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. 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.
