2
$\begingroup$

Suggested by Willie Wong, I reformulate my question posed Reference request : differential equation with operator valued unknowns to a concrete case.

I wish to emphasise that I look for the references on infinite-dimensional Lindblad equation that are written for mathematicians.

Let $\mathbf H:=L^2(\mathbb R^d)$ be a Hilbert space, and denote by $\mathcal L(\mathbf H)$ the collection of linear operators (that may not be bounded) on $\mathbf H$. For every $A\in \mathcal L(\mathbf H)$, denote by $D(A)$ the domain of $A$. Consider the Lindblad equation

\begin{equation} \mathrm{d} \rho_t = -\mathrm{i}\left[-\frac 12 \Delta + V,\rho_t\right]\mathrm{d} t + \left(K\rho_tK - \frac{1}{2}K^2\rho_t-\frac{1}{2}\rho_tK^2 \right)\mathrm{d} t, \end{equation}

where $K, V$ stand for the multiplication operator defined by the functions $k, v:\mathbb R^d\to\mathbb R$, i.e. for $\varphi\in \mathbf H$, $K(\varphi)\in \mathbf H$ is defined by

$$K(\varphi)(x):=k(x)\varphi(x),$$

and $[\cdot,\cdot]$ denotes the commutator, i.e. for $A, B\in \mathcal L(\mathbf H)$, $[A,B]:=AB-BA$. We look for the unknown $\rho$ living in the so-called trace class, i.e. $\rho_t$ is Hermitian, $\rho_t\ge 0$ and $trace(\rho_t)=1$ for all $t\ge 0$.

Question : What is the known results on the wellposedness of this equation?

If $\rho_t$ can be identified by some kernel, denoted by $\rho(t,x,y)$, the above Lindblad equation rewrites as the following PDE on $\mathbb R_+\times \mathbb R^d \times\mathbb R^d$:

\begin{eqnarray*} \partial_t \rho(t,x,y) &=& -\mathrm{i}\Big(-\frac 12 (\Delta_{x}-\Delta_{y})\rho(t,x,y) +\big( v(x)-v(y)\big) \rho(t,x,y)\Big)\\ && - \frac{1}{2}\Big(k(x)- k(y) \Big)^2 \rho(t,x,y), \end{eqnarray*}

Question: Under which conditions this PDE is well posed (admit a unique solution in the trace class)?

When $k\equiv 0$, it is known that $\rho(t,x,y)=\varphi(t,x)\overline\varphi(t,y)$, where $\varphi$ solves some Schrödinger equation, see e.g. Energy estimation of density operator to von Neumann equation

$\endgroup$
6
  • $\begingroup$ What I'm saying is a standard trick. Your equation for the evolution operator $U(t)$ can be written as $\dot{U} = K(U) + D(U)$ where $K$ is the Hamiltonian part and $D$ the dissipative one. Now write $U(t)=U_0(t)V(t)$ where $U_0$ solves the Hamiltonian part: $U_0(t)=\exp(t K)$. Now you get the following, time dependent equation for $V$: $\dot{V} = D_I (t) (V)$, where $D_{I}(t)= U_0^\dagger(t) D U_0(t)$. $\endgroup$ Commented Jan 20 at 17:26
  • $\begingroup$ In any case your original generator should generate a contraction semi-group for which you can invoke Hille-Yoshida. One paper that could be a starting point is: Avron et al Adiabatic Theorems for Generators of Contracting Evolutions, Commun. Math. Phys. 314, 163–191 (2012) $\endgroup$ Commented Jan 20 at 17:39
  • $\begingroup$ @lcv Thanks a lot. I have one further question: Using this mild formulation (with $U_0$) and the fixed-point argument, can we always guarantee the solution $\rho$ always belongs to the trace class? $\endgroup$ Commented Jan 21 at 8:04
  • $\begingroup$ The Lindblad equation is trace preserving so if you can prove existence of the semigroup and start with a trace class state you will stay trace class. $\endgroup$ Commented Jan 24 at 16:09
  • $\begingroup$ @lcv For me it is clear that the Hermitian property and unit trace are preserved by the Lindblad equation. However, why the semidefinite positivity is also preserved? $\endgroup$ Commented Jan 28 at 12:47

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.