Let us consider a entangled state for the signal-idler system in its Schmidt decomposition form \begin{align} \vert \psi \rangle_{SI} = \sum_{\alpha} \sqrt{p_{\alpha}} \vert w_{\alpha} \rangle_S \vert v_{\alpha} \rangle_I, \end{align} where $\langle v_{i} \vert v_j \rangle = \langle w_i \vert w_j \rangle = \delta_{ij}$. How to prove the following equality? \begin{align} {\rm Tr}_S [s^\dagger b- s b^\dagger, \vert \psi \rangle \langle \psi \vert \otimes \rho_B] &= \sum_{ij} \sqrt{p_i p_j} \vert v_i \rangle \langle v_j \vert \otimes [\langle w_j \vert s^\dagger \vert w_i \rangle b - \langle w_j \vert s \vert w_i \rangle b^\dagger, \rho_B], \end{align} where $s^\dagger$ is the creation operator for $\vert \psi \rangle_{SI}$ and $b^\dagger$ is the creation operator for $\rho_B$.
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$\begingroup$ I feel like to answer this you need to specify the boson-number decomposition of the states $|w_\alpha\rangle_S$. Otherwise how would you know the result of multiplying by $s,s^\dagger$ and tracing? $\endgroup$– glS ♦Jul 30 at 22:56
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$\begingroup$ @glS I don't think so: the $s$ and $s^\dagger$ operators are left uncalculated at the end. They are only supposed to apply to the system S, not I. $\endgroup$– Quantum MechanicAug 1 at 13:13
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$\begingroup$ @QuantumMechanic far enough; I was thinking of finding a more explicit expression but the question here was in fact different $\endgroup$– glS ♦Aug 1 at 18:45
1 Answer
This can be computed directly by using the partial trace $Tr_S(\rho)=\sum_j\langle \omega_j|\rho|\omega_j\rangle$. I will drop the subscripts S and I and use the labels $\omega$ and $v$ to specify the appropriate state.
\begin{align} {\rm Tr}_S [s^\dagger b- s b^\dagger, \vert \psi \rangle \langle \psi \vert \otimes \rho_B] &= \sum_{j}\langle \omega_j|[(s^\dagger b- s b^\dagger)\vert \psi \rangle \langle \psi \vert \otimes \rho_B-\vert \psi \rangle \langle \psi \vert \otimes \rho_B(s^\dagger b- s b^\dagger)]|\omega_j\rangle\\ &=\sum_{j}\langle \omega_j|(s^\dagger b- s b^\dagger)\vert \psi \rangle \langle v_j \vert \sqrt{p_j}\otimes \rho_B-\sqrt{p_j}\vert v_j \rangle \langle \psi \vert \otimes \rho_B(s^\dagger b- s b^\dagger)|\omega_j\rangle\\ &=\sum_{ij} (\langle\omega_j|s^\dagger|\omega_i\rangle b- \langle\omega_j|s|\omega_i\rangle b^\dagger)\sqrt{p_i}|v_i\rangle \langle v_j \vert \sqrt{p_j}\otimes \rho_B-\sqrt{p_j}\vert v_j \rangle \langle v_i| \sqrt{p_i} \otimes \rho_B(\langle\omega_i|s^\dagger|\omega_j\rangle b- \langle\omega_i|s|\omega_j\rangle b^\dagger)\\ &=\sum_{ij}\sqrt{p_i p_j}|v_i\rangle \langle v_j \vert \otimes[(\langle\omega_j|s^\dagger|\omega_i\rangle b- \langle\omega_j|s|\omega_i\rangle b^\dagger) \rho_B - \rho_B(\langle\omega_j|s^\dagger|\omega_i\rangle b- \langle\omega_j|s|\omega_i\rangle b^\dagger)]\\ &=\sum_{ij}\sqrt{p_i p_j}|v_i\rangle \langle v_j \vert \otimes[(\langle\omega_j|s^\dagger|\omega_i\rangle b- \langle\omega_j|s|\omega_i\rangle b^\dagger), \rho_B]. \end{align} The only trick was to collect terms by swapping the indices $i\leftrightarrow j$ in the final terms for the second-last equality.
