5
$\begingroup$

I seem to read a lot of times that some materials called this $\mathcal{L}$ in the equation(Lindblad master equation) below as the generator: $$ \mathcal{L} \rho=-i[H, \rho]+\sum_{\alpha}\left(V_{\alpha} \rho V_{\alpha}^{\dagger}-\frac{1}{2}\left\{V_{\alpha}^{\dagger} V_{\alpha}, \rho\right\}\right). $$ I only know the generator in group theory means the minimum component that can span the whole group by the multiplication action, but I can't see what does the generator there means?

$\endgroup$
1
  • 1
    $\begingroup$ to add some context to the notation, you'll find this notation used this way in the context of Lie theory and functional analysis. The general Lie-theoretic result is that given a continuous homomorphism $f:\mathbb R\to G$ from the additive group of real numbers to some matrix Lie group $G$, there must be some $X\in\mathrm{End}(G)$ such that $f(t)=\exp(tX)$ for all $t\in\mathbb R$. This $X$ would be the "generator" of the homomorphism. $\endgroup$
    – glS
    Aug 19, 2021 at 6:23

1 Answer 1

5
$\begingroup$

The idea is that $$ \frac{d}{dt}\rho=\mathcal{L}\rho\qquad \Leftrightarrow\qquad \rho(t)=\exp(t\mathcal{L})\rho(0). $$ In that sense, the Lindbladian $\mathcal{L}$ generates evolution through $$\rho(dt)\approx \rho(0)+dt \mathcal{L}\rho(0),$$ which is the same sense as an infinitesmal generator of a Lie group. You might be familiar with something like a rotation of a vector $\mathbf{u}(0)$ by some rotation angle $|\mathbf{r}|$ about some axis pointing in the direction $\mathbf{r}$: $$R(\mathbf{r})\mathbf{u}(s)=\exp(i \mathbf{J}\cdot\mathbf{r})\mathbf{u} (0)\qquad\Leftrightarrow\qquad \mathbf{u}(d\mathbf{r})\approx \mathbf{u}(d\mathbf{r})+i \left(\mathbf{J}\cdot d\mathbf{r}\right) \,\mathbf{u}(0);$$ then, the operators $\mathbf{J}=(J_1,J_2,J_3)$ are the generators of rotations.

The idea is explicitly explained in Lindblad's original paper, from which the name ``generators of quantum dynamical semigroups'' derives, where the relevant fact is that $\lim_{t\to 0}\rho(t)=\rho(0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.