This post is a continuation of "The Spin Chronicles (Part 3): Spin frames". It is purely technical. This previous post ended with:
"In other words, when you rotate an element of O(V) by an SU(2)
matrix A, the corresponding lab element rotates, too, by and SO(3)
matrix R(A). It’s like a cosmic interlock between spinors and
real-world rotations."
Today we will see how this can be realized, and, in fact, is realized, in quantum mechanics. I should say that I am not happy with this way of dealing with spin, but, at the moment, this is the best I can do. What I present here is just one way of dealing with spinors. There are other formulations possible, but, in fact, they are all equivalent and equally unsatisfactory.
We will treat V as a Hilbert space, with its scalar product (u|v).
Note: Here, more in the spirit of quantum physics we will
use the notation (u|v) instead of (u,v) for the scalar product.
Once we have a Hilbert space, we have linear operators acting on this space. Let us denote the space of linear operators by L(V). Then L(V) becomes itself a Hilbert space if we define the scalar product by
(A|B) = Tr(A*B),
where (*) denotes the Hermitian conjugation, and "Tr" stands for
"trace". In an orthonormal basis operators are represented by matrices,
and (*) becomes the ordinary Hermitian conjugation of matrices (complex
conjugate and transpose), while trace becomes the sum of diagonal
elements of the matrix. Notice that (A|B)* = (B|A), where now (*) is
applied to a complex number, and denotes the ordinary complex
conjugation, usually denoted by a "bar" over that number. But "bar" is
easy in Latex, and not really supported on the web page. So I will stick
with the (*) notation.
In L(V) we have, in particular, Hermitian operators. The set of all Hermitian operators is a real vector space. We denote it by H(V). For Hermitian A,B the scalar product (A|B) is real. So H(V) is a real vector space, with a real scalar product. In order to see what kind of scalar product it is, we need a basis in H(A), So, let us construct one. First of all, given any two unit vectors u,v in V, where where "unit" means (u|u) = (v|v) = 1,we denote by |u)(v| the operator in L(V) defined as
|u)(v| w = (v|w) u.
Then (|u)(v|)*= |v)(u|
Then |u)(u| is the orthogonal projection operator on u. It acts, by definition, as
|u)(u| v =(u|v)u
It is Hermitian (thus in H(V)), and idempotent (that means ( |u)(u| )2 = |u)(u|. Its eigenvalues are 1 and 0. Its eigenvectors are u (for eigenvalue 1), and 0 for any non-zero vector orthogonal to u). Sometimes it is easier yo write Pu instead of |u)(u|. So we have Pu=Pu*, and Pu2 = Pu, (Pu|Pu) = 1. We will construct a basis in H(V) from a basis in V. But first let us talk some "physics". Let us think of "spin in a given direction" as a binary quantity. It can be "up" or "down". So, we want to associate with its "measurement" a Hermitian operator with eigenvalues +1 and -1, +1 for "up" and -1 for "down". But which one? The answer, as far as I know, is this: "anyone", but it should be done in a "consistent" way. We can do it as follows: first choose an orthonormal basis e1,e2 in V. Then we have the following (easy to prove) Theorem:
Theorem: Every Hermitian operator with eigenvalues ±1
can be uniquely represented as σ(n) = n1σ1+n2σ2+n3σ3,
where
σ1,σ2,σ3
are the Pauli matrices
and n is a unit vector in R3. Conversely, if n is a unit vector in R3, then σ(n) represents a Hermitian operator with eigenvalues ±1 in the basis e1,e2. Moreover, if A is any matrix from SU(2), then
A σ(n) A* = σ( R(A)n ).
Now we associate spin "up" in "z" direction in the lab with n
=(0,0,1). The formula above allows us to consistently associate
Hermitian operator, namely σ(n),
with "spin up" in any direction.
Why is that not quite satisfactory? The problem is that we have chosen the particular Pauli matrices above. Why these and not other? If U is any unitary matrix, then the matrices U σi U* would do as well (with a modified R(A)) . This seems to tell us that these "spin directions" are in a space that looks and behaves as "our space", but it is not "our space". It is a mystery that needs to be understood.
We will continue these considerations in the next post.
P.S. 19-10-24 This is my reply to the comment Bjab October 19, 2024 at 10:11 AM. O(V) was defined in the previous post as the set of special orthonormal frames. Taking C2 for V, as in the example in that post, for special orthonormal frame we require
1) (ei|ej) = δij,
2) ε(ei,ej) = εij.
where
(u|v) = u1* v1 + u2* v2,
=u*v,
ε(u,v) = u1 v2 - u2 v1.= uT ε v,
where ε on the right
hand side denotes the matrix
| 0 | 1 |
| -1 | 0 |
One special orthonormal basis is e1 = (1,0)T, e2 = (0,1)T. If e'1,e'2 is another special orthonormal frame, they are related by some invertible complex matrix A: e'i = ej Aji. Then from 1) it follows that the matrix must be unitary, while from 2) it follows that
ε = AT ε A,
which implies that det(A) = 1. Therefore A is SU(2). Conversely, any such A transforms special orthonormal frame into a special orthonormal frame. Thus the space of all special orthonormal frames has the same dimension as SU(2), that is 3.
I don't feel it yet.
ReplyDeleteHow many dimensions does the space O(V) have?
ReplyDeleteHow many dimensions does spinor have?
Delete"Thank you. You are right."
DeleteI don't know where I'm right. I asked the questions just to verify my assumptions.
Thank you. You are right. Something is wrong. I will think it over, and fix it tomorrow.
DeleteSpinor is a unit vector in V. So, it has 4-1=3 real dimensions. O(V) has the same dimension as U(V) -the space of unitary operators. It has 2^2=4 complex dimensions, 8 real dimensions. But I will make it all better tomorrow. Thank you for making me to rethink the problem.
Delete"Spinor is a unit vector in V. So, it has 4-1=3 real dimensions."
DeleteThat's what I thought.
" O(V) has the same dimension as U(V) -the space of unitary operators. It has 2^2=4 complex dimensions, 8 real dimensions."
DeleteWell, I thought that unitary operators (SU(2)) have 3 real dimensions. And I thought that O(V) has 6 real dimensions.
But I wrote a complete nonsense about the dimension of O(V). They are special orthonormal frames. The dimension of O(V) is the same as SU(2), that is 3.
Delete@Bjab. I thought it over. There is nothing wrong with what I wrote in the post, but I overlooked the fact that I need to introduce the 3D space of traceless Hermitian operators, which is important for non-relativistic discussion of spin. So I will do it in the next post.
DeleteДля наглядности действия групп U(2), SU(2) можно использовать модель с собственными движениями тора и вращениями тора в евклидовом пространстве. Дело в том, что композиция, состоящая из произведения трёх матриц, первая и последняя из которых в общем случае разные диагональные матрицы с элементами e^(i\varphi_1), e^(i\varphi_2), а средняя матрица принадлежит группе SO(2), является элементом группы U(2). Если e^(i\varphi_1)* e^(i\varphi_2)=1, то это будет элемент группы SU(2). Аналогично строится и SU(n). Но главный вопрос в том, как надо вращать тор. Я предлагаю вращать его вокруг оси, которая является диаметром некоторой окружности Вилларсо.
DeleteИгорь, прежде чем предлагать что-то новое, сначала исправьте старое.
DeleteIf you introduce your own operation of multiplication of vector fields, you should first define it - otherwise you mislead the readers. So, do it. Explain in your paper the relation between vector field and matrices, explain how it differs from Noviko's book. Explain how you got your matrices A1,A2,A3, and how you got your vector fields from these matrices. And do it all clearly and mathematically precisely. Do it!
Аркадиуш, я подумаю над вашим советом. Хотя, по большому счёту, никакие матрицы не нужны, достаточно того определения произведения векторных полей, которое там уже имеется.
Delete"того определения произведения векторных полей, которое там уже имеется."
DeleteAre you sure Igor? Where it is defined in your paper? This "определение произведения векторных поле"? I do not see it.
Эту строку вы видите?
Deletewith respect to the multiplication operation of vector fields
\begin{equation}
A\ast B = \nabla_{B}A = \sum_{i=1}^{n}\nabla_{B}a^{i}\partial x_{i} =
\sum_{i=1}^{n}b^{j}\frac{\partial a^{i}}{\partial x_{j}}\partial x_{i}
\end{equation}
Now I see it. Thank you. What are X,Y,Z in (1.1.7)? Numbers or functions? Is your "3-dimensional unit sphere" invariant under the action of Γ?
DeleteЭто числа, которые задают выбор динамической системы \Gamma (ориентацию векторного поля особенности в 4-мерном пространстве). Наверно, можно ещё попробовать задать динамическую систему, составленную из суперпозиции векторных полей особенностей, имеющих разную ориентацию и разные центры в 4-мерном пространстве.
Delete"Is your "3-dimensional unit sphere" invariant under the action of Γ?"
Не понял вопроса.
Не понял вопроса.
DeleteAre the trajectories of Γ on your "3-dimensional unit sphere"?
Да, на 3-мерной, но не на 2-мерной X^2+Y^2+Z^2=1
DeleteBut I think that ∂ϕx, ∂ϕy, ∂ϕz are tangent to the 3D sphere, and ∂ρ is orthogonal to all three of them. Am I mistaken?
DeleteВсё правильно, но на единичной 3-сфере ортогональное поле зануляется.
DeleteBy your definition:
Delete∂ρ = x1∂x1 + x2∂x2 + x3∂x3 + x4∂x4
This field does not vanish on the unit 3-sphere!
А вы посмотрите на \Gamma Там стоит \rho\partial\rho
Delete\rho=\log(r)
DeleteSo, on your unit sphere Γ = ∂ϕ , yes? This should be your (1.1.8). Why then are you introducing ∂ρ, if you do not need it?
DeleteДинамическая система задана не на 3-сфере, а в 4-мерном евклидовом прочтранстве - в пространстве эволюционирующей 3-сферы.
DeleteIn your paper you write: "and
Deletetherefore a dynamical system with an inertial manifold on a
3-dimensional unit sphere is given by the equation". You do not have "evolving sphere". And in order to write "evolving sphere", you would have to first define it. So, please, do it.
Or, simply, write "in R^4.
DeleteAre you sure you want "+ ρ∂ρ" in Γ and not " - ρ∂ρ"?
DeleteСейчас на даче работаю. Отвечу чуть позже.
DeleteЯ тут немного подправил
DeleteIn turn, in order to transfer a particle-like dynamical system into the Euclidean space of an evolving 3-sphere $\mathbb{R}^{4}$, it must be borne in mind that on $S^{3}$ there are three linearly independent and orthogonal to each other (in the metric of a 4-dimensional Euclidean space) linear vector fields
Кроме того, в самом начале есть предложение
Going beyond the framework of classical mechanics, we are starting from the representation of a particle as a material point in physical space (i.e., a geometric point having its own mass), moved on to representing a particle as a feature of the vector velocity field of matter particles closed by their own current lines on a 3-dimensional sphere, and having its own angular velocity in the 4-dimensional Euclidean space of the evolving 3-sphere $\mathbb{R}^{4}= r\ast S^{3}$, where the radius of the sphere $r$ and the evolutionary parameter $\tau$ are related by the dependence $\tau =\log(r)$. In other words, from now on, the inertial manifold of the dynamical system of 4-dimensional space, formed by limit cycles closing on the 3-sphere, is the prototype of a material point, and due to the fact that absolute time is a function of the polar coordinate of space $\mathbb{R}^{4}$, the dynamical system is a vector velocity field matter in evolution.
Что касается замены плюса на минус, то я не против, но это ни на что не влияет, поскольку формально стрелу времени можно и развернуть (в эволюции сфера может как расширяться так и сжиматься).
How you define "limit cycles"?
DeleteПредельным циклом я называю особенность векторного поля на плоскости с замкнутой траекторией, то есть это замкнутая кривая вблизи которой траектории не замкнуты.
Delete@Ark
ReplyDelete"There is nothing wrong with what I wrote in the post"
Perhaps yes.
But, please, explain me why you think that O(V) has 3 real dimensions.
Let me explain my reasoning that O(V) has 6 dimensions.
When we are dealing with a plane (two-dimensional space), the rectangular coordinate system is one-dimensional. (One angle of rotation of such a system).
When we are dealing with a three-dimensional space, the rectangular coordinate system is three-dimensional. (Two angles of rotation selecting the direction of one coordinate plus one angle of rotation guiding the other two coordinates of such a system).
When we are dealing with a four-dimensional space, the rectangular coordinate system is six-dimensional. (Three angles of rotation selecting the direction of one coordinate plus three angles of rotation guiding the other three coordinates of such a system).
OK. I will write an explanatory P.S. Writing things down helps me to think over the details. And the details is where the devil often hides.
Delete@Ark
DeleteThank you for explanatory P.S.
I don't see the catch in it.
On the other hand
I found Wiki's article:
https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space
where there is something lie this:
"Group structure of SO(4)
SO(4) is a noncommutative compact 6-dimensional Lie group."
This is true. But our transformations are 1) complex linear and 2) "special".
DeleteCorrect me if I'm wrong.
DeleteU(2) is four-dimensional.
SU(2) is three-dimensional.
So here specialization reduces dimension by one.
O(3) is three-dimensional.
SO(3) is also three-dimensional.
So here specialization does not reduce dimension.
What about your O(V)?
What reduces its dimensionality to 3 from 6 and by how many?
1) complex linearity?
2) "special" speciality (what is it exactly (epsilon bi-form?)?
Let me use Mathematica matrix notation. We have 4x4 real matrix. Let us write it as 2x2 block matrix form X as
DeleteX={{A,B},{C,D}}. Then, in our 4D real space we have imaginary "i" 4x4 matrix, say i={{0,I},{-I,0}}. Then iX=Xi implies D=A, C=-B.
Thus X={{A,B}},-B,A}}.
So complex linearity divides the number of parameters by 2. To consider what happens to orthogonality condition it is convenient to consider the Lie algebra level - infinitesimal transformations. At the infinitesimal level X^TX=I becomes X^T=-X. This implies A^T=-A, B^T=B. Thus we have 1 parameter for A, and 2+1=3 parameters for B. This way we get complex unitary matrices. The requirement of this complex unitary matrix being of determinant one takes away one parameter.
But, starting with the next post, I will start approaching spinors in a different way, via Clifford algebra. All will be much more transparent, I think.
DeleteThank you. I feel that's OK.
DeleteBut I don't understand the following sentensces:
"At the infinitesimal level X^TX=I becomes X^T=-X."
and
"The "special", determinant 1, becomes infinitesimal trace zero. So B must have trace 0."
Suppose we have a path of orthogonal matrices O(t), with O(0) = I - a path through the identity matrix. Define X=dO(t)/dt at t=0. We have, for each t, O(t)^TO(t)=I for each t. Diffrentiate this equation at t=0. You get X^T+X=0, or X^T=-X. These X are called the Lie algebra of the group. Conversely, if X^T=-X, then we can define O(t)=exp(tX), and we get a path of orthogonal matrices through I.
DeleteNow take complex unitary 2x2 matrices. Do the same as before. You get that X* = -X, so X must be anti-Hermitian. But if you write U(t) as {{a(t),b(t)},{c(t),d(t)}}, the determinant 1 becomes a(t)d(t)-b(t)c(t)=0. Differentiate at t=0, and rtake into account the fact that b(0)=c(0)=0, a)0)=d(0)=1. You will get d a/dt+d d/dt =0 at t=0. That is X must have trace 0.
In general the number of parameters (the dimension) of the group is the linear dimension of its Lie algebra - the space of infinitesimal generators - derivatives of t the paths through identity.
Может быть, стоит также указать, что exp(tX^T+tX)=exp(tX)^T exp(tX)=I, поскольку tX^T+tX=0
DeleteI understand that for O(t)^TO(t)=I we get X^T+X=0 but I don't understand what you wrote:
Delete"At the infinitesimal level X^TX=I becomes X^T=-X."
I also don't understand :
"the determinant 1 becomes a(t)d(t)-b(t)c(t)=0"
It was unfortunate that I have used the same letter X for both X and the one-parameter subgroup exp(tX) generated by X. Like if there was problem with the alphabet. My fault. Skip this part of my comment. I apologize.
DeleteSimilarly I used a,b,c,d for their derivatives at t=0. Another very bad abuse of notation.
Delete"Similarly I used a,b,c,d for their derivatives at t=0."
DeleteI don't think so. Now I think that you wanted to write instead of
"the determinant 1 becomes a(t)d(t)-b(t)c(t)=0"
"the determinant 1 means a(t)d(t)-b(t)c(t) = ONE"
You are right. I first wrote an even sillier comment that I have removed. It was still in my mind. Working on too many projects at the same time. But I can't edit my comments once they re posted, so I can't fix this error.
DeleteSo I see that ALL the issues presented in the last posts concerned ONLY three-dimensional spaces. Therefore, dealing with spinor is like kneading three-dimensional (of course) yeast dough.
DeleteYes, indeed. And the same will be with the Clifford algebra approach - it will be adapted to three dimensions. It is easy to grasp the idea for such a simple and, at the same time, realistic case.
Delete