11.5: Schrödinger's cat, Wigner's friend and Laplace's demon

Contents:
11.5.1: Schrödinger's cat and Wigner's friend

Let us explain Schrödinger's cat paradox in the Schrödinger picture.

Problem 11.14 [Schrödinger's cat]

 $(a):$ Suppose we put a cat in a cage with a radioactive atom, a Geiger counter, and a poison gas bottle; further suppose that the atom in the cage has a half-life of one hour, a fifty-fifty chance of decaying within the hour. If the atom decays, the Geiger counter will tick; the triggering of the counter will get the lid off the poison gas bottle, which will kill the cat. If the atom does not decay, none of the above things happen, and the cat will be alive.

Here, we have the following question:
 $(b):$ Asuume that, after one hour, you look at the inside of the box. Then, do you know whether then the cat is dead or alive after one hour, (= $60^{60}$ seconds ) ? Of course, we say that it is half-and-half whether the cat is alive. However, our problem is Clarify the meaning of "half-and-half"

$\fbox{Note 11.1}$ [Wigner's friend]: Instead of the above (b), we consider as follows.
 $(b)':$ after one hour, Wigner's friend look at the inside of the box, and thus, he know whether the cat is dead or alive after one hour, (= $60^{60}$ seconds ). And further, after two hours, Wigner's friend informs you the fact. How is the cat?
This problem is not difficult. That is because the linguistic interpretation says that "the moment you measured" is out of the description of quantum language. Recall the spirit of the linguistic world-view (i.e., Wittgenstein's words) such as \begin{align} \color{magenta}{\mbox{ The limits of my language mean the limits of my world}} \end{align} and \begin{align} \color{magenta}{\mbox{ What we cannot speak about we must pass over in silence. }} \end{align}

11.5.2: The usual answer without quantum language

Put ${\mathbf q}=(q_{11}, q_{12}, q_{13}, q_{21},q_{22}, q_{23}, \dots, q_{n1},q_{n2}, q_{n3} ) \in {\mathbb R}^{3n}$. And put

\begin{align} \nabla_i^2 = \frac{\partial^2}{\partial q_{i1}^2} + \frac{\partial^2}{\partial q_{i2}^2} + \frac{\partial^2}{\partial q_{i3}^2} \end{align} Consider the quantum system basic structure: \begin{align} [{\mathcal C}(H) \subseteq B(H) \subseteq B(H)] \qquad \mbox{ ( where,$H=L^2 ( {\mathbb R}^{3n}, d {\mathbf q} )$ )} \end{align}

And consider the Schrödinger equation (concerning $n$-particles system):

\begin{align} \left\{\begin{array}{ll} i \hbar \frac{\partial}{\partial t} \psi ( {\mathbf q}, t ) = \Big[ \sum_{i=1}^n \frac{- \hbar^2}{2m_i} \nabla_i^2 + V( {\mathbf q}, t ) \Big] \psi ( {\mathbf q}, t ) \\ \\ \psi_0({\mathbf q})=\psi ( {\mathbf q}, 0 ): \mbox{initial condition} \end{array}\right. \tag{11.24} \end{align}

where,$m_i$ is the mass of a particle $P_i$, $V$ is a potential energy.

If we believe in quantum mechanics, it suffices to solve this Schrödinger equation (11.24). That is,

 $(A_1):$ Assume that the wave function $\psi( \cdot, 60^2 )=U_{0,60^2} \psi_0$ after one hour (i.e., $60^2$ seconds) is calculated. Then, the state $\rho_{60^2}$ $(\in {\mathcal Tr}_{+1}^p (H) )$ after $60^2$ seconds is represented by \begin{align} \rho_{60^2}= | \psi_{60^2} \rangle \langle \psi_{60^2} | \tag{11.25} \end{align} (where, $\psi_{60^2} = \psi(\cdot , 60^2 )$).

Now, define the observable ${\mathsf O}= (X=\{\mbox{life}, \mbox{death}\}, 2^X, F)$ in $B(H)$ as follows.

 $(A_2):$ that is,putting \begin{align} & V_{\mbox{life}} (\subseteq H ) = \Big\{ u\in H \;|\; "\mbox{ the state } \frac{|u\rangle \langle u |}{\|u\|^2}" \Leftrightarrow \mbox{"cat is alive"} \Big\} \\ & V_{\mbox{death}} (\subseteq H )=\mbox{the orthogonal complement space of }V_{\mbox{life}} \\ & \qquad \qquad \qquad = \{ u \in H \;|\; \langle u, v \rangle = 0 \;\;(\forall v \in V_{\mbox{life}} ) \} \end{align} define $F(\{\mbox{life}\})(\in B(H) )$ is the projection of the closed subspace $V_{\mbox{life}}$ and $F(\{\mbox{death}\})=I - F(\{\mbox{life}\})$,
Here,

 $(A_3):$ consider the measurement ${\mathsf M}_{B(H)}({\mathsf O}=(X, 2^X, F), S_[\rho_{60^2]})$. The probability that a measured value $\left[\begin{array}{ll} \mbox{life} \\ \mbox{death} \end{array}\right]$ is obtained is given by \begin{align} \left[\begin{array}{ll} {}_{{\mathcal Tr}(H)} \Big(\rho_{60^2}, F(\{\mbox{life}\}) \Big){}_{B(H)}=\langle \psi_{60^2},F(\{\mbox{life}\})\psi_{60^2} \rangle=0.5 \\ {}_{{\mathcal Tr}(H)} \Big(\rho_{60^2}, F(\{\mbox{death}\}) \Big){}_{B(H)}=\langle \psi_{60^2},F(\{\mbox{death}\})\psi_{60^2} \rangle=0.5 \end{array}\right] \end{align}

Therefore,we can assure that \begin{align} & \psi_{60^2}= \frac{1}{\sqrt 2}( \psi_{\mbox{life}}+ \psi_{\mbox{death}}) \tag{11.26} \\ (\mbox{where, } & \psi_{\mbox{life}}\in V_{\mbox{life}},\|\psi_{\mbox{life}}\|=1 \quad \psi_{\mbox{death}}\in V_{\mbox{death}},\|\psi_{\mbox{death}}\|=1 ) \nonumber \end{align}

Hence, we can conclude that
 $(A_4):$ the state (or, wave function) of the cat (after one hour ) is represented by (11.26), that is, $\frac{ \mbox{"Fig.$(\sharp_1)$"} + \mbox{ "Fig. ($\sharp_2$)"} }{\sqrt{2}}$

And,
 $(A_5):$ After one hour (i.e, to the moment of opening a window), It is decided "the cat is dead" or "the cat is vigorously alive." That is, \begin{align} & \mbox{"half-dead"} \Big(=\frac{1}{2}(|\psi_{\mbox{life}}+ \psi_{\mbox{death}}\rangle \langle \psi_{\mbox{life}}+ \psi_{\mbox{death}}|)\Big) \\ & \xrightarrow[{\mbox{ the collapse of wave function}}]{{ \mbox{ to the moment of opening a window }}} \left\{\begin{array}{ll} \mbox{"alive"} (=|\psi_{\mbox{life}}\rangle \langle \psi_{\mbox{life}}|) \\ \\ \mbox{"dead"} (=|\psi_{\mbox{death}}\rangle \langle \psi_{\mbox{death}}|) \end{array}\right. \end{align}
$\square \quad$
In quantum language, the quantum decoherence is permitted. That is, we can assume that
 $(B_1):$ the state $\rho'_{60^2}$ after one hour is represented by the following mixed state \begin{align} \rho'_{60^2} = \frac{1}{2} \Big( | \psi_{\mbox{life}} \rangle \langle \psi_{\mbox{life}} | + | \psi_{\mbox{death}} \rangle \langle \psi_{\mbox{death}} | \Big) \end{align} That is, we can assume the decoherent causal operator $\Phi_{0, 60^2}: B(H) \to B(H)$ such that \begin{align} (\Phi_{0, 60^2})_* ( \rho_0) = \rho'_{60^2} \end{align}
Here,consider the measurement ${\mathsf M}_{B(H)}({\mathsf O}=(X, 2^X, F), S_[\rho'_{60^2}])$, or, its Heisenberg picture ${\mathsf M}_{B(H)}(\Phi_{0, 60^2}{\mathsf O}=(X, 2^X, \Phi_{0, 60^2}F), S_[\rho'_{0}])$. Of course we see:
 $(B_2):$ The probability that a measured value $\left[\begin{array}{ll} \mbox{life} \\ \mbox{death} \end{array}\right]$ is obtained by the measurement \\ ${\mathsf M}_{B(H)}(\Phi_{0, 60^2}{\mathsf O}=(X, 2^X, \Phi_{0, 60^2}F), S_[\rho'_{0}])$ is given by \begin{align} \left[\begin{array}{ll} {}_{{\mathcal Tr}(H)} \Big(\rho_{0}, \Phi_{0, 60^2}F(\{\mbox{life}\}) \Big){}_{B(H)}=\langle \psi'_{60^2},F(\{\mbox{life}\})\psi_{60^2} \rangle=0.5 \\ {}_{{\mathcal Tr}(H)} \Big(\rho_{0}, \Phi_{0, 60^2}F(\{\mbox{death}\}) \Big){}_{B(H)}=\langle \psi'_{60^2},F(\{\mbox{death}\})\psi_{60^2} \rangle=0.5 \end{array}\right] \end{align}

Also,"the moment of measuring" and "the collapse of wave function" are prohibited in the linguistic interpretation, but the statement (B$_2$) is within quantum language.
$\square \quad$

11.5.4: Summary (Laplace's demon)
Summary 11.17 [Schrödinger's cat in quantum language]
Here, let us examine
Answer11.15 :(A$_5$) v.s. Answer11.16 :(B$_2$)
 $(C_1):$ the answer (A$_5$) may be unnatural, but it is an argument which cannot be confuted,
On the other hand,
 $(C_2):$ the answer (B$_2$) is natural. but the non-deterministic time evolution is used.
Since the non-deterministic causal operator (i.e., quantum decoherence) is permitted in quantum language, we conclude that
 $(C_3):$ Answer11.16:(B$_2$) is superior to Answer11.15:(A$_1$)

For the reason that the non-deterministic causal operator (i.e., quantum decoherence) is permitted in quantum language, we add the following.
 $\bullet$ If Newtonian mechanics is applied to the whole universe, Laplace's demon appears. Also, if Newtonian mechanics is applied to the microworld, chaos appears. This kind of supremacy of physics is not natural, and thus, we consider that these are out of "the limit of Newtonian mechanics"
And,
 $\bullet$ when we want to apply Newton mechanics to phenomena out of "the limit of Newtonian mechanics", we often use the stochastic differential equation (and Brownian motion). This approach is called "dynamical system theory", which is not physics but metaphysics. \begin{align} & \underset{\mbox{ physics}}{ \fbox{Newtonian mechanics} } \xrightarrow[\mbox{linguistic turn}]{\mbox{out of the limits}} \underset{\mbox{ metaphysics}}{ \fbox{dynamical system theory; statistics} } \\ &{} \end{align}
In the same sense, we consider that quantum mechanics has "the limit".That is,
 $\bullet$ Schrödinger's cat is out of quantum mechanics.
And thus,
 $\bullet$ When we want to apply quantum mechanics to phenomena out of "the limit of quantum mechanics", we often use the quantum decoherence. Although this approach is not physics but metaphysics, it is quite powerful. \begin{align} \underset{\mbox{ physics}}{ \fbox{quantum mechanics} } \xrightarrow[\mbox{linguistic turn}]{\mbox{out of the limits}} \underset{\mbox{ metaphysics}}{ \fbox{quantum language} } \end{align}

 $\fbox{Note 11.2}$ If we know the present state of the universe and the kinetic equation (=the theory of everything), and if we calculate it, we can know everything (from past to future). There may be a reason to believe this idea. This intellect is often referred to as Laplace's demon. Laplace's demon is sometimes discussed as the super realistic-view (i.e., the realistic-view over which the degree passed). Thus, we consider the following correspondence: \begin{align} \underset{\mbox{ physics}}{ \fbox{Newtonian mechanics} } \xrightarrow[\mbox{super realistic-view}]{\mbox{out of the limits}} \underset{\mbox{ physics?}}{ \fbox{Laplace's Demon} } \end{align} This should be compared with the formula (11.19).