# Show the linearity of $(\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U(I_{A}\otimes I_B\otimes |0_{C}\rangle\otimes |0_{D}\rangle)$

Suppose a composite system $$AB$$ initially in an unknown quantum state $$\rho$$ is brought into contact with a composite system $$CD$$ initially in some standard state $$|0\rangle$$, and the two systems interact according to a unitary interaction $$U$$. After the interaction we discard systems $$A$$ and $$D$$, leaving a state $$\rho\:'$$ of the system $$BC$$. Show that the map $$\mathcal{E}(\rho)=\rho\:'$$ satisfies $$\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$$ for some set of linear operators $$E_k$$ from the state space of system $$AB$$ to the state space of system $$BC$$, and such that $$\sum_kE_k^\dagger E_k=I$$.

Overall state of the system after the interaction is $$U(\rho\otimes|0\rangle\langle 0|)U^\dagger$$

Let $$|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$$ be orthonormal basis for the state space of the systems $$A,B,C,D$$, respectively.

Discarding systems $$A$$ and $$D$$, i.e., tracing out system $$A$$ first then tracing out the system $$D$$ obtains

\begin{align} \mathcal{E}(&\rho)=tr_D\Big[tr_A\big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big]\Big]\\ &=tr_D\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big]\\ &=\sum_q(I\otimes I\otimes I\otimes \langle d_q|)\\ &\times\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big](I\otimes I\otimes I\otimes |d_q\rangle)\\ &=\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle) \end{align}

Expanding $$\rho\otimes|0\rangle\langle 0|$$ into the products

\begin{align} \rho\otimes|0\rangle\langle 0|&=\rho_{AB}\otimes|0_{CD}\rangle\langle 0_{CD}|=\color{blue}{(\rho_{AB}\otimes I_{CD})(I_{AB}\otimes|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(\rho_{AB}I_{AB}\otimes I_{CD}|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\rho_{AB}\otimes |0_{CD}\rangle.1)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)(\rho_{AB}\otimes 1)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho_{AB}}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho}(I_{AB}\otimes\langle 0_{CD}|) \end{align} Substituting back into the expression, \begin{align} \mathcal{E}(&\rho)=\\\\ =&\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\rho(I_{AB}\otimes\langle 0_{CD}|)U^\dagger(|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle)\\\\ =&\sum_{m,q} E_{m,q}\rho E_{m,q}^\dagger \end{align} where $$E_{m,q}=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)$$

\begin{align} &\color{red} {\text{How do I show that this is a linear operator from the state space of AB to the state}}\\ &\color{red} {\text{ space of BC ?}} \end{align} My Attempt

If $$|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$$ be orthonormal basis for the state space of the systems $$\mathbb{C}^{i},\mathbb{C}^{j},\mathbb{C}^{k},\mathbb{C}^{l}$$ respectively, then $$|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle$$ is an orthonormal basis for the combined system $$\mathbb{C}^{i}\otimes\mathbb{C}^{j}\otimes\mathbb{C}^{k}\otimes\mathbb{C}^{l}=\mathbb{C}^{ijkl}$$. \begin{align} &\color{gray}{\text{Lemma: If V and W are vector spaces with basis |v_i\rangle and |w_j\rangle respectively. Then }}\\ &\color{gray}{|v_i\rangle\otimes_{outer}|w_j\rangle=|v_i\rangle\langle w_j| \text{form a basis for the vector space V\otimes_{outer}W.}} \end{align} Therefore, $$(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$$ for a basis for the space $$\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$$.

$$\therefore U\in\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$$ can be written as a linear combination of the basis $$(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$$.

i.e., $$U=\sum_{m',n',p',q'}\nu_{m',n',p',q'} (|a_{m'}\rangle\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes|d_{q'}\rangle)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes\langle c_{p''}|\otimes\langle d_{q''}|)\\ =\sum_{m',n',p',q'}\nu_{m',n',p',q'} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|$$

And $$|0_{CD}\rangle$$ being a standard state of the system $$CD$$ can be represented in terms of the orthonormal basis vectors as, $$|0_{CD}\rangle=\sum_{p,q}\eta_{p,q} |c_p\rangle\otimes|d_q\rangle$$

Substituting back into the expression for $$E_{m,q}$$ obtains \begin{align} &E_{m,q}=\\ &=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &\times\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &\times(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ \\ &=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \langle a_{m}|a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|c_{p'''}\rangle\otimes\langle d_{q}|d_{q'}\rangle\langle d_{q''}|d_{q'''}\rangle\\ \\ &=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \delta_{m,m'}\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\delta_{p'',p'''}\otimes\delta_{q,q'}\delta_{q'',q'''}\\ \\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} \langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\otimes 1\\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1.\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle.1\otimes 1.1)\\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes 1)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes 1\otimes 1)\\\end{align}

$$\implies E_{m,q}$$ is a sum of operators from $$AB$$ to $$BC$$.

If I understand correctly, you want to show that $$E_{m,q}=(\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U(I_{A}\otimes I_B\otimes |0_{C}\rangle\otimes |0_{D}\rangle)$$ is a linear operator from $$AB$$ to $$CD$$.
For this, you can just expand the identity operators in the expression. E.g. $$I_B=\sum_i |i_B\rangle\!\langle i_B|$$ for any orthonormal basis $$\{|i_B\rangle\}_i$$ for the space $$B$$, and same for the other identities. You'll end up with an expression of the form $$E_{m,q} = \sum_{ijk\ell} c_{ijk\ell} (|i_B\rangle\otimes|j_C\rangle)(\langle k_A|\otimes \langle\ell_B|)$$ for some set of numbers $$c_{ijk\ell}$$. This is a linear operator between the relevant spaces.