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I've tried two quantum computing textbooks "QUANTUM COMPUTING From Linear Algebra to Physical Realizations" and "quantum information and quanutum computing" , and most only have a lot of discussion on single quantum systems and less on composite systems. So I have a question.

Is $$(|{{w}_{1}},{{U}_{2}}{{w}_{3}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}},{{U}_{2}}{{w}_{3}}|=\langle {{w}_{1}},{{w}_{3}}|U_{2}^{\dagger }$$ correct? ${w}_{i}$ is a quantum state, ${U}_{i}$ is an operator.

How to solve the expectation of a composite system $$A=(4|{{w}_{1}},{{w}_{2}}\rangle +|{{w}_{1}},{{U}_{1}}{{w}_{2}}\rangle +|{{U}_{2}}{{w}_{1}},{{w}_{2}}\rangle +|{{w}_{1}},{{U}_{3}}{{w}_{2}}\rangle +|{{U}_{4}}{{w}_{1}},{{w}_{2}}\rangle )$$ and calculate its expectation $\langle A|A\rangle $ on the standard orthonormal basis?

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First line is not entirely correct, it should read something like : $$\langle {{w}_{1}},{{U}_{2}}{{w}_{3}}|=\langle {{w}_{1}},{{w}_{3}}|I \otimes U_{2}^{\dagger }$$ to know to which qubit the matrix $U_2^\dagger$ applies, where $I$ is the identity.

We proceed by taking all the $U$s out the kets $$|A\rangle=(4|{{w}_{1}},{{w}_{2}}\rangle +I\otimes U_1|{{w}_{1}},{{w}_{2}}\rangle +U_2\otimes I|{{w}_{1}},{{w}_{2}}\rangle +I\otimes U_3|{{w}_{1}},{{w}_{2}}\rangle +U_4\otimes I|{{w}_{1}},{{w}_{2}}\rangle )=(4I\otimes I +I\otimes U_1 +U_2\otimes I +I\otimes U_3 +U_4\otimes I)|{{w}_{1}},{{w}_{2}}\rangle $$

to calculate its "expectation". We have

$$\langle A|A\rangle=\langle{{w}_{1}},{{w}_{2}}|(4I\otimes I +I\otimes U^\dagger_1 +U^\dagger_2\otimes I +I\otimes U^\dagger_3 +U^\dagger_4\otimes I)(4I\otimes I +I\otimes U_1 +U_2\otimes I +I\otimes U_3 +U_4\otimes I) | {{w}_{1}},{{w}_{2}}\rangle $$

If $U_i$ is unitary then $U^\dagger_i U_i=I$, also $\langle{{w}_{1}},{{w}_{2}}|{{w}_{1}},{{w}_{2}}\rangle=1$, then you can expand the product and work out a more compact form.

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  • $\begingroup$ For example $(I\otimes U_{q2}^{\dagger })\times ({{U}_{k1}}\otimes I)$, when tensor operations and multiplication operations occur at the same time, who should compute first?$$(I\otimes U_{q2}^{\dagger })\times ({{U}_{k1}}\otimes I)=(I\times {{U}_{k1}})\otimes (U_{q2}^{\dagger }\times I)?$$ $\endgroup$ Aug 24, 2022 at 18:24
  • $\begingroup$ That is equivalent to $(I\otimes U^\dagger_{i})(U_{j}\otimes I)=U_j\otimes U^\dagger_i$ $\endgroup$
    – Mauricio
    Aug 25, 2022 at 2:07
  • $\begingroup$ Got it! Thank you a lot. $\endgroup$ Aug 25, 2022 at 2:15

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