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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 20 Jan 2019
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      I'm starting a series on "Quantum Mechanics Basics". Feel free to comment with any questions about the posted material. Be sure to stay tuned and enjoy!pic.twitter.com/NHmeD3kR4H

      16 replies 107 retweets 470 likes
      Show this thread
    2. Herb da Hegelian‏ @HerbertHitchens 20 Jan 2019
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      Replying to @InertialObservr

      What is an “observable” or should I just say “observer” in QM?

      1 reply 0 retweets 2 likes
    3. 〈 Berger | Dillon 〉‏ @InertialObservr 20 Jan 2019
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      Replying to @HerbertHitchens

      Another good question. An observable is simply a dynamical value that one can measure (e.g. momentum, position, spin, angular momentum etc.). These are represented by hermitian operators because that enforces that the eigenvalues (what we measure) always be real

      2 replies 1 retweet 15 likes
    4. Surge‏ @SCUngureanu 23 Oct 2019
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      Replying to @InertialObservr @HerbertHitchens

      But there are other ways (functions) of producing real values from states |\Psi> . Why are all observables linear operators?

      1 reply 0 retweets 0 likes
    5. Raphael J F Berger‏ @rjf_berger 23 Oct 2019
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      Replying to @SCUngureanu @InertialObservr @HerbertHitchens

      There is a dicussion about non selfadjungated but CP inversion invariant operators, they can have also real eigenvalues. In fact real eigenvalues are actually a consequence of energy conservation. But if a system recieves as much energy as it releases you get CP invariance and

      2 replies 0 retweets 1 like
      〈 Berger | Dillon 〉‏ @InertialObservr 23 Oct 2019
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      Replying to @rjf_berger @SCUngureanu @HerbertHitchens

      Are u sure it’s not PT invariant operators? I’ve seen papers on PT invariant operators that generalize hermitian ones

      7:19 AM - 23 Oct 2019
      1 reply 0 retweets 0 likes
        1. Raphael J F Berger‏ @rjf_berger 23 Oct 2019
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          Replying to @InertialObservr @SCUngureanu @HerbertHitchens

          Sure you are rigth, time momentum, of course. Sorry, I was confused!

          0 replies 0 retweets 0 likes
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