Definition of inner product of Hilbert–Schmidt operator

Question

Wikipedia (Hilbert–Schmidt operator) states the following statement:

$ \langle A,B\rangle _{\text{HS}}=\operatorname {Tr} (A^{*}B)=\sum\limits _{i}\langle Ae_{i},Be_{i}\rangle .$

The second equation confuses me a lot. According to my computation, $\operatorname {Tr} (A^{*}B)=\sum\limits _{i}\langle (A^{*}B)e_{i},e_{i}\rangle=\langle Be_{i},Ae_{i}\rangle=\overline{\langle Ae_{i},Be_{i}\rangle},$ distinctive to the Wikipedia one.

Hope for an answer!

score 1 · Accepted Answer · answered Aug 12 '22 at 16:04

1

It's about the convention for the inner product. The Wikipedia article seems to be using the physics convention where the first component is the conjugate-linear one, while you are using the math convention where it is the second component that is conjugate-linear.

answered Aug 12 '22 at 16:04

Martin Argerami

205,756

The problem seems to be that https://en.wikipedia.org/wiki/Trace_class has it the other way around. For this definition we should have $\langle A, B \rangle_{\text{HS}} = \mathrm{Tr}(B^{*}A)$. – unwissen Aug 12 '22 at 16:24
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Indeed. $ \ \ $ – Martin Argerami Aug 12 '22 at 16:30

Definition of inner product of Hilbert–Schmidt operator

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