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

• Quantum probability for outcome \(m\) for entangled state
  \[p_m=\langle\psi_{AB}|E_{Am}\otimes\mathbb{1}_{AB}|\psi_{AB}\rangle=\text{tr}_AE_{Am}\rho_A\neq\langle\psi_{AB}|E_{Am}|\psi_{AB}\rangle \\ \rho_A=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}|\]

• 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 ensemble \(\{p_b,|\psi_{Ab}\rangle\}\)
• \(B\) can prepare different ensemble by performing a different measurement
• If \(B\) doesn't inform \(A\) which measurement he takes, the marginal state is the same

⭐️⭐️⭐️ Same mixed quantum state of A may have different ensemble preparations by B

⭐️⭐️⭐️ In fact, there are infinitely many ensembles 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 \(V\)
  • \(V^\dagger V=\mathbb{I}\)

Ensemble v.s. Mixed state

• Ensemble: if \(A\) knows \(b\) then the basis in \(B\) does matter
  \[\rho_{Ab}=|\psi_{Ab}\rangle\langle\psi_{Ab}|\]
  • \(A\) is given an unknown state \(|\psi_{Ab}\rangle\) with probability \(p_b\) from known ensemble
  • Coherence/purity of \(A\) is maintained by access of complete outcome \(b\) on \(B\)
• Mixed state: 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 unknown state from unknown ensemble
  • Decoherence of \(A\) by entanglement with inaccessible/discarded system \(B\)

⭐️⭐️⭐️ Decoherence = 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 canonical/Schmidt decomposition \[|\psi_{AB}\rangle=\sum_k\sqrt{\lambda_k}|k_A\rangle|k_B\rangle\text{ with ONBs }\{|k_A\rangle\}, \{|k_B\rangle\}\]
nonzero eigenvalues of \(\rho_A\) and \(\rho_B\) are always equal because \(|\psi_{AB}\rangle\) is pure.

• 🧐 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.
  • \[|\psi_{AB}\rangle=(\mathbb{1}_A\otimes U_B)|\psi_{AB}'\rangle\]

Purification

• \[\text{marginal }\rho_A=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}|\text{ purification}\]
  \[ \begin{aligned} &\text{entangled pure state} &|\psi_{AB}\rangle=\sum_b\sqrt{p_b}|\psi_{Ab}\rangle|b_B\rangle \\ &\text{pure ensemble} &\{p_b,|\psi_b\rangle\langle\psi_b|\} \\ &\text{mixed state} &\rho_A=\sum_bp_b|\psi_b\rangle\langle\psi_b|=\text{tr}_B|\psi_{AB}\rangle\langle\psi_{AB}| \end{aligned} \]

• \(\text{entangled pure state }\)\(\xrightarrow[basis \{|b_B\rangle\}]{Discard}\) \(\text{ pure ensemble }\) \(\xrightarrow[basis \{outcomes\ b\}]{Discard}\) \(\text{ mixed state}\)

• \(\text{mixed state}\) \(\xrightarrow[into\ ensembles]{Discompose}\)\(\text{ pure ensemble }\)\(\xrightarrow[ensembles]{Purify}\) \(\text{ entangled pure state }\)

⭐️⭐️⭐️ No information without disturbance