11.4: Quantum Zeno effect

This section is extracted from

 $(\sharp):$ S. Ishikawa, "A Measurement Theoretical Foundation of Statistics," arXiv:1308.5469 [quant-ph] 2014
11.4.1:Quantum decoherence: non-deterministic sequential causal operator

Let us start from the review of $\S$ 10.6.2 (quantum decoherence). Consider the quantum basic structure:

\begin{align} [{\mathcal C}(H) \subseteq B(H) \subseteq {B(H)}] \end{align}

Let ${\mathbb P}=[P_n ]_{n=1}^\infty$ be the spectrum decomposition in $B(H)$, that is,

\begin{align} P_n \mbox{ is a projection, and}, \sum_{n=1}^\infty P_n =I \end{align}

Define the operator $(\Psi_{\mathbb P})_*: {\mathcal Tr}(H) \to {\mathcal Tr}(H)$ such that

\begin{align} (\Psi_{\mathbb P})_* (|u \rangle \langle u |) = \sum_{n=1}^\infty |P_n u \rangle \langle P_n u | \quad (\forall u \in H) \end{align} Clearly we see \begin{align} \langle v, (\Psi_{\mathbb P})_* (|u \rangle \langle u |) v \rangle = \langle v, (\sum_{n=1}^\infty |P_n u \rangle \langle P_n u |) v \rangle =\sum_{n=1}^\infty |\langle v, |P_n u \rangle |^2 \ge 0 \qquad (\forall u,v \in H ) \end{align} and, \begin{align} & \mbox{Tr}((\Psi_{\mathbb P})_* (|u \rangle \langle u |)) \\ = & \mbox{Tr} (\sum_{n=1}^\infty |P_n u \rangle \langle P_n u |) = \sum_{n=1}^\infty \sum_{k=1}^\infty |\langle e_k , P_n u \rangle|^2 = \sum_{n=1}^\infty \| P_n u \|^2 = \|u\|^2 \qquad (\forall u \in H ) \end{align} And so, \begin{align} (\Psi_{\mathbb P})_* ({\mathcal Tr}_{+1}^p(H)) \subseteq {\mathcal Tr}_{+1}(H) \end{align} Therefore,
 $(\sharp):$ $\Psi_{\mathbb P} (=((\Psi_{\mathbb P})_*)^*): B(H) \to B(H)$ is a causal operator, but it is not deterministic.

In this note, a non-deterministic (sequential) causal operator is called a quantum decoherence.

Example 11.11 [Quantum decoherence in quantum Zeno effect] Furthermore consider a causal operator $(\Psi_S^{\Delta t})_* : {\mathcal Tr}(H) \to {\mathcal Tr}(H)$ such that

\begin{align} (\Psi_S^{\Delta t})_* (|u \rangle \langle u |) = |e^{-\frac{i {\cal H} \Delta t}{\hbar}}u \rangle \langle e^{-\frac{i {\cal H} \Delta t}{\hbar}} u | \quad (\forall u \in H) \end{align}

where the Hamiltonian ${\mathcal H}$ (cf. (10.23) ) is, for example, defined by

\begin{align} {\mathcal H} = \Big[ \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial q^2 } + V( {q}, t ) \Big] \end{align}

Let ${\mathbb P}=[P_n ]_{n=1}^\infty$ be the spectrum decomposition in $B(H)$, that is, for each $n$, $P_n \in B(H)$ is a projection such that

\begin{align} \sum_{n=1}^\infty P_n =I \end{align}

Define the $(\Psi_{\mathbb P})_*: {\mathcal Tr}(H) \to {\mathcal Tr}(H)$ such that

\begin{align} (\Psi_{\mathbb P})_* (|u \rangle \langle u |) = \sum_{n=1}^\infty |P_n u \rangle \langle P_n u | \quad (\forall u \in H) \end{align}

Also, we define the Schrödinger time evolution $(\Psi_S^{\Delta t})_* : {\mathcal Tr}(H) \to {\mathcal Tr}(H)$ such that

\begin{align} (\Psi_S^{\Delta t})_* (|u \rangle \langle u |) = |e^{-\frac{i {\cal H} \Delta t}{\hbar}}u \rangle \langle e^{-\frac{i {\cal H} \Delta t}{\hbar}} u | \quad (\forall u \in H) \end{align}

where ${\mathcal H}$ is the Hamiltonian (10.22). Consider $t=0,1$. Putting $\Delta t = \frac{1}{N}$, $H=H_0=H_1$, we can define the $(\Phi_{0,1}^{(N)})_*: Tr(H_0) \to Tr(H_1)$ such that

\begin{align} (\Phi_{0,1}^{(N)})_* =((\Psi_S^{1/N})_* (\Psi_{\mathbb P})_*)^N \end{align}

which induces the Markov operator $\Phi_{0,1}^{(N)} : B(H_1) \to B(H_0)$ as the dual operator $\Phi_{0,1}^{(N)} =((\Phi_{0,1}^{(N)})_*)^*$. Let $\rho=|\psi \rangle \langle \psi |$ be a state at time $0$. Let ${\mathsf{O}_1}{\; :=} (X, {\cal F}, F)$ be an observable in $B(H_1)$. Then, we see

\begin{align} \overset{\rho=|\psi \rangle \langle \psi |}{\fbox{$B(H_0)$}} \xleftarrow[\Phi_{0,1}^{(N)}]{} \underset{ {\mathsf{O}_1}{\; :=} (X, {\cal F}, F) }{\fbox{$B(H_1)$}} \end{align}

Thus, we have a measurement:

\begin{align} {\mathsf{M}}_{B(H_0)} (\Phi_{0,1}^{(N)} {\mathsf{O}_1}, S_{[\rho]}) \end{align}

$\big($ or more precisely, ${\mathsf{M}}_{B(H_0)} (\Phi_{0,1}^{(N)}{\mathsf{O}}{\; :=} (X, {\cal F}, \Phi_{0,1}^{(N)}F),$ $S_{[|\psi \rangle \langle \psi |]})$ $\big)$. Here, Axiom 1 ( $\S$2.7) says that

 $(A):$ the probability that the measured value obtained by the measurement belongs to $\Xi (\in {\cal F})$ is given by \begin{align} \mbox{Tr}(| \psi \rangle \langle \psi | \cdot \Phi_{0,1}^{(N)}F(\Xi)) \tag{11.21} \end{align}
Now we will explain "quantum Zeno effect" in the following example.

Example 11.12 [Quantum Zeno effect] Let $\psi \in H$ such that $\|\psi \|=1$. Define the spectrum decomposition

\begin{align} {\mathbb P}=[ P_1 (=|\psi \rangle \langle \psi |), P_2(=I-P_1) ] \tag{11.22} \end{align} And define the observable ${\mathsf{O}_1}{\; :=} (X, {\cal F}, F)$ in $B(H_1)$ such that

\begin{align} X=\{ x_1 , x_2 \}, \qquad {\cal F}=2^X \end{align} and \begin{align} F(\{x_1 \})=|\psi \rangle \langle \psi |(=P_1), F(\{x_2 \})=I- |\psi \rangle \langle \psi |(=P_2), \end{align} Now we can calculate (11.21)(i.e., the probability that a measured value $x_1$ is obtained) as follows. \begin{align} (11.21) &= \langle \psi, ((\Psi_S^{1/N})_* (\Psi_{\mathbb P})_*)^N (|\psi \rangle \langle \psi |) \psi \rangle \nonumber \\ & \ge |\langle \psi , e^{-\frac{i {\cal H} }{\hbar N}}\psi \rangle \langle \psi , e^{\frac{i {\cal H} }{\hbar N}}\psi \rangle|^N \nonumber \\ & \approx \Big(1 - \frac{1}{N^2} \big( || (\frac{ {\cal H} }{\hbar }) \psi ||^2 - |\langle \psi, (\frac{ {\cal H} }{\hbar }) \psi \rangle |^2 \big) \Big)^N \to 1 \nonumber \\ & \qquad \qquad \qquad \qquad ( N \to \infty) \tag{11.23} \end{align}

Thus, if $N$ is sufficiently large, we see that

\begin{align} {\mathsf{M}}_{B(H_0)} (\Phi_{0,1}^{(N)} {\mathsf{O}_1}, S_{[|\psi \rangle \langle \psi |]}) & \approx {\mathsf{M}}_{B(H_0)} (\Phi_I {\mathsf{O}_1}, S_{[|\psi \rangle \langle \psi |]}) \end{align} \begin{align} & = {\mathsf{M}}_{B(H_0)} ( {\mathsf{O}_1}, S_{[|\psi \rangle \langle \psi |]}) \end{align}

Hence, we say, roughly speaking in terms of the Schrödinger picture, that

the state $|\psi \rangle \langle \psi |$ does not move. 11.13: Remark The above argument is motivated by B. Misra and E.C.G. Sudarshan. However, the title of their paper: "The Zeno's paradox in quantum theory" is not proper. That is because

 $(B):$ the spectrum decomposition ${\mathbb P}$ should not be regarded as an observable (or moreover, measurement).

The effect in Example 11.12 should be called "brake effect" and not "watched pot effect".