Spin Chronicles Part 35: Rotating vectors

 Anna asked " why is the spin in m-direction unitarily equivalent to sigma3 and what does it mean?"

Spinning - Rotations in action

This question was asked in a comment to the previous post, and here I will propose my answer. There is more than one way of answering possible, and I will chose just one, that relates to previous discussions.

Let us address a more general question: given any two unit vectors m and n in V, why is the spin in m-direction unitarily equivalent to the spin in n-direction? 

Let us analyze the question first. What exactly do we mean by that? What is "spin in m-direction"? The fact that we are asking about "unitary equivalence" indicates that we have in mind operators, not state vectors. Which operators?

Introduction

Given a unit vector m in V, we consider it as an element of the Clifford algebra Cl(V). Then it acts, in particular, as an operator L(m) of left multiplication by m on Cl(V) . But Cl(V) is 4-dimensional complex, and when we say "unitarily equivalent", we mean in a 2-dimensional complex space. So, we probably have in mind some irreducible representation of CL(V), perhaps one provided by one of its non-trivial left ideals. But we can start answering our question even before specifying the representation. And this is what we will do now. Cl(V) is a Hilbert space, the scalar product <v,w> is defined as the scalar part of v*w. We know that L(u*) = L(u)*, where on the right-hand-side we have Hermitian conjugate of L(u) with respect to the <v,w> scalar product. The spin operator in m-direction is then L(m), the spin operator in n-direction is L(n). In matrix representation, if we select an oriented orthonormal basis e¹,e²,e³  in V, and represent the basic vectors by the three Pauli matrices,  L(n) would be represented by the matrix σ(n) = n₁σ₁+n₂σ₂+n₃σ₃. But we do not have to use matrix representation yet. We can proceed on an algebraic level, and descend to a particular representation only at the very end. It will be more "geometrical" this way. Instead of using almost mindless matrix multiplication we are going to use scalar products and vector products, which have rather simple geometrical meaning. Well, we will also use the exponential, but this will be  just a compact way of using sin and cos functions. It is a longer way, but it gives some satisfaction.

In the past, using n²=1 (if n is a unit vector), we calculated, within Cl(V),  exp(itn), with the result

exp(itn) = cos(t) + in sin(t).

But we do not have to calculate anything, we can just define u(t,n):

u(t,n) = cos(t) + in sin(t).

Then u(t,n) is in Cl(V), and we can verify that u(t,n) is unitary:

u(t,n)u(t,n)* = u(t,n)*u(t,n) = 1.

But the U(t,n)=L(u(t,n)) is a unitary operator acting on the Hilbert space Cl(V).

U(t,n)U(t,n)* = U(t,n)*U(t,n) = I (the identity operator on Cl(V)).

We will use this fact in what follows, remembering that it holds for any unit vector n and any real t.

Back to the original question

Let us return to the original question. We have two unit vectors m and n. There are two cases here. The first, generic case, is when m and n are not parallel. There are two exceptions from this generic case: m=n and m=-n. We will discuss first the generic case.

If m is not parallel to n, then the cross product m⨯n is non-zero. In fact we have

||m⨯n|| = |sin(θ)|,

where θ is the angle between these vectors. The vector k=m⨯n/|sin(θ)| is then in V, and of unit norm. Thus

u(t,k) = cos(t) + i k sin(t)

is unitary. We can use our formula for the Clifford product to calculate u(t,k) m u(t,k)*. This is a simple exercise with cross products. The result is:

Exercise 1. Calculate the result.

Exercise 2. Check that with t = θ/2 (or t = -θ/2, I am not sure which, since I did not yet do these calculations!) we get

u(t,k) m u(t,k)* = n.

So m and n are unitarily equivalent in Cl(V). But then, since L is a *-representation, we have that U(t,k) gives us unitary equivalence of L(m) and L(n) acting on Cl(V). This is in four complex dimensions. How to descend to two? Simple: choose two-dimensional left ideal. For instance choose e¹,e²,e³, and choose the left ideal determined by p = (1+e³)/2, as we have done before. But any other choice will do as well. Since it is a left-ideal, it is invariant under the action of  U(t,k) = L(u(t,k)). And U(t,k) being unitary in the whole Cl(V), is also unitary within any invariant subspace.

What remains is the exceptional case of m = -n.  This exceptional case can be handled even simpler. Let k be any unit vector in the plane perpendicular to n. Let u = kn (Clifford algebra product). Then

unu* =(kn)n(nk) =knk = -n.

Moreover, u is unitary in Cl(V).

Exercise 3. Verify this last statement.

Decscending from 4 to 2 dimensions
(left ideal) works as before.

Spin Chronicles Part 33: Consciousness and conscious choices

 Once upon a time, guided by a whisper of intuition—or perhaps a playful nudge from fate—we set out on a journey. At first, our quest seemed clear: to uncover the mysteries of the enigmatic spinors. We had a map (or so we thought) and a destination in mind. But as we wandered deeper into the unknown, the wide road faded into a meandering trail, and the trail became a wisp of a path. Before we knew it, we were in a forest—dense, shadowy, and alive with secrets.

The forest wasn’t on the map, but here we were. And while our grand quest felt like a distant memory, the forest itself had other lessons to teach. At first, we worried: how would we ever find our way? But then we noticed the sweetness in the air, the earthy scent of moss, and the rustling leaves whispering ancient songs. We saw plump berries, glistening with dew—some delicious, others mysterious. And then, as if out of a dream, a gentle roe deer emerged, its soft eyes urging us to follow. It led us to a crystal-clear lake, where the water was cool and refreshing, as though the forest itself offered us a blessing.

... the forest itself offered us a blessing.

Forests, after all, are not just places to get lost; they are places to be found. They nourish the soul, if only we stop to look. So we paused, took a deep breath, and began to notice both the towering trees and the soft carpet of the forest floor. In this moment of stillness, we remembered our original quest. Yes, we were here to understand spinors, but perhaps the forest—the journey—was as important as the destination.

This particular forest is called Geometric Algebra A. It is a simple land, yet rich with wonder. To truly know it, we must not just walk its trails but see its beauty, smell its air, touch its textures, and listen to its tales. Some stories are soft whispers; others roar like waterfalls. This is one of those stories, told by the forest itself.

So, dear traveler, let us begin.

We are in geometric algebra A. It is simple, but it has a rich structure. We need to feel this structure by sight, by smell and by touch. We need to be able to hear the stories it says to us, sometimes silently, sometimes in a really loud voice. So this is on of these stories.

A is simple. In algebra saying this has a precise meaning: an algebra is simple if it has no non-trivial two-sided ideals. A two-sided ideal is a subalgebra that is at the same time left and right ideal. We did not consider two-sided ideals yet (and we will not in the future), but it is not difficult to show that A is indeed simple. But we did consider left ideals, and those of a particular form. To construct such an ideal we select a direction (unit vector) n in V, from this we construct p, with p=p*=pp:

p=(1+n)/2

Then we define, let us call it  I_n:

I_n = {u: up=u}.

This is a left ideal.

Now, the defining equation up=u is equivalently written as un=u (convince yourself that this is indeed the case!). Since n² =1 (n is a unit vector), and n*=n, it follows that n, considered as an operator acting on A from the right, has two possible eigenvalues +1 and -1. So the equation  un=u means that I_n consists of eigenvectors of n belonging to the eigenvalue +1. This is our left ideal under consideration.

The first thing we notice is that p itself is an element of I_n. But that is not all I_n. I_n is a complex two-dimensional space. Thus there are two linearly independent (even mutually orthogonal) vectors in I_n.

Thus we proceed as follows: we choose an oriented orthonormal basis e¹,e²,e³ in V in such a way that e³ coincides with n. Then e² and e³ are perpendicular to n. Then we define a basis E₁,E₂ in I_n by choosing:

E₁ = p
E₂ = (e¹ - ie²)/2,

Then magic happens: in this basis the left action of e¹,e²,e³ on  I_n is given exactly by the three Pauli matrices!

We came to the lake in a forest and it is time to fore out the thinking machine in our brains. We have arrived naturally at Pauli matrices, which is very rewarding. Except for the fact that there is nothing "natural" in this process! First we had to select a direction n in otherwise completely isotropic space V. This cannot be deterministic. No deterministic process can lead to breaking a perfect symmetry. It can be done only by a conscious choice. So consciousness is entering here (or it can be a random choice, but then consciousness is needed to define what precisely a "random choice" is). In practice the choice of a reference direction, and of an orthonormal basis is being made by a conscious "observer" (or by a machine programmed by a conscious "observer"). You can, of course, replace "observer" by "experimental physicist" or "an engineer", but that will not change the idea.

Then we decided to define E₂ the way we did above. Another application of consciousness. Thus, temporarily, I am associating the right action of the algebra on itself, the action of p in up=u, with consciousness. It is not needed for further considerations, but it something that should be thought about: we have left and right actions of operators on our Hilbert space. In ordinary quantum theory only left actions are being considered, What can be the meaning of right actions, if any? But let us abandon philosophy and return to math.

We have the basis E_α (α=1,2) in I_n, we have the basis eⁱ (i=1,2,3) in V and they are related by Pauli matrices σ_i by the following relation

eⁱ E_α = E_β (σ_i)_βα.       (*)

Notice that I write the right hand side by putting the basis vectors first, and coefficients after. That has the advantage that matrices transforming the components are transposed to those transforming the basis vectors. This way I do not have to transpose anything.

But what happens if we replace our basis E_α by some other orthonormal basis in I_n? Then the whole beauty and simplicity of Pauli matrices will be spoiled. And we like Pauli matrices so much! And here is the place to demonstrate the power of the desire. We want Pauli matrices, whatever the cost would be! So, we start thinking. When there is a desire, there must be a way! So we start looking at our equation (*) from a different point of view. The elements E_α  and eⁱ are related (or "correlated") by the Pauli matrices. If change E_α, perhaps eⁱ also need to be changed so, that the correlation stays the same? We try our great idea of saving our love - the sigmas. The result is condensed in the following statement:

Proposition. There is one and only one way to have Eq. (*), with Pauli matrices in it, valid for all orthonormal bases in I_n. It goes as follows: if E_α is replaced by E'_α related to E_α by a 2 by 2 unitary matrix A of determinant 1 (element of SU(2)):

E'_α = E_β A_βα,

then e_i must be replaced by e'i, related to e_i by a real orthogonal 3 by 3 matrix R(A) (an element of SO(3)):

e'_i = e_j R(A)_ji,

where the relation between A and R is

Aσ_iA* = σ_j R(A)_ji.

Proof. Left as a straightforward, but needing use of indices,  exercise.

And this way we have accomplished something that was left unexplained in October 16 post  Part 3: Spin frames.

What we see now is that this correlation between spin  frames Eα and orthonormal frames e_i is not so "natural" at all. It requires certain arbitrary human-made choices. It has little to do with the "true state of affairs". Spin frames and orthonormal frames are two different realities. Yes, they can be "correlated", but this correlation is artificial. So, the question remains: what are spinors? Elements of a left ideal? But which one? And why this one, and not some other?

Exercise 1. Do the calculations needed to prove the Proposition.

Exercise 2. For any x in A denote by Ax the set

Ax = {ax: a in A}

Show that Ax is a left ideal. Show that it is the smallest left ideal containing x. With I_n and p =(1+n)/2  show that I_n = Ap. Why Ap is not the same as An? What is An?

Exercise 3a.  If u in A is invertible, it cannot be contained in any of the  I_n's.

Exercise 4. Show that the * operation transforms every left ideal into a right ideal, and conversely.

Exercise 5. If Il1 and Il2 are two left ideals, is their intersection also a left ideal? If Il is a left ideal and Ir a right ideal, is their intersection a two-sided ideal?