Sunday, October 6, 2024

Navigating the Quantum Maze: Spin State Vectors and the Magic of Projection

Welcome to the Spin Vector Universe

Today, we embark on a cosmic adventure, learning how to navigate through a space where the points are none other than spin ½ state vectors. These vectors, much like secret agents, carry both the “observable information” that we can detect and a hefty dose of hidden data that remains mysterious—at least for now. Who knows what tomorrow’s physicists will uncover about these enigmatic vectors? Perhaps one day, even consciousness itself (and even  "psi" phenomena) will be written into the math.

These vectors, much like secret agents, carry both the “observable information”...

That’s right, physicists may finally crack the "consciousness code," but until then, let's stick to the basics—like juggling four numbers in a multidimensional space.


The Joy of Four Numbers

We initially presented the state vector as a pair of points on a two-dimensional sphere. But why settle for simplicity when we can dive into complexity? We now represent it by four real numbers: X, Y, Z, and W. And if you’re keeping score, the squares of these four beauties add up to 1:

X² + Y² + Z² + W² = 1

But wait, there's more! We can rewrite these numbers using angular coordinates φ, θ, and ψ. Let’s get fancy:

  • X = sin(φ/2) cos(ψ)
  • Y = sin(φ/2) sin(ψ)
  • Z = cos(φ/2) cos(ψ+θ)
  • W = cos(φ/2) sin(ψ+θ)

The ranges of variation of these angles are:

  •  φ from 0 to π. (latitude)
  • θ, ψ from 0 to 2 π.

 Cue the applause from the programmers in the room! Of these three angles, only θ and φ are observable in our space; ψ remains the mysterious guest who refuses to show their face.

NoteIn quantum mechanics we usually present the spin vector as a column of two complex numbers, the "Pauli spinor":

X + iY

Z + iW


A Tour of Spherical Spaces

To keep things simple (or at least as simple as quantum mechanics gets), here’s a quick refresher:

  • X² + Y² = 1: Points on the plane (X, Y) lie on a one-dimensional circle.
  • X² + Y² + Z² = 1: Points in three-dimensional space lie on a two-dimensional sphere.
  • X² + Y² + Z² + W² = 1: Points with coordinates (X, Y, Z, W) lie on a… wait for it… three-dimensional sphere embedded in four-dimensional space.

Wrap your head around that one! But don’t worry, mathematicians have our backs. Using stereographic projection, they can help us visualize these four-dimensional wonders in our humble three-dimensional world.


Stereographic Projections: Cartography for the Quantum Age

Remember those flat maps of the Earth that distort everything around the poles? That’s stereographic projection in action. We can apply the same trick to our three-dimensional sphere, squishing it down into a more manageable three-dimensional space. Here’s the cartographer’s recipe:

  • If X, Y, Z are the coordinates on a 2-sphere, the projection onto a plane gives:
    • x = X / (1 - Z)
    • y = Y / (1 - Z)


To up the ante for our four-dimensional friends:

  • For X, Y, Z, W, the projection into 3D space becomes:
    • x = X / (1 - W)
    • y = Y / (1 - W)
    • z = Z / (1 - W)

What Do We Gain? What Do We Lose?

Through stereographic projection, we lose just one point: the North Pole, or in this case, the point where W = 1. When W = 1, X, Y, and Z all vanish into thin air, leaving us with a single state vector that escapes to infinity. But don’t worry, we’ll survive this minor loss.

On the plus side, we get to visualize the structure of spin state vectors in all their three-dimensional glory, and the angles between intersecting lines remain preserved. Cartographers everywhere are cheering in solidarity.


The Aesthetics of Spin Space

Is it worth painting a picture of the spin state vector space? Absolutely! 


Both for its educational value and sheer aesthetic beauty. We can craft these images with relative ease (shout-out to the programmers) and even delve into the lovely world of Villarceau circles, which have graced the stairs of Strasbourg Cathedral for centuries. Talk about a deep cut!

In our next post, we’ll dive into how to create such images and translate them into physicist-friendly terms. For now, imagine a torus—a mathematical donut—and the intricate circles that define its geometry. Doesn’t quantum physics just make you hungry?


PS: For Your Viewing Pleasure

If you’re curious about more stunning images and videos (probably better than the ones I can whip up), check out the “Dimensions” website: Dimensions. Just a heads-up, though—they skip the spin physics. You’ll have to come back here for that!


And there you have it—our journey through the quirky, brain-bending world of spin state vectors. Let’s call it the appetizer for the main course, which will be served in the coming posts. Until then, keep your angles sharp and your vectors spinning! 

P.S.07-10-24 13:26 In the meantime I am working on my paper about positive and negative energy conformal universe. Have just created a nice illustration for 1+1 spacetime dimensions. Matter and antimatter are separated by "conformal infinity". Here it is:

Universe as a yummy donut
P.S. 07-10-24 20:38 Minus for Igor - From "Современные геометрические структуры и поля", p. 279:




P.S. 09-10-24 10:56 My new post will be delayed, as I am learning how to use the 3D graphics program that I have discovered yesterday.


Hopefully I will be able to figure it out later today.
P.S. 09-10-24 19:07:56 Finally I have managed to use this software and draw what I wanted to draw. Still lacking Villarceau circles though.
Explanations will follow tomorrow, with a new post.



54 comments:

  1. What are those tori a projection of?
    What are those black lines on the tori a projection of?

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    1. Thanks for asking! All this will be coming in next posts. The last picture in this post is just a teaser.

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    2. A very provocative and intriguing teaser, for sure! Thanks!

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    3. Perhaps I will start revealing some secrets sooner. But I need to test available free 3d graphing software, so that a Reader can start making plots of elements of the teaser herself.

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  2. Если в вашей конформной Вселенной используются окружности Вилларсо, то, на всякий случай, учтите, что они не совпадают с (1,1)-торическими узлами. Кстати, про случайные блуждания вы не забыли?

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    1. I did not forget. But still I would like to know how did you get (1.1.10)? And what is the relation of (x1,x2,x3,x4) to quaternions?

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    2. Чуть позже. То был на свадьбе, то теперь на работе.

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  3. Алгебра линейных векторных полей изоморфна матричной алгебре, поэтому (1.1.10) соответствуют матричным образующим алгебры кватернионов. Например, первое векторное поле соответствует матрице A = \diag[\antidiag[1,-1], \antidiag[1,-1]], где A^2 = \diag[-1,-1,-1,-1]. В свою очередь, координаты (x1,x2,x3,x4) имеют точно такое же отношение к кватернионам как координаты, сумма квадратов которых равна радиусу трёхмерной сферы. Иначе говоря, только в таких координатах будет ковариантной алгебра линейных векторных полей, изоморфная алгебре кватернионов.

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    1. "Алгебра линейных векторных полей изоморфна матричной алгебре"

      Where is this isomorphism? Can you write it down? Or provide a reference?

      "координаты (x1,x2,x3,x4) имеют точно такое же отношение к кватернионам..."

      Can you provide this relation explicitly?

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    2. When you say "algebra", do you mean "Lie algebra"? If so, why are you so imprecise? Matrices we can multiply, but we can't multiply vector fields!

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    3. Нет, я не имел в виду алгебры Ли. А векторные поля вполне себе умножаются. Например, на плоскости (x,y), используя понятие производная компоненты векторного поля B по направлению векторного поля A, получаем определение производной векторного поля B по направлению векторного поля A, а именно, \nabla_{A} B = \nabla_{A} b_{x}\partial x + \nabla_{A} b_{y}\partial y

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    4. Строгое изложение вопроса о соответствии алгебры векторных полей и матричной алгебры смотрите в книге
      С.П. Новиков, И.А. Тайманов, Современные геометрические структуры и поля, Москва, МЦНМО, 2005.

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    5. Very nice book! Very nice indeed. But I checked, and I am not able to find in this book the isomorphism you are talking about!

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    6. Странно, но (по памяти) там целая глава (или параграф) о линейных векторных полях.

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    7. страница 279

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    8. Indeed there is a whole paragraph about linear vector fields, but about Lie algebra, with the commutator, NOT as an algebra with just multiplication.

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    9. Параграф 8.3 Векторные поля

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    10. Lie algebra, Igor. Commutators. Not multiplications.

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    11. Это всего лишь потому, что авторы позволяют себе говорить о ковариантных алгебрах. Однако, как мне помнится, о соответствии алгебры линейных векторных полей и алгебры Mat(n, R) они всё же упоминают.

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    12. они всё же упоминают. Yes yes, but it is about Lie algebras, not about "algebras". Theorem 8.9. It is about commutators, not about simple multiplications.

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    13. An interesting coincidence: working on my paper I just while ago came onto the matrices diag[\antidiag[1,-1], \antidiag[1,-1]] and \antidiag[1,-1] for my own reasons while playing with transformations in 6d space!

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    14. Хорошо, пусть авторы книги и говорят только о соответствии коммутаторов векторных полей и коммутаторов матриц. На ваш взгляд, разве нет соответствия произведения линейных векторных полей и произведения матриц? Если есть, то есть и соответствие алгебры линейных векторных полей и полной матричной алгебры.

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    15. When we restrict ourselves to invariance only only under affine transformations, perhaps it can be done. I will check it.

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    16. If, following the book, we denote by T_A the vector field corresponding to the matrix A, and if we use your prescription of multiplying vector fields, we obtain
      T_{AB} = -T_B T_A, NOT an isomorphism of algebras. But on commutators we get isomorphism of Lie algebras.

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    17. Igor, using your own prescription I decoded the three matrices responsible for (1.1.10). They make no relation to quaternions, even though square of each gives -1. Here they are:
      Ax = {{0, 1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, -1, 0}};
      Ay = {{0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 0, 0, 0}, {0, -1, 0, 0}};
      Az = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}};
      For instance Ax and Az anticommute, but Ax and Ay commute, Ay and Az commute.

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  4. Сначала возьмите произвольные линейные преобразования.

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    1. Обратимые, естественно.

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    2. Откуда вылез минус? Посмотрите на простых примерах. Там нет минуса.

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    3. See P.S. 0710-24 20:38 under the note. Minus is necessary so that the authors of this book can get the Lie algebra isomorphism.

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    4. So, look at my comment October 8, 2024 at 3:17 PM, and tell: how did you get (1.1.10)? What it is supposed to represent?

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    5. Если использовать обозначение без минуса A*B = \nabla_B(A), то искомый изоморфизм имеет место быть.

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    6. That's better. But how did you get (1.1.10)?

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    7. Под вашим постом с матрицами я ответил как можно исправить ошибку, чтобы все образующие линейные векторные поля антикоммутировали.
      А вообще, мне приходилось конструировать линейные векторные поля нечётномерных сфер. Если интересно, то полистайте последнюю главу книги. Там как раз и ответ на ваш прямой вопрос.

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    8. "Под вашим постом с матрицами я ответил ..."

      Где именно этот ответ?

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    9. Мне не сложно повторить свой ответ. Предлагалось заменить A_y на (A_x)*(A_z).

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    10. Igor, what you wrote doesn't make any sense. Can you answer these simple questions:
      1) how did you get (1.1.10)?
      2) what are your matrices that are supposed to anticommute, and how they relate to your vector fields (1.1.10)?

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    11. Ну вот, я старался, строил векторные поля сфер, связывал их с антикоммутирующими матрицами, а оказывается, что это не имеет никакого смысла. Зачем тогда вы спрашиваете как я получил эти (1.1.10)? Вопрос то простой, но не могу я ответить просто.

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    12. Igor, if you are not able to answer simple questions about formulas in your paper, your paper is only adding to the noise, it is misleading the readers, you are wasting other people time. Don't you care?

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    13. Мне не всё равно, но эти формулы во введении, а там невозможно сразу всё объяснить. Большая часть пояснений в основном тексте. Что касается формул (1.1.10), то из контекста понятно, что это векторные поля, касательные к 3-сфере, а то, что представляющие их матрицы должны антикоммутировать для того, чтобы векторные поля были ортогональны друг к другу я там, действительно, не отметил. А получить их можно и эмпирически.

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    14. Igor, you do not really care. I am asking you: write down these matrices. Then we will check if they anticommute or not. And we will check if they indeed corresponds to vector fields in (1.1.10). Is that too difficult for you?

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    15. Аркадиуш, я обязательно исправлю ошибки, на которые вы мне указали. С 20.00 на работе у меня только мобильный интернет (на телефоне), поэтому все формулы переносятся на завтра.

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    16. Arkadiusz

      At 5:23 Igor wrote:
      Предлагалось заменить A_y на (A_x)*(A_z).

      So now:
      Ax = {{0, 1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, -1, 0}};
      Ay = {{0, 0, 1, 0}, {0, 0, 0, -1}, {-1, 0, 0, 0}, {0, 1, 0, 0}};
      Az = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}};

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    17. @Bjab These matrices anticommute, but they do not correspond to the vector field 1.1.10).

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    18. Maybe because there are typos in (1.1.10).

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    19. Anyway they are some permutation of gamma matrices:
      https://en.wikipedia.org/wiki/Gamma_matrices

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    20. @Bjab Maybe. But that is what Igor needs to tell us. I think one stil needs to change signs of all three matrices if they are supposed to represent quaternions.

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    21. "... one stil needs to change signs of all three matrices if they are supposed to represent quaternions."

      Or change succession

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    22. Wikipedia for instance quoptes two versions of representing quaternions by real 4x4 matrices

      https://en.wikipedia.org/wiki/Quaternion

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    23. Пару слов о связи матриц и линейных векторных полей. Если у нас имеется векторное поле X= (X_1,X_2,X_3,X_4), то из него с помощью матрицы действительных чисел A можно получить векторное поле AX. Если мы хотим получить касательное к S^3 линейное векторное поле, то нам надо взять такую матрицу A, что A^2=-I, где I - матричная единица, поскольку только такая матрица обеспечит нам условие ортогональности (AX,X)=0. Вместе с тем, если мы хотим получить другое векторное поле A'X, ортогональное полю AX, то есть чтобы (AX,A'X)=0, необходимо чтобы их матрицы антикоммутировали, то есть чтобы AA'=-A'A. Для S^3 найдутся три таких ортогональных друг другу векторных поля, а именно:
      A_1=(-X_{2},X_{1},-X_{4},X_{3})
      A_2=(-X_{4},-X_{3},X_{2},X_{1})
      A_3=(-X_{3},X_{4},X_{1},-X_{2})
      Если я там наврал в (1.1.10), то уж извините, виноват, исправлюсь.

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    24. Да, Игорь, ты наврал, но Господь Бог простит тебя и даже поможет тебе. Через час-два в новой заметке в блоге.

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  5. ∂ρ = x∂x + y∂y (1.1.2)

    How does such thing come out?

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    Replies
    1. This follows from the principle of least action for the rotational motion of matter. (1.3.1) - (1.3.5)
      https://www.researchgate.net/publication/377842836_Mathematical_Notes_on_the_Nature_of_Things_fragment

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    2. Arkadiuszu
      Do you understand the above Igor's explanation?

      Shouldn't it be not
      ∂ρ = x∂x + y∂y (1.1.2)
      but
      ρ∂ρ = x∂x + y∂y (1.1.2)

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    3. @Bjab Sure, if only for dimensional reasons.

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Thank you for your comment..

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