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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 $B'$?

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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 $\dim A=3\ne 2=\dim A'$.

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  • $\begingroup$ 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? $\endgroup$
    – Mohan
    Dec 22, 2023 at 23:14
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    $\begingroup$ 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. $\endgroup$ Dec 23, 2023 at 7:34
  • $\begingroup$ Thank you for your kind answer! $\endgroup$
    – Mohan
    Dec 25, 2023 at 3:02

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