Given two quantum system $A$ and $B$, their Hilbert space $H_{A}$ and $H_{B}$, we can form a tensor product $H_{A}\otimes H_{B}$ to describe the larger system $A$ with $B$. Of course the operators act on $H_{A}$ and $H_{B}$ have to be tensor product as well. Assume all quantum observables are bounded(which is not true, but my focus is not here). We denote the space of bounded linear operator act on $H_{A}\otimes H_{B}$ be $L(H_{A}\otimes H_{B})$, then $L(H_{A}\otimes H_{B})$ is a Banach space with some norm. As far as I know, the choice of norm for tensor product of norm space is not unique. Are all these norms physically equivalent?
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2what do you physically need a norm on $L(H_A\otimes H_B)$ for, anyway? I.e. when you ask whether they are "physically equivalent", what statements are you thinking of where it might matter? – ACuriousMind May 19 '21 at 19:25
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@ACuriousMind I don't know if this is correct or not, but one formalism of QM is to use C* algebra, so we have to define a norm on space of operators. – Ken.Wong May 19 '21 at 19:44
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6$H_A \otimes H_B$ is a Hilbert space in its own right so that it has its Hilbert norm which, in turn, induces the usual operator norm in $L(H_A\otimes H_B)$. That norm is compatible with the tensor product structure. The problem you consider arises when one refers to a pair of abstract C*-algebras... – Valter Moretti May 19 '21 at 19:55
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Related : Total spin of two spin-1/2 particles. – Frobenius May 21 '21 at 00:11