Timeline for Why does Riesz's Representation Theorem apply in quantum mechanics?
Current License: CC BY-SA 4.0
17 events
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| Feb 6, 2021 at 15:26 | comment | added | Yemon Choi | I have missed a lot of the history of the discussion on this question but isn't some of it addressed by noting that the space $S_1(H)$ of trace-class operators on a general Hilbert space H is in duality with the space $B(H)$ of all bounded operators on $H$, and while this duality resembles the self-duality between $S_2(H)$ and its complex-conjugate, these three spaces are all distinct in general. The former duality is not a special case of the Riesz-Frechet duality theorem that I think you are referring to | |
| Feb 2, 2021 at 19:47 | comment | added | jjcale | See also books.google.de/… | |
| Feb 2, 2021 at 19:38 | comment | added | jjcale | By the argument of Martin Argerami in math.stackexchange.com/questions/77820/… there exists a pure state that maps all compact operators to 0. So the answer to Question 1 is no. | |
| Feb 2, 2021 at 4:10 | history | became hot network question | |||
| Feb 1, 2021 at 23:25 | vote | accept | Andrew NC | ||
| Feb 1, 2021 at 23:24 | history | edited | LSpice | CC BY-SA 4.0 |
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| Feb 1, 2021 at 23:20 | answer | added | Pedro Lauridsen Ribeiro | timeline score: 12 | |
| Feb 1, 2021 at 22:15 | answer | added | Nik Weaver | timeline score: 14 | |
| Feb 1, 2021 at 22:07 | comment | added | Buzz | Why do you think that the collection of all positive linear functions on $A$ is the actual object of interest? | |
| Feb 1, 2021 at 21:59 | comment | added | Andrew NC | No, I'm not concerned with the interpretation of states as probability measures, which I am aware of and understand perfectly. What I am concerned with is that a state is a priori a positive funcitonal taking 1 to 1 defined on the $C^*$ algebra, which is a Banach space, but potentially not a Hilbert space. Then the question is: why are physics textbooks treating it as if the $C^*$ algebra is a Hilbert space? This seems to make sense only if you restrict to the Hilbert Schmidt operators among the $C^*$ algebra. | |
| Feb 1, 2021 at 21:33 | comment | added | Christian Remling | Are you perhaps confusing two different Riesz representation theorems (not saying you are, just a hunch)? If the $C^*$ algebra is commutative, so $A=C(K)$, then the Riesz representation theorem (the one about measures being the dual of $C(K)$) says that the states are exactly the prob measures. What we are dealing with here is a non-commutative version of this, so perhaps it's natural to use the same words. | |
| Feb 1, 2021 at 21:09 | history | edited | Andrew NC | CC BY-SA 4.0 |
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| Feb 1, 2021 at 20:42 | history | edited | Andrew NC | CC BY-SA 4.0 |
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| Feb 1, 2021 at 20:42 | comment | added | Andrew NC | No. It is clear that a positive operator of unit trace is Hilbert Schmidt, and that it defines a positive linear functional on $A$ taking $1$ to $1$. It is not clear why any positive linear function on $A$ is determined by its restriction to the sub-algebra of $A$ made up of only the Hilbert Schmidt operators. The reason to normalize, i.e. look at unit trace operators, is irrelevant to the question. | |
| Feb 1, 2021 at 20:37 | comment | added | Carlo Beenakker | Isn't Q1 answered in the Wikipedia definition of a state: a positive linear operator of unit trace, and hence by definition a Hilbert-Schmidt operator. The restriction to unit trace is a matter of normalization, in the same sense that a probability distribution is normalized to unity. Pure states satisfy the additional requirement that the operator squares to itself. | |
| Feb 1, 2021 at 20:24 | history | edited | Andrew NC | CC BY-SA 4.0 |
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| Feb 1, 2021 at 20:07 | history | asked | Andrew NC | CC BY-SA 4.0 |