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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 30 Oct 2019
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      Quick Introduction To Quantum Entanglementpic.twitter.com/9aDo9lAuNC

      11 replies 69 retweets 338 likes
    2. Nikita Lisitsa‏ @lisyarus 12 Dec 2019
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      Replying to @InertialObservr

      From mathematician's point of view entanglement is exactly the fact that not all tensors are decomposable. It totally clicked me once I realized that.

      2 replies 0 retweets 2 likes
    3. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Dec 2019
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      Replying to @lisyarus

      hmm what do you mean? this is the singlet representation, an irreducible representation, of su(2) x su(2)

      2 replies 0 retweets 1 like
    4. Nikita Lisitsa‏ @lisyarus 12 Dec 2019
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      Replying to @InertialObservr

      I mean the following: given a system with Hilbert state X and another with Y, the joint state space is X⨂Y. A state there X & Y are independent would look like a⨂b (a decomposable tensor) for some a∊X and b∊Y, but not all elements of X⨂Y are decomposable.

      1 reply 0 retweets 3 likes
    5. Nikita Lisitsa‏ @lisyarus 12 Dec 2019
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      Replying to @lisyarus @InertialObservr

      In general, elements of X⨂Y are linear combinations of decomposable tensors, just like in your post.

      1 reply 0 retweets 3 likes
    6. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Dec 2019
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      Replying to @lisyarus

      I’m seeing what you’re saying, but I thought that a tensor product of reps can always be decomposed into a direct sum of its irreps.. or is that just SU(N)?

      2 replies 0 retweets 0 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 12 Dec 2019
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      Replying to @InertialObservr @lisyarus

      tensor product of irreps* I mean

      11:27 PM - 12 Dec 2019
      1 reply 0 retweets 0 likes
        1. Nikita Lisitsa‏ @lisyarus 12 Dec 2019
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          Replying to @InertialObservr

          Tensor product of any reps (or irreps) is still a representation, and for compact groups (e.g. SO(N), SU(N), Spin(N), finite groups) any representation decomposes into irreps. (technically I believe you need the reps to be strongly continuous, though)

          0 replies 0 retweets 2 likes
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