If I have derived a master equation (e.g. in the Lindblad form) and solved for the density matrix, $\rho(t)$ I can get the mean value of an operator, A as:

$ <A> = \mathrm{Tr}A\rho $.

But this is just the mean value. Suppose I want an equation for the operator in the Heisenberg picture

$ A_H (t) $

For this I can derive quantum-Langevin equations (aka. Heisenberg-Langevin equations see e.g. API p.340ff) which gives this in terms of the Hamiltonian for the full system including the reservoir OR in terms of quantum Langevin forces.

Is there a way to go from the density matrix in the Schrödinger picture to the operator in the Heisenberg picture (in the case of an open system like described above)?

API: Cohen-Tannoudji et al., Atom Photon Interactions, 1998 & 2004.


For a linear master equation as the one described by the Lindbladian there is no problem in passing from the Schrödinger to the Heisenberg picture, apart from an eventual loss of regularity wrt time.

This is because there is a duality chain between the "set of normal states" $V_*$ (to be precise the normal states are positive elements with norm one of this predual space), the von Neumann algebra of observables $V$ and the states $V^*$ (again the positive elements with norm one): $$V^*=(V)^*=(V_*)^{**}\;,$$ where the $*$ operation stands for the topological dual.

So, suppose you are given a (strongly-continuous) semigroup (Lindblad semigroup) $(e^{\,\cdot\,\mathscr{L}})_*: \mathbb{R}_+\times V_*\to V_*$, that to a time $t\geq 0$ and a normal state (density matrix) $\rho\in V_*$ associates another normal state $(e^{\,t\,\mathscr{L}})_*\,\rho$.

By the aforementioned map we can define the evolution $e^{\,\cdot\,\mathscr{L}}: \mathbb{R}_+\times V\to V$ on observables $V$ easily by duality: $$\forall A\in V,\rho\in V_*, t\geq 0,\quad \rho(e^{\,t\,\mathscr{L}}A)=(e^{\,t\,\mathscr{L}})_*\,\rho(A)\; .$$

The semigroup $e^{\,\cdot\,\mathscr{L}}$ is called the dual semigroup of $(e^{\,\cdot\,\mathscr{L}})_*$. The only care that has to be taken is that this dual semigroup is, in general, less regular than the original one: if the original one is strongly continuous, the dual is a priori only weakly continuous. There is however a subspace $V^{\dagger}$ of the observables where the restriction of the dual semigroup is again strongly continuous. Nevertheless you have defined your Heisenberg evolution, starting from the Schrödinger one (and you can do the same thing to obtain the evolution for general states as well).

Remark. Here I have considered the Lindblad generator $(\mathscr{L})_*$ as a linear operator on the states (density matrices); its explicit form is the usual $$(\mathscr{L})_*\rho= -i[H,\rho]+\frac{1}{2}\sum_j\Bigl([U_j\rho,U_j^{\dagger}]+[U_j,\rho U_j^{\dagger}]\Bigr)\; .$$ The generator for the observables semigroup $\mathscr{L}$ (acting in $V^\dagger$ where it can be defined) is simply $$\mathscr{L}X= i[H,X]+\sum_j\Bigl(U_j^\dagger X U_j -\tfrac{1}{2}\{U^\dagger_jU_j,X\}\Bigr)\; .$$

  • 1
    $\begingroup$ I am not sure that you understood my question and I definitely did not understand your answer :) I wanted to know if there was an expression for $A_H (t)$ given $\rho (t)$. I already understood that the two formalisms should correspond to the same dynamics. $\endgroup$ Jun 19 '15 at 16:47
  • 1
    $\begingroup$ @HansHarhoff The expression for $A(t)$ is simply $A(t)=e^{t\mathscr{L}}A$, with $\mathscr{L}$ given by the last equation (it does not get much more explicit than that except in very few cases). This is, roughly speaking, a direct consequence of $\rho(t)=(e^{t\mathscr{L}})_*\rho$ and duality between states and observables. $\endgroup$
    – yuggib
    Jun 19 '15 at 17:48
  • $\begingroup$ Ah, I see. The math was a bit heavy for me so this clarification helps. $\endgroup$ Jun 19 '15 at 18:45
  • 1
    $\begingroup$ @HansHarhoff no, I mean another semigroup regularity, roughly speaking its continuity properties with respect to $t$. An usual requirement (but there are others) is that the semigroup is strongly continuous when acting on states: $(e^{(\cdot)\mathscr{L}})_*\rho: \mathbb{R}_+ \to V_*$ is a continuous function for every $\rho\in V_*$. This continuity property, however, is not preserved when taking the dual. This means that in general we can only say that $\rho(e^{i(\cdot)\mathscr{L}}A):\mathbb{R}_+\to \mathbb{C}$ is a continuous function for all $\rho\in V_*$, $A\in V$ (weak continuity). $\endgroup$
    – yuggib
    Jul 1 '15 at 21:25
  • 1
    $\begingroup$ @HansHarhoff However there is a subspace of the dual called the adjoint space $V^\dagger\subseteq V$, where $e^{t\mathscr{L}}\lvert_{V^\dagger}:V^\dagger\to V^\dagger$ it is again strongly continuous wrt $t$: $e^{(\cdot)\mathscr{L}}\lvert_{V^\dagger} X:\mathbb{R}_+ \to V^\dagger$ is continuous for any $X\in V^\dagger$. Physically this is not irrelevant, because continuity properties are related to the possibility of properly define the generator (Lindbladian) $\mathscr{L}$, as the derivative of the semigroup calculated in $t=0$. $\endgroup$
    – yuggib
    Jul 1 '15 at 21:30

Yuggib's answer is correct. Here is my elaboration for how to apply it in my case (Quantum optics):

@yuggib's answer (and for my own future reference): Ah I see. So I can derive the Heisenberg-Langevin equations from that like this: $ A(t) = e^{\mathscr{L}t} A(0) $

where $\mathscr{L}$ is the Lindbladian of the relevant master equation (up to a sign change for the commutator):

$ \dot{\rho} = \mathscr{L} \rho $

we should be able to obtain

$ \dot{A}(t) = \mathscr{L} A(t) $

by differentiating the first equation.

Here the Lindbladian working on the operator space is the dual of the Lindbladian working on states density matrices and is given by Lindblad 1976, eq. 4.2 and 4.3:

$ \mathscr{L} X = i [H,X] + \frac{1}{2} \sum_j \bigg( U_j^\dagger X U_j - \frac{1}{2} \big\{ U_j^\dagger U_j, X \big\} \bigg)$

An application example

I want to test this by finding the derivative of the $n=a^\dagger a $ operator in the Heisenberg picture by using the expression for the derivative of an operator.

My system is a harmonic oscillator with decay terms $V_j = a $ and for the purposes of the example I assume $H=0$ to simplify the expression.

We then have (note that in this equation all operators are time dependent (but taken at the same time):

$\dot{n} = \mathcal{L} n = \kappa(a^\dagger na - \frac{1}{2} a^\dagger a n - \frac{1}{2} n a^\dagger a) = \frac{1}{2} \kappa(a^\dagger na - a^\dagger a n + a^\dagger na - n a^\dagger a) = \frac{1}{2} \kappa(a^\dagger na - a^\dagger a n + a^\dagger na - n a^\dagger a) = -\kappa n $

Which is the expected behavior. Note that some authors have factor of two in this Lindblad equation which is due to a factor of $\sqrt{2}$ in the definition of the decay operators.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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