Convergence of vacuum state of Klein Gordon field in a box

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Discussion Overview

The discussion revolves around the convergence of the vacuum state of the Klein-Gordon field in a box, particularly focusing on the interpretation of the field operator as an operator-valued distribution rather than a conventional operator. Participants explore the implications of this interpretation and its mathematical foundations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that the introduction of the functions \chi_{1} and \chi_{2} is meant to redefine the notion of convergence of the operator \Phi(t,{\bf{x}}).
  • Others argue that \Phi(t,{\bf{x}}) is not an operator, and that it only yields an operator when integrated against a function, which leads to different operators based on the choice of function.
  • A participant seeks clarification on the nature of \mathcal{F}_{s}^{0}(\mathcal{H}), which is described as a Fock space that accommodates varying particle numbers.
  • There is a discussion about the analogy between the quantum field operator and the Dirac delta function, emphasizing that the quantum field operator acts as a distribution.
  • One participant requests resources for further reading on the connection between quantum fields and distributions.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the quantum field operator as a distribution rather than a conventional operator. However, there remains some uncertainty regarding the implications of this interpretation and the specific motivations behind the use of the functions \chi_{1} and \chi_{2}.

Contextual Notes

The discussion touches on the mathematical complexities of operator-valued distributions and their convergence properties, but does not resolve the underlying assumptions or definitions that may affect the interpretation.

Who May Find This Useful

This discussion may be useful for those interested in quantum field theory, operator theory, and the mathematical foundations of distributions in physics.

maverick280857
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Hi

I've been reading through the book "Quantum Field Theory: A Tourist Guide for Mathematicians" by George B. Folland. On page 101, he describes the construction of a scalar field "in a box" [itex]\mathbb{B}: \left[-\frac{1}{2}L,\frac{1}{2}L\right]^3[/itex] in [itex]\mathbb{R}^3[/itex]. Here [itex]\bf{p}[/itex] lies in the lattice:

[tex]\Lambda = \left[\frac{2\pi}{L}\mathbb{Z}\right]^3[/tex]

the field being given by

[tex]\Phi(t,{\bf{x}) = \sum_{\Lambda}\frac{1}{\sqrt{2\omega_{p}L^3}}(e^{ip\cdot x - i\omega_{p}t}A_{p} + e^{-ip\cdot x + i\omega_{p}t}A_{p}^{\dagger})[/tex]

Suppose we apply the operator [itex]\Phi(t,{\bf{x}})[/itex] to the vacuum state:

[tex]\Phi(t,{\bf{x}})|0,0,\ldots\rangle = \sum_{\Lambda}\frac{1}{\sqrt{2\omega_{p}L^3}}e^{-ip\cdot x + i\omega_{p}t}|0,0,\ldots,0,1,0,\ldots\rangle[/tex]

(with the 1 in the [itex]p^{th}[/itex] place).

The author goes on to say that

The square of the norm of this alleged vector is [itex]\sum_{\Lambda}1/2(|{\bf{p}}|^2+m^2)[/itex], which is infinite (by comparison to [itex]\int_{\mathbb{R}^3}d{\bf{p}}/(|{\bf{p}}|^2+m^2) = 4\pi\int_{0}^{\infty} r^{2}dr/(r^2 + m^2)[/itex])...

The way out is to interpret [itex]\Phi[/itex] as a operator-valued distribution rather than an operator-valued function.

I get it so far...

That is, [itex]\Phi[/itex] is the linear map that assigns to each compactly supported [itex]C^{\infty}[/itex] function [itex]\chi_{1}[/itex] on [itex]\mathbb{R}[/itex] and each [itex]C^{\infty}[/itex] [itex]\Lambda[/itex]-periodic function [itex]\chi_{2}[/itex] on [itex]\mathbb{R}^3[/itex] the operator

[tex]\int_{\mathbb{B}}\int_{\mathbb{R}}\Phi(t,{\bf{x}})\chi_{1}(t)\chi_{2}({\bf{x}})dt d{\bf{x}} = \sum_{\Lambda}\frac{1}{\sqrt{2\omega_{p}L^3}}\left[\hat{\chi}_{1}(\omega_{p})\hat{\chi}_{2}(-p)A_{p} + \hat{\chi}_{1}(-\omega_{p})\hat{\chi}_{2}(p)A_{p}^{\dagger}\right][/tex]

with the obvious interpretation of the Fourier coefficients [itex]\hat{\chi}_{1}[/itex] and [itex]\hat{\chi}_{2}[/itex]. The rapid decay of these coefficients as [itex]|{\bf{p}}|\rightarrow\infty[/itex] guarantees that this series converges nicely as an operator on the finite-particle space [itex]\mathcal{F}_{s}^{0}(\mathcal{H})[/itex].

I have a few questions here:

1. Have we introduced the two functions [itex]\chi_{1}[/itex] and [itex]\chi_{2}[/itex] as a means to redefine the notion of convergence of the operator [itex]\Phi(t,{\bf{x}})[/itex]?

2. If yes, what is the motivation for such a definition?

3. What is [itex]\mathcal{F}_{s}^{0}(\mathcal{H})[/itex] really?

Thanks.
 
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maverick280857 said:
I have a few questions here:

1. Have we introduced the two functions [itex]\chi_{1}[/itex] and [itex]\chi_{2}[/itex] as a means to redefine the notion of convergence of the operator [itex]\Phi(t,{\bf{x}})[/itex]?
No, the point is that [itex]\Phi(t,{\bf{x}})[/itex] is not an operator. The only way to get an operator from it is to "smear" it by integrating it against a function. This results in a different operator for every choice of function. Here the author is just using two different functions to smear in time and space.

maverick280857 said:
3. What is [itex]\mathcal{F}_{s}^{0}(\mathcal{H})[/itex] really?
Fock space. Basically get the Hilbert space for a Klein-Gordon particle, attach the Hilbert space for two particles and so on, until you have a Hilbert space which can deal with any particle number. It's all the Hilbert spaces for different particle numbers put together.
 
DarMM said:
No, the point is that [itex]\Phi(t,{\bf{x}})[/itex] is not an operator. The only way to get an operator from it is to "smear" it by integrating it against a function. This results in a different operator for every choice of function. Here the author is just using two different functions to smear in time and space.

Can you please elaborate on that?
 
maverick280857 said:
Can you please elaborate on that?
The quantum field operator isn't an operator. It simply gives you operators when you integrate it against a function. Just like the dirac delta isn't a function, but gives you a number when integrated against a function.
 
DarMM said:
The quantum field operator isn't an operator. It simply gives you operators when you integrate it against a function. Just like the dirac delta isn't a function, but gives you a number when integrated against a function.

Ok, so its a distribution. Where can I read more about the connection between quantum fields and distributions? More like an expository article/book..
 
http://home.uchicago.edu/~seifert/geroch.notes/ .
 
Last edited by a moderator:
Thanks, I'll take a look at these.
 

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