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Creation Operator is not a densely defined operator...

  1. Aug 23, 2015 #1
    Hi everyone,

    I am currently preparing myself for my Bachelor thesis in local quantum field theory. I was encouraged by my advisor to read the books of M. Reed and Simon because of my lag of functional analysis experience but I have quite often problems understand the “obvious” conclusions.

    For example:

    Why is the creation operator not a densely defined operator? And how do I proof that formally correctly? I am not asking for a step by step solution but I have absolutely no idea how to start.

    Reed, Simon define the creation operator as:

    $$(a^\dagger(p)(\psi))^{(n)}(k_1,...k_n) = \frac{1}{\sqrt{n}} \sum_{l=1}^n \delta(p-k_l) \psi^{(n-1)}(k_1,...,k_{l-1},k_k,...,k_n)$$

    as the adjoint of the annihilation operator :

    $$(a(p)\psi)^{(n)}(k_1,...k_n) = \sqrt{n+1} \psi^{(n+1)}(p,k_1,...k_n)$$

    with $$\psi$$ a Schwartz function.

    I really appreciate any hints.
     
  2. jcsd
  3. Aug 23, 2015 #2

    dextercioby

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    Who told you the operator was not densely defined?
     
  4. Aug 23, 2015 #3
    The book of Simon and Reed. (Fourier Analysis chapter X.7)
    "The adjoint of the operator a(p) is not a densely defined operator since it is given formally by" the expression I posted before. There is no more explanation given. They merely mention that it is possible to define $$a^\dagger$$ as a quadratic form.
     
  5. Aug 25, 2015 #4

    naima

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    I see (books google) that the chapter is about self adjointness.
    Relativist fields need creation and annihilation operators. Not only creator operations.
     
  6. Aug 25, 2015 #5

    A. Neumaier

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    According to the formulas given by you, the annihilation operator is densely defiened, but the creation operator moves the states in the alleged domain out of the Hilbert space because of the delta function, hence is not defined on this domain. (This is probably what is really claimed in your quoted piece of text.) However, smeared creation operators are densely defined since smearing removes the delta function.
     
    Last edited: Aug 26, 2015
  7. Aug 25, 2015 #6

    George Jones

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    In other words, a creation "operator" is actually an operator-valued distribution. See Chapter 10 "Creation and Annihilation Operators" in the book "Quantum Field Theory II" by Zeidler.
     
  8. Aug 25, 2015 #7

    dextercioby

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    As Arnold remarked, the „trick” comes from the delta-Dirac which „explodes” for p equal to any of the k-s, therefore the equality means something only in terms of distributions, i.e. putting integrals and smearing test functions. Actually, the true operators are defined simply by omitting one of the arguments. These operators are defined on pages 208-209 of R-S, Vol.II. For those of you who do not have R-S and understand German, the notes attached by Prof. Keyl are gold. For those who cannot understand German, use the other notes by Prof. Luecke.
     

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    Last edited: Aug 25, 2015
  9. Aug 26, 2015 #8
    I thank all of you.
    The attached notes are a great help. I really appreciate it.
     
  10. Sep 26, 2015 #9

    naima

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  11. Sep 26, 2015 #10
    I think i see what you are stuck on. I am not sure if my response will help you or complicate the issue so i am rather reluctant to post.

    δ(p−kl)
    δ is not a function of (p−kl) according to the usual mathematical definition of a function which requires a function to have a definite value for each point in its domain.



    as above, the range of the integration should contain the origin
     
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