11.7: quantum eraser experiment

Let us explain quantum eraser experiment. This section is extracted from

 $(\sharp):$ S. Ishikawa, The double-slit quantum eraser experiments and Hardy's paradox in the quantum linguistic interpretation, arxiv:1407.5143[quantum-ph],( 2014)

11.7.1: Tensor Hilbert space

Let ${\mathbb C}^2$ be the two dimensional Hilbert space, i,e., ${\mathbb C}^2$ $=$ $\Big\{ \left[\begin{array}{l} z_1 \\ z_2 \\ \end{array}\right] \;| \; z_1, z_2 \in {\mathbb C} \Big\}$. And put

\begin{align} e_1 =\left[\begin{array}{l} 1 \\ 0 \\ \end{array}\right], \qquad e_2 =\left[\begin{array}{l} 0 \\ 1 \\ \end{array}\right] \end{align}

Here, define the {observable ${\mathsf O}_x=(\{-1, 1 \}, 2^{\{-1, 1 \}}, F_x )$ in $B({\mathbb C}^2)$} such that

\begin{align} F_x(\{ 1 \} ) = \frac{1}{2} \left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \\ \end{array}\right], \quad F_x(\{ -1 \} ) = \frac{1}{2} \left[\begin{array}{ll} 1 & -1 \\ -1 & 1 \\ \end{array}\right], \end{align}

Here, note that

\begin{align} & F_x(\{ 1 \} )e_1 = \frac{1}{2}(e_1 + e_2 ), \quad F_x(\{ 1 \} )e_2 = \frac{1}{2}(e_1 + e_2 ) \\ & F_x(\{ -1 \} )e_1 = \frac{1}{2}(e_1 - e_2 ), \quad F_x(\{ -1 \} )e_2 = \frac{1}{2}(-e_1 + e_2 ) \end{align}

Let $H$ be a Hilbert space such that $L^2({\mathbb R})$. And {let ${\mathsf O}=(X, {\mathcal F}, F )$ be an observable in $B(H)$}. For example, consider the position observable, that is, $X={\mathbb R}$, ${\mathcal F}={\mathcal B}_{{\mathbb R}}$, and

\begin{align} [F(\Xi)](q)= \left\{\begin{array}{ll} 1 \quad & ( q \in \Xi \in {\mathcal F} ) \\ 0 \quad & ( q \notin \Xi \in {\mathcal F} ) \end{array}\right. \end{align}

Let $u_1$ and $u_2$ $( \in H)$ be orthonormal elements, i.e., $\| u_1 \|_H=\| u_2 \|_H=1$ and $\langle u_1 , u_2 \rangle =0$. Put

\begin{align} u= \alpha_1 u_1 + \alpha_2 u_2 \end{align}

where $\alpha_i \in {\mathbb C}$ such that $| \alpha_1|^2 + |\alpha_2|^2=1$. Furthermore, define $\psi \in {\mathbb C}^2 \otimes H$ $($ the tensor Hilbert space of ${\mathbb C}^2$ and $H$) such that

\begin{align} \psi = \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2 \end{align}

where $\alpha_i \in {\mathbb C}$ such that $| \alpha_1|^2 + |\alpha_2|^2=1$.

11.7.2: Interference

Consider the measurement: \begin{align} {\mathsf M}_{B({\mathbb C}^2 \otimes H )} ({\mathsf O}_x \otimes {\mathsf O} , S_{[| \psi \rangle \langle \psi |]} ) \tag{11.27} \end{align} Then, we see:
 $(A_1):$ the probability that a measured value $(1, x ) (\in \{-1,1\} \times X )$ belongs to $\{1\} \times \Xi$ is given by
\begin{align} & \langle \psi, (F_x(\{ 1 \}) \otimes F(\Xi) ) \psi \rangle \\ = & \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, (F_x(\{ 1 \} \otimes F(\Xi) )) ( \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2 ) \rangle \\ = & \frac{1}{2} \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, \alpha_1 (e_1+e_2) \otimes F(\Xi) u_1 + \alpha_2 (e_1+ e_2) \otimes F(\Xi) u_2 \rangle \\ = & \frac{1}{2} \Big( |\alpha_1|^2 \langle u_1 , F(\Xi)u_1 \rangle + |\alpha_2|^2 \langle u_2 , F(\Xi)u_2 \rangle + \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle + {\alpha}_1 \overline{\alpha}_2 \langle u_2 , F(\Xi)u_1 \rangle \Big) \\ = & \frac{1}{2} \Big( |\alpha_1|^2 \langle u_1 , F(\Xi)u_1 \rangle + |\alpha_2|^2 \langle u_2 , F(\Xi)u_2 \rangle + 2 \mbox{[Real part]} ( \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle ) \Big) \end{align}

where the interference term (i.e., the third term) appears.
Define the probability density function $p_1$ by

\begin{align} \int_\Xi p_1 (q) dq =\frac{\langle \psi, (F_x(\{ 1 \}) \otimes F(\Xi) ) \psi \rangle}{\langle \psi, (F_x(\{ 1 \}) \otimes I ) \psi \rangle} \quad (\forall \Xi \in {\mathcal F} ) \end{align}

Then, by the interference term (i.e., $2 \mbox{[Real part]} ( \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle )$ ), we get the following graph.

Also, we see:

 $(A_2):$ the probability that a measured value $(-1, x ) (\in \{-1,1\} \times X )$ belongs to $\{-1\} \times \Xi$ is given by
\begin{align} & \langle \psi, (F_x(\{ -1 \}) \otimes F(\Xi) ) \psi \rangle \\ = & \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, (F_x(\{ -1 \} \otimes F(\Xi) )) ( \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2 ) \rangle \\ = & \frac{1}{2} \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, \alpha_1 (e_1-e_2) \otimes F(\Xi) u_1 + \alpha_2 (-e_1+ e_2) \otimes F(\Xi) u_2 \rangle \\ = & \frac{1}{2} \Big( |\alpha_1|^2 \langle u_1 , F(\Xi)u_1 \rangle + |\alpha_2|^2 \langle u_2 , F(\Xi)u_2 \rangle - \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle - {\alpha}_1 \overline{\alpha}_2 \langle u_2 , F(\Xi)u_1 \rangle \Big) \\ = & \frac{1}{2} \Big( |\alpha_1|^2 \langle u_1 , F(\Xi)u_1 \rangle + |\alpha_2|^2 \langle u_2 , F(\Xi)u_2 \rangle - 2 \mbox{[Real part]} ( \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle ) \Big) \end{align}

where the interference term (i.e., the third term) appears. Define the probability density function $p_2$ by

\begin{align} \int_\Xi p_2 (q) dq =\frac{\langle \psi, (F_x(\{- 1 \}) \otimes F(\Xi) ) \psi \rangle}{\langle \psi, (F_x(\{ -1 \}) \otimes I ) \psi \rangle} \quad (\forall \Xi \in {\mathcal F} ) \end{align} Then, by the interference term (i.e., $-2 \mbox{[Real part]} ( \overline{\alpha}_1\alpha_2 \langle u_1 , F(\Xi)u_2 \rangle )$ ), we get the following graph.
11.7.3: No interference

Consider the measurement: \begin{align} {\mathsf M}_{B({\mathbb C}^2 \otimes H )} ({\mathsf O}_x \otimes {\mathsf O} , S_{[| \psi \rangle \langle \psi |]} ) \tag{11.29} \end{align} Then, we see
 $(A_3):$ the probability that a measured value $(u, x ) (\in \{1, -1\} \times X )$ belongs to $\{1,-1\} \times \Xi$ is given by
\begin{align} & \langle \psi, (I \otimes F(\Xi) ) \psi \rangle \\ = & \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, (I \otimes F(\Xi) ) ( \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2 ) \rangle \\ = & \langle \alpha_1 e_1 \otimes u_1 + \alpha_2 e_2 \otimes u_2, \alpha_1 e_1 \otimes F(\Xi) u_1 + \alpha_2 e_2 \otimes F(\Xi) u_2 \rangle \\ = & |\alpha_1|^2 \langle u_1 , F(\Xi)u_1 \rangle + |\alpha_2|^2 \langle u_2 , F(\Xi)u_2 \rangle \end{align} where the interference term disappears.
Define the probability density function $p_3$ by \begin{align} \int_\Xi p_3 (q) dq = \langle \psi, (I \otimes F(\Xi) ) \psi \rangle \quad (\forall \Xi \in {\mathcal F} ) \end{align}

Since there is no interference term, we get the following graph.

Remark 11.17

Note that \begin{align} \underset{\mbox{no interference}}{\fbox{(A$_3$)}} = \underset{\mbox{interferences are canceled}}{\fbox{(A$_1$)+(A$_2$)}} \end{align}