11.6: "Whereof one cannot speak, thereof one must be silent"
This section is extracted from
$(\sharp):$  S. Ishikawa, The doubleslit quantum eraser experiments and Hardy's paradox in the quantum linguistic interpretation, arxiv:1407.5143[quantumph],( 2014) 
In the conventional quantum mechanics, the question: "particle or wave?" may frequently appear. However, this is a foolish question. On the other hand, the argument about the "particle vs. wave" is clear in quantum language. As seen in the following table, this argument is traditional:
In the table 11.1, Newtonian mechanics (i.e., mass point $\leftrightarrow$ state) may be easiest to understand. Thus, "particle" and "wave" are not confrontation concepts.
Concerning "particle or wave", we have the following statements:
$(A_1):$  "Particle or wave" is a foolish question. 
$(A_2):$  Wheeler's delayed choice experiment is related to the question "particle or wave" 
$(A_3):$  How is Wheeler's delayed choice experiment described in terms of quantum mechanics? 
11.6.2: Preparation
Let us start from the review of $\S$2.10 (de Broglie paradox in $B({\mathbb C}^2)$) Let $H$ be a two dimensional Hilbert space, i.e., $H={\mathbb C}^2$. Consider the basic structure
\begin{align} [B({\mathbb C}^2)\subseteq B({\mathbb C}^2 ) \subseteq B({\mathbb C}^2 )] \end{align}Let $f_1, f_2 \in H$ such that
\begin{align} f_1 = \left[\begin{array}{l} 1 \\ 0 \end{array}\right], \qquad f_2 = \left[\begin{array}{l} 0 \\ 1 \end{array}\right] \end{align} Put \begin{align} u=\frac{f_1 +f_2}{{\sqrt 2}} \end{align}Thus, we have the state $\rho = u \rangle \langle u $ $(\in {\frak S}^p(B({\mathbb C}^2)))$. Let $U (\in B({\mathbb C}^2 ))$ be an unitary operator such that
\begin{align} U = \left[\begin{array}{ll} 1 & 0 \\ 0 & e^{i\pi/2 } \end{array}\right] \quad \end{align} and let $\Phi: B({\mathbb C}^2) \to B({\mathbb C}^2) $ be the homomorphism such that \begin{align} \Phi(F) = U^* F U \qquad (\forall F \in B({\mathbb C}^2) ) \end{align}Consider two observable ${\mathsf O}_f=(\{1,2\}, 2^{\{1,2\}}, F)$ and ${\mathsf O}_g=(\{1,2\}, 2^{\{1,2\}}, G)$ in $B({\mathbb C}^2 )$ such that
\begin{align} & F(\{1\}) = f_1 \rangle \langle f_1  , \quad F(\{ 2 \}) = f_2 \rangle \langle f_2  \end{align} \begin{align} & G(\{1\}) = g_1 \rangle \langle g_1  , \quad G(\{ 2 \}) = g_2 \rangle \langle g_2  \end{align} where \begin{align} & g_1 = \frac{f_1  f_2 }{\sqrt{2}}, \qquad g_2 = \frac{f_1 + f_2 }{\sqrt{2}} \end{align}11.6.3: de Broglie's paradox in $B({\mathbb C}^2)$ (No interference)
Now we will explain, by the Schrödinger picture, Figure 11.4(1) as follows.
The photon P with the state $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ ( precisely, $\rho= u \rangle \langle u $ ) rushed into the halfmirror 1,
$(B_1):$  the $f_1$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ passes through the halfmirror 1, and goes along the course 1. And it is reflected in the mirror 1, and goes to the photon detector $D_1$. 
$(B_2):$  the $f_2$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ rebounds on the halfmirror 1 (and strictly saying, the $f_2$ changes to ${\sqrt{1}}f_2$, we are not concerned with it ), and goes along the course 2. And it is reflected in the mirror 2, and goes to the photon detector $D_2$. 
This is, by the Heisenberg picture, represented by the following measurement:
\begin{align} {\mathsf M}_{B({\mathbb C}^2)} (\Phi {\mathsf O}_f, S_{[\rho]} ) \tag{11.27} \end{align} Then, we see:$(C):$  the probability that $\left[\begin{array}{l} \mbox{ a measured value }1 \\ \mbox{ a measured value }2 \end{array}\right]$ is obtained by $ {\mathsf M}_{B({\mathbb C}^2)} (\Phi {\mathsf O}_f, S_{[\rho]} ) $ is given by \begin{align} \left[\begin{array}{l} \langle Uu, F(\{1\})Uu \rangle \\ \langle Uu, F(\{2\}) Uu \rangle \end{array}\right] = \left[\begin{array}{l}  \langle Uu, f_1 \rangle^2 \\  \langle Uu, f_2 \rangle^2 \end{array}\right] = \left[\begin{array}{l} \frac{1}{2} \\ \frac{1}{2} \end{array}\right] \tag{11.28} \end{align} 
By the analogy of Section 11.2 ( The projection postulate ), Figure 11.4(1) is also described as follows. That is, putting $e_1= \begin{bmatrix} 1 \\ 0 \end{bmatrix} $ and $e_2= \begin{bmatrix} 0 \\ 1 \end{bmatrix} $ $( \in {\mathbb C}^2 )$, we have the observable ${\mathsf O}_E=(\{1,2\},2^{\{1,2\}}, E)$ in $B({\mathbb C}^2)$ such that $E( \{1 \})=e_1 \rangle \langle e_1$ and $E( \{1 \})=e_1 \rangle \langle e_1$. Hence,
11.6.4: MachZehnder interferometer (Interference)
Next, consider the following figure:
Now we will explain, by the Schrödinger picture, {Figure 11.4(2)} as follows. The photon P with the state $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ ( precisely, $\rho= u \rangle \langle u $ ) rushed into the halfmirror 1,
$(D1):$  the $f_1$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ passes through the halfmirror 1, and goes along the course 1. And it is reflected in the mirror 1, and passes through the halfmirror 2, and goes to the photon detector $D_1$. 
$(D2):$  the $f_2$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ rebounds on the halfmirror 1 (and strictly saying, the $f_2$ changes to ${\sqrt{1}}f_2$, we are not concerned with it ), and goes along the course 2. And it is reflected in the mirror 2, and further reflected in the halfmirror 2, and goes to the photon detector $D_2$. 
This is, by the Heisenberg picture, represented by the following measurement: $ {\mathsf M}_{B({\mathbb C}^2)} (\Phi^2 {\mathsf O}_g, S_{[\rho]} ) $ Then, we see:
$(E):$  the probability that $\left[\begin{array}{l} \mbox{ a measured value }1 \\ \mbox{ a measured value }2 \end{array}\right]$ is obtained by $ {\mathsf M}_{B({\mathbb C}^2)} (\Phi^2 {\mathsf O}_g, S_{[\rho]} ) $ is given by \begin{align} \left[\begin{array}{l} \langle u, \Phi^2 G(\{1\})u \rangle \\ \langle u, \Phi^2 G(\{2\}) u \rangle \end{array}\right] =& \left[\begin{array}{ll}  \langle u, UUg_1\rangle ^2 \\  \langle u, UU g_2 \rangle^2 \end{array}\right] = \left[\begin{array}{l} 1 \\ 0 \end{array}\right] \tag{11.29} \end{align} 
11.6.5: Another case
Consider the following Figure 11.4(3).
Now we will explain, by the Schrödinger picture, Figure 11.4(3) as follows.
The photon P with the state $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ ( precisely, $\rho= u \rangle \langle u $ ) rushed into the halfmirror 1,
$(F_1):$  the $f_1$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ passes through the halfmirror 1, and goes along the course 1. And it reaches to the photon detector $D_1$. 
$(F_2):$  the $f_2$ part in $u=\frac{1}{\sqrt{2}}( f_1+f_2 )$ rebounds on the halfmirror 1 (and strictly saying, the $f_2$ changes to ${\sqrt{1}}f_2$, we are not concerned with it ), and goes along the course 2. And it is again reflected in the mirror 1, and further reflected in the halfmirror 2, and goes to the photon detector $D_2$. 
This is, by the Heisenberg picture, represented by the following measurement:
\begin{align} {\mathsf M}_{B({\mathbb C}^2)} (\Phi^2 {\mathsf O}_f, S_{[\rho]} ) \tag{11.30} \end{align}Therefore,we see the following:
$(F_3):$ 
The probability that
$\left[\begin{array}{l} \mbox{measured value }1 \\ \mbox{measured value }2 \end{array}\right]$
is obtained by
the measurement
$
{\mathsf M}_{B({\mathbb C}^2)} (\Phi^2 {\mathsf O}_f, S_{[\rho]} )
$
is given by
\begin{align}
\left[\begin{array}{l}
\mbox{Tr}(\rho \cdot \Phi^2 F(\{1\}) )
\\
\mbox{Tr}(\rho \cdot \Phi^2 F(\{2\}) )
\end{array}\right]
=
\left[\begin{array}{l}
\langle UUu, F(\{1\})UUu \rangle
\\
\langle UUu, F(\{2\}) UUu \rangle
\end{array}\right]
=
\left[\begin{array}{l}
 \langle UUu, f_1 \rangle^2
\\
 \langle UUu, f_2 \rangle^2
\end{array}\right]
=
\left[\begin{array}{l}
\frac{1}{2}
\\
\frac{1}{2}
\end{array}\right]
\end{align}
Therefore,if the photon detector $D_1$ does not react, it is expected that the photon detector $D_2$ reacts. 
The above argument is just Wheeler's delayed choice experiment. It should be noted that the difference among Examples in $\S$11.6.3 (Figure 11.4(1)) $\S$11.6 (Figure 11.4(3)) is that of the observables (= measuring instrument ). That is,
$\quad$  $ \left\{\begin{array}{ll} \mbox{ $\S$11.6.3 (Figure 11.4(1)) } \quad & \xrightarrow[\mbox{ Heisenberg picture}]{} \Phi {\mathsf O}_f \\ \mbox{ $\S$11.6.4 (Figure 11.4(2)) } \quad & \xrightarrow[\mbox{ Heisenberg picture}]{} \Phi^2 {\mathsf O}_g \\ \mbox{ $\S$11.6.5 (Figure 11.4(3)) } \quad & \xrightarrow[\mbox{ Heisenberg picture}]{} \Phi^2 {\mathsf O}_f \end{array}\right. $ 
$(H):$ 
Wheeler's delayed choice experiment
can not be described paradoxically in quantum language.
Thus, it should be recalled Wittgenstein's famous words:
"Whereof one cannot speak, thereof one must be silent" 
However, it should be noted that the nonlocality paradox (i.e., "there is some thing faster than light") is not solved even in quantum language.
$\fbox{Note 11.3}$ 
What we want to assert in this book may be the following:
