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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 22 Oct 2019
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      Replying to @danijarh @JasonHise64

      there’s a lot of non-Standard terminology there..

      1 reply 0 retweets 1 like
    2. 〈 Berger | Dillon 〉‏ @InertialObservr 22 Oct 2019
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      Replying to @InertialObservr @danijarh @JasonHise64

      There is no probability of measuring phi.. only its eigenvalue

      1 reply 0 retweets 2 likes
    3. Jason Hise‏ @JasonHise64 22 Oct 2019
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      Replying to @InertialObservr @danijarh

      Every different technical field invents different nomenclature for what are essentially the same concepts. 'Non-standard terminology' assumes a preferred frame of reference! :D Brb, googling 'one-hot'.

      1 reply 0 retweets 3 likes
    4. Jason Hise‏ @JasonHise64 22 Oct 2019
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      Replying to @JasonHise64 @InertialObservr

      pic.twitter.com/gK5iVpEDgC

      2 replies 0 retweets 2 likes
    5. Danijar Hafner‏ @danijarh 22 Oct 2019
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      Replying to @JasonHise64 @InertialObservr

      Haha, yeah I mean a vector of the standard basis where one element is 1 and all others are 0. Does it make sense to measure in the standard basis if we can only measure eigenvalues though?

      1 reply 0 retweets 2 likes
    6. 〈 Berger | Dillon 〉‏ @InertialObservr 22 Oct 2019
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      Replying to @danijarh @JasonHise64

      I guess I’m just not sure what you mean by measure in the standard basis .. the measurement is a physical process and is basis independent and is represented by a unitary operator acting on the projective Hilbert space

      2 replies 0 retweets 1 like
    7. 〈 Berger | Dillon 〉‏ @InertialObservr 22 Oct 2019
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      Replying to @InertialObservr @danijarh @JasonHise64

      Terminology 🤧

      1 reply 0 retweets 1 like
    8. Jason Hise‏ @JasonHise64 22 Oct 2019
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      Replying to @InertialObservr @danijarh

      Side effect of CS foundation is thinking in terms of data structures used to represent concepts in memory. In practice this means vectors always need a basis. In this case, a dot product can become a fast lookup if 𝜓 is represented using the operator's eigenvectors as a basis.

      1 reply 0 retweets 3 likes
    9. 心言自由 (XinYaanZyoy)‏ @XinYaanZyoy 22 Oct 2019
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      Replying to @JasonHise64 @InertialObservr @danijarh

      Can we make these vectors independent of basis? So that we can define any basis space whenever we want for these vectors in order to get output accordingly! These basis space may be a data structure, these vectors then gives components based on what data structure we impose!

      1 reply 0 retweets 1 like
    10. Jason Hise‏ @JasonHise64 22 Oct 2019
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      Replying to @XinYaanZyoy @InertialObservr @danijarh

      At some point you have to choose specific numeric values to physically store in each 4-byte floating point slot representing a component of your vector. You can certainly transform this vector into other spaces, but it has to be stored in *some* space to be stored at all.

      1 reply 0 retweets 3 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 22 Oct 2019
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      Replying to @JasonHise64 @XinYaanZyoy @danijarh

      Makes sense I just never think about that

      10:33 PM - 22 Oct 2019
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      • ℙ rojective_orbits Jason Hise 心言自由 (XinYaanZyoy)
      1 reply 0 retweets 3 likes
        1. Jason Hise‏ @JasonHise64 22 Oct 2019
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          Replying to @InertialObservr @XinYaanZyoy @danijarh

          Basis choices have huge effects on performance. If you have a hierarchical scene graph with each node defined in the local space of its parent, and objects in separate branches need to interact, you can waste tons of time multiplying matrices just to get things in the same space!

          0 replies 0 retweets 1 like
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