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8.14 Generalized Measurements and Open Quantum Systems

Elements of Quantum Mechanics: Open systems

Generalized Measurement

Two types of Evolution

  1. Projective von-Neumann Measurement \(\{\Pi_x\}_{x\in[m]}\)
    Initial state: \(|\psi\rangle\)
    Outcome \(x\) occur with probability \(P_x = \langle\psi|\Pi_x|\psi\rangle\)
    Post-Measurement state: \(\frac{\Pi_x|\psi\rangle}{\sqrt{P_x}} := |\psi_x\rangle\)

    image​

  2. Unitary Evolution
    ​image​

Then combining (1) + (2)

Capture_20250820_143859 where \(P_x = \langle\psi|U^*\Pi_x U|\psi\rangle\), \(|\phi_x\rangle := \frac{1}{\sqrt{P_x}}\Pi_x U|\psi\rangle\)

Here we transform \(|\psi\rangle \longrightarrow |\phi_x\rangle, P_x\)
Let's \(M_x := \Pi_x U\)(no longer projection and unitary), then \(|\phi_x\rangle = \frac{1}{\sqrt{P_x}}M_x|\psi\rangle\), \(\sum_{x\in[m]} M_x^*M_x = I\), \(P_x = \langle\psi|M_x^*M_x|\psi\rangle\) since \(\Pi_x\) is hermitian (\(\Pi_x = \sum|\phi_x\rangle\langle\phi_x|\))

Generally

image​

Here we also get \(|\psi_x\rangle = \frac{1}{\sqrt{P_x}} M_x |\psi\rangle\) and \(P_x\)​

Then \(\sum_x P_x = \sum_x \langle\psi|M_x^* M_x|\psi\rangle\)

Of course \(\sum_{x}P_{x}=1,\forall|\psi\rangle\implies\langle\psi|\sum_{x}M_{x}^{*}M_{x}|\psi \rangle=1,\forall|\psi\rangle\implies\sum_{x}M_{x}^{*}M_{x}=I\)

Definition

A collection of m complex matrices \(\{M_x\}_{x\in[m]} \in \mathbb{C}^{d \times d}\) is called a generalized measurement if \(\sum_{x\in[m]} M_x^* M_x = I^A\) with \(A \cong \mathbb{C}^d\)

Theorem

Let \(\{M_x\}_{x\in[m]}\) be a generalized measurement.
Then, there exists system \(R\) of dimension \(m\), and a unitary matrix \(U^{RA}: RA \rightarrow RA\) (\(U\) is \(md \times md\)).
For all \(|\psi\rangle \in A\), image
We have \(|\psi_y^A\rangle = \frac{1}{\sqrt{P_y}} M_y |\psi\rangle\) with \(P_y := \langle\psi|M_y^* M_y|\psi\rangle\) \(= ||M_y|\psi\rangle||^2\)

Proof

Every unitary matrix \(U^{RA}\) can be written as \(U^{RA} = \sum_{y,y' \in [m]} |y\rangle\langle y'|^R \otimes \Lambda_{yy'}^A\) where \(\Lambda_{yy'}\) - complex matrices. Because \(U^{RA}=\sum_{x,x^{\prime},y,y^{\prime}}C_{xx^{\prime}yy^{\prime}}|y\rangle\langle y^{\prime} |^{R}\otimes|x\rangle\langle x^{\prime}|^{A}=\sum_{y,y^{\prime}}|y\rangle\langle y^{\prime}|\otimes\left(\sum_{x,x^{\prime}}C_{xx^{\prime}yy^{\prime}}|x\rangle\langle x^{\prime}|\right)\rightarrow\Lambda_{yy^{\prime}}\), then \(U^{RA} = \begin{pmatrix}\Lambda_{11} & \Lambda_{12} & \cdots & \Lambda_{1m} \\\Lambda_{21} & \Lambda_{22} & \cdots & \Lambda_{2m} \\\vdots & \vdots & \ddots & \vdots \\\Lambda_{m1} & \Lambda_{m2} & \cdots & \Lambda_{mm}\end{pmatrix}\)

Then we define \(\Lambda_{11} := M_1, \Lambda_{21} := M_2 \quad \forall y \in [m] \quad \Lambda_{y1} := M_y \quad \sum_{y=1}^m M_y^* M_y = I\)
\(U^{*}U = \begin{bmatrix} \Lambda_{11}^{*} & \Lambda_{21}^{*} & \cdots & \Lambda_{m1}^{*} \\ & \vdots & & \\ \end{bmatrix} \begin{bmatrix} \Lambda_{11} & \\ \Lambda_{21} & \cdots \\ \vdots & \\ \Lambda_{m1} & \end{bmatrix}= \begin{pmatrix} \sum_{y=1}^{m}\Lambda_{y1}^{*}\Lambda_{y1}= I & * & * \\* & * & * \\* & * & * \end{pmatrix}=I\)
For the first block\(\begin{bmatrix} a_{1}^{*} & a_{2}^{*} & a_{3}^{*} \\ b_{1}^{*} & b_{2}^{*} & b_{3}^{*} \\ \vdots & \vdots & \vdots \\ - & - & - \end{bmatrix} \begin{bmatrix} a_{1} & b_{1} & \cdots \\ a_{2} & b_{2} & \cdots \\ a_{3} & b_{3} & \cdots \end{bmatrix} = \begin{bmatrix} 1 & 0 & * \\ 0 & 1 & * \\ * & * & \ddots \end{bmatrix}\)

Then the first column is orthonormal. Then we can complete the columns to all orthonormal columns by G.S. process.

Thus such dimension and matrix exists

image

\(t_0 : |1\rangle^R |\psi^A\rangle\)​

\(t_{1}:U^{RA}(|1\rangle^{R}|\psi^{A}\rangle)=\sum_{y}|y\rangle^{R}\otimes\Lambda_{y1} ^{A}|\psi\rangle\) \(=\sum_{y^{\prime}}|y^{\prime}\rangle^{R}\otimes M_{y^{\prime}}|\psi\rangle:=|\phi ^{RA}\rangle\)

\(t_2:\)Probability for outcome \(y \in [m]\) is: \(\Pi_{y}=|y\rangle\langle y|\), \(P_{y}=\langle\phi^{RA}|\Pi_{y}^{R}\otimes I^{A}|\phi^{RA}\rangle=\langle\psi|M_{y} ^{*}M_{y}|\psi\rangle\)

\(|\psi_{y}^{RA}\rangle=\frac{1}{\sqrt{P_{y}}}\underbrace{\Pi_{y}^{R}}_{|y\rangle\langle y|}\otimes I^{A}|\phi^{RA}\rangle=\frac{1}{\sqrt{P_{y}}}|y\rangle^{R}\otimes M_{y} |\psi\rangle=|y\rangle_{R}\otimes\left(\frac{1}{\sqrt{p_{y}}}M_{y}|\psi\rangle\right )\)

Homework:

  1. Given \(\{M_x\}_{x\in[m]}, \{N_y\}_{y\in[n]}\) - Generalized measurements. Prove that \(\{M_x N_y\}_{x,y}\) is also a generalized measurement.

    Proof:
    We know \(\sum_{i=1}^m M_i^* M_i = I, \sum_{i=1}^n N_i^* N_i = I\)
    NTP: \(\sum_{i=1}^m \sum_{j=1}^n (M_i N_j)^* (M_i N_j) = I \iff \sum_{i=1}^m \sum_{j=1}^n N_j^* M_i^* M_i N_j = I\)
    ​\(\iff \sum_{j=1}^n N_j^* (\sum_{i=1}^m M_i^* M_i) N_j = I \iff \sum_{j=1}^n N_j^* \cdot I \cdot N_j = I \quad \checkmark\)

  2. Given \(M_{0}=a|+\rangle\langle0|,M_{1}=b|0\rangle\langle+|\). Find conditions on \(a,b \in \mathbb{C}\) s.t. \(\exists M_2\) with \(\{M_0, M_1, M_2\}\) are Generalized Measurements.

    Proof:
    We know \(M_0^*M_0 + M_1^*M_1 + M_2^*M_2 = I\)
    Then \(\bar{a}|0\rangle\langle+|\cdot a|+\rangle\langle0|+\bar{b}|+\rangle\langle0|\cdot b|0\rangle\langle+|+M_{2}^{*}M_{2}=I\)
    Then \(||a||^{2} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{||b||^{2}}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + M_{2}^{*}M_{2} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
    Then \(M_2^*M_2 = \begin{pmatrix} 1-||a||^2 & 0 \\ 0 & 1 \end{pmatrix} - \frac{||b||^2}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\)
    Then \(M=M_{2}^{*}M_{2} = \begin{pmatrix} 1-||a||^{2} - \frac{||b||^2}{2} & -\frac{||b||^2}{2} \\ -\frac{||b||^2}{2} & 1-\frac{||b||^2}{2} \end{pmatrix}\) since \(M\) is hermitian and positive, then the eigenvalue should be non-negative

    \(\det(xId-M) = \det \begin{pmatrix} x-1+\|a\|^2+\frac{\|b\|^2}{2} & -\frac{\|b\|^2}{2} \\ -\frac{\|b\|^2}{2} & x-1+\frac{\|b\|^2}{2} \end{pmatrix}\)
    ​\(= (x-1+\frac{\|b\|^2}{2})^2 + \|a\|^2(x-1+\frac{\|b\|^2}{2}) - \frac{\|b\|^4}{4}\)
    ​\(= x^2+1+\frac{\|b\|^4}{4} -2x + x\|b\|^2 - \|b\|^2 + \|a\|^2x - \|a\|^2 + \frac{\|a\|^2\|b\|^2}{2} - \frac{\|b\|^4}{4}\)
    ​\(= x^2 + x(-2+\|b\|^2+\|a\|^2) + 1 - \|b\|^2 - \|a\|^2 + \frac{\|a\|^2\|b\|^2}{2}\)
    ​\(= 0 \Rightarrow x = \frac{2-\|b\|^2-\|a\|^2 \pm \sqrt{\|a\|^4+\|b\|^4}}{2}\)

    Then \(x \geq 0 \Rightarrow \frac{2 - ||b||^2 - ||a||^2 - \sqrt{||a||^4 + ||b||^4}}{2} \geq 0\)

    Then \(2 - ||b||^2 - ||a||^2 \geq \sqrt{||a||^4 + ||b||^4} \geq 0\)

    Then \(4 + ||b||^4 + ||a||^4 - 4||b||^2 - 4||a||^2 + 2||b||^2 ||a||^2 \geq ||a||^4 + ||b||^4\)

    Then \(||a||^2 (||b||^2 - 2) \geq 2||b||^2 - 2\)

    Since \(P_0 = \langle 0 | M_0 | 0 \rangle = |a|^2 \leq 1\), the same, \(|b|^2 \leq 1\), then \(||a||^2 \leq \frac{2||b||^2 - 2}{||b||^2 - 2}\)

    Thus \(0 \leq ||b||^2 \leq 1\), \(0 \leq ||a||^2 \leq \min \left\{ 1, \frac{2||b||^2 - 2}{||b||^2 - 2} \right\}\)

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