Quantum Asymptotic Equipartition

Question

From Information Theory, we have the Asymptotic Equipartition Property, which can be proved by the Weak Law of Large Number:

$\log P(x^n)=\log \prod\limits_{i=1}^{n} P(x_i)=\sum\limits_{i=1}^{n} \log P(x_i)=n\frac{1}{n}\sum\limits_{i=1}^{n} \log P(x_i) \overset{LLN}{=}n\mathrm{E}[\log P(X)]=-nH(X)$

However, when I encountered the quantum version...

Suppose the quantum information source emits n quantum states:

$\left|\psi_1\right> \otimes...\otimes\left|\psi_n\right>=\left|\psi\right>^{\otimes n}$

The density operator of these states is $\boldsymbol{\rho}^{\otimes n}$.

How do I calculate the asymptotic probability of observing $\left|\psi\right>^{\otimes n}$ ?

I tried to calculate:

$\log \Big[ Tr\big( \left|\psi\right>^{\otimes n}\left<\psi\right|^{\otimes n} \boldsymbol{\rho^{\otimes n}} \big) \Big]$

but I faced some problems, and failed...

Answer 1

I found a clear explanation [Property 15.1.3 (Equipartition)] in http://arxiv.org/abs/1106.1445

in which states that

Let \begin{equation} \Pi_{X^n}^\delta=\sum\limits_{x^n\in T_\delta^{x^n}}\left|x^n\right>\left<x^n\right| \end{equation} be the projection operator onto the typical subspace $T_\delta^{x^n}$

Also, since the density matrix is, \begin{equation} \rho_{X^n}=\sum\limits_{x^n\in\mathcal{X}^n}P_{X^n}(x^n)\left|x^n\right>\left<x^n\right| \end{equation}

we have \begin{equation} \Pi_{X^n}^\delta\rho_{X^n}\Pi_{X^n}^\delta=\sum\limits_{x^n\in T_\delta^{x^n}}P_{X^n}(x^n)\left|x^n\right>\left<x^n\right| \end{equation}

Recall that the typicality states that \begin{equation} 2^{-n(H(X)+\delta)}\leq P_{X^n}(x^n) \leq 2^{-n(H(X)-\delta)} \end{equation} where $H(X)=-E[\log P_X(X)]=-Tr(\rho_A \log \rho_A)$ (note that the basis of $\rho_A$ are orthonormal)

Therefore we arrive at

\begin{equation} 2^{-n(H(X)+\delta)}\Pi_{X^n}^\delta\leq \Pi_{X^n}^\delta\rho_{X^n}\Pi_{X^n}^\delta \leq 2^{-n(H(X)-\delta)}\Pi_{X^n}^\delta \end{equation} which is the desired result.