Eric P. Hanson

March 21, 2019

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?

At each time step, exactly one ball switches sides.

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

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

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.

10 timesteps per frame, 600 total timesteps.

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

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!

*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.

- 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 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)

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.

*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

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!

*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.

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.

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).]