Talk:Creation and annihilation operators

message: Hi to all. First I have to say that my english is very poor, sorry for that. I think that can be usefull put the form of the raising and lowering operartors as, for example 
: J_{+}=J_{x}+iJ_{y}
: J_{-}=J_{x}-iJ_{y} —The preceding unsigned comment was added by 193.144.179.151 (talk • contribs) .
Perhaps it would be a good idea to include them...but somehow we need to say that those come from angular momentum operators, and you can't ladder up to infinity. --HappyCamper 19:55, 19 July 2006 (UTC)
replies: text: yes you can, but you get 0 pretty soon --MarSch 11:42, 30 January 2007 (UTC)
thread_title:  An advice 

message: "Interestingly enough, no actual function actually represents the |0> state"
There's some weird stuff being said in this article. This statement is the weirdest. States are elements of a vector space of states. If you're using a representation where the states are wavefunctions, then |0> is a wavefunction like any other. If you're using some other more abstract representation where |0> isn't a 'function', then the other states aren't functions either. |0> isn't a different kind of thing from the other states. It's just another vector in some vector space.
And that stuff about |> and |0> makes no sense to me. It doesn't correspond to any notation I've met before, and I spent quite a few years doing QM and QFT. It's standard to call the lowest energy state |0>. There's no need for a |>. Sigfpe 23:16, 5 September 2007 (UTC)
thread_title:  Vacuum State 

message: Is the term "Bose operator" synonymous to creation and annihilation operator?
136.152.170.235 18:47, 30 October 2007 (UTC)
thread_title:  Untitled 2 

message: I changed the properties of the annihilation operator back to a|n> = (n)|n-1> etc.
for some reason someone converted it to the "quantum" sqrt(n) behavior. this is not what is used for reaction diffusion.
-RZ  —Preceding unsigned comment added by 62.219.245.34 (talk) 12:58, 25 November 2009 (UTC)
replies: text: Maybe I'm missing something but the creation/annihilation operators don't seem to check out for the reaction diffusion system. The current versions don't seem to satisfy the commutator relationship if you work things through. The previous (quantum) forms do check out, (but are apparently wrong?). It seems like the commutator relationship works out if the constant for the creation operator is (n/(n+1)) rather than (n+1), but I have no clue if this is really correct or canonical.— Preceding unsigned comment added by 2a00:23c7:85a2:500:4d4a:7e8a:b088:4f91 (talk) 17:42, 31 May 2021 (UTC)
replies: text:  Note: I have switched the reaction-diffusion operators to the definitions found in Gunnar Pruessner's lecture slides (now added as a new citation). These seem to obey the commutator relationship. (see: Talk:"Definition of a_i in "Creation and annihilation operators for reaction-diffusion equations")— Preceding unsigned comment added by 2a00:23c7:85a2:500:4d4a:7e8a:b088:4f91 (talk) 18:01, 31 May 2021 (UTC)
thread_title:  Reaction diffusion 

message: In the introduction, it says that an operator creates a particle. So if we are speaking about mathematical objects (operators), I guess the word particle also refers to another mathematical object (or the model and the "physical system" are not being properly distinguished, which is confusing in my opinion). What is the definition of particle in the model? What does it mean to create a particle??--190.188.3.11 (talk) 20:52, 4 August 2010 (UTC)
replies: text: It is, in fact, not "another mathematical object", but a physical one, a most-important one, and the closing sentence in the introduction, that on "2nd quantization", should be noted; I repeat, it should definitely be noted. - Regards, Meier99 (talk) 08:44, 12 January 2012 (UTC)
replies: text: I repeat, it should definitely be noted. Meaning that it is a whole 'nother story. Still, what does it mean to say that the operator creates a particle? This idea is taken for granted, but not actually explained in the article as far as I have read -- only the raising and lowering functions, which are really cool but are easy to explain by the efficacy of solving the eigenvalue problem (leaving aside deeper thoughts from representation theory). 178.38.76.15 (talk) 18:17, 21 November 2014 (UTC)
thread_title: Problem with intro

message: The section entitled Generalized creation and annihilation operators  makes little sense.
(1) If f is an element of the Hilbert space H, then a(f) is an element of H, not an operator on H.
(2) a is never defined or restricted in any way. Surely some conditions on a and on f are required in order to arrive at anything useful. Or is f an arbitrary, variable element?
(3) It is never explained why or how a and a† might "create" or "annihilate" anything. The terminology is just "slid into" like a kind of catechism, with no motivation. Ditto for how a†a might "count" anything. 178.38.76.15 (talk) 18:04, 21 November 2014 (UTC)
replies: text: I improved the presentation re: (1) and (2) considerably. a(f) is actually an algebra element labelled by f, and a is a linear embedding of the original Hilbert space H in an algebra with a conjugation operator *. It is similar but not identical to the construction of a Weyl algebra and a Clifford algebra. The difference is only that a skew-symmetric (Hermitian symmetric) inner product is used, instead of a truly symmetric one. a(f) becomes an operator only after a representation for the *-algebra is chosen. Missing from the exposition, but ought to be there: take the Banach space completion of the algebra with * to get a C* algebra.84.226.185.221 (talk) 09:46, 21 September 2015 (UTC)
thread_title:  Generalized creation and annihilation operators  

message: I attempted to define ai. Did I do it right? 
Also, it was not clear to me what space ψ is in. Is it really a tensor product of Hilbert spaces? If so, do  ai† and ai+1 (and other combinations) commute, and does the commutator make sense? 178.38.76.15 (talk) 00:15, 22 November 2014 (UTC)
replies: text: The discussion of 'non-quantum' nature and different normalization of creation & annihilation operators that follows (i.e. the action of a and a^\dagger without square roots) seems to be wrong, i.e. the commutation relations are not satisfied if one omits the square roots. Actually the whole section about reaction-diffusion equations seems to lower the quality of the wikipedia page.  — Preceding unsigned comment added by 86.49.105.240 (talk) 10:57, 18 May 2021 (UTC)
replies: text: This is correct, the normalization previously matched the quantum versions, but was changed to the new (wrong) values circa 2009 (see discussion topic "Reaction diffusion"). These slides by Gunnar Pruessner define the constants as 1 for the creation operator and n for the annihilation operator. This works. I'm going to change it and add the lecture slides as a citation.— Preceding unsigned comment added by 2a00:23c7:85a2:500:4d4a:7e8a:b088:4f91 (talk) 17:50, 31 May 2021 (UTC)
thread_title:  Definition of a_i in "Creation and annihilation operators for reaction-diffusion equations"