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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. Anomaly Canceller‏ @litgenstein Feb 21
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      Interesting Q someone asked: quantum electrodynamics has an obvious classical theory it corresponds to, but quantum chromodynamics and electroweak theory don’t—why’s that? Possible answer: there’s electromagnetism in daily life, but no strong or electroweak force

      12 replies 9 retweets 130 likes
      Show this thread
    2. Sean Carroll‏Verified account @seanmcarroll Feb 21
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      Replying to @litgenstein

      There definitely is a classical Lagrangian for those theories. It’s just that quantum effects are very important in the presence of confinement and SSB, not so much when everything’s perturbative.

      2 replies 2 retweets 47 likes
    3. Anomaly Canceller‏ @litgenstein Feb 21
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      Replying to @seanmcarroll

      Good stuff! Do you think the classical lagrangian is unique? Or that it depends on how you take the classical limit?

      3 replies 0 retweets 16 likes
    4. 〈 Berger | Dillon 〉‏ @InertialObservr Feb 21
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      Replying to @litgenstein @seanmcarroll

      In non perturbative QFTs there actually exist observables for which we don't have Lagrangians .. Do you mean unique modulo gauge symmetries & conserved global symmetries? Even the Dirac equation gives the same physics depending on if you use +-m

      4 replies 0 retweets 11 likes
    5. Eric Aspling‏ @AsplingEric Feb 21
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      Replying to @InertialObservr @litgenstein @seanmcarroll

      Kinda curious what kind of observable we could find in a non-perturbative QFT that doesn't have a Lagrangian. Maybe something from QCD?

      1 reply 0 retweets 0 likes
      〈 Berger | Dillon 〉‏ @InertialObservr Feb 21
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      Replying to @AsplingEric @litgenstein @seanmcarroll

      consider the operator e^(-1/φ^2)

      3:34 PM - 21 Feb 2020
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      • Eric Aspling Anomaly Canceller
      1 reply 0 retweets 2 likes
        1. New conversation
        1. Eric Aspling‏ @AsplingEric Feb 21
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          Replying to @InertialObservr @litgenstein @seanmcarroll

          Honestly, I'd have no idea how to treat that operator to get any reasonable number. Therefore, imma take your word for it! Thank you!

          1 reply 0 retweets 0 likes
        2. 〈 Berger | Dillon 〉‏ @InertialObservr Feb 21
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          Replying to @AsplingEric @litgenstein @seanmcarroll

          we treat all functions of field operators as their taylor series expansion (hint)

          1 reply 0 retweets 0 likes
        3. 5 more replies

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