### Question 1.

In Notes 8, why does equation (1.3) follow hold? We invert \(I_d \otimes R\) and \(\Lambda \otimes I_n\). Can you describe why please? Is it because the identity matrix commutes with everything?

Good question! The short answer is yes. In this part, we start with \[
\Lambda \otimes \operatorname{id}_k ( |\varphi_i\rangle \langle\varphi_i|),
\] and use a relation from Notes 7 to write \(|\varphi_i\rangle = I_d\otimes R_i\, |\Omega\rangle\) for some matrix \(R_i\). Substituting this, we have \[
\Lambda \otimes \operatorname{id}_k ( I_d\otimes R_i\, |\Omega\rangle \langle\Omega| \,I_d \otimes R_i^\dagger),
\](1) which (1.3) claims is equal to \[
I_d\otimes R_i\, (\Lambda \otimes \operatorname{id}_k ( |\Omega\rangle \langle\Omega| ) )I_d \otimes R_i^\dagger.
\](2) Intuitively, we can see this because \(I_d\otimes R_i\) only acts non-trivially on the second system, and \(\Lambda\otimes \operatorname{id}_k\) only acts non-trivially on the first system. But how can we show this algebraically? We can define a linear map \(\mathcal{R}_i : A\mapsto R_i A R_i^\dagger\). Then (1) becomes \[\begin{aligned}
(\Lambda \otimes \operatorname{id}_k) \circ (\operatorname{id}_d \otimes \mathcal{R}_i) (|\Omega\rangle \langle\Omega|) &= (\Lambda \circ \operatorname{id}_d) \otimes (\operatorname{id}_k \circ \mathcal{R}_i ) \, (| \Omega \rangle\langle \Omega |) \\
&=(\operatorname{id}_d \circ \Lambda ) \otimes (\mathcal{R}_i \circ \operatorname{id}_k) \, (| \Omega \rangle\langle \Omega |)\\
&=\operatorname{id}_d \otimes \mathcal{R}_i \circ ( \Lambda \otimes \operatorname{id}_k) \, (| \Omega \rangle\langle \Omega |)\\
&=I_d \otimes R_i\, ( \Lambda \otimes \operatorname{id}_k \, (| \Omega \rangle\langle \Omega |) ) \, I_d \otimes R_i^\dagger
\end{aligned}\] which is (2). So it just follows from the rules of composition^{1} and tensor product, along with the fact that, e.g. \(\operatorname{id}_k \circ \mathcal{R}_i = \mathcal{R}_i = \mathcal{R}_i \circ \operatorname{id}_k\). That is, composing a map with the identity map does nothing.

In practice, one usually doesn’t write out full proofs for facts like these, because it comes up a lot and it’s the same argument every time. But it’s important to see how it works to know that the intuition about “which system does it act on non-trivially” really holds up.

Note that these superoperators \(\mathcal{R}_i\) are just linear maps like \(R_i\), except on a larger Hilbert space. So composition of superoperators respects tensor product just as matrix multiplication does, since they are in fact the same operation, just on different sized Hilbert spaces.↩