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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. 〈 Berger | Dillon 〉‏ @InertialObservr 18 Dec 2019
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      Is empty space 'empty'? Fields extend throughout all space. So photons moving through space 'know' about other fields (like the e⁻ field) This 'knowing' is called field interaction, & the leading order correction to the vacuum in E&M is a virtual electron, anti-electron pairpic.twitter.com/1wFEus8KB0

      18 replies 101 retweets 516 likes
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    2. 〈 Berger | Dillon 〉‏ @InertialObservr 18 Dec 2019
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      In Quantum Field Theory, 'classical' empty space is the lowest order term in the expansion of the full vacuum A photon in 'classical' empty space will propagate with no interactions Saying 'empty space is not empty' is just to say that the quantum 'loop' corrections are nonzeropic.twitter.com/iY12RonriR

      3 replies 24 retweets 101 likes
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    3. Rogier Brussee‏ @RogierBrussee 19 Dec 2019
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      Replying to @InertialObservr

      Isn't in QFT, the "classical" vacuum just the ground state in Hilbertspace of the Hamiltonian constructed from the fields with all interactions higher than quadratic set to zero? (I wish I understood that Hilbert space with an action of fields concretely, say as some L_2)

      2 replies 0 retweets 0 likes
    4. udit gupta‏ @guptadagger 20 Dec 2019
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      Replying to @RogierBrussee @InertialObservr

      These operators always exist in any Poincare-invariant QFT. In an interacting gapped QFT, we say the vacuum has "virtual" particles because the Hamiltonian now doesn't commute at different times but we still require the vacuum to have a zero energy eigenvalue.

      1 reply 0 retweets 1 like
    5. Rogier Brussee‏ @RogierBrussee 20 Dec 2019
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      Replying to @guptadagger @InertialObservr

      Mmmh let me brood on that. My fundamental problem is "the state" assumes a (projectivised) Hilbert space on which the field you want to quantise acts. A representation of the Poincaré group (or more generally the isometries of the Lorentzian manifold) should come for free.

      1 reply 0 retweets 0 likes
    6. udit gupta‏ @guptadagger 21 Dec 2019
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      Replying to @RogierBrussee @InertialObservr

      I assumed the canonical quantization scheme which does infact assume a Hilbert space with a particular inner product and existence of an algebra of linear operators (in a representation of the Poincare group) that act on the Hilbert space.

      1 reply 0 retweets 0 likes
    7. Rogier Brussee‏ @RogierBrussee 22 Dec 2019
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      Replying to @guptadagger @InertialObservr

      Canonical quantisation is (for a mathematician) a deeply unsatisfactory procedure. "promote Fourier components (on 3-space or prob. forward light cone) to operators on Hilbert space with equal time canonical commutation relations".🤢 And depends on Lagrangian, and x-t split!?

      2 replies 0 retweets 1 like
      〈 Berger | Dillon 〉‏ @InertialObservr 22 Dec 2019
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      Replying to @RogierBrussee @guptadagger

      is there a more fundamental thing going on when we ‘promote’ operators?

      5:33 AM - 22 Dec 2019
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        2. Rogier Brussee‏ @RogierBrussee 22 Dec 2019
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          Replying to @InertialObservr @guptadagger

          Obviously. It means that you _have_ actually constructed a Hilbert space with a unitary representation of the Poincare group and a *-representation of an algebra constructed from "the (cotangent?) space of fields" and its all equivariant and suitably continuous. Just how?

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        1. Rogier Brussee‏ @RogierBrussee 23 Dec 2019
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          Replying to @InertialObservr @guptadagger

          These notes by Prakash Panangaden come fairly close to what I mean for the simplest case real Klein Gordon: https://www.cs.mcgill.ca/~prakash/Qft/lecture6.pdf … (see also lecture 3, 7, 8)

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