11.5: Schrödinger's cat, Wigner's friend and Laplace's demon
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 halflife of one hour, a fiftyfifty 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. 
$(b):$ 
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 halfandhalf whether the cat is alive. However, our problem is Clarify the meaning of "halfandhalf" 
$\fbox{Note 11.1}$ 
[Wigner's friend]:
Instead of the above (b), we consider as follows.

11.5.2: The usual answer without quantum language
Answer 11.15 [The ordinary answer to Problem11.14(i.e.,
the 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
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 $F(\{\mbox{death}\})=I  F(\{\mbox{life}\})$, 
$(A_3):$  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,

$(A_5):$  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{"halfdead"} \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} 
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} 
$(B_2):$  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.
11.5.4: Summary (Laplace's demon)
Here, let us examine
$(C_1):$  the answer (A$_5$) may be unnatural, but it is an argument which cannot be confuted, 
$(C_2):$  the answer (B$_2$) is natural. but the nondeterministic time evolution is used. 
$(C_3):$ 

For the reason that the nondeterministic 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" 
$\bullet$  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} 
$\bullet$  Schrödinger's cat is out of quantum mechanics. 
$\bullet$  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 realisticview (i.e.,
the realisticview
over which the degree passed).
