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 $A_j⊗B_j$?


1 Answer 1


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.


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