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michael_nielsen's profile
michael_nielsen
michael_nielsen
michael_nielsen
@michael_nielsen

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michael_nielsen

@michael_nielsen

Searching for the numinous. Co-purveyor of https://quantum.country/ 

San Francisco, CA
michaelnielsen.org
Joined July 2008

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    1. michael_nielsen‏ @michael_nielsen Jan 12
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      A new & peculiar essay, on a (very!) unusual approach I've been using to deepen my understanding of mathematics: http://cognitivemedium.com/srs-mathematics 

      31 replies 110 retweets 536 likes
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    2. Per Vognsen‏ @pervognsen Jan 12
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      Replying to @michael_nielsen

      My favorite way to deepen my understanding of mathematics is to repeatedly re-learn topics when I inevitably forget the details over the years. :)

      1 reply 0 retweets 1 like
    3. Per Vognsen‏ @pervognsen Jan 12
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      Replying to @pervognsen @michael_nielsen

      Incidentally: Any operator T can be written in the form T = A + iB where A and B are self-adjoint operators. Then T* = A - iB, so T T* = A^2 - B^2 + [A, B] and T* T = A^2 - B^2 + [B, A]. So T T* = T* T iff [A, B] = [B, A] = 0 iff A and B commute.

      1 reply 0 retweets 2 likes
    4. Per Vognsen‏ @pervognsen Jan 12
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      Replying to @pervognsen @michael_nielsen

      That lets you deduce the spectral theorem for normal operators from the spectral theorem for self-adjoint operators and simultaneous diagonalization for commuting operators. That's been my preferred proof for a while, but everyone has their pet favorites for these basic theorems.

      1 reply 0 retweets 2 likes
    5. michael_nielsen‏ @michael_nielsen Jan 12
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      Replying to @pervognsen

      Amusingly, that's one of the proofs I discovered (and forgot) on my walk. The proof of diagonalizability for self-adjoint matrices is essentially obvious. The commuting => simultaneous basis needs a little more care, though.

      1 reply 0 retweets 0 likes
      michael_nielsen‏ @michael_nielsen Jan 12
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      Replying to @michael_nielsen @pervognsen

      (I did _not_ think at all about operators, though.)

      11:57 PM - 12 Jan 2019
      1 reply 0 retweets 0 likes
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        1. michael_nielsen‏ @michael_nielsen Jan 13
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          Replying to @michael_nielsen @pervognsen

          Inductive proofs, like that in my essay, are harder to make work in infinite dimensions!

          1 reply 0 retweets 0 likes
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