Quantum Asymptotic Equipartition 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...
 A: 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.
