What’s a Spin Axis, Anyway?
This post was inspired by a question from Bjab, who asked: "What do you mean by ‘spin axis’?" Well, buckle up because I’m about to explain that! I’m not entirely thrilled with the answer I have right now, but hey—it works (kind of). Hopefully, in the near future, I'll come up with something that satisfies both Bjab and my restless curiosity. For now we have to be happy with the standard concept of the spin structure - representing the universe as a set of interlocked rotations - described below.
Where Do Spinors Live?
Imagine a strange and mysterious land—let’s call it V—a two-dimensional complex vector space. It’s not just any space, though. It has "structure." Picture it equipped with a hermitian non-degenerate scalar product that we’ll denote by , and a bilinear antisymmetric non-degenerate form we’ll call . Sounds fancy, right? These two should be related in a specific way, and here’s how:
There exists a special kind of basis for this space, which we’ll call “special orthonormal.” It satisfies:
- ,
- ,
“Special” indeed. The first natural question is: Can this magical basis even exist?
The Magic of C²: Relax, It’s All Possible
Yes, it exists! Here’s why. Let’s choose any basis in V. Using this, we can identify V with (our familiar two-dimensional complex space). In , the scalar product is defined as:
where the star (*) represents complex conjugation. As for , we define it as:
Now, using a standard basis , , it’s easy to check that our special orthonormal basis conditions hold. So, crisis averted! We don’t need to worry about whether this basis exists—it does. Phew.
Oh, and here’s a fun fact: if is a special orthonormal basis, then is also a special orthonormal basis. Isn’t symmetry beautiful?
Back to the Mysterious World of V
So, we’ve got our scalar product, our fancy , and we’re cruising in V—home to the enigmatic spinors. But are there other “special orthonormal” bases? And if so, how many?
Let’s say is another basis. There exists some matrix such that:
(Yes, that’s a summation over “j”.)
Now, here’s the neat part: will also be special orthonormal if and only if the matrix is unitary with determinant 1. Matrices like this form a group called SU(2).
By the way, SU(2) is isomorphic to the group of unit quaternions. I won’t dive into that now, but it’s a pretty cool connection between algebra and geometry.
The Spinor Connection
Let’s denote by O(V) the space of all these special orthonormal bases. The group SU(2) acts on O(V) from the right, both transitively and freely. The elements of O(V) are what we call spinor bases, and the elements of V are simply spinors. If is a spinor, and you change the basis using an SU(2) matrix , the components of the spinor change according to this formula:
Got it? Good! Now, let’s connect this with something real, like the lab—where actual experiments happen (sometimes with coffee stains on the data sheets).
What About Our Lab?
To bridge this to the physical world, we need some more algebra (don’t worry, just a little more). Quaternions are often used for handling 3D rotations, but since we’re sticking with matrices, here’s what we do in quantum mechanics:
For each point in , associate the following 2x2 Hermitian matrix, :
Now, for any SU(2) matrix , it turns out:
where is a 3x3 real rotation matrix, an element of the special orthogonal group SO(3). In fact, the map from and are two SU(2) elements, then if and only if .
Hence, SU(2) is a double cover of SO(3). It’s like the universe giving you two chances to rotate.
Note: With quaternions it would look more "natural". To each x we would associate a pure imaginary quaternion
q(x) = xi+yj+zk,
and for each unit quaternion u, we would have
uq(x)u* = q(R(u)x),
where R(u) is in SO(3).
The Spin Structure: So What About That Spin Axis?
Let’s bring this back to the real world—our lab, where we have Cartesian coordinates. Any two right-handed orthogonal systems (say, different setups in the same room) are related by a unique rotation matrix from SO(3). The set of all such frames forms a space O(L), which SO(3) acts on.
Now, here’s the kicker: nature somehow maps each element of O(V) (the spinor world) to an element of O(L) (our lab’s coordinate system) in a way that respects the rotations. We call this map the spin structure.
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.
Note (for experts): This is a baby version of a spin structure, good for a 3D flat space or, more generally, for a space with a distant parallelism. But this baby version is all that we need for now.
What’s Next?
All this brings us closer to answering the question: what do I mean by a spin axis? Stay tuned, because we’ll tackle that mystery in the next post! We will also take under the loop the spin structure - the most important point here.
P.S. I am experimenting with the mathematical notation on this blog. Sometimes I mess up, then correct what can be corrected.
P.S. I have uploaded the pdf version of this post to academia.edu.
P.S. 18-10-24 9:33 Igor Bayak (1.1.6)
Re experiments.
ReplyDeleteIn the paragraph from the words "In fact, the map" everything started to fall apart. What a person sees is different from what can be copied to the clipboard. This is very strange. The engine cannot handle processing, perhaps due to syntax errors?
It ate whole ordinary words, causing complete lack of meaning.
Thank you. That probably means that is better to show just pdf?
DeleteWorking directly with the HTML code is a nightmare for me!
DeleteI understand what I did wrong this time. Tomorrow's post should be better.
Delete"I understand what ... ".
DeleteThat's great.
Thanks for the PDF, there're just small differences compared to what can be seen in the post, but it makes the text much more readable.
DeleteIn A Linear Algebraic Introduction to Quaternions file you shared the link to several posts ago, it said that Hamilton claimed and fervently defended that the quaternion product uq itself was the rotation operation and not the "conjugation" uqu*, like written in the post and what seemed to be closer to correct interpretation after some evaluation by hand.
During the "disentanglement" some interesting things popped up that might relate to the spin structure and spin axis measurement topics at hand, and all comments on the texts below and eventual corrections are highly appreciated and welcomed.
A bit of a background for the discussion, so we're all on the same page, sort of.
A quaternion q=(q0+iq1), where q0 and q1 are its real/scalar and imaginary/vector parts, respectively, has a polar form and unique decomposition as,
q = ||q|| * (cosϕ_q + i*n_q*sinϕ_q) = ||q|| * exp(i*n_q*ϕ_q) = ||q|| * U_q,
where ||q||^2 = q0^2 + q1^2 is its norm, angle ϕ_q defined as cosϕ_q = q0/||q|| and sinϕ_q = |q1|/||q||, n_q = q1/|q1| is its unit vector or axis, and U_q = q/||q|| = exp(i*n_q*ϕ_q) is its unique unit quaternion, ||U_q||=1.
In that respect, a pure imaginary vector quaternion q'= (0+i*q1),
q' = i*q1 = i*|q1|*n_q1 = |q1| * (cosϕ_q1 + i*n_q1*sinϕ_q1) = |q1| * exp(i*n_q1*ϕ_q1) = |q1|*U_q1,
shares the same unit vector axis with its general/origin quaternion q, n_q1 = n_q, which means that q can be written as q = q0 + i*n_q*|q1| = q0 + |q1|*U_q1, where U_q1 = i*n_q is imaginary unit vector of q1 or its unique unit vector/imaginary quaternion. In other words, all quaternions with their axis equal to n_q have the same U_q1.
Using polar decomposition of q1, it's easily seen that ϕ_q1=pi/2, as cosϕ_q1=0 and sinϕ_q1=1 regardless of the value |q1| or direction of axis n_q, meaning that every pure imaginary vector quaternion q' has the form,
q' = i*q1 = |q1|*U_q1 = |q1| * exp(i*n_q*pi/2).
Defining the vector product, that is the product of two pure imaginary vector quaternions q' and p' as,
q' * p' = i*q1 * i*p1 = i^2 * q1 * p1 = i^2 * (q1.p1 - i*(q1×p1)),
where "." and "×" are Gibbs' defined and usually known dot and cross vector products in vector analysis, respectively, for a product of two general quaternions q=(q0+q') and p=(p0+p') we obtain,
q * p = q0*p0 - q1.p1 + i*q0*p1 + i*p0*q1 + i*(q1×p1)
= q0*p0 - |q1|*|p1|*cos(ϕ_(q1,p1)) + q0*|p1|*U_p1 + p0*|q1|*U_q1 + i*|q1|*|p1|*sin(ϕ_(q1,p1))*n⊥(q1,p1),
where ϕ_(q1,p1) is the angle between q1 and p1 lying in the plane spanned by those vectors, and n⊥(q1,p1) is the unit vector pointing in the direction, using right-hand rule, perpendicular to that plane.
Now that we hopefully all have the basic ropes in our hands and stuff under our belts, lets see in following comment what can be deduced from multiplication of sort of collinear quaternions, that is those quaternions sharing the same imaginary unit 3d vector or axis, n_q, like q, U_q, q' and of course U_q1.
Lets start with multiplication of pure imaginary quaternions, like U_q1 and q' for example. Since we're looking at quaternions with the same axis n_q, we don't need to worry about the fact that quaternion product is not commutative because that comes into play only for those quaternions with non-collinear imaginary 3d vector axes.
DeleteApplying U_q1 to q', that is to q1 itself, we get,
U_q1 * q' = i*n_q * i*n_q*|q1| = |q1| * U_q1 * U_q1 =
= i^2 * |q1| (n_q*n_q) = -|q1|*(n_q.n_q - i*(n_q×n_q)) = -|q1|,
which can be interpreted, taking into consideration that ϕ_q1=pi/2 and thus q'=U_q1*|q1| = |q1|*exp(i*n_q*pi/2), in a sense that U_q1 or imaginary 3d unit vector "pulls" from the 4th dimension of reals/scalars, which is perpendicular to imaginary 3d space R^3, the axis n_q perpendicular to that dimension, as for the pi/2 angle, and when applied again on the same axis, it then "pushes" it towards for pi/2 in the same direction of original action. Since all imaginary vector quaternions in general and their axes in particular are perpendicular to the real 4th dimension of scalars, it's natural that all pure imaginary vector quaternions have its ϕ_q'=pi/2 regardless of their specific axes n_q. That also suggests, as there are infinitely many possibilities to draw an axis through the point, that there might be infinitely many real directions or scalar "dimensions" of 4th dimensional character perpendicular to our imaginary 3d space.
When U_q1 is applied again, acting or rotating the axis n_q in the same direction as the one in which it got pulled out from the 4th real dimension, the axis then points in the opposite direction in the 4th scalar dimension, as for the minus sign at the end, that is the scalar or real got rotated once for pi/2 to become imaginary vector i*q1=i*|q1|n_q, and then again for 90° to make a complete angle of pi and so become sort of its original reflection or opposite, -|q1|. In that sense, multiplication of imaginary vectors with themselves, yields q'*q' = |q1|^2 * U_q1*U_q1 = -|q1|^2.
The situation clears itself a bit more when we take a look at what U_q1 does when acting on its origin general quaternion q,
U_q1 * q = i*n_q * (q0 + q') = q0 * i*n_q - |q1| = -|q1| + i*n_q*q0,
indicating that while it "pushed" imaginary vector q1 into the 4th dimensional scalar in the opposite direction, at the same time it "pulled" the scalar quaternion part q0 from the real 4th dimension into imaginary 3d space, pointing in its characteristic imaginary 3d direction n_q. In mathematical jargon, we could say that it sort of exchanged the values of real and imaginary parts, if not for that minus sign, but when applied again, as indicated in previous passage, it turns its origin general quaternions completely to their opposites,
U_q1 * U_q1 * q = -1 * q.
Other three possible combinations of multiplying general with their vector quaternions, that is q*q1, U_q*U_q1 and U_q*q1, can be easily evaluated/deduced using polar decomposition, and all three, that is all four of them in total, have identical resulting unit quaternion of the form U_q*U_q1, so lets see its exact form on the example above,
DeleteU_q1 * q = -|q1| + i*n_q*q0 = ||q|| * exp(i*n_q*pi/2) * exp(i*n_q*ϕ_q) =
= ||q|| * (-|q1|/||q|| + i*n_q * q0/||q||) = ||q|| * exp(i*n_q*(pi/2 + ϕ_q)) =
= ||q||*(-sin(ϕ_q) + i*n_q*cos(ϕ_q)) = ||q||*(cos(ϕ_q + pi/2) + i*n_q*sin(ϕ_q + pi/2),
where, for the first term in the third line, relations for cos(ϕ_q) and sin(ϕ_q) from previous comment were used. Like discussed in the passage above, action of a unit vector on its original general quaternion just "adds" an additional pi/2 to its original characteristic angle ϕ_q which is evaluated with respect to the real 4th scalar dimension, going counter-clockwise as indicated by the relation for its real part, Re(q) = q0 = ||q||*cos(ϕ_q).
In that sense, Hamilton was not actually completely wrong when he argued that multiplication of quaternions acts as a rotation; it sort of does, but not only strictly in imaginary 3d space R^3, but in relation to complete 4 dimensional quaternion space, including also the real 4th dimension of scalars.
To complete this part of multiplication of "collinear" quaternions, and to drive the point/discussion home that this multiplication sort of speaks about rotation with respect to the 4th scalar dimension, lets check an example of a general quaternion product. Using the expression from the previous comment, for q*q we get,
q * q = q0^2 - |q1|^2 + 2*q0*|q1| * n_q*i = ||q||^2 * U_q * U_q =
= ||q||^2 * (q0^2/||q||^2 - |q1|^2/||q||^2 + 2*(q0*|q1|/||q||^2)) * n_q*i =
= ||q||^2 * (cos^2(ϕ_q) - sin^2(ϕ_q) + 2*cos(ϕ_q)*sin(ϕ_q) * n_q*i) =
= ||q||^2 * (cos(2*ϕ_q) + sin(2*ϕ_q) * n_q*i) = ||q||^2 * exp(i*n_q*(2*ϕ_q)),
showing that the angle of quaternion squared, or its unit quaternion squared U_q*U_q, is in fact 2*ϕ_q, that is application of a unit quaternion to its origin general quaternion "moves" the origin quaternion for an additional value of its characteristic angle ϕ_q, "from" the 4th scalar dimension and "into" the imaginary 3d space R^3. It can be reasonably deduced from the discussion above that this movement or rotation "from" 4th dimension, perpendicular to 3d space R^3, turns into movement "to" when the overall value of rotation angle value reaches pi/2.
Next comes the general application of quaternion multiplication, like an arbitrary (U_q * p) for example, where axes n_q and n_p are non-collinear, but before diving into that a bit more complex part, our imaginary 3d world takes its "toll", so its time for a dinner and an energy refill, sort of. :)
As the approach of gradually building our action elements and in such a way observe what's been going on with each multiplication application, there seems to be no significant need for any dramatic change of course. So, lets start with the usual application of a given unit vector, U_q1 = i*n_q, onto an arbitrary pure imaginary vector, p' = i*|p1|*n_p = |p1|*U_p1,
DeleteU_q1 * p' = i*n_q * i*n_p*|p1| = -1*|p1|*(n_q.n_p - i*(n_q×n_p)) =
= -|p1|*(cos(ϕ_(n_q,n_p)) - i*sin(ϕ_(n_q,n_p))*n⊥(n_q,n_p)) =
= -|p1|*cos(ϕ_(q,p)) + i*|p1|*sin(ϕ_(q,p))*n⊥(q,p),
where ϕ_(n_q,n_p) = ϕ_(q,p) is the angle between n_q and n_p, determined in the plane spanned by them, and i*n⊥(n_q,n_p) = i*n⊥(q,p) is the imaginary unit vector pointing in the direction perpendicular to that plane with its orientation determined by the right-hand rule.
Knowing what happened with the product U_q1*q', we might interpret the action of U_q1 to p' in the following way.
Only thing that the U_q1 is sort of certain of, is that its direction is the orthogonal one with respect to real 4th dimension and that it acts by sending or rotating its quaternion product imaginary partner, whatever it might be, in the perpendicular direction with respect to itself. In that sense, part of the product partner that aligned with its imaginary direction n_q, that is |p1|*cos(ϕ_(q,p)), deserved to go to their sort of mutual 4th dimensional scalar home, just like its very own |q1| in the product U_q1*q'. Part of the p' partner that showed sort of misalignment with n_q, |p1|*sin(ϕ_(q,p)), got rotated in the only direction that's sure to be perpendicular to them both, as U_p1 = exp(i*n_p*pi/2) also performs a rotation for 90°, that is to direction orthogonal to the plane spanned by n_q and n_p, that is by them both.
Since p', that is U_p1, in fact acts in the same manner, just with respect to its own imaginary axis n_p, exchanging places in the product just changes the orientation of i*n⊥(q,p) to -i*n⊥(q,p), which might be poetically summarized that a left product partner shakes with its right hand, while the actor that multiplies from the right, seals the deal with its left hand.
By now, it seems rather straightforward to deduce what happens when general origin quaternion p=(p0+p') replaces its pure imaginary part p' in the product, as only difference between them is the addition of a scalar part p0,
U_q1 * p = i*n_q * (p0 + i*n_p*|p1|) =
= -|p1|*cos(ϕ_(q,p)) + i*|p1|*sin(ϕ_(q,p))*n⊥(q,p) + i*n_q*p0,
that is being the real of 4th dimension, p0 got the same action applied as q0 in previous comments and ended up pointing in the direction of the imaginary unit vector i*n_q.
To sort of complete the mathematical part of the story, here are the expressions still remaining to be discussed,
DeleteU_q * p' = (cos(ϕ_q) + i*n_q*sin(ϕ_q)) * i*n_p*|p1| = |p1|*cos(ϕ_q) * i*n_p - |p1|*sin(ϕ_q) * (n_q*n_p) =
= -|p1|*sin(ϕ_q) * cos(ϕ_(q,p)) + i*n_p * |p1|*cos(ϕ_q) + i*n⊥(q,p) * |p1|*sin(ϕ_q) * sin(ϕ_(q,p),
and finally,
U_q * p = (cos(ϕ_q) + i*n_q*sin(ϕ_q)) * (p0 + i*n_p*|p1|) =
= p0*cos(ϕ_q) - |p1|*sin(ϕ_q)*cos(ϕ_(q,p)) + i*n_p*|p1|*cos(ϕ_q) +
+ i*n⊥(q,p)*|p1|*sin(ϕ_q)*sin(ϕ_(q,p) + i*n_q*p0*sin(ϕ_q).
As it's a bit late already, and busy working day awaits tomorrow, just a quick closing remark on the fly about possible application of presented things in this comment series vis-a-vis measurement of unknown axis n_p.
In the expression for the product (U_q1*p), the imaginary vector part is composed of two terms, and the second, i*n_q*p0, points in a given imaginary unit n_q direction. By changing or rotating the axis n_q, and measuring the overall imaginary vector component, which is basically the only part of the quaternions and their products that we know how to deal with physically in a lab, in the direction of n_q, we can determine the angle ϕ_(q,p) and thus the axis n_p up to the collinearity with respect to n_q.
When the imaginary vector component in the n_q direction reaches its maximum value while rotating n_q, it's an indication that sin(ϕ_(q,p)) = 0 or as close as it can be to 0, meaning that ϕ_(q,p) is also closest it can be to 0, because the other imaginary vector component is pointing to n⊥(q,p) and the existence of its non-zero value can only diminish the overall imaginary vector value in the direction of a given n_q.
Good night.
В квантовой механике применяются комплексные волновые функции, но для прояснения алгебры этих функций можно обойтись и без комплексных чисел. Именно об этом ваш пример с кватернионами. Наверно, дополнительно можно было бы привести пример с 4-мерными действительными кососимметрическими матрицами. Но с точки зрения физики важно дать интерпретацию этих матриц.
ReplyDeleteНу раз сказал а, то уже скажу и бэ. На заметку физикам: эрмитовы матрицы, чисто мнимые кватернионы и кососимметрические матрицы изоморфны алгебре Ли векторных полей Киллинга 4-мерного евклидова пространства, которая эквивалентна алгебре Ли линейных векторных полей, касательных к трёхмерным сферам с центром в нулевой точке.
DeleteDo you get (1.1.6) from your matrices? Or you get something else?
DeleteВы меня спросили, что я ещё (кроме алгебры кватернионов) получаю из этих матриц и я вам ответил, что ничего. Однако, это не означает, что из алгебры кватернионов нельзя получить алгебру Ли. Наоборот каждой алгебре Клиффорда соответствует своя алгебра Ли. Во введении об этом я просто не пишу.
DeleteSo, how did you get (1.1.6)? Where from?
DeleteЯ их просто аккуратно выписал из последней главы (заметки) книги, где конструируются векторные поля сфер с помощью группы зеркальных симметрий.
DeleteYou write before (1.1.6) : " orthogonal to each other (in the
Deletemetric of a 4-dimensional Euclidean space). Can you show me how do they are "orthogonal to each other"? Show me the orthogonality of the first two fields. They do not want to be orthogonal for me. How does it happen that they are orthogonal for you?
В каждой точке X_1X_2X_3X_4 скалярное произведение векторных полей (которое вы не признаёте) равно X_2X_4-X_1X_3-X_4X_2+X_3X_1=0
Delete"скалярное произведение (которое вы не признаёте)"
DeleteWhere did you find such a "scalar product"? Is it your own invention? How exactly you define your "metric of a 4-dimensional Euclidean space"?
AB=A_1B_1+...+A_4B_4 в ортонормированном базисе. Не верю, что это для вас новость.
Deleteφx has components (x2, -x1, x4, -x3)
Deleteφy has components (x3, x4, -x1, x2)
If I use your definition, I get
x2 x3 - x1 x4 -x1 x4 -x2 x3 = -2 x1 x4 instead of 0.
Из предыдущей заметки
Delete∂φ_x=A_x(x_1,x_2,x_3,x_4)^{T}(∂x_1,∂x_2,∂x_3,∂x_4) =
(-x_2,x_1,-x_4,x_3)
∂φ_y=A_y(x_1,x_2,x_3,x_4)^{T}(∂x_1,∂x_2,∂x_3,∂x_4) =
(-x_4,-x_3,x_2,x_1)
SO(3) aka SU(2) in multivector form is XZ, YZ, XY which has a transverse-longitudinal structure of sorts. For quaternions aka Cl(2) it would be X, Y, XY plus the scalar aka also a transverse-longitudinal (plus scalar/timelike) structure. Spinors are kind of vector-like in the vector-half spinor-half spinor triality sense at 8-dim which 4-dim might be a dimensional reduction from.
ReplyDeleteI tend to think of everything in a multivector sense and via Tony Smith I have cellular automata pictures in mind as an analogy for multivectors. The transverse-longitudinal-timelike structure of spacetime is plastered all over the place for multivectors.
"For each point
ReplyDeletex=(x1,x2,x3) in R3, associate the following 2x2 Hermitian matrix,
σ(x):
σ(x)=(x3 x1−ix2, x1+ix2 −x3)"
Could we use Anti-Hermitian matrix instead?
Like iσ(x) ?
We could, indeed. It is sufficient to multiply by "i" or by "-i". Then we would have "quaternion algebra." But here, for applications in QM, I prefer Hermitian matrices.
ReplyDeleteАркадиуш, позвольте полюбопытствовать,- вы сами рисуете картинки к своим постам?
ReplyDeleteNo. It is the computer that is painting them. I am only providing instructions to the computer what to paint and how.
DeleteА у меня первое поле -X_2,X_1,-X_4 а второе -X_4,-X_3,X_2,X_1
ReplyDeleteLook at your (1.1.6) - P.S. 18-10-24 9:33 (I put a screenshot at the end of this post). Do you see -X2 as the first component?
DeleteАркадиуш, прошу прощения. Похоже там осталась старая (ошибочная) версия этих уравнений.
DeleteSo, fix it. And come back after fixing it.
DeleteБыстро не получится, я сейчас на работе.
DeleteИсправил.
Delete"the basic vector fields ∂ϕx, ∂ϕy, ∂ϕz form the algebra of quaternions"
DeleteThey do not form the algebra of quaternions. For instance ∂ϕx squared is not -1.
0-1 0 0
Delete1 0 0 0
0 0 0-1
0 0 1 0
=A
A^2=diag(-1,-1,-1,-1)
You wrote "vector fields", not "matrices". Your vector fields do not form the algebra of quaternions. This is a serious mathematical error!
DeleteОни образуют алгебру кватернионов относительно операции умножения векторных полей $A\ast B = \nabla_{B}A$, поскольку в результате такого умножения векторных полей получается векторное поле, представленное матрицей, равной произведению матриц, представляющих векторные поля сомножителей. И в чём тут ошибка?
DeleteIf you introduce your own operation of multiplication of vector fields, you should first define it - otherwise you mislead the readers. Are you misleading them on purpose?
DeleteAnd second: quaternion algebra is 4-dimensional (thus the name "quaternions"), and you have only three vector fields. They do not "form" the algebra of quaternions, they do not even "form" a basis in this algebra. This is another serious mathematical error. Fix these errors, then come back.
Аркадиуш, спасибо за уточнение. Я по наивности полагал, что это общепринятое определение, а словом "образуют" заменил фразу "являются образующими". Ещё раз спасибо, буду исправлять.
DeleteWhen correcting, don't forget this: If you introduce your own operation of multiplication of vector fields, you should first define it.
DeleteИсправил как умел.
Delete