# On inner product for operators on $\scr H \otimes H$

Given two linear operators $$A \in {\scr H}_A$$ and $$B \in {\scr H}_B$$, their inner product $$A \cdot B \in {\scr H}_A \otimes {\scr H}_B$$ is defined to be: $$A \cdot B := \sum_{j} A_j \otimes B_j$$

Question. Is there a general rule for computing the inner product between operators, both of them living in the tensor product? I mean how to compute $$A \cdot B = (A_1⊗A_2)⋅(B_1⊗B_2)$$ for the generic linear operators $$A, B \in \scr H_1 \otimes \scr H_2$$.

According to the previous definition, it should give: $$\sum_j(A⊗B)_j⊗(C⊗D)_j$$ but I'm in trouble understanding what does it mean, if it means anything at all. Is $$(A⊗B)_j$$ simply $$A_j⊗B_j$$?

There is no general notion of inner products between operators. However, if $$\mathbf A:=(A_1,A_2,A_3)$$ and $$\mathbf B:=(B_1,B_2,B_3)$$ are collections of operators, then $$\mathbf A \cdot \mathbf B \equiv \sum_i A_i \circ B_i$$ is a relatively common shorthand, where $$A_i \circ B_i$$ means "apply $$B_i$$ and then apply $$A_i$$."
Extending this idea to the tensor product space $$\mathcal H \otimes \mathcal H$$, if we let $$\mathcal A_i := A_i \otimes \mathbb I$$ and $$\mathcal B_i := \mathbb I \otimes B_i$$, we would find $$\mathcal A\cdot \mathcal B := \sum_i(A_i\otimes \mathbb I)\circ (\mathbb I\otimes B_i) = \sum_i(A_i \otimes B_i)$$
with the latter equality following straightforwardly by noting the action of $$\mathcal A\cdot \mathcal B$$ on any product state:
$$(\mathcal A \cdot \mathcal B)(\psi\otimes \phi) = \left[\sum_i (A_i\otimes \mathbb I )\circ(\mathbb I \otimes B_i)\right](\psi\otimes \phi) = \sum_i (A_i\otimes\mathbb I)(\psi\otimes B_i\phi)$$ $$= \sum_i (A_i\psi\otimes B_i \phi) = \left[\sum_i (A_i\otimes B_i) \right](\psi\otimes \phi)$$
In standard notation we usually drop the operator composition symbol $$\circ$$ and just write $$A\circ B \equiv AB$$; I have kept it in for pedagogical clarity.