> ⭐️⭐️⭐️ The essense of quantum theory is the probability of **state purification**.

- Quantum probability for outcome $m$ for <font color=red>entangled state</font>
$$p_m=\langle\psi_{AB}|E_{Am}\otimes\mathbb{1}_{AB}|\psi_{AB}\rangle=\text{tr}_AE_{Am}\rho_A\neq\langle\psi_{AB}|E_{Am}|\psi_{AB}\rangle \\
\rho_A=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}|$$

- projective measurement on $B$ in *complete basis* $\{|b_B\rangle\}$ for $B$
$$P_b=\mathbb{1}_{A}\otimes|b_B\rangle\langle b_B|$$

- 🧐 *How does a mixed state evolve when system $A$ evolves unitarily on its own?*
- Unitarily evolve only subsystem $A$
$$|\psi_{AB}\rangle\mapsto(U_{A}\otimes\mathbb{1}_B)|\psi_{AB}\rangle$$
- $$\rho_A\mapsto\text{tr}_B(U_{A}\otimes\mathbb{1}_B)|\psi_{AB}\rangle\langle\psi_{AB}|(U_{A}^\dagger\otimes\mathbb{1}_B)=U_A\rho_A U_A^\dagger$$

- 🧐 *How does a mixed state change by a projective measurement on system $A$ only?*
- $$|\psi_{AB}\rangle\mapsto\frac{1}{\sqrt{p_m}}(P_{Am}\otimes\mathbb{1}_B)|\psi_{AB}\rangle\\
p_m=\langle\psi_{AB}|P_{Am}\otimes\mathbb{1}_B|\psi_{AB}\rangle$$
- $$\rho_A\mapsto \frac{1}{p_m}P_{Am}\rho_AP_{Am}$$

### Density operator
1. Hermicity: $\rho_A^\dagger=\rho_A$
2. Positivity (implies Hermicity): $\mathbb{R}\ni\langle\varphi|\rho_A|\varphi\rangle\geq0, \forall\ |\varphi\rangle$
3. Unit trace: $\text{tr}_A \rho_A=1$

- 🧐 *Are state-vector projections precisely the pure states?*
- ⭕️ $\rho=|\psi\rangle\langle\psi|$ is a pure state.
- $\rho^2=\rho$

### Ensemble of pure state
- commuting measured outcome $b$ with $A$
- $A$ is given a state from <font color=blue>ensemble $\{p_b,|\psi_{Ab}\rangle\}$</font>
- $B$ can prepare <font color=red>different ensemble</font> by performing a different measurement
- If $B$ doesn't inform $A$ which measurement he takes, the marginal state is <font color=green>the same</font>
> ⭐️⭐️⭐️ <font color=green>Same</font> mixed quantum state of A may have <font color=red>different ensemble</font> preparations by B

> ⭐️⭐️⭐️ In fact, there are <font color=red>infinitely many ensembles</font> for one non-pure density operator !
> they can be **non-orthogonal basis**.
> $$\sum_bp_b|\psi_{Ab}\rangle\langle\psi_{Ab}|=\rho_A=\sum_bp_{b'}|\psi_{Ab}'\rangle\langle\psi_{Ab}'|$$

- 🧐 *How are all ensemble decompositions of the same mixed state related ?*
- Two basis related by an unitary $V$
- $V^\dagger V=\mathbb{I}$

### <font color=blue>Ensemble v.s. Mixed state</font>
- <font color=blue>Ensemble</font>: if $A$ **knows** $b$ then the basis in $B$ does matter
$$\rho_{Ab}=|\psi_{Ab}\rangle\langle\psi_{Ab}|$$
- $A$ is given an <font color=green>unknown state</font> $|\psi_{Ab}\rangle$ with probability $p_b$ from <font color=red>known ensemble</font>
- <font color=green>Coherence/purity</font> of $A$ is maintained by *access of complete* outcome $b$ on $B$
- <font color=blue>Mixed state</font>: if $A$ **doesn't know** $b$, then basis in $B$ **does not matter**
$$\rho_{A}=\sum_bp_b|\psi_{Ab}\rangle\langle\psi_{Ab}|$$
- $A$ is given an <font color=green>unknown state</font> from <font color=red>unknown ensemble</font>
- <font color=red>Decoherence</font> of $A$ by *entanglement* with *inaccessible/discarded* system $B$
> ⭐️⭐️⭐️ <font color=red>Decoherence</font> = Loss of purity ~ entanglement + inaccessible information

- 🧐 *Can a* **given mixed state** $\rho_A$ *of some system $A$ be prepared from some entangled bipartite state $|\psi_{AB}\rangle$ by discarding some system $B$?*
- ⭕️ If you pick a **sufficient large** system $B$

- 🧐 *Can a* **given mixed state** $\rho_A$ *of some system $A$ be prepared from some entangled bipartite state $|\psi_{AB}\rangle$ by discarding* **a fixed system** *$B$?*
- ⭕️ If you have a same Hilbert space size $B$ as system $A$ (or larget). $$\text{dim }\mathcal{H}_A=\text{dim }\mathcal{H}_B$$
- A *canonical purified state* $|\psi_{AB}\rangle$ can be constructed directly from $$|\psi_{AB}\rangle=\sum_k\sqrt{\lambda_k}|k_A\rangle|k_B\rangle$$
where $\rho_A=\sum_k\lambda_k|k_A\rangle\langle k_A|$ and choose $\{|k_B\rangle\}$ for system $B$
- $\{|k_B\rangle\}$ are indeed **orthogonal** and can be normalized to ONB for $B$

> ⭐️⭐️⭐️ Every pure bipartite state $|\psi_{AB}\rangle$ has a <font color=red>*canonical/Schmidt decomposition*</font> $$|\psi_{AB}\rangle=\sum_k\sqrt{\lambda_k}|k_A\rangle|k_B\rangle\text{ with ONBs }\{|k_A\rangle\}, \{|k_B\rangle\}$$
> nonzero eigenvalues of $\rho_A$ and $\rho_B$ are always <font color=blue>equal because $|\psi_{AB}\rangle$ is pure.</font>

- 🧐 *How are the purifications of the same mixed state related ?*
- Purifications of the same state $\rho_A$ are related by a *unitary* on the purification syste.
- $$|\psi_{AB}\rangle=(\mathbb{1}_A\otimes U_B)|\psi_{AB}'\rangle$$

### <font color=green>Purification</font>
- $$\text{marginal }\rho_A=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}|\text{ purification}$$
$$
\begin{aligned}
&\text{entangled pure state} &|\psi_{AB}\rangle=\sum_b\sqrt{p_b}|\psi_{Ab}\rangle|b_B\rangle \\
&\text{pure ensemble} &\{p_b,|\psi_b\rangle\langle\psi_b|\} \\
&\text{mixed state} &\rho_A=\sum_bp_b|\psi_b\rangle\langle\psi_b|=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}|
\end{aligned}
$$
- $\text{entangled pure state }$<font color=red>$\xrightarrow[basis \{|b_B\rangle\}]{Discard}$</font> $\text{ pure ensemble }$ <font color=red>$\xrightarrow[basis \{outcomes\ b\}]{Discard}$</font> $\text{ mixed state}$

- $\text{mixed state}$ <font color=green>$\xrightarrow[into\ ensembles]{Discompose}$</font>$\text{ pure ensemble }$<font color=green>$\xrightarrow[ensembles]{Purify}$</font> $\text{ entangled pure state }$

> ⭐️⭐️⭐️ No information without disturbance