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Suppose $A$ is a Hilbert-Schmidt operator on a Hilbert space $\mathcal{H}$, then when is it true that

$$\langle Ax, x \rangle_{\mathcal{H}} = \langle A, x \otimes x \rangle_{\text{HS}}, \quad \forall \, x \in \mathcal{H}$$

I know from the definition of Hilbert-Schmidt inner product and the definition of tensor product that

$$ \langle A, x \otimes x \rangle_{\text{HS}} = \sum_{j \in J} \langle A e_j, \langle x, e_j \rangle_{\mathcal{H}} x \rangle_{\mathcal{H}} $$ where $\{e_j\}_{j \in J}$ is an arbitrary ONB of $\mathcal{H}$. However, I am unable to simplify further. Any help would be appreciated.

If the above result is incorrect, in general, I am interested in going from an expression of the form $\langle Ax, x \rangle_{\mathcal{H}}$ to an expression of the form $\langle A, x \otimes x \rangle_{\text{HS}}$. How could I do that?

Also, I would really appreciate some references where I can look into these kind of results in more detail.

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    $\begingroup$ Shouldn't this follow simply from a resolution of the identity? You have outer products of $e_j$ in there which sum to $I$. $\endgroup$ Commented Apr 4, 2021 at 4:08
  • $\begingroup$ I am sorry, but I am very new to functional analysis. Could you be more detailed about what you mean by the outer product of $e_j$ and how it follows from resolution of the identity? Let me read up on resolution of the identity till you reply. $\endgroup$ Commented Apr 4, 2021 at 4:15
  • $\begingroup$ @CameronWilliams forgot to tag you. $\endgroup$ Commented Apr 4, 2021 at 4:28

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What you need to do is use an orthonormal basis whose first element is $x/\|x\|$. Then $$ \langle A,x\otimes x\rangle_{\rm HS}=\operatorname{Tr}(A(x\otimes x))=\langle A(x\otimes x)\tfrac{x}{\|x\|},\tfrac{x}{\|x\|}\rangle=\langle Ax,x\rangle, $$ since $(x\otimes x)x=\|x\|^2\,x$.

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  • $\begingroup$ This is brilliant! Just to confirm one thing though: $\langle A , x \otimes x \rangle_{\text{HS}}$ should equal $\text{Tr}(A^* (x \otimes x))$, right? and therefore we have $\langle A , x \otimes x \rangle_{\text{HS}} = \langle A^* x , x \rangle$. Or is it true that $A$ being Hilbert-Schmidt must also be self-adjoint? $\endgroup$ Commented Apr 4, 2021 at 4:45
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    $\begingroup$ No, $\langle A,x\otimes x\rangle$ is $\operatorname{Tr}((x\otimes x)^*A)$. And $x\otimes x$ is selfadjoint. $\endgroup$ Commented Apr 4, 2021 at 4:55
  • $\begingroup$ Okay, must be different conventions then? The Wikipedia page says $\langle A, B \rangle_{\text{HS}} = \text{Tr}(A^* B)$. $\endgroup$ Commented Apr 4, 2021 at 4:59
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    $\begingroup$ It's about convention. If you want to use the physicist's convention, then you also need to use it in the definition of the trace. $\endgroup$ Commented Apr 4, 2021 at 5:05

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