8.19 Mixed Quantum States and POVM

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Density Matrices - Mixed Quantum States

Assume the physical system is open, it is described by a density matrix \(\rho^A:A\to A,\,\rho\geq 0,\text{Tr}\rho^A=1\)

That is the information about the physical system is encoded in a density matrix

Any measurement is described by Generalized Measurement

We apply a Generalized Measurement: \(\{M_x\}_{x\in[m]}\) and \(\sum_{x\in[m]}M^*_xM_x=I^A\)

Physical process, transform \(\rho\) to \(\sigma_x:=\frac{1}{p_x}M_x\rho M_x^*\)(post-measurement state) where \(p_x=\text{Tr}[M_x^*M_x\rho]\)

This tells us how to extract the information from system (many times we get partial information)

And \(x\) is the outcome of the measurement

Suppose we have a pure state \(|\psi^{AB}\rang\), example: \(|\psi^{AB}\rang =\sqrt{\frac{1}{3}}|00\rang+\sqrt{\frac{2}{3}}|11\rang\)

We know any purification can be written as \(|\psi^{AB}\rangle=\sqrt{\rho}\otimes V^{A\to B}|\Omega^{A\tilde{A}}\rangle\)(Proof. In this case, the bases of Bob and Alice are same, thus \(V=I\)), \(\rho^{A}:=\text{Tr}_{B}[|\psi^{AB}\rang\lang\psi^{AB}|]\)

Example: \(\rho^{A}=\text{Tr}_{B}[|\psi^{AB}\rangle\langle\psi^{AB}|]=\frac{1}{3}|0\rangle\langle 0|+\frac{2}{3}|1\rangle\langle1|\), \(|\psi^{AB}\rangle=\sqrt{\rho}\otimes I|\Omega^{A\tilde{A}}\rangle\)(\(\checkmark\))

Suppose Alice performs a generalized measurement described by \(\{M_x\}_{x\in[m]}\) in

And suppose \(x\) occurs with probability \(p_{x}=\langle\psi^{AB}|M_{x}^{*}M_{x}\otimes I^{B}|\psi^{AB}\rangle\) (after outcome \(x\) occurred)

\(=\langle\Omega^{A\tilde{A}}|(\sqrt{\rho}\otimes V^{*})(M_{x}^{*}M_{x}\otimes I^{B})(\sqrt{\rho}\otimes V)|\Omega^{A\tilde{A}}\rangle\)

\(=\langle\Omega^{A\tilde{A}}|(\sqrt{\rho}M_{x}^{*}M_{x}\sqrt{\rho}\otimes V^{*}V) |\Omega^{A\tilde{A}}\rangle=\langle\Omega^{A\tilde{A}}|(\sqrt{\rho}M_{x}^{*}M_{x} \sqrt{\rho}\otimes I)|\Omega^{A\tilde{A}}\rangle\)

Since this, \(=\text{Tr}\left\lbrack\sqrt{\rho}M_{x}^{*}M_{x}\sqrt{\rho})\right\rbrack=\text{Tr} \left\lbrack\rho M_x^*M_x\right\rbrack\)

The post measurement state: After outcome \(x\) occurred
\(|\phi_{x}\rangle=\frac{1}{\sqrt{p_{x}}}M_{x}\otimes I|\psi^{AB}\rangle=\frac{1}{\sqrt{p_{x}}} (M_{x}\sqrt{\rho}\otimes V)|\Omega^{A\tilde{A}}\rangle\), then \(\sigma_{x}=\text{Tr}_{B}[|\phi_{x}\rang\lang \phi_{x}|]=\frac{1}{p_{x}}M_{x}\rho M_{x}^{*}\) since we only want the sub system of Alice
\(\sigma_{x}=\text{Tr}_{B}[|\phi_{x}\rangle\langle\phi_{x}|]=\text{Tr}_{B}[\frac{1}{\sqrt{p_{x}}} (M_{x}\sqrt{\rho}\otimes V)|\Omega^{A\tilde{A}}\rangle\frac{1}{\sqrt{p_{x}}}\langle \Omega^{A\tilde{A}}|(\sqrt{\rho}M_{x}^{*}\otimes V^{*})]\)
\(=\frac{1}{\sqrt{p_{x}}}\text{Tr}_{B}[(M_{x}\sqrt{\rho}\otimes V)|\Omega^{A\tilde{A}} \rangle\langle\Omega^{A\tilde{A}}|(\sqrt{\rho}M_{x}^{*}\otimes V^{*})]\frac{1}{\sqrt{p_{x}}} =\frac{1}{\sqrt{p_{x}}}M_{x}\sqrt{\rho}\text{Tr}_{B}[V\Omega^{A\tilde{A}}V^{*}]\sqrt{\rho} M_{x}^{*}\frac{1}{\sqrt{p_{x}}}\)

\(=\frac{1}{p_{x}}M_{x}\rho\text{Tr}_{\tilde{A}}[\Omega^{A\tilde{A}}]M_{x}^{*}=\frac{1}{p_{x}} M_{x}\rho M_{x}^{*}\) since \(\text{Tr}_{\tilde{A}}[\Omega^{A\tilde{A}}]=I\)

A classical - quantum state(cq-state)

When \(x\) remains both unknown and unrecorded, as discussed previously, the complete characterization of the system is encapsulated by the density operator \(\rho^A = \sum_{x \in [m]} p_x |\psi_x \rangle \langle \psi_x |\).

When \(x\) is recorded within the classical system \(X\) using the mapping \(x\to|x\rangle\langle x|\), the description of the system adopts a classical-quantum state, represented by \(\rho^{XA}=\sum_{x\in[m]}p_{x}|x\rang\lang x|^{X}\otimes \rho^{A}_{x}\) where \(X\) is classical system and \(A\) is quantum system

\(\{p_{x},\rho_{x}^{A}\}_{x\in[m]}\) is ensemble of states where we define \(\rho^A:=\sum_xp_x\rho_x^A\), \(x\to|x\rangle\langle x|\)

\(\{p_{x}, \sigma_{x}^{A}\}\) describe it with \(\sigma^{XA}:=\sum_{x}p_{x}|x\rangle\langle x|\otimes\sigma_{x}^{A}\) \(=\sum_{x}|x\rangle\langle x|^{X}\otimes M_{x}\rho M_{x}^{*}\) where \(\rho^A \xrightarrow{Measurement} \sigma^{XA}\) and since we substitute \(p_x \sigma_x^A = M_x \rho^A M_x^*\)

Separable State

\(\rho^{AB}:A\otimes B\to A\otimes B\)

\(\rho^{AB}:=\sum_{x\in[m]}p_{x}\omega_{x}^{A}\otimes\tau_{x}^{B}\) is separable state where \(p_x\) is probability

A convex combination of vectors in \(\R^n\) is \(\sum^m_{x=1}p_x\vec v_x\) for vectors \(\vec v_1,\vec v_2,\vec v_3,...,\vec v_m\) and \(p_x\geq 0,\sum^m_{x=1}p_x=1\) (It's linear combination and coefficients is probability)

Remark: Given a separable state, you can write it with rank 1 \(\omega\) and \(\tau\)

That is \(\rho=\sum_{y}p_{y}\psi_{y}\otimes\phi_{y}\), \(\omega_{x}^{A} = \sum_{y} \gamma_{y|x}\psi_{yx}^{A}\) and \(\tau_{x}^{B} = \sum_{z} s_{z|x}\phi_{zx}^{B}\) since \(\omega,\tau\) are density matrix, they can be decomposed as pure states

\(\rho = \sum_{x,y,z}\underbrace{\gamma_{yx} s_{z|x} p_x}_{\lambda_\alpha}\psi_{yx} \otimes \phi_{zx}\) \(= \sum_{\alpha}\lambda_{\alpha}\psi_{\alpha}\otimes \phi_{\alpha}\) where \(\sum_{\alpha} \lambda_{\alpha} = 1\)

Show that: \(\Phi_+^{AB} = |\Phi_+\rangle \langle \Phi_+|\), and \(|\Phi_{+}^{AB}\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)\) \(\neq |\Psi^A\rangle|\Phi^B\rangle\)

Proof

Assume \(|\Phi_+^{AB}\rangle = |\Psi^A\rangle |\Phi^B\rangle\), where \(|\Psi^A\rangle = \alpha |0\rangle + \beta |1\rangle, \quad |\Phi^B\rangle = \gamma |0\rangle + \delta |1\rangle,\) with \(|\alpha|^2 + |\beta|^2 = 1\) and \(|\gamma|^2 + |\delta|^2 = 1\)

Then \(\alpha \gamma = \frac{1}{\sqrt{2}}\), \(\alpha \delta = 0\), \(\beta \gamma = 0\), \(\beta \delta = \frac{1}{\sqrt{2}}\). No solution

A quantum state is said to be entangled if it is not separable

A referee samples a number \(x\) with a probability distribution \(p_{x}\) (e.g. roll a possibly biased dice, or flip a coin) and sends the number \(x\) to Alice and Bob who are spatially separated.
Based on this value, Alice prepares the state \(\rho_{x}^{A}\), and Bob prepares the state \(\tau_{x}^{B}\). Then, if Alice and Bob forget the value of \(x\), but still know the distribution \(p_{x}\) from which \(x\) was sampled, the state of their
composite system becomes \(\rho^{AB}:=\sum_{x\in[m]}p_{x}\omega_{x}^{A}\otimes\tau_{x}^{B}\)

Positive operator valued measured(POVM)

If we don't care about the post measurement state, we only focus on probabilities

Since \(p_{x}= \text{Tr}[\rho M_{x}^{*}M_{x}]\), then we define \(\Lambda_{x}:= M_{x}^{*}M_{x}\qquad \{\Lambda_{x}\}_{x \in [m]}- POVM\)

If \(\sum_{x\in[m]}\Lambda_{x}=I,\Lambda_{x}\ge0\), then it's a POVM

And \(0 \le \Lambda_x \le I\) which means \(I - A_x \ge 0\)(Positive semi-definite)

No Post-Measurement State

To every generalized measurement there exists a unique POVM that corresponds to it via the relation \(\Lambda_x = M_x^* M_x\).

However, for every POVM there are many quantum measurements corresponding to it.