# 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 composition1 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.
1. 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.