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.