Timeline for Density matrix formalism
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27 events
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| Oct 9, 2023 at 20:32 | comment | added | Tobias Fünke | @ValterMoretti Ah, I see --sorry for the confusion. I've now formulated a question, since the confusion won't end. Thanks again so far for your help. | |
| Oct 9, 2023 at 19:26 | comment | added | Valter Moretti | No, I simply misunderstood you. You are considering a certain weak * topology on the set of the mixed states. You called it also ultraweak topology, but the ultraweak topology is not the same topology, it is a certain weak* topology on $B(H)$ and not on the set of mixed the states. | |
| Oct 9, 2023 at 18:24 | comment | added | Tobias Fünke | As you can read, the assertion in my math SE question is indeed false and someone gave a nice counter example. But if I understood you correctly, you say that one can make sense of weak*-compactness of the set of density matrices if properly stated/modified? Sorry to have bothered you with this; it seems I am still confused about all the different topologies etc. involved. | |
| Oct 9, 2023 at 16:13 | comment | added | Tobias Fünke | @ValterMoretti I used (or at least tried to) the notion introduced e.g. in this thread. The same is used in a paper by Lieb: Density Functionals for Coulomb Systems, 1983, proof of remark after theorem 3.3. He writes (I will paraphrase the words): The dual of compact operators is the trace class operators and a $\gamma$ trace-class takes a compact $A$ to $\mathrm{Tr}\gamma A$, and a sequence $\gamma_n$ of trace-class op. converges in the $w^*$-top. iff $\mathrm{Tr}_n\gamma A \to \mathrm{Tr}\gamma A$ for all $A$. | |
| Oct 9, 2023 at 15:36 | comment | added | Valter Moretti | The only problem I see with your statement is that the ultraweak topology is on $B(H)$ and not on the space of trace class operators. Could you write very clearly the statement ? | |
| Oct 9, 2023 at 15:21 | comment | added | Tobias Fünke | Hi, thanks for the reference. Unfortunately I could not find the relevant passage -- I hope I did not overlook it. I actually found a reference: Theory of Quantum Information with Memory. M. H. Chang, section 2.4; but I could not follow the proof. I have tried to prove it myself, but I got stuck with a sub-proof. If you are interested and have time, I'd highly appreciate if you could have a look here; sorry in advance if this is a trivial point I miss. | |
| Oct 8, 2023 at 2:46 | comment | added | Valter Moretti | Dear Tobias, I suggest to have a look at Busch’s (and coauthors) book on quantum measurement, 2nd edition. | |
| Oct 7, 2023 at 21:06 | comment | added | Tobias Fünke | Dear Valter, I have a question: IIRC, the set of density operators is (for an infinite-dimensional Hilbert space) compact in the corresponding weak-* topology (ultraweak topology), which I think follows from the Banach-Alaoglu theorem. However, I cannot find a suitable source. Do you happen to know some? Thanks in advance. BTW: Great answer, helped me a lot some time ago. | |
| Dec 28, 2019 at 19:59 | comment | added | Valter Moretti | I discuss the two approaches in my last book "fundamental mathematical structures in quantum theory".... | |
| Dec 28, 2019 at 17:39 | comment | added | user199113 | @ValterMoretti Thanks! Could you please provide me a reference (or references) from where I can learn these two approaches that you mention? | |
| Dec 28, 2019 at 17:30 | comment | added | Valter Moretti | The only other approach is the algebraic one where states are normalized positive functionals on the C*-algebra of observables. In finite dimension the to approaches coincide, in infinite dimentions the second is more general and therein there are states that are not represented by statistical operators (the non-normal states). | |
| Dec 28, 2019 at 17:26 | comment | added | Valter Moretti | @S.D. In the standard approach to quantum theory states are probability measures on the lattice of orthogonal projectors on the Hilbert space of the system. Gleason's theorem proves that those measures are exactly the positive trace-class unit-trace operators. | |
| Dec 28, 2019 at 12:22 | comment | added | user199113 | @ValterMoretti I see. Well, then I suppose my question is why quantum states (density operators) necessarily have well-defined traces? Is it a physical/experimental observation or merely a mathematical idealization to simplify the math to model quantum systems? (I guess a possible answer is that the sum of probabilities $p_k$ must add up to a finite value i.e. 1). BTW a google search returns some results for "unbounded density operator". | |
| Dec 28, 2019 at 7:32 | comment | added | Valter Moretti | Well, the notion of trace is only well-defined for (some) bounded oprators. | |
| Dec 28, 2019 at 2:28 | comment | added | user199113 | Is there any specific reason why density operators must be bounded? | |
| Dec 18, 2019 at 7:08 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Jun 23, 2014 at 16:23 | comment | added | Daniel Hogg | Thank you for your clear and informative response. I have been reading over it and the links you suggested with great interest. | |
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| Jun 22, 2014 at 18:04 | vote | accept | Daniel Hogg | ||
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| Jun 22, 2014 at 15:54 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:45 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:39 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:34 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:25 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:19 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Jun 22, 2014 at 15:10 | history | answered | Valter Moretti | CC BY-SA 3.0 |