 Answer 1


The statement is almost true, although one has to redefine both $H_S$ and $H_{SB}$ to make it work. You need to also make use of the standard assumption of factorising initial conditions, $$ \rho(0) = \rho_S(0)\otimes \rho_B,$$ where $\rho_B$ is the bath reference state, which must commute with the bath Hamiltonian, i.e. $[\rho_B,H_B]=0$. Often (but not always) the bath state is taken to be thermal $$ \rho_B = \frac{\mathrm{e}^{-\beta H_B}}{Z}.$$ Using the cyclicity of the partial trace, you can show that the offending term is equal to $$\mathrm{Tr}_B\left\lbrace[H(t),\rho_S(0)\otimes \rho_B]\right\rbrace = [\mathrm{Tr}_B\{H(t)\rho_B\},\rho_S(0)]\},$$ where the operator $\mathrm{Tr}_B\{H(t)\rho_B\}$ acts only in the Hilbert space of the system.

In order to remove this term, define the shifted Hamiltonians $$H_{SB}' = H_{SB} - \mathrm{Tr}_B\{H_{SB}\rho_B\},$$ $$H_A'= H_{A} + \mathrm{Tr}_B\{H_{SB}\rho_B\},$$ so that $H = H_A' + H_B + H_{SB}'$ is unchanged. Then, carrying out the derivation as before using the new Hamiltonians, you should be able to prove that $$\mathrm{Tr}_B\left\lbrace[H'(t),\rho_S(0)\otimes \rho_B]\right\rbrace = 0, $$ where $$H'(t) = \mathrm{e}^{\mathrm{i}(H_A'+H_B)t}H_{SB}'\mathrm{e}^{-\mathrm{i}(H_A'+H_B)t}.$$ In practice, one finds that common interaction Hamiltonians often already satisfy the desired property $\mathrm{Tr}_B\{H_{SB}\rho_B\}=0$ for states $\rho_B$ that are diagonal in the reservoir energy eigenbasis.


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