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?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.