Talk:Self-adjoint operator

message: Hey there! 
What about having the distinction between Hermitian operators and Self adjoint operators added to this otherwise very good presentation... se for example some common lectures like http://www.hep.caltech.edu/~fcp/math/differentialEquations/linDiffEq.pdf
Thanks   — Preceding unsigned comment added by 89.242.21.31 (talk) 07:30, 29 April 2012 (UTC)
replies: text: It is already written: "Bounded symmetric operators are also called Hermitian." Boris Tsirelson (talk) 08:15, 29 April 2012 (UTC)
thread_title: Untitled

message: This is probably obvious to most experts, but I think that the notation < | > should be defined before it is used in the first section.  I added a clarification tag to this.  —Preceding unsigned comment added by BradLipovsky (talk • contribs) 02:12, 11 September 2010 (UTC)
thread_title: Notation Unclear

message: A symmetric operator on a hilbert space can have no eigenvalues? Isn't 0 an eigenvalue?
: Not in general. If it's invertible, for instance 0 is not an eigenvalue.--CSTAR 07:54, 30 December 2006 (UTC)
replies: text: To mu knowledge there isn't a proof policy.  This is supposed to be an encyclopedia, not a textbook or reference.--CSTAR 07:54, 30 December 2006 (UTC)
replies: text:  There are a couple of math wiki projects, though still in their infancy. It would be worth linking to them once they are up and running: Swim, VDASH. Drevicko (talk) 06:04, 23 March 2010 (UTC)
thread_title: Proof Needed?

message: As i know sigma-finiteness is not needed for a multiplication operator to be densely defined. Enlighten me or i will delete the remark.
: Hmmm. I think you're right.--CSTAR 21:32, 26 February 2006 (UTC)
thread_title:  necessary? 

message: The article states:
:The following is an alternate characterization of symmetric operator: A densely defined operator A is symmetric iff
:: A \subseteq A^*.
What the heck does this subset notation mean?? linas 21:46, 18 March 2006 (UTC)
replies: text: This is standard terminology for operators. One is an extension of the other.--CSTAR 21:47, 18 March 2006 (UTC)
replies: text: Do we have an article somewhere that defines this? I'm thinking of this in terms of topology, that the one is defined in a topology that is finer than the other. But rather than guessing, I'd like to send the gentle reader to an actual definition. linas 23:09, 18 March 2006 (UTC)
replies: text: It means the graph is a subset. -lethe talk [ +] 03:29, 19 March 2006 (UTC)

text: So should I edit the article to say "...  G(A) \subseteq G(A^*). This is often written as  A \subseteq A^*, with the use of subset notation implicitly implying that the relation holds for the graph of the operator." I'll make this edit now; if its wrong, back it out please.   linas 05:19, 19 March 2006 (UTC)
replies: text: That language sounds funny to me.  It's not "often written as".  It's the definition of extension of operators. -lethe talk [ +] 05:23, 19 March 2006 (UTC)
replies: text: So where is this defined? Read my first post, wherein I ask: "what the heck does this mean"? We need a definition of this notation. Anyone who does not already know the answer won't be able to glean the meaning from the current wording. Note redlinks: Graph of an operator, operator extension. linas 20:48, 19 March 2006 (UTC)
replies: text:  In the usual set-theoretic formalization of function, a function is identical to its graph .--CSTAR 20:55, 19 March 2006 (UTC)

text:  It looks OK now, mathematically, although I don't particularly like the way the inline tex looks, aesthetically.--CSTAR 05:14, 20 March 2006 (UTC)
replies: text: I use a font size about the same size as the image fonts, and also enable MathML in my WP preferences. I have hopes for BlahTeX. linas 15:11, 20 March 2006 (UTC)

text:  BTW further down, I had used \mathrm{perp} instead of \perp (maybe when I added that \perp wasn't working.) --CSTAR 05:16, 20 March 2006 (UTC)
replies: text: Will fix shortly. linas 15:11, 20 March 2006 (UTC)
thread_title: Alternate characterization

message: I don't think this article is the right place to define an extension of an operator.  It should go in unbounded operator.  -lethe talk [ +] 08:51, 20 March 2006 (UTC)
replies: text: I'd suggest its own unique article, e.g. extension of symmetric operators. That section and everything beyond it certainly does take the discussion of operators to a whole new niveau. linas 15:38, 20 March 2006 (UTC)
replies: text: Well, I might agree with you on that.  I think there might be enough to say about extensions to symmetric operators to merit its own article, but I'm not sure.  In any event, the definition of the closabillity of operators, and also of extension of operators, makes sense for general unbounded operators, not just symmetric operators, therefore that definition should go there.  Then we don't have to clutter this article with definitions.  As for your complaint that it's too "operator theoretic", well, the distinction between self-adjointness and symmetry of operators is itself a bit operator theoretic.  Anyone who wants to read this article will have to get at least a little of the terminology of unbounded operators. I'm going to take a look at that article now. -lethe talk [ +] 15:52, 20 March 2006 (UTC)
replies: text: I'm looking at closed operator now, and I don't really like it.  It's not really set-up to talk about general unbounded or non-everywhere-defined operators, which makes it difficult to have the definitions there that I want.  I don't think I have the energy to do a rewrite at the moment.  I'm starting to think that I'd like to see an article about linear operators on general TVSes.  It could talk about the different kinds of boundedness, and when boundedness implies continuity, and operators not everywhere-defined, graphs, extensions, and closures.  All of these make sense in general TVSes.  What do you think?  This might be nice because like eg Oleg doesn't like to have stuff about spaces that are not normable in a lot of places, but those spaces still need to have a home. -lethe talk [ +] 16:08, 20 March 2006 (UTC)
replies: text:  No reason not to. They occur in nature.--CSTAR 20:47, 20 March 2006 (UTC)
replies: text:  PS. I don't like the idea of breaking up the article.  I rather like the idea of having a lot of stuff on s.a. operators in one place.--CSTAR 20:50, 20 March 2006 (UTC)

text: I'm afraid I can't add anything. My understanding of these topics is incomplete and spotty; I don't have the vision  to suggest an overall structure. I know little/nothing about operator extensions beyond what this article says, and  a breif skim of some papers on rigged hilbert space and related topics. On the other hand, I do have a tried-n-true method of writing: write a few paragraphs, put them where they seem to fit best at this point in time. Write a few more. After a while, one realizes that the first few paragraphs would be better if moved somewhere else .. or expanded or deleted ... rinse, and repeat. As Charles Matthews put it, the result is "organic". linas 20:25, 20 March 2006 (UTC)
thread_title: Extension of symmetric operators

message: These days, whenever I see the words "Cayley transform" I think "Riemann surface/Fuchsian group/j-function/etc.". Can anything interesting be said along these lines? Or is this a crazy non-question? linas 15:38, 20 March 2006 (UTC)
replies: text: Not much more than is already on the page to which it is linked (somewhere in the article). The Cayley transform is a fractional linear transformation; the namw used for operators is mostly used by analogy, although H \oplus H does have a symplectic structure, but not too much should be made of this fact, at least not in this article.--CSTAR 20:45, 20 March 2006 (UTC)
replies: text: OK thanks then. I've come up with a remarkable little problem, then; I dare say no more. I haven't had this much fun playing at solving things in quite a while. linas 00:11, 21 March 2006 (UTC)
thread_title: Crazy question

message: This article states, in the section on the spectral theorem, that
: Theorem. Any multiplication operator is a (densely defined) self-adjoint operator. Any self-adjoint operator is unitarily equivalent to a multiplication operator.
By contrast, the article spectral theorem qualifies this further, stating that any bounded self-adjoint operator is unitarily equivalent to a multiplication operator. So, does an operator need to be bounded for this theorem to hold, or not?  linas 23:05, 18 March 2006 (UTC)
replies: text: No. It's true for arbitrary self-adjoint operators.--CSTAR 23:41, 18 March 2006 (UTC)
replies: text: Can I get you to review the article on spectral theorem, and remove the "bounded" qualifier from it? linas 00:05, 19 March 2006 (UTC)
replies: text:  There is already a short section on the unbounded case. Isn't that good enough? --CSTAR 00:49, 19 March 2006 (UTC)
replies: text: OK, right. I guess I skipped reading the last paragraph, which circles right back to this article. I've been feeling confused today. linas 05:07, 19 March 2006 (UTC)
thread_title:  Multiplication operators 

message: "partially defined isometry" is perculiar terminology. is it used at all in literature? as pointed out later in the same paragraph (extension of symmetric operators), one customarily extend by continuity and then by zero to get something defined everywhere. "unique partially defined linear operator", also rather uncommon language, should be taken to mean the continuous extension is unique anyhow. i didn't change it since i didn't see anything incorrect, just somewhat unusual and perhaps unnecessary. Mct mht
: Why is this a problem? Partially defined function is a perfectly good concept and standard terminology.--CSTAR 23:00, 23 April 2006 (UTC)
replies: text: Do you mean if they are unitarily equivalent or if not at least similar? The answer is no to both questions, because spectrum (or more strongly spectral measure) does not deal with spectral multiplicity.--CSTAR 16:24, 12 January 2007 (UTC)
thread_title:  partial isometry 

message: CSTAR actually, you can. it's not trivial but it's been done. i just thought it should be mentioned Mct mht 23:26, 23 April 2006 (UTC)
replies: text: above comment is re possibility of convergence of operator integral in strong operator topology. Mct mht 23:28, 23 April 2006 (UTC)
replies: text:  OK I believe you, but it's news to me.--CSTAR 23:30, 23 April 2006 (UTC)
replies: text:  Could you provide a reference for how this is done? I'm still pretty skeptical.--CSTAR 13:17, 24 April 2006 (UTC)
replies: text:  CSTAR, ok, it is rather obscure. i haven't found an explicit reference to a paper. it's done using vector valued measures, according to Reed and Simon. convergence in the weak operator topology looks to be the most convenient and common approach taken. since all the well-known texts seem to take that approach, i am going to remove the SOT note. Mct mht 01:19, 27 April 2006 (UTC)

text:  This is an ancient post, but I would like to mention that I am taking a graduate course where we are currently discussing operator integrals in the strong topology. Maybe screwy things can happen in the cases we haven't covered yet though.--91.11.110.253 (talk) 22:05, 13 January 2012 (UTC)
thread_title:  operator integral 

message: Is it just me or is this article confusing symmetric operators with Hermitian ones?  It even goes so far as to say that they are synonymous, and fails to mention complex conjugates.  This seems odd to me since for matrices, "adjoint" means "take the transpose, and also take the complex conjugate of each entry", hence a "self-adjoint" matrix would be a Hermitian rather than a symmetric one.
Am I missing some subtle point, or should the article be changed? Nathaniel Virgo (talk) 22:53, 26 June 2010 (UTC)
replies: text: The article is fine in this point the way it is, but I am not sure how best to explain. A common definition for symmetric operators in a complex Hilbert Space is that a operator A is called symmetric if <Ax,x> is real for all x in the Hilbert space. This is merely a convention, but it is called symmetric because it implies that you can move an operator from the right side of an inner product to the other without changing it's value. It is slightly unfortunate however that this doesn't concide with the common definition/notation for operators on finite dimensional spaces. One thing to note is that for real Hilbert spaces, the same definition doesn't suffice. And actually the article doesn't cover this, but as soon as one allows operators to be defined only on a subset of the Hilbert space, then self adjoint and symmetric cease to be the same thing. In the case of operators defined on a dense subset of the entire space, for example, there are many examples of symmetric operators which are not self-adjoint.--91.11.110.253 (talk) 21:57, 13 January 2012 (UTC)
thread_title:  Symmetric != Hermitian 

message: About the ongoing edit war , , , ,  I fail to see, what is the fuss? All I see in the article (rather than its wikitext) is, few tiny distinctions in vertical spacing. Could the combatants explain here, what is the  essense of the conflict? Boris Tsirelson (talk) 07:15, 21 April 2018 (UTC)
thread_title:  Edit war, why? 

message: The section on Hermitian operators restricts the definition to "bounded operators". But, especially in quantum mechanics and quantum field theory, where this terminology is used, most operators under consideration are not bounded operators. Physicists often don't bother with boundednes and use the terminology Hermitian anyway. The preceding part of the article discusses unbounded operators, so is there a reason for the qualifier here? I feel it might confuse an uninitiated reader. Aaroneisenberg (talk) 09:30, 2 December 2021 (UTC)
thread_title:  Hermitian operators