# When do we lose correlations under partial Markovian evolution?

March 21, 2019

## Strategy

• Classical analog: what kind of correlations are we talking about? How can we lose them?

• What is entanglement?

• (Briefly:) What quantum evolutions cause a complete loss of entanglement in finite time?

## Ehrenfest model (classical)

At each time step, exactly one ball switches sides.

## Coupled system

Well-known invariant distribution: $\pi = \frac{1}{2^N}\sum_{n=0}^N{N \choose n} (\delta _n \otimes \delta _{N-n})$

Perfectly correlated!

## Loss to the environment

Loss of correlations! Becomes $\delta _0 \otimes p$ for $p = \frac{1}{2^N}\sum_{n=0}^N{N \choose n} \delta _{N-n}$.

## Loss to the environment

Loss of correlations! Becomes $\delta _0 \otimes p$ for $p = \frac{1}{2^N}\sum_{n=0}^N{N \choose n} \delta _{N-n}$.

## Alternate dynamics: Birth-death

At each timestep, prob. $\frac{1}{10}$ for a ball to be born (until 20); prob. $\frac{1}{10}$ for ball to die (until 0). Otherwise nothing happens.

## Alternate dynamics: Birth-death

10 timesteps per frame, 600 total timesteps.

Loss of correlations! Asymptotically converges to $u \otimes p$ for $u$ uniform.

## More generally: irreducible + aperiodic

Initial joint distribution: $\pi = \frac{1}{2^N}\sum_{n=0}^N{N \choose n} (\delta _n \otimes \delta _{N-n})$.

Let $P$ be irreducible and aperiodic with invariant distribution $q$. Then \begin{aligned}\relax \pi ( P^k \otimes \operatorname{id}) &= \frac{1}{2^N}\sum_{n=0}^N{N \choose n} ( (\delta _n P^k) \otimes \delta _{N-n}) \\ & \rightarrow \frac{1}{2^N}\sum_{n=0}^N{N \choose n} ( q \otimes \delta _{N-n}) = q \otimes p \end{aligned}
where $p = \frac{1}{2^N}\sum_{n=0}^N{N \choose n} \delta _{N-n}$. Decouples!

## But we don’t always lose correlations!

Summary of the classical case

If we let one part of a joint system evolve independently, the two may decouple.

That decoupling can take infinitely long

Under periodic evolution, the joint system may not decouple.

## Classical $\textcolor{purple}{\to}$ quantum

• Probability distributions $p \in \mathbb{R}^d$ $\textcolor{purple}{\to}$ quantum states $\rho \in M_{d\times d}$.
• $p_i \ge 0 \textcolor{purple}{\to}\rho \succeq 0$,
• $\sum _i p_i = 1 \textcolor{purple}{\to}\operatorname{tr}[\rho ] = 1$.
• Markov transition matrix $P$ $\textcolor{purple}{\to}$ $\Phi$ linear, completely positive, trace-preserving map.
• $P$ maps $\mathbb{R}^d$ to $\mathbb{R}^d$ $\textcolor{purple}{\to}$ $\Phi$ maps $M_{d\times d}$ to $M_{d\times d}$
• Joint distribution $\sum_i \lambda_i p_i \otimes q_i$ (convex combination of product distribution) $\textcolor{purple}{\to}$ joint state $\rho \in M_{d\times d} \otimes M_{d\times d}$.
• Key difference: it can be that $\rho \in M_{d\times d} \otimes M_{d\times d}$ cannot be expressed as $\sum_i \lambda_i \rho _i \otimes \sigma _i$ for $\rho _i \succeq 0$ and $\sigma _i \succeq 0$, and a probability distribution $\{\lambda_i\}$. Entanglement!

## Entanglement

Entanglement is a non-classical correlation between the two systems. It is a bit of a strange type of correlation.

• “No-signalling”; you can’t use entanglement alone to send messages

• “Superdense coding” One can communicate 2n bits of classical information by only transmitting n bits of data, using n pre-shared “bits of entanglement” (n pairs of entangled qubits)

## Which states are entangled?

3D cross section of two-qubit quantum states.

• Green: all states.

• Blue: classical states.

• Inside green, outside blue: entangled states

Avron, Joseph, and Oded Kenneth. 2019. “An Elementary Introduction to the Geometry of Quantum States with a Picture Book.” http://arxiv.org/abs/1901.06688.

## Magic square game

Task: Two isolated, cooperative players Alice and Bob try to construct a magic square

Rows must have an even number of $-1$’s

Columns must have an odd number of $-1$’s

## Winning the game

Alice and Bob are judged probabilistically:

• the referee asks Alice for one specific row, and Bob for one specific column (e.g. top row, second column)

• Alice’s row must have an even number of $-1$’s, and Bob’s column an odd number, and they must agree on the intersection.

• That’s it! They win or lose the game.

• Classically, the best they can do is win $8/9$ percent of the time

• If they have access to pre-shared entanglement, they can win every time!

## How do we lose quantum entanglement?

Classically…

If we let one part of a joint system evolve independently, the two may decouple.

That decoupling can take infinitely long

Under periodic evolution, the joint system may not decouple.

## Irreducible + aperiodic quantum evolutions

• Similar to property to the classical case: $\Phi ^n(\rho ) \rightarrow \sigma$ for any $\rho$.

• Just like in the classical case, this gives convergence of the joint state to a product state (here, $\sigma \otimes I/d$ where $I/d$ is the identity matrix )

• It turns out, product states of full support are in the interior of the set of separable states.

• Then the convergence $\Phi ^n(\rho ) \rightarrow \sigma$ gives that there is a finite time for any joint state to become separable under this evolution.

## Which evolutions destroy all entanglement in finite time?

• Irreducible + aperiodic

• Direct sums thereof

• Any map such that after some number iterations, it is a direct sum of irreduce + aperiodic maps.

• That’s it!$^*$

$*$: up to a faithfulness assumption, namely that there exists an invariant state with full support (possibly non-unique).

[Joint work with Cambyse Rouzé and Daniel Stilck França (https://arxiv.org/abs/1902.08173).]