Theory of the Decoherence Effect in Finite and Infinite Open Quantum Systems Using the Algebraic Approach

Abstract

Quantum mechanics is the greatest revision of our conception of the character of the physical world since Newton. Consequently, David Hilbert was very interested in quantum mechanics. He and John von Neumann discussed it frequently during von Neumann’s residence in Göttingen. He published in 1932 his book Mathematical Foundations of Quantum Mechanics. In Hilbert’s opinion it was the first exposition of quantum mechanics in a mathematically rigorous way. The pioneers of quantum mechanics, Heisenberg and Dirac, neither had use for rigorous mathematics nor much interest in it. Conceptually, quantum theory as developed by Bohr and Heisenberg is based on the positivism of Mach as it describes only observable quantities. It first emerged as a result of experimental data in the form of statistical observations of quantum noise, the basic concept of quantum probability.

Appendix

Sketch of Proof of Theorem 7

The proof is organized by successively proving the following three properties.

P1 :

$x \in L^{\infty }(\pi _{\omega _{0}}(A))$ if and only if $\left [\pi _{\omega _{0}}(A),\left [\pi _{\omega _{0}}(A),x\right ]\right ] = 0$.

P2 :

$L^{\infty }(\pi _{\omega _{0}}(A)) \cap L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0})) = \mathbb{C}{\!1\!\!1}$.

P3 :

$\mathfrak{M} = \mathbb{C}{\!1\!\!1}$ if and only if $\ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})' = \mathbb{C}{\!1\!\!1}$.

Given these properties the proof is easily carried through. If $x \in \ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})'$ then $[\pi _{\omega _{0}}(A),[\pi _{\omega _{0}}(A),x]] = 0$ and [H _S, x] = 0. Consequently, by P1 we get $x \in L^{\infty }(\pi _{\omega _{0}}(A))$, and hence $[\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}),x] = [H_{\mathrm{S}},x] = 0$, i.e., $x \in L^{\infty }(\pi _{\omega _{0}}(A)) \cap L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0})) = \mathbb{C}{\!1\!\!1}$ according to P2. In this way we obtained $\ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})' = \mathbb{C}{\!1\!\!1}$, i.e., $\mathfrak{M} = \mathbb{C}{\!1\!\!1}$ according to P3. To complete the proof we therefore need to establish properties P1, P2 and P3.

Proof of P1 : :

We start by proving the direction “ ⇐ ”. Let us define the derivation $\delta _{x}(\cdot ) = \mathrm{i}[\cdot,x]$. If $[\pi _{\omega _{0}}(A),[\pi _{\omega _{0}}(A),x]] = 0$ then $[\pi _{\omega _{0}}(A),x] \in L^{\infty }(\pi _{\omega _{0}}(A))$ as $L^{\infty }(\pi _{\omega _{0}}(A))$ is m. a. s. a. (see the proof of P2 below). Let P be any polynomial, then

$$\displaystyle{\delta _{x}(P(\pi _{\omega _{0}}(A))) = \mathrm{i}[P(\pi _{\omega _{0}}(A)),x] = \mathrm{i}[\pi _{\omega _{0}}(A),xP'(\pi _{\omega _{0}}(A))] \in L^{\infty }(\pi _{\omega _{ 0}}(A)).}$$

This means that $\delta _{x}(L^{\infty }(\pi _{\omega _{0}}(A))) \subseteq L^{\infty }(\pi _{\omega _{0}}(A))$ since δ _x is continuous in the weak operator topology. But $L^{\infty }(\pi _{\omega _{0}}(A))$ is abelian, thus δ _x (E) = 0 for any projector E from the domain of the derivation δ _x . On the other hand, δ _x is defined on the whole algebra and in particular L ^∞(π(A)) is contained in its domain. Hence

$$\displaystyle{\delta _{x} \upharpoonright _{L^{\infty }(\pi _{\omega _{ 0}}(A))} = 0.}$$

In particular, $[\pi _{\omega _{0}}(A),x] = -\mathrm{i}\delta _{x}(\pi _{\omega _{0}}(A)) = 0$. But $L^{\infty }(\pi _{\omega _{0}}(A))$ is a m. a. s. a., hence $x \in L^{\infty }(\pi _{\omega _{0}}(A))$. The proof of the converse is obvious.

Proof of P2 : :

Let $C_{3} \subseteq \mathcal{A}$ be C*-algebra generated by the set

$$\displaystyle{\{\sigma _{i_{1}} \otimes \cdots \otimes \sigma _{i_{n}}\otimes{\!1\!\!1} \otimes \cdots \,,\ i_{k} = 0,3\ \text{for}\ k = 1,\ldots,n,\ n = 1,2,\ldots \}.}$$

Then $\pi _{\omega _{0}}(C_{3})''$ is a m. a. s. a. If we substitute σ ₃ by σ ₁ we can define C ₁ in a similar fashion and get another m. a. s. a. $\pi _{\omega _{0}}(C_{1})''$. Evidently, $\pi _{\omega _{0}}(C_{3})'' \cap \pi _{\omega _{0}}(C_{1})'' = \mathbb{C}{\!1\!\!1}$. Now the choice of the sequences $\{h_{l}\}_{l=1,2,\ldots }$ and $\{a_{l}\}_{l=1,2,\ldots }$ in the statement of the Theorem ensures that $L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0})) =\pi _{\omega _{0}}(C_{3})''$, and $L^{\infty }(\pi _{\omega _{0}}(A)) =\pi _{\omega _{0}}(C_{1})''$ as is proven below. Let us introduce some notation. As the Cantor set $\mathcal{C}$ is homeomorphic to {0, 1}^×∞ we shall use the representation of elements $\omega \in \mathcal{C}$ by

$$\displaystyle{\omega \equiv \{ i_{1},i_{2},\ldots \}\quad \text{with}\ i_{1},i_{2},\ldots \in \{ 0,1\}.}$$

We say that two elements ω ₀ and ω ₁ constitute a pair of adjoint points if

$$\displaystyle\begin{array}{rcl} \omega _{0}& =& \{i_{1},i_{2},\ldots,i_{m},0,1,1,1,\ldots \}\quad \text{and} {}\\ \omega _{1}& =& \{i_{1},i_{2},\ldots,i_{m},1,0,0,0,\ldots \} {}\\ \end{array}$$

for some non-negative integer m. Let $\mathcal{C}_{0}$ be the set of pairs of adjoint points. Moreover, let $\Phi:\pi _{\omega _{0}}(C_{3})\longrightarrow C(\mathcal{C})$ denote the Gelfand–Naimark isomorphism, where $C(\mathcal{C})$ is the set of all continuous functions defined on Cantor set. Let us consider the unique extension of $\Phi$ to the normal isomorphism $\Psi:\pi _{\omega _{0}}(C_{3})''\longrightarrow L^{\infty }(\mathcal{C},\mu )$, where $\mu =\prod \mu _{0}$ with the measure μ ₀ on {0, 1} defined by $\mu _{0}(\{0\}) =\mu _{0}(\{1\}) = \frac{1} {2}$.

Finally, $C_{0}(\mathcal{C})$ is the set of all continuous functions f such that f(ω ₀) = f(ω ₁) for some pair of adjoint points ω ₀ and ω ₁. Let $h = \Phi (\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}))$, then

$$\displaystyle{h(\{i_{1},i_{2},\ldots \}) =\sum _{ l=1}^{\infty }(-1)^{i_{l} }h_{l}.}$$

We can evaluate the difference at the pair of adjoint points to obtain

$$\displaystyle{h(\omega _{0}) - h(\omega _{1}) = 2{\Bigl (h_{m+1} -\sum _{l=m+2}^{\infty }h_{ l}\Bigr )} \geq 0,}$$

so the function h takes different values to different points, except perhaps a pair of adjoint points of $\mathcal{C}_{0}$. Repeating the standard argument we get $C_{0}(\mathcal{C}) \subseteq \Phi (C^{{\ast}}({\!1\!\!1},\pi _{\omega _{0}}(H_{\mathrm{S}}^{0})))$, where C ^∗(D) denotes the smallest C*-algebra generated by the set D.

Let $P_{i_{1}i_{2}\cdots i_{m}} =\pi _{\omega _{0}}(P_{i_{1}} \otimes P_{i_{2}} \otimes \cdots \otimes P_{i_{m}}\otimes{\!1\!\!1} \otimes \cdots \,)$ be one of the generating projectors of π(C ₃). Then

$$\displaystyle{ P_{i_{1}i_{2}\cdots i_{m}} \in L^{\infty }(\pi _{\omega _{ 0}}(H_{\mathrm{S}}^{0})) }$$

(1.16)

for any i ₁, i ₂, …, i _m  ∈ { 0, 1}, and m = 1, 2, …. Consequently, $\pi _{\omega _{0}}(C_{3}) \subseteq L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}))$ and hence $\pi _{\omega _{0}}(C_{3})'' \subseteq L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}))$. We give the proof of (1.16) only in the special case of the projector P ₀ (m = 1, i ₁ = 0) to avoid the complex notation of the general case. Let {f _n } be the sequence in $C_{0}(\mathcal{C})$ given by

$$\displaystyle{f(\{i_{1},i_{2},\ldots \}) = \left \{\begin{array}{@{}l@{\quad }l@{}} 1 \quad &: i_{1} = 0 \\ \frac{1} {2} +\sum _{ l=1}^{\infty }(-1)^{i_{n}+l} \frac{1} {2^{l+1}}\quad &: i_{1} = 1,\ \sum _{l=2}^{n}i_{ l} = 0 \\ 0 \quad &: i_{1} = 1,\ \sum _{l=2}^{n}i_{ l}\not =0\end{array} \right.,}$$

where $0 \leq f_{n+1} \leq f_{n} \leq 1$ for n = 1, 2, …. As $f_{n} \in C_{0}(\mathcal{C})$, there exists $F_{n} \in C^{{\ast}}({\!1\!\!1},\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}))$ such that $f_{n} = \Phi (F_{n})$ and moreover $0 \leq F_{n+1} \leq F_{n} \leq{\!1\!\!1}$ for all n = 1, 2, …. Hence F _n  → F in the strong operator topology. The map $\Psi$ is normal, hence $1 - \Psi (F) =\sup (1 - f_{n}) = 1 - \Psi (P_{0})$, where we have used μ({1, 0, 0, …}) = 0. Consequently, $P_{0} = F \in L^{\infty }(\pi _{\omega _{0}}(H_{\mathrm{S}}^{0}))$. This ends the proof of P2.

Proof of P3 : :

We start by proving the direction “ ⇐ ”. Using (1.11) we deduce that $\delta _{H_{\mathrm{S}}}(\mathfrak{M}) \subseteq \mathfrak{M}$, so we can introduce another derivation defined by

$$\displaystyle{\delta _{1} \equiv \delta _{H_{\mathrm{S}}} \upharpoonright _{\mathfrak{M}}.}$$

The derivation δ ₁ is inner, i.e., $\delta _{1}(\cdot ) = \mathrm{i}[H_{1},\cdot ]$ for some Hermitian operator $H_{1} \in \mathfrak{M}$ (see, e.g., [24]). In particular, for each spectral projector $P \in L^{\infty }(H_{1})$ we get $P \in \ker L_{\mathrm{D}}$. On the other hand, $[H_{\mathrm{S}},P] = -\mathrm{i}\delta _{1}(P) = [H_{1},P] = 0$, hence $P \in L^{\infty }(H_{\mathrm{S}})'$. Summarizing, we have $P \in \ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})' = \mathbb{C}{\!1\!\!1}$. This means that $H_{1} =\lambda{\!1\!\!1}$, and consequently $[H_{\mathrm{S}},x] = -\mathrm{i}\delta _{1}(x) = [\lambda{\!1\!\!1},x] = 0$ for any $x \in \mathfrak{M}$. We conclude that $\mathfrak{M} \subseteq \ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})' = \mathbb{C}{\!1\!\!1}$.

For the direction “ ⇒ ” notice that for a projector $P \in L^{\infty }(H_{\mathrm{S}})'$ we have [H _S, P] = 0 and hence $L_{\mathrm{D}} \circ \delta _{H_{\mathrm{S}}}^{m} = 0$ for any m = 1, 2, …. This means that $P \in \mathfrak{M}$ if and only if $P \in \ker L_{\mathrm{D}}$. Consequently, if $P \in \ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})'$ then $P \in \mathfrak{M} = \mathbb{C}{\!1\!\!1}$. This in turn means that $P ={\!1\!\!1}$ or P = 0. Hence $\ker L_{\mathrm{D}} \cap L^{\infty }(H_{\mathrm{S}})' = \mathbb{C}{\!1\!\!1}$.

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