Talk:Creation and annihilation operators

| > versus |0>

So I edited the last section replacing | > with |0> before I read the section above explaining the need for the distinction. However, after reading that bit I still don't understand the distinction. It seems it would be important only if you had a infinite (-∞, ∞) versus semi-infinite [constant, ∞) set of energies, in which case you'd be hosed anyways. --Laura Scudder 23:46, 4 April 2005 (UTC)

Actually you brought forward something that I didn't consider before! We should explain this subtlety about the infinite/semi-infinite set of energies available to the system. The | > state is only needed in the case where the energies are semi-infinite, such as the quantum harmonic oscillator. In the quantum harmonic oscillator, it's necessary to define |0> as the "ground" state of the system, which arguably already has 1 quanta of energy in it, and | > the completely "null" state where there's absolutely no energy whatsoever in the oscillator. It's necessary to explain why the annihilation operation a|0> = 0. It's because it really is a short form for a|0> = 0| > = 0, since 0 is the eigenvalue for the |0> state. However, admittingly, this is somewhat of a subtle point that isn't often mentioned, although in Wikipedia, there is a reference to the "zero ket" in quantum harmonic oscillator about halfway down the page. I'll attempt to add this to the page - please feel free to modify it if necessary :) HappyCamper 14:43, 5 April 2005 (UTC)

I've never heard of this distinction before and I am in quantum optics. As the article stands right now, this is confusing. In the quantum harmonic oscillator, the |0> state is known as the ground state, lowest energy eigenstate, zero ket, or the vacuum state. The point is there are no states with absolutely zero energy - there are always vacuum fluctuations. The eigenvalue equation should read a|0> = 0|0> where 0 is the eigenvalue. If you have a reference for the |> state please post it. Otherwise I'll remove this section of the article.J S Lundeen 15:46, 16 June 2005 (UTC)

This distinction is made for example, in some areas of quantum chemistry. See for example, page 93 of:

• Szabo A., Ostlund N.S., Modern Quantum Chemistry - Introduction to Advanced Electronic Structure Theory, Dover Publications Inc., Mineola, New York, 1996.
• It also has ISBN 0-486-69186-1
• It can also be found with the Library of Congress code: QD462.S95 1996.

This reference I cite here is one of the classics of the quantum chemistry field. I agree that the writeup at the bottom is not quite clear. The paragraph at the bottom is an attempt to explain this confusing subtlety. Please feel free to modify it to make it more rigourous. If this |> state is absent from quantum optics, perhaps it would be judicious to point out this distinction in the article. One thing which might complicate this is that both |0> and |> are present in quantum chemistry. --HappyCamper 16:47, 16 June 2005 (UTC)

Thanks. Now that I ask around - it appears that this concept does arise in quantum optics also. Something to do with Bogolubov transformations. My interest has been piqued and I'm going to try to get a better understanding of this.J S Lundeen 20:25, 18 June 2005 (UTC)

Hey, no problem! Glad you're interested. They're also related to Fock spaces too. If you're into programming, I usually think of the |> as the "null state" and |0> as the "zero state". I'm going to add this comment here into the article. --HappyCamper 21:46, 18 June 2005 (UTC)

Is this ``|>`` simply the null vector? --MarSch 11:40, 30 January 2007 (UTC)

If so then I propose to use the conventional notation 0.--MarSch 10:54, 2 March 2007 (UTC)

An advice

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)

yes you can, but you get 0 pretty soon --MarSch 11:42, 30 January 2007 (UTC)

Matrix representation

I thought I understood this stuff, until people started claiming it could be represented by matrices. What does "in the number basis" mean? Can you give an example of say |0102> in this "basis"? 163.1.146.129 21:14, 1 March 2007 (UTC) (signs... Thor2023 21:22, 1 March 2007 (UTC))

Well, let's say you wanted to find the matrix representation of the creation operator. You can make use of the relationship

$A_{ij}=\langle \psi _{i}|{\hat {a}}^{\dagger }|\psi _{j}\rangle$

in which the bras and kets are the wavefunctions of the quantum harmonic oscillator. A_ij is the matrix. HTH. (Hmm...langle and rangle don't seem to work? ) --HappyCamper 04:49, 2 March 2007 (UTC)

Right - this gives you the matrices as given in the article. This is for the case of a single particle in a harmonic potential. But in the more general QFT case - where there are multiple particles in each mode - raising and lowering actually mean something else (adding a particle to a state, not moving a particle to a higher state). Is it possible to create a matrix representation of this situation - as for example Weinberg's book seems to claim you can? I want the form of the matrix eqns (if they exist) that express what I would write as a^+_1 |00>=|10> a^+_2 |00>=|01> a^+_1 |10>~|20> etc. Thor2023 14:24, 2 March 2007 (UTC)

(${\hat {a}}_{1}^{\dagger }|00>=|10>{\hat {a}}_{2}^{\dagger }|00>=|01>{\hat {a}}_{1}^{\dagger }|10>=|20>$) Thor2023 14:25, 2 March 2007 (UTC)

Oops, I misread your question. I'm fairly certain the end result is simply the tensor product of the original matrices. However, generally you try to avoid having to construct this matrix at all, because the dimensionality grows exponentially. You can also think of it this way. In the example of the 2 mode case, your basis states are

$|\psi \rangle =|MN\rangle =|M\rangle \otimes |N\rangle .$

Now, let's say you only want to put at most b quanta in a particular state. b is a fixed number that you pick in advance, say 10. Then, define, say

$|Q\rangle =|Mb+N\rangle$, and of course another $|P\rangle =|Mb+N\rangle$

as your basis states. This is simply a clever way of indexing the original basis. So, now you can easily compute the matrix elements via

$O_{pq}=\langle P|{\hat {O}}|Q\rangle$

where O is your operator of choice. However, this becomes quite problematic for many modes. If you have say 10 modes and 10 quanta, you are dealing with a square matrix that has a side with 10¹⁰ elements already, so very likely some other tricks are involved in the QFT problem you wish to solve. --HappyCamper 19:29, 2 March 2007 (UTC)

Gotcha - so you find some way to map your whole state onto a single number, then you can just do it as for the single particle case. In fact, you probably don't need the maximum number of quanta constraint (so long as you don't mind infinite matrices) as there are various tricks you could play to map multiple integers onto one. Thanks Thor2023 19:48, 4 March 2007 (UTC)

The cap is so that you could actually construct the matrix in a finite computer. Also, the labeling is sort of arbitrary. You could say, number the states like this:

• 0 |00> -- states with 0 total quanta distributed among the modes
• 1 |10> 2 |01> -- states with 1 total quanta distributed among the modes
• 3 |20> 4 |11> 5 |02> -- states with 2 total quanta distributed among the modes

and continue the pattern. So, as an example, the state that you label "|5>" actually means "|02>". The advantage here is that you have all the lower states confined in a more "localized" part of the matrix. As for the "tricks", I was thinking more of things like squeeze operators and bogoliubov transformations. It doesn't sound like a good idea to map multiple states onto one though. The labeling is just a nice way of keeping track of things. --HappyCamper 07:52, 5 March 2007 (UTC)

Vacuum State

"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)

Untitled 2

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

Reaction diffusion

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)

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)

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)

Problem with intro

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)

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)

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)

Generalized creation and annihilation operators

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)

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)

Definition of a_i in "Creation and annihilation operators for reaction-diffusion equations"

I attempted to define a_i. 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 a_i^† and a_i+1 (and other combinations) commute, and does the commutator make sense? 178.38.76.15 (talk) 00:15, 22 November 2014 (UTC)

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)

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)

Reaction diffusion formula

From the article:

The reaction term can be deduced by noting that $n$ particles can interact in $n(n-1)$ different ways, so that the probability that a pair annihilates is $\lambda n(n-1)dt$, yielding a term

$\lambda \sum (a_{i}a_{i}-a_{i}^{\dagger }a_{i}^{\dagger }a_{i}a_{i})$

where number state n is replaced by number state n-2 at site i at a certain rate.

I couldn't make this work out algebraically. 84.226.185.221 (talk) 09:53, 21 September 2015 (UTC)

I've updated the operator definition, and I think it works now. See my comment under Talk:"Definition of a_i in "Creation and annihilation operators for reaction-diffusion equations", based on Gunnar Pruessner's lecture slides. Using these operator definitions, we find:

$(a_{i}a_{i}-a_{i}^{\dagger }a_{i}^{\dagger }a_{i}a_{i})|n\rangle =n(n-1)\left[|n-2\rangle -|n\rangle \right]$

This seems correct to me.— Preceding unsigned comment added by 2a00:23c7:85a2:500:4d4a:7e8a:b088:4f91 (talk) 18:13, 31 May 2021 (UTC)