8.21 Exercise

Exercise 2.41. Show that for any \(\rho \in \mathcal{L}(AB)\) and any two matrices \(\eta, \zeta \in \mathcal{L}(B)\),
\(\operatorname{\mathrm{Tr}}_{B}\left[\left(I^{A}\otimes\eta^{B}\right)\rho^{AB}\left (I^{A}\otimes\zeta^{B}\right)\right]=\operatorname{\mathrm{Tr}}_{B}\left[\left(I^{A} \otimes\zeta^{B}\eta^{B}\right)\rho^{AB}\right]\)

Proof

拆开\(\rho\) 然后用\(\text{Tr}(AB)=\text{Tr}(BA)\), partial trace 不能换

For any pure bipartite state as in \(|\psi^{AB}\rangle = M_{\psi} \otimes I^{B} |\Omega^{\tilde{B}B}\rangle\), there is a unique reduced density matrix \(\rho_{\psi}^{A}\). On the other hand, for any density matrix \(\rho \in \mathfrak{D}(A)\) there are many bipartite pure states \(|\psi^{A B}\rangle\) with the same reduced density matrix \(\rho\) (see Exercise 2.44). - Exercise 2.44. Let \(\rho \in \mathcal{D}(A)\) be a density matrix.

Show that \(\sqrt{\rho}\otimes I^{\tilde{A}}|\Omega^{A\tilde{A}}\rangle\) is a purification of \(\rho\).

See notes on lectures 2. Show that \(|\psi^{AB}\rangle \in AB\) and \(|\phi^{AC}\rangle \in AC\) (assuming \(|B| \leq |C|\) ) are two purifications of \(\rho \in \mathcal{D}(A)\) if and only if there exists an isometry matrix \(V: B \to C\) such that \(|\phi^{AC}\rangle = I^{A}\otimes V^{B \to C}|\psi^{AB}\rangle.\)

Proof

\(\Rightarrow\)) See notes on lectures

\(\Leftarrow\)) Assume \(|\psi^{AB}\rangle\) is a purification of \(\rho\), then \(\text{Tr}_B[|\psi^{AB}\rang\lang \psi^{AB}|]=\rho\)

Since \(|\phi^{AC}\rangle = I^{A}\otimes V^{B \to C}|\psi^{AB}\rangle\), then \(\text{Tr}_{B}(|\phi^{AC}\rangle\langle\phi^{AC}|)=\text{Tr}_{B}\left(I^{A}\otimes V^{B\to C}\psi^{AB}\cdot\left(I^{A}\otimes V^{B\to C}\right)^{*}\right)=\text{Tr} _{B}\left(I^{A}\otimes I\psi^{AB}\right)=\text{Tr}_{B}(\psi^{AB})=\rho\) 3. Use Part 2 to provide alternative (simpler!) proof of the claim in Exercise 2.27.

Exercise 2.27. Let \(\left\{\left|\psi_{x}\right\rangle, p_{x}\right\}_{x \in[m]}\) and \(\left\{\left|\phi_{y}\right\rangle, q_{y}\right\}_{y \in[n]}\) be two ensembles of quantum states in A with \(m \geq n\). Show that they correspond to the same density matrix
\(\rho:=\sum_{x \in[m]}p_{x}\left|\psi_{x}\right\rangle \langle\psi_{x}|= \sum_{y \in[n]}q_{y}| \phi_{y}\rangle\left\langle\phi_{y}\right|\) if and only if there exists an \(m \times n\) isometry matrix \(V=\left(v_{x y}\right)\) (i.e. \(V^{*} V=I_{n}\)) such that \(\sqrt{p_{x}}\left|\psi_{x}\right\rangle=\sum_{y \in[n]} v_{x y} \sqrt{q_{y}}\left|\phi_{y}\right\rangle \quad \forall x \in[m] .\)

Proof

Using Schmidt Decomposition

Define purifications on orthonormal bases \(\{|y\rangle\}_{y=1}^n\) and \(\{|x\rangle\}_{x=1}^m\):
\(|\Psi^{AB}\rangle=\sum_{y=1}^n \sqrt{q_y}\,|\phi_y\rangle^A\otimes|y\rangle^B,\quad|\Phi^{AC}\rangle=\sum_{x=1}^m \sqrt{p_x}\,|\psi_x\rangle^A\otimes|x\rangle^C.\)
Then easily \(\operatorname{Tr}_B|\Psi\rangle\langle\Psi|=\sum_{y}q_y|\phi_y\rangle\langle\phi_y|,\quad\operatorname{Tr}_C|\Phi\rangle\langle\Phi|=\sum_{x}p_x|\psi_x\rangle\langle\psi_x|.\)

\(\Rightarrow\)) Since \(n=|B|\le |C|=m\), by Part 2 there exists an isometry \(V:B\to C\) such that
\(|\Phi^{AC}\rangle=(I^A\otimes V^{B\to C})\,|\Psi^{AB}\rangle.\)
Since this, then \(V^{B\to C}=\sum_{x\in[m]}\sum_{y\in [n]}v_{xy}|x\rang \lang y|\), then \(V|y\rangle=\sum_{x=1}^m v_{xy}\,|x\rangle\).
Then \(\sum_{x=1}^{m} \sqrt{p_{x}}\,|\psi_{x}\rangle\otimes|x\rangle=\sum_{y=1}^{n} \sqrt{q_{y}} \,|\phi_{y}\rangle\otimes V|y\rangle=\sum_{x=1}^{m} \Big(\sum_{y=1}^{n} v_{xy}\sqrt{q_{y}} \,|\phi_{y}\rangle\Big)\otimes|x\rangle\)
Then \(\sqrt{p_{x}}\,|\psi_x\rangle=\sum_{y=1}^n v_{xy}\sqrt{q_{y}}\,|\phi_y\rangle\quad\forall x\in[m]\)

\(\Leftarrow\)) If there exists an \(m\times n\) isometry \(V=(v_{xy})\) with the above relations, then
\(\sum_{x}p_{x} |\psi_{x}\rangle\langle\psi_{x}|= \sum_{x}\sum_{y,y'}v_{xy}v_{xy'} ^{*}\sqrt{q_{y}q_{y'}}\,|\phi_{y}\rangle\langle\phi_{y'}|= \sum_{y,y'}(V^{\ast} V )_{y'y}\sqrt{q_{y}q_{y'}}\,|\phi_{y}\rangle\langle\phi_{y'}|= \sum_{y}q_{y} |\phi _{y}\rangle\langle\phi_{y}|\) since \(V^\ast V=I_n\).
Hence the ensembles yield the same density matrix. - Exercise 3.1. Show that if \(\{M_x\}_{x \in [m]}\) and \(\{N_y\}_{y \in [n]}\) are two generalized measurements, then \(\{M_x N_y\}\) is also a generalized measurement. Use this to show that a sequence of generalized measurements can be simulated by a single generalized measurement.

See lecture notes - Exercise 3.2. Show that any collection of \(m\) operators \(\{M_x\}_{x\in[m]} \subseteq \mathcal{L}(A, B)\) that satisfy \(\sum_{x \in[m]}M_{x}^{*}M_{x}=I^{A}\) can also be realized as a generalized measurement as depicted in Figure 3.1: .
Hint: Consider a unitary operator \(U : RA \rightarrow R'B\) where the reference systems \(R\) and \(R'\) are such that \(|RA| = |R'B|\).

‍ - Exercise 3.4. Consider \(d\) (rank 1) operators \(\{M_x = |\psi_x\rangle\langle\phi_x|\}_{x \in [d]}\) in \(\mathcal{L}(\mathbb{C}^d)\), where \(\{|\psi_x\rangle\}_{x \in [d]}\) and \(\{|\phi_x\rangle\}_{x \in [d]}\) are some normalized states in \(\mathbb{C}^d\). Show that \(\{M_x\}_{x \in [d]}\) is a generalized measurement if and only if \(\{|\phi_x\rangle\}_{x \in [d]}\) is an orthonormal basis of \(\mathbb{C}^d\).

Proof

\(\Rightarrow\)) We know \(\sum_{x\in[d]}M_{x}^{*}M_{x}=I^{A}\Rightarrow\sum_{x\in[d]}|\phi_{x}\rangle\langle \psi_{x}|\psi_{x}\rangle\langle\phi_{x}|=I^{A}\Rightarrow\sum_{x\in[d]}|\phi_{x}\rangle \langle\phi_{x}|=I^{A}\)
But we know this is a iff proposition, then \(\{|\phi_x\rangle\}_{x \in [d]}\) is an orthonormal basis of \(\mathbb{C}^d\).

\(\Leftarrow\)) We know \(\{|\phi_x\rangle\}_{x \in [d]}\) is an orthonormal basis of \(\mathbb{C}^d\). Same reason, \(\sum_{x\in[d]}|\phi_{x}\rangle\langle\phi_{x}|=I^{A}\)
Since \(\{|\psi_x\rangle\}_{x \in [d]}\) are normalized, then \(\sum_{x\in[d]}|\phi_{x}\rangle\langle\psi_{x}|\psi_{x}\rangle\langle\phi_{x}|=I^{A}\)
Then \(\sum_{x\in[d]}M_{x}^{*}M_{x}=I^{A}\) - Exercise 3.9. [Polar Decomposition]

Show that for any \(n \times n\) complex matrix A there exists an \(n \times n\) unitary matrix \(U\) such that \(A = U|A| \quad \text{where} \quad |A| := \sqrt{A^*A}.\)
Show that for \(A\) and \(U\) as above, \(\max_U \text{Tr} [AU] = \text{Tr}[|A|],\) where the maximum is over all unitary matrices \(U\).
Exercise 3.10. Let \(\{\Lambda_x\}_{x \in [m]}\) be a POVM in Pos(A). Show that a generalized measurement \(\{M_x\}_{x \in [m]} \subset \mathcal{L}(A)\) corresponds to the POVM \(\{\Lambda_x\}_{x \in [m]}\) if and only if there exists \(m\) unitary matrices, \(\{U_x\}_{x \in [m]}\), in \(\mathcal{L}(A)\) such that \(M_{x}=U_{x}\sqrt{\Lambda_{x}}\)

Hint: Use the polar decomposition of a complex matrix.

Proof

\(\Rightarrow\)) We have \(\Lambda_x=M_x^*M_x\) where \(\Lambda_{x}\ge 0 \quad \text{and}\quad \sum_{x \in [m]}\Lambda_{x}= I^{A}\)

By polar decomposition, exists \(U_x\) s.t. \(M_{x}=U_{x}|M_x|=U_x\sqrt{M_x^*M_x}=U_x\sqrt{\Lambda_x}\)

\(\Leftarrow\)) \(M_{x}^{*}M_{x}=\sqrt{\Lambda_{x}}^{*} U_{x}^{*}\cdot U_{x}\sqrt{\Lambda_{x}}=(\sqrt{\Lambda_{x}} )^{2}=\Lambda_{x}\) - Exercise 3.11. Consider the following POVM in \(\text{Pos}(\mathbb{C}^2)\)

\(\Lambda_{1} = a|1\rangle\langle 1|, \quad \Lambda_{2} = b|-\rangle\langle -|, \quad \Lambda_{3} = I - \Lambda_{1} - \Lambda_{2} \quad a,b \in \mathbb{R}.\)

Find all the possible values of \(a\) and \(b\) for which the set \(\{\Lambda_1, \Lambda_2, \Lambda_3\}\) is a POVM.

We need to ensure \(\Lambda_{x}\ge 0 \quad \text{and}\quad \sum_{x \in [m]}\Lambda_{x}= I^{A}\)

\(\Lambda_{1}= \begin{pmatrix} 0 & 0 \\ 0 & a \end{pmatrix}\), then \(a\geq 0\). \(\Lambda_{2}= \begin{pmatrix} \frac{b}{2} & -\frac{b}{2} \\ -\frac{b}{2} & \frac{b}{2} \end{pmatrix}\), then \(b\geq 0\)

\(\Lambda_{3}= \begin{pmatrix} 1-\frac{b}{2} & \frac{b}{2} \\ \frac{b}{2} & 1-a-\frac{b}{2} \end{pmatrix}\), then \(b\leq 2,b\leq 2-2a,\,\,1-b-a+\frac{ab}{2}\geq 0\) and \(b\neq 2\)
Then \(a\leq \frac{2-2b}{2-b}\), thus \(0\leq a\leq \frac{2-2b}{2-b},\,0\leq b\leq 1\) 2. Which values of \(a\) and \(b\) that you found in Part 1 correspond to a rank 1 POVM (i.e. all the POVM elements have rank 1)?

\(0<b<1,a=\frac{2b-2}{2-b}\)

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