Talk:Fock space

Help!

Could somebody knowledgeable in this area take a look at the article. The sentence "WARNING: Fock space only describes noninteracting quantum fields. See Haag's theorem." is strange, but I don't know what to do about it. Thanks! Oleg Alexandrov 02:34, 9 Mar 2005 (UTC)

"(to describe many species of particles, made the tensor products of as many different Fock spaces)." I assume that "make" is meant, and not "made." Since I don't know that, I'll let someone else fix it.

Why there are two different types of phi??? --77.176.71.6 (talk) 17:25, 18 December 2009 (UTC)

Definition

In the definition section it says:

$|0\rangle$ is a vector of length 1, called the vacuum state and $\,a_{0}\in \mathbb {C}$ is a complex coefficient,

$|\psi _{1}\rangle \in H$ is a state in the single particle Hilbert space,

But it doesn't say anywhere whether

$|\psi _{2}\rangle$ is a different state in the same single particle Hilbert space or if it is a state in a two-particle Hilbert state.

Can somebody who knows clarify the article? — Preceding unsigned comment added by 129.6.107.65 (talk) 15:44, 25 June 2014 (UTC)

The 10th line is wrong.It is instead (-1)^{\pi{ij}}. — Preceding unsigned comment added by 183.63.97.18 (talk) 12:04, 30 August 2016 (UTC)

Bottom two paragraphs

I can't seem to make head or tail out of the last two paragraphs, so I have moved them over here instead:

WARNING: Fock space only describes noninteracting quantum fields. See Haag's theorem. However, if the model has a mass gap, the asymptotic past and asymptotic future states can be described by a Fock space. Fock space does not describe finite temperature physics as well.

While Fock space is appropriate for free massive particles with finite energy (i.e. zero temperature) because for a collection of these particles, finite total energy and finite particle number mean the same thing, it is no longer appropriate for massless particles because an infinite number of them can still have finite energy. See soft photon

Some of the terminology seems out of place, and sounds extremely foreign in the context of quantum chemistry. Perhaps it would make more sense in quantum electrodynamics? --HappyCamper 02:02, 19 November 2005 (UTC)

The context doesn't have to be quantum chemistry. Fock spaces are used in quantum field theory in general. I dunno what the guy is talking about though. I think this is an example of people from different fields talking past each other - for instance I've seen Fock space used to derive temperature-dependent stuff, but then again I'm a condensed matter guy. These kinds of conflict are somewhat common on physics-related pages, I wonder if there's some kind of resolution mechanism. Nvj 20:57, 29 May 2006 (UTC)

Is all the notation correct within this article, why these different phi for single particle states? and one could better describe the indices, because this is where many of the things are that one needs to understand. why are the indices sometimes in the bracket, sometimes out of the bracket?

I think the caveats arising from Haag's theorem are very important and should be restored. 86.147.126.104 (talk) 03:49, 30 March 2023 (UTC)

Hilbert space

As far as I understand it, it is not true that the Fock space is a Hilbert space. It is not possible to have a scalar product between states with a different number of particles.

For every k, the k-particle space is a Hilbert space. But the direct sum of all these spaces is not.

FelixP (talk) 15:12, 16 November 2009 (UTC)

Well, as far as I know and also according to Wikipedia, this is possible ([[1]]) Joasiak (talk) 17:17, 29 November 2009 (UTC)

Ok thanks, I undid it. I had misinterpreted something in the chemistry liturature. FelixP (talk) 01:24, 2 December 2009 (UTC)

Relation to Bargmann-Fock space

This article could do with a discussion of how Fock spaces are related to the Bargmann-Fock space i.e.

${\mathcal {F}}^{2}(\mathbb {C} ^{n})=\{f\colon \mathbb {C} ^{n}\to \mathbb {C} \mid \Vert f\Vert _{{\mathcal {F}}^{2}(\mathbb {C} ^{n})}<\infty \}$, where $\Vert f\Vert _{{\mathcal {F}}^{2}(\mathbb {C} ^{n})}:=\int _{\mathbb {C} ^{n}}\vert f(\mathbf {z} )\vert ^{2}e^{-\pi \vert \mathbf {z} \vert ^{2}}\,d\mathbf {z}$

Perhaps the relationship is just "there isn't one (apart from the name)", but even mentioning this would be useful. But if, as I suspect, they are different views (quantum mechanical / pure mathematical) of the same thing, a description of this would be very useful. 128.86.179.98 (talk) 15:30, 21 June 2011 (UTC)

According to the one answer to this related question, the two are isomorphic, and "the creation and anihilation operators are just the multiplication a_j = z_j and the derivation a*_j = d/dZ_j and consequently, the theory of several complex variables can be used for the analysis on this space". Obviously that page isn't a reliable source itself, but the author also gives a couple of useful references. 128.86.179.98 (talk) 15:48, 21 June 2011 (UTC)

Convergence Issue

The following quote, at the end of the Definition section, is confusing:

"The convergence of this infinite sum is important if $F_{\nu }(H)$ is to be a Hilbert space. Technically we require $F_{\nu }(H)$ to be the subspace of $\bigoplus _{n=0}^{\infty }S_{\nu }H^{\otimes n}$ which consists of all vectors with finite norm (where the norm is defined by the inner product as $\|\Psi \|_{\nu }:={\sqrt {\langle \Psi |\Psi \rangle _{\nu }}}$)."

The direct sum requires that all but finitely many coefficients are zero, so every vector is a sum of finitely many vectors. Since $S_{\nu }H^{\otimes n}$ is a Hilbert space, every vector in $\bigoplus _{n=0}^{\infty }S_{\nu }H^{\otimes n}$ has finite norm. The simple way to fix this would be to eliminate the paragraph entirely, since at the moment it's irrelevant and confusing. However, I may be misinterpreting this. If the direct sum should instead be a direct product, then some sense can be made of the statement. In particular, then the statement would be necessary. Not being a professional physicist, I don't know which is correct, but I suspect the latter space is the one that is intended. Could someone who knows about this sort of thing edit the page (or reply here to tell me why it's correct as is)? 129.15.139.211 (talk) 01:44, 13 December 2011 (UTC)

Okay, ignore the above. Direct sums in the category of Hilbert Spaces are defined differently from arbitrary modules. The quoted paragraph is certainly unnecessary then, but it's not incorrect and not really confusing. I'll leave it up to the regular editors whether any change should be made. Btw this is the same person as above even though my ip is different. 68.97.39.154 (talk) 12:02, 13 December 2011 (UTC)

Changed the formulation such that the Hilbert direct sum is a subspace of the (algebraic) direct _product_ of spaces with finite norm--RogierBrussee (talk) 18:47, 31 January 2012 (UTC)

Missing coefficient in defnition?

In the defnition, we read that:

A typical state in $F_{\nu }(H)$ is given by

$|\Psi \rangle _{\nu }=|\Psi _{0}\rangle _{\nu }\oplus |\Psi _{1}\rangle _{\nu }\oplus |\Psi _{2}\rangle _{\nu }\oplus \ldots =a_{0}|0\rangle \oplus |\psi _{1}\rangle \oplus \sum _{ij}a_{ij}|\psi _{2i},\psi _{2j}\rangle _{\nu }\oplus \ldots$.

Why is there no coefficient for $|\psi _{1}\rangle$? I suggest the following (with an $a_{1}$):

$|\Psi \rangle _{\nu }=|\Psi _{0}\rangle _{\nu }\oplus |\Psi _{1}\rangle _{\nu }\oplus |\Psi _{2}\rangle _{\nu }\oplus \ldots =a_{0}|0\rangle \oplus a_{1}|\psi _{1}\rangle \oplus \sum _{ij}a_{ij}|\psi _{2i},\psi _{2j}\rangle _{\nu }\oplus \ldots$.

——Mcasariego (talk) 13:21, 19 August 2015 (UTC)Mcasariego

PR stunt

The comment about the error in the end of section "Wave Function Interpretation" does not respect the Wikipedia rule stating that articles should not contain original work. — Preceding unsigned comment added by 130.233.206.195 (talk) 07:23, 8 August 2016 (UTC)

Simple example?

There is quite a bit of mathematical abstraction here, but no concrete example, which can make it hard for newbies to get a sense of what's going on.

Abstract, generalized discussions are useful as a reference for experts, but inappropriate as an introduction. Wikipedia is not a textbook, so I'm not requesting exercises and worked examples, but a single intuitive example would make this page useful rather than useless as it is. — Preceding unsigned comment added by 129.219.43.70 (talk) 20:19, 10 November 2016 (UTC)