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Convergence of vacuum state of Klein Gordon field in a box

  1. Jun 3, 2009 #1

    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

    I get it so far...

    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?

  2. jcsd
  3. Jun 3, 2009 #2


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    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.

    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.
  4. Jun 3, 2009 #3
    Can you please elaborate on that?
  5. Jun 4, 2009 #4


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    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.
  6. Jun 4, 2009 #5
    Ok, so its a distribution. Where can I read more about the connection between quantum fields and distributions? More like an expository article/book..
  7. Jun 4, 2009 #6


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    http://home.uchicago.edu/~seifert/geroch.notes/ [Broken].
    Last edited by a moderator: May 4, 2017
  8. Jun 4, 2009 #7
    Thanks, I'll take a look at these.
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