3
$\begingroup$

Background

Shannon's source coding theorem tells us the following. We shall consider a binary alphabet for simplicity. Suppose Alice has $n$ independent and identically distributed instances of a random variable $X\in \{0, 1\}$. Let us call this string $M$. There exists an encoding scheme that allows her to compress this into $m = H(X)$ bits, where $H(X)$ is the Shannon entropy, such that she can later decode it to some $\tilde{M}$. The error she makes will vanish as $n\rightarrow \infty$ i.e. in that limit $\tilde{M} = M$.

Schumacher compresssion is the correct quantum analogue of this. We shall consider qubits for simplicity. Suppose Alice has $n$ independent copies of some qubit state $\rho$ i.e. she holds $\rho^{\otimes n}$. There exists an encoding scheme that allows her to compress these registers into $m = S(\rho)$ qubits, where $S(\rho)$ is the von Neumann entropy, such that she can later decode it to $\tilde{\rho}^n$. As $n\rightarrow\infty$, the error will vanish and $\tilde{\rho}^n = \rho^{\otimes n}$.

The quantum case is interesting because $\rho^{\otimes n}$ may be entangled with some inaccessible system $R$. The encoding and decoding scheme should work such that these correlations are preserved.

Question: What does the encoded state look like?

In the classical case, the encoded string is necessarily a uniformly randomly chosen $m$ bit string. Were this not the case, Alice could compress it further. What is the correct analogue of this result for the quantum case? What we know is that this state must still retain correlations with the inaccessible register $R$ and must have von Neumann entropy of exactly $m$ bits.

My guess

  1. Is it a uniformly random $m$ qubit pure state? Since the state has to maintain entanglement with $R$ in general, this cannot be the answer.
  2. A uniformly randomly chosen $m$ qubit mixed state? This seems more plausible and the set of choices depends on how much entanglement the input state $\rho$ shares with $R$.
$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.