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
    Commented 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 $\lim_{t\to 0}\rho(t)=\rho(0)$.

$\endgroup$

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