1
$\begingroup$

In the Heisenberg picture we have the evolution of the operator in time given by: $$A(t)=U^+A(0)U$$

I was looking into the theory of open quantum systems where we introduce the concept of a quantum dynamical map. So, my question is in which all cases can we write the above evolution of the operator as the action of a quantum dynamical map? That is, $$A(t)=\Phi*A(0)$$ where $\Phi$ is a quantum dynamical map which is a mapping within the space of operators.

Can we always find a quantum dynamical map which has an explicit form? Can we always find an explicit form for such a map for closed systems or systems with some noise(described by a stochastic function included in the Hamiltonian of the system)?

I'm new to this subject and sorry if I'm asking something stupid. Also, I would like hints rather than any complete answers as I would like to put some thought into it myself.

$\endgroup$
  • $\begingroup$ Comments would be too long, I will post a complete answer...However it may be so technical that it works out as a collection of hints ;-P $\endgroup$ – yuggib May 3 '16 at 9:02
2
$\begingroup$

The most effective way to think about such problems is by means of operator algebras.

The best way to treat systematically quantum observables (and their associated unitary operators) is to collect them in a C$^*$-algebra, roughly speaking a Banach algebra where observables have a norm, can be summed and multiplied (with some additional technical conditions). This point of view fits perfectly to describe bounded operators on Hilbert spaces, but for unbounded operators it is not possible to define a norm (for closed unbounded operators it is possible to define a metric, however the space is not closed). The most used workaround - as far as I know - is to say that an unbounded self-adjoint operator (observable) is affiliated to a C$^*$-algebra of operators if all its spectral projections belong to the algebra.

Nevertheless, consider the easiest C$^*$-algebra, the algebra $L(\mathscr{H})$ of all linear bounded operators on the Hilbert space $\mathscr{H}$, endowed with the usual operator norm. The quantum states of this algebra of observables are the functionals of the topological dual $L(\mathscr{H})^*$ with norm one and positive, i.e. that give positive values when positive observables are evaluated. Now $L(\mathscr{H})$ has also a predual, the trace class operators $I^1(\mathscr{H})$, such that $L(\mathscr{H})=I^1(\mathscr{H})^*$. Hence it is a W$^*$-algebra, and the positive and norm one elements of $I^1(\mathscr{H})$ are called normal states. These normal states are none other than the usual quantum mechanical states: the density matrices.

Now the lengthy preamble is finished, and let think about the structure of the usual unitary dynamics for an isolated system. This is given by a self-adjoint (usually unbounded) operator $H$ that is affiliated to $L(\mathscr{H})$ and that generates the dynamics via the unitary group $(e^{-itH})_{t\in\mathbb{R}}$. Clearly, such group induces an evolution on the normal states $I^1(\mathscr{H})$, i.e. a group of linear continuous maps $(e^{-it L_0})_{t\in\mathbb{R}}\subset L\bigl(I^1(\mathscr{H})\bigr)$ that acts as follows: $$e^{-itL_0}\varrho=e^{-itH}\varrho e^{itH}\; .$$ This a so-called $C^0$-group, i.e. it is continuous with respect to the strong topology of $I^1(\mathscr{H})$ for any $t\in\mathbb{R}$.

The group $(e^{-itL_0})_t$ induces, by duality, a group on the observables $L(\mathscr{H})$ however this new group is in general only ultraweakly continuous - i.e. continuous wrt the ultraweak topology $\sigma\bigl(L(\mathscr{H}),I^1(\mathscr{H})\bigr)$. This is not so important, however this ultraweakly continuous group is strongly continuous on a suitable subspace of $L(\mathscr{H})$ called the adjoint subspace. Let's denote by $(e^{-itL_0^*})_t$ this dual group, whose action is $$e^{-itL_0^*}A=e^{itH}A e^{-itH}\; .$$

To sum up, even for the usual closed systems with unitary evolution it is possible to define the dynamical maps on both the space of states and on the space of observables, using the solution of the Schrödinger equation. The continuity properties of this map however change in the duality between states and operators, and usually there is stronger continuity in the space of states than in the space of observables.

The next situation where it is possible to define an explicit one-parameter dynamical map is for open Markovian systems. In this case, the Markov property is reflected in the fact that we still have a semigroup structure for the evolution: for any positive times $t,s\geq 0$, the evolution semigroup $(e^{-itL})_{t\geq 0}$ satisfies $$e^{-itL}e^{-isL}=e^{-i(t+s)L}\; .$$ If in addition we require it to be norm continuous and that map states into states, i.e. that it preserves positivity and trace, we obtain the Lindblad superoperator, whose generator $L$ necessarily has a very specific form. The norm continuity is preserved by duality, so in this case we have a norm-continuous evolution map in both the space of states and of observables. This is only a semigroup, i.e. it is defined only for positive times because in general an open evolution is irreversible. This can be extended to strongly continuous Markov dynamical maps on states, and by duality ultraweakly continuous dynamical maps on observables. It is rather hard to show that an effective system where the degrees of freedom of the environment have been traced out exhibits the Markov property, nevertheless it can be proved in some very special situation, and it is thought to be true in many reasonable physical systems (e.g. cavity QED, hot spin injections,...).

The most difficult situation that I am aware of that still allows for a dynamical map treatment is the case where the system has memory, but still it can be described by a two-parameter semigroup, i.e. a semigroup where the generator $L(t)$ depends explicitly on time. It is very hard to define the corresponding two parameter evolution $T(t,s)$, and it is required that the generator satisfies rather technical assumptions. Even more difficult (in fact I know no result in this sense) is to prove that such effective dynamics is obtained tracing out the degrees of freedom of the environment.


All the discussions above are limited to linear maps. If we allow the map to be non-linear, we may be able to describe other types of effective evolutions, that essentially would be a generalization to the space of states of the non-linear Schrödinger equations such as Hartree or Gross-Pitaevskii. How such non-linear evolution maps of states would then be translated to observables is not clear, since by effect of the non-linearity it is not possible to exploit the duality between the space of states and of observables.

$\endgroup$
  • $\begingroup$ Sir, Thank you for your detailed answer though I'm not quite familiar with operator algebras. Consider the following situation of a qubit exposed to a white noise. It turns out that given the noise is piece wise constant at equal time-slices we can form a quantum dynamical map(which turns out to be a transfer matrix) for the evolution of the generators of SU(2). $\endgroup$ – Rajath Krishna R May 3 '16 at 9:29
  • $\begingroup$ Since the transfer matrix can be always diagonalized we have exact solution for the evolution of these generators and hence enabling us to get the time evolution of any operator in this 2-D hilbert space. So, is there a general way to see whether this is possible for example for a 3-state system? Do you recommend me studying operator algebras to get a proper idea of what exactly is happening and then try to look into these problems? $\endgroup$ – Rajath Krishna R May 3 '16 at 9:29
  • $\begingroup$ Finite dimensional systems are rather easy, so probably it is not worth to study operator algebras if you're just interested in them. Probably it would be more interesting for you to take a look at random matrix theory; but I'm not an expert on the field. It seems however that it can indeed be related to non-equilibrium dynamics of finite dimensional systems. $\endgroup$ – yuggib May 3 '16 at 9:35
  • $\begingroup$ So, to make my thinking more clear. We have the 8-Gell-Mann matrices(along with the identity) as generators of SU(3) and we have the unitary operators $U$ which takes each of these 8 Gell-Mann matrices $\lambda_i(0)$ where $i={1,2,3,..,8}$ to $\lambda_i(t)$. I was thinking whether we can explicitly find a quantum dynamical map(which would be a 8X8 matrix) which does the same. $\endgroup$ – Rajath Krishna R May 3 '16 at 9:43
  • $\begingroup$ Every linear operator in a finite dimensional space can be written as a matrix. And since you say the space has nine generators, it is nine-dimensional (provided they're linearly indpendent). Therefore the linear operator $\lambda\mapsto U^*(t)\lambda U(t)$ can be also written as a $9\times 9$ matrix that to a vector $(a_0,\dotsc,a_8)$ (of coefficients of the generators) associates $(a_0(t),\dotsc,a_8(t))$ with the rule $$\sum_{j=0}^8 a_j(t)\lambda_j= U^*(t) \bigl(\sum_{j=0}^8 a_j\lambda_j\bigr) U(t)\; .$$ $\endgroup$ – yuggib May 3 '16 at 10:00

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.