# Tensor products question

In the course of doing some quantum mechanical calculation I encountered an object of the form $$Tr(\cdots)Tr(\cdots).$$ In order to simplify this, I use the fact that $$Tr(A)Tr(B)=Tr(A\otimes B).$$

Now, let us focus on the object inside the trace and forget where it came from. Suppose that both $$A$$ and $$B$$ are unitaries that acts on a linear chain of $$N$$ sites, each site has local Hilbert space of dimension $$q$$. Now suppose that both $$A,B$$ acts trivially on the chain, in the sense that they do not not contain any interactions. In physics notation, I would write it as $$A= \prod_{r=1}^N A_r$$ and everyone would understand that each $$A_r$$ only acts on one site, whereas the proper meaning of this is actually $$A = A_1\otimes A_2\otimes ... \otimes A_N$$.

Now, I would like to believe that I can write $$A\otimes B = \prod_r (A_r\otimes B_r)$$ but I am not sure how to show that using basic properties, for example https://archive.siam.org/books/textbooks/OT91sample.pdf.

On the other hand, the expression $$A\otimes B = A_1\otimes A_2\otimes ... \otimes A_N \otimes B_1 \otimes B_2\otimes ... \otimes B_N$$ doesnt quite make sense to me, and I do not think I can arrive to the previous expression from this one.

What is the proper way of making sense of these "2 kind" of tensor products and how to properly show this kind of relations?

Let's set $$N = 3$$ for clarity. First, it's worth noting that the decomposition $$A = A_1 \otimes A_2 \otimes A_3$$ doesn't make sense for most operators. For example, if $$A$$ represents the total energy, or the total spin, or anything else like that, then as long as it doesn't involve any interaction between distinct sites, it has the form $$A = A_1 \otimes I_2 \otimes I_3 + I_1 \otimes A_2 \otimes I_3 + I_1 \otimes I_2 \otimes A_3$$ where $$I_i$$ is the identity operator on the $$i^{\text{th}}$$ site.
Anyway, if $$A$$ is a unitary and the $$A_i$$ represent a unitary acting separately on each site, then $$A = A_1 \otimes A_2 \otimes A_3$$ makes perfect sense. Now, one quantity that makes sense is the composition of two such operators, $$AB = (A_1 B_1) \otimes (A_2 B_2) \otimes (A_3 B_3).$$ What you seem to want instead is the tensor product of $$A$$ and $$B$$, which doesn't have a direct physical interpretation. For example, if $$A$$ and $$B$$ are defined on a Hilbert space of dimension $$D = d^n$$, then $$A \otimes B$$ is an operator on a $$D^2 = d^{2n}$$-dimensional vector space, not the original Hilbert space. So it is formally completely correct that $$A \otimes B = A_1 \otimes A_2 \otimes A_3 \otimes B_1 \otimes B_2 \otimes B_3$$ which follows trivially from the associative property of the tensor product, but this has no physical meaning in terms of the original lattice sites like $$AB$$ would; physically $$A \otimes B$$ would act on two copies of the lattice.
• $A$ is a unitary. Then I think what I wrote is correct, no? Commented Jan 25, 2020 at 22:13
• Then is it correct to say that $A\otimes B = A_1\otimes B_1 \otimes A_2 \otimes B_2 \otimes A_3 \otimes B_3$, like I did in the question? Commented Jan 25, 2020 at 22:47