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Insights Mathematical Quantum Field Theory - Observables - Comments

  1. Nov 19, 2017 #1

    Urs Schreiber

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  2. jcsd
  3. Nov 19, 2017 #2

    strangerep

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    Typo near "linear off-shell observables": $$A(\Phi_1+\Phi_2) ~ A(\Phi_1)+A(\phi_2)$$ I.e., "=" sign is missing, and 2nd phi is lower case.

    And further down... "field field histories"

    Also: "...in that that subset of field histories..."
     
    Last edited: Nov 19, 2017
  4. Nov 20, 2017 #3

    Urs Schreiber

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    Thanks! All fixed now.
     
  5. Nov 21, 2017 #4

    strangerep

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    Another typo: hiorizontal
     
  6. Nov 21, 2017 #5

    Urs Schreiber

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    Thanks again! Also fixed now.
     
  7. Jan 28, 2018 #6
    I have troubles with the comment about observables on fermionic fields (under the definition of observables)
    1. Why ##Obs = [\Gamma_\Sigma(E_{odd}) , \mathbb{C}]## has only 1 point?

    I assume that when we talk about "points", we're using ##Obs(\mathbb{R}^0) ##. However,
    $$
    \begin{align*}
    Obs(\mathbb{R}^0) &= \{\mathbb{R}^0 \rightarrow Obs \} \\
    &= \{\mathbb{R}^0 \rightarrow [\Gamma_\Sigma(E) , \mathbb{C}] \}\\
    &= \{\mathbb{R}^0 \times \Gamma_\Sigma(E) \rightarrow \mathbb{C} \} \\
    &= \{\Gamma_\Sigma(E) \rightarrow \mathbb{C} \}
    \end{align*}
    $$
    (this is true for both ##E = E_{even}## and ##E=E_{odd}##.) We know that ##\Gamma_\Sigma(E_{odd})## has only 1 "point" (the zero-section), and so hand-wavvily I would assume that there are several morphisms going from that unique "point" to ##\mathbb{C}## which correspond to several "points" in ##Obs = [\Gamma_\Sigma(E_{odd}) , \mathbb{C}]##.

    2. What is the map ##(\theta \mapsto \theta A): \mathbb{R}^{0|1} \rightarrow Obs ##? (The notation seems suggesting but I could not yield any conclusion.)

    I am inclined to think this is an extension/consequence of the correspondence between a morphism ##\psi_{(-)}:\mathbb{R}^{0|1} \rightarrow \Gamma_\Sigma(E_{odd}) ## and a morphism ##\phi_{(-)}:\mathbb{R}^{0} \rightarrow \Gamma_\Sigma(E_{even}) ##, but I haven't been successful to see how.
     
    Last edited: Jan 28, 2018
  8. Jan 28, 2018 #7

    Urs Schreiber

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    Thanks for catching this. Indeed this comment was wrong as stated. I was trying to say that these global points of the space of field histories of the Dirac field always yield the zero linear observable for the fermions (i.e cannot observe ##\mathbf{\Psi}^\alpha(x)##), but of course they do see the even powers, such as ##\mathbf{\Psi}^\alpha(x) \cdot \mathbf{\Psi}^\beta(x)##. I have corrected the statement in that comment and added a new example that says this in some detail, now here
     
  9. Jan 28, 2018 #8
    Thanks! I think I got what you meant. Then perhaps still in that comment there're a few typos:

    1. The plot ##\mathbb{R}^{0|1} \rightarrow Obs## should be specified by ## c\mapsto \theta \mathbf{\Psi}^\alpha##.
    2. (in the line below) The bosonic-observable component should be ## \mathbf{\Psi}^\alpha ## instead of ##\Pi##.

    I think it is useful to add that the bosonic observable which originally is of even degree now is regarded as of odd-degree (so that ##\theta \mathbf{\Psi}^\alpha ## is of even degree), and this fact is made possible by Prop. 3.51. Also, I think it is very nice to keep the mentioning of "the zero linear observable" if we insist on thinking of the observables as "points".
     
    Last edited: Jan 28, 2018
  10. Jan 29, 2018 #9

    Urs Schreiber

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    Right, thanks again. Should be fixed now.
     
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