# unitary that transforms one Hilbert space to another Hilbert space

Let $$H = A \otimes B$$. If there exists a unitary operator $$U$$ that transforms the Hilbert space $$H$$ into another Hilbert space $$H' = A' \otimes B'$$ (meaning that $$U$$ maps each basis of $$H$$ to each basis of $$H'$$), does this imply that $$U$$ necessarily rotates the subsystems, i.e. mapping $$A$$ to $$A'$$ and $$B$$ to $$B'$$?

No. Suppose $$A$$ and $$B'$$ are qutrits, $$A'$$ and $$B$$ are qubits and $$U$$ is the SWAP unitary that sends $$|ab\rangle$$ to $$|ba\rangle$$ for every $$a\in\{0,1,2\}$$ and $$b\in\{0,1\}$$. Clearly, SWAP transforms $$H=A\otimes B$$ into $$H'=A'\otimes B'$$, but it certainly does not map $$A$$ to $$A'$$, because $$\dim A=3\ne 2=\dim A'$$.
• You're definitely right. What if $\dim(A) = \dim(A')$ and $\dim(B) = \dim(B')$? I believe the statement holds in that case, is it? Dec 22, 2023 at 23:14
• Yes, it does. In this case, pick a product basis $|ab\rangle$ in $A\otimes B$ and impose the new tensor product structure $A'\otimes B'$ by defining $|a'b'\rangle:=U|ab\rangle$. More formally, one way to define the tensor product is via a bilinear map $T: A\times B\ni(|a\rangle,|b\rangle)\mapsto|a\rangle\otimes|b\rangle\in A\otimes B$ which is linearly disjoint, i.e. sends every pair of linearly independent sets to a linearly independent set. It is a simple exercise to check that if $T$ is a linearly disjoint bilinear map then so is $U\circ T$ where $\circ$ denotes function composition. Dec 23, 2023 at 7:34