Thursday, October 3, 2024

Untangling the Mysteries of Spinors: A Wild Ride Through 4D Geometry (Who Said Math Wasn't Fun?)

Minimizing the Mystery (Kind of)

In my spinor notes, I started with the humble attempt to strip away some of the mystique. You see, a spinor is just a pair of points in "internal space." Nothing too spooky, right? Well, actually, it kind of is spooky—but more on that later. 

Well, actually, it kind of is spooky—but more on that later. 

First, let's take a closer look at this pair of points. I even drew them in a
previous note (pretty handy, right?).



Now, let's get serious—or as serious as one can get when doodling 4D geometry. Today, we'll draw these points more precisely, name a few angles, and write out some formulas that will make your head spin (pun intended). To this end let us define vectors a and b ending at the respective points as follows: 


View from Above: The Pair of Points

Looking from above, our pair of points looks like this:

Fig. 1

And looking from the side:

Fig.2

Feel free to click and enlarge the image, though honestly, it's not that exciting—unless you're an angle enthusiast (no judgment).

In Figure 1, I marked the angle ψ between the red point (the one hanging out closer to the South Pole) and the X-axis of our inner space. The angle θ is the one between the red and blue points, which is kind of a big deal because it's what shows up in the outer space as the longitude of the spin state. Fancy, right?

In Figure 2, I’ve marked φ, which is the latitude shared by both points. We’re measuring it from the North Pole, because unlike geography, where latitude is measured from the equator, in math we like things to be... a bit more north-centric. Makes things more convenient—if you're into convenience, that is.


The Dance of Phi and Psi: Trigonometry Comes to Play

So here’s a cool tidbit: trigonometry tells us (thank you, middle angle theorem) that point "a" makes an angle of φ/2 with the vertical axis. I decided to make life a bit easier by setting the radius of my sphere to ½. That means the diameter is 1, and thanks to our old buddy Pythagoras, we get the equation a² + b² = 1.

This angle φ also makes its way into the outer space, representing the latitude of the spin state. So what we're really dealing with here is geometry pretending to be physics.


Circles, Vectors, and... Math?

At the top and bottom of the sphere, I've drawn two circles and two vectors (you know, to keep things interesting). The vector at the bottom has a phase ψ and a length a. Meanwhile, the vector at the top is just trying to be fancy with its phase ψθ and length b.

When you superimpose these two planes—like the mathematical equivalent of stacking pancakes—you get this:

Fig. 3


So, we have two circles, one with radius a and one with radius b. The important part? a² + b² = 1. That’s it! You’ve got yourself a spin state vector—a spinor, if you will.


The Coordinates: Let’s Break It Down

Let’s describe this using coordinates. For the vector a, let's use X and Y for its plane coordinates. For b, we'll use Z and W. Simple enough, right?

So:

  • a = (X, Y)
  • b = (Z, W)

But here's the catch: remember, these vectors are hanging out on two planes—one at the top of the sphere and one at the bottom. They’re superimposed, so you’re seeing them both at once.


Time for Some Equations

From Figure 3, we can deduce:

  • X = a cos(ψ)
  • Y = a sin(ψ)
  • Z = b cos(ψ + θ)
  • W = b sin(ψ + θ)

And from Figure 2, we know:

  • a = cos(φ/2)
  • b = sin(φ/2)

Bringing it all together, we get:

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

And voila! X² + Y² + Z² + W² = 1. That’s your whole spinor, folks—four numbers whose squares add up to one. The first pair, (X, Y), is vector a, and the second pair, (Z, W), is vector b. Easy, right?


The Fourth Dimension: Not As Scary As It Sounds

At this point, the mathematically inclined among you might recognize that X² + Y² + Z² + W² = 1 is the equation of a three-dimensional sphere in four-dimensional space! Don’t freak out about the extra dimension. If it makes you feel better, just think of it as two plane vectors, a and b, whose lengths always add up to 1. Phew, problem solved!

If you’re still feeling brave, you can even project this 4D sphere onto 3D space using something called a stereographic projection, which basically compresses it into something we can visualize. We’ll dive into that graphic wizardry in the next note.


The Algebraic Approach: Complex Numbers to the Rescue

But wait, there’s more! We can also turn our vectors into two complex numbers. Here’s how:

  • a+ = X + iY
  • a- = Z + iW

And just like that, we jump straight into the algebraic description used in quantum mechanics. Don’t worry, we’ll cover this in more depth in future notes, where quantum math takes over and everything gets extra... well, quantum-y.

Stay tuned!


21 comments:

  1. The descriptions a, b in Fig. 1 do not correspond to the descriptions a, b in Fig. 2.
    Not enough letters?

    ReplyDelete
    Replies
    1. I have assed one sentence before "View from Above: The Pair of Points". It should help.

      Delete
    2. What is called a (b) in Fig. 1?

      What is called a (b) in Fig. 2?

      (a in Fig 1. is near red point,
      a in Fig 2. is near blue point. Red point is absent.)

      Delete
  2. Why do your theta and phi angles not correspond to the ISO standard latitude and longitude names? (They are swaped.)

    ReplyDelete
    Replies
    1. Because I have a somewhat rebelious nature? Maybe....

      Delete
    2. Your naming of angles is the old nomenclature of mathematicians (in contrary to physicists and ISO standard.)

      So your soul is more of a mathematician than a physicist.

      Delete
  3. Replies
    1. It's not Friday, but I will be busy tomorrow morning. Perhaps I will change completely to Tuesday, Thursday, Saturday evening. Will fix a,b at some point tomorrow. Thank you!

      Delete
  4. Для конструирования образа спинора вам понадобился взгляд с двух ракурсов на две точки на сфере. Это вполне коррелирует с моим представлением унитарной группы как композиции движений на классической сфере.

    ReplyDelete
    Replies
    1. I wish I would understand what I am doing! Random walk - that is what I am doing.... But I will try to understand your point of view.

      Delete
    2. I still do not understand your (1.1.10) in your "Zametki". I do not understand where it comes from, and I do not understand your notation. You have three different vector fields in R^4, not just one vector field. And you have confusing notation on the left hand side. Can you explain, Igor? You mentioned quaternions before, but I do not see what (1.1.10) has to do with quaternions.

      Delete
    3. Производная каждого из этих трёх векторных полей по собственному направлению равна векторному полю -\partial\rho, что соответствует основному соотношению алгебры кватернионов
      j^2=j^2=k^2=-1

      Delete
    4. I don't understand. What are x1,x2,x3,x4? What is their relation to quaternions? What is "Производная каждого из этих трёх векторных полей по собственному направлению"? How did you get (1.1.10)? Can you answer precisely?

      Delete
  5. Здесь очень неудобно писать формулы, но если вы посмотрите на формулу (1.5) по ссылке
    https://www.researchgate.net/publication/282661319_Applications_of_the_local_algebras_of_vector_fields_to_the_modelling_of_physical_phenomena
    то поймёте о чём я говорю.

    ReplyDelete
    Replies
    1. (1.5)? Are you sure? It does not explain anything to me. Moreover (1.4) and (1.5) are not covariant definitions. Only (1.6) - the Lie bracket, is covariant.

      Delete
    2. Если вас волнует ковариантность, то пусть координаты будут ортонормальными в соответствующей метрике.

      Delete
    3. There is no metric in Sec. 1 of this paper. There is an "affine manifold", without any definition what this term means for you. Moreover, that does not answer my question what (1.5) has to do with (1.1.10) in Zametki?

      Delete
    4. Похоже мой ответ потерялся, поэтому повторюсь. Действительно, векторные поля задаются в аффинном пространстве, но если мы потребуем сохранения алгебры векторных полей, то на преобразование координат наложится требование соответствия соответствующей метрике. Об это говорится во второй секции. А вопрос про (1.1.10) я так и не понял. Повторите, пожалуйста, его ещё раз, но другими словами.

      Delete

    5. Я не понимаю. Что такое x1, x2, x3, x4? Какое отношение они имеют к кватернионам? Что значит «Производная каждого из этих трёх векторных полей по собственному направлению»? Как вы получили (1.1.10)? Можете ли вы ответить точно?

      Delete

Thank you for your comment..

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