The Spin Chronicles (Part 24): Conformal dig

 Continuing from Part 23,  in the grand tradition of curious explorers, let’s keep digging..

We dig up diamonds by the score

A thousand rubies, sometimes more

Though we don't know what we dig 'em for

We dig dig dig a-dig dig

A thousand rubies, sometimes more

Though we don't know what we dig 'em for

We are left with exp(tX) for yet unexplored candidates for X, namely e₁, e₂, ie₁, ie₂. Here the calculations can become cumbersome, and, in fact, this what they are. But they do not have to be such if we are clever enough. I was not that clever at the beginning, so I lost quite some time, until I remembered something vaguely, and an enlightenment suddenly occurred to me. After that I checked, and rechecked, and here comes you rabbit from the hat - the Eureka great discovery!  The trick is extremely simple ..... once you know it!

We have these e₁, e₂, ie₁, ie₂ that we want to exponentiate. But why exactly THEM? They are part of the "Lie algebra" (you don't have to know what the Lie algebra is), they satisfy the fundamental requirement X+ ν(X) = 0, but so is any linear combination of them! So, let's take these linear combination, but in a clever way. And what is this "clever way" I am telling you right away, now.

Here are the linear combinations that work for us:

1. e¹ + ie²

2. e² - ie¹

3. e¹ - ie²

4. e² + ie¹.

We will take 1/2 of them. Today, we’ll focus on 1 and 2. Combinations 3 and 4? I leave those as a rewarding exercise for you, dear Reader.

Now, what is so special about these linear combinations? What is this "great discovery"? Take, for instance

X = (e¹ + ie²)/2

We have that e¹ and e² anti-commute and their squares are 1, while i² = -1. As the result X² = 0. Check it! As the result the whole series for the exponential exp(tX) terminates after the linear term:

exp(tX) = 1 + tX. = 1 + t(e¹ + ie²)/2

Then τ(X) = (e¹ - ie²)/2,

and

τ(exp(tX)) = 1 + t(e¹ - ie²)/2.

Thus we need to calculate

e'⁰ = (1 + t(e¹ + ie²)/2) e⁰ (1 + t(e¹ - ie²)/2),

e'¹ = (1 + t(e¹ + ie²)/2) e¹ (1 + t(e¹ - ie²)/2),

e'² = (1 + t(e¹ + ie²)/2) e² (1 + t(e¹ - ie²)/2),

e'³ = (1 + t(e¹ + ie²)/2) e³ (1 + t(e¹ - ie²)/2).

It’s a straightforward algebraic exercise—boring, yes, but manageable. I used the powerful and free computer algebra system "Reduce" to automate these tedious steps. If you’re not using tools like this, you’re missing out! Anyway, you tell the program the rules, and then it does the job. My rules were

Operator e$
Noncom e$
for all i let e(i)*e(i)=1$
for all i,j such that i>j let e(i)*e(j)=-e(j)*e(i)$
let e(1)*e(2)=i*e(3)$
let e(2)*e(3)=i*e(1)$
let e(1)*e(3)=-i*e(2)$

Note: Alternatively, one can play with the Pauli matrices instead.

In my mind I set e⁰=1, as it is an exception. Then you give it expressions to calculate, and then, without even a blink of the eye, it spits out the results. You tell the program to write them to a file. Here they are, rewritten using our notation, where I manually replaced the scalar part by e⁰:

e'⁰ = (1+t²/2) e⁰ + te¹ + (t²/2)e³,
e'¹ = te⁰ + e¹ + te³,
e'² = e²,
e'³ = -(t²/2)e⁰ - te¹ + (1-t²/2)e³.

The transformation matrix, in Mathematica notation, is

{{1 + t^2/2,  t,  0,  t^2/2},
{t,  1,  0,  t},
{0,  0,  1, 0},
{-t^2/2,  -t,  0,  1 - t^2/2}}

The coordinates transformation matrix is the transposed matrix,

m={{1 + t^2/2, t, 0, -(t^2/2)},
{t, 1, 0, -t},
{0, 0, 1, 0},
{t^2/2, t, 0, 1-t^2/2}}

We are almost done. We recall from Part 18 the formula for embedding the x¹=x, x²=y plane into the null cone of Minkowski space:

x ⟼ ζ={ζ₀,ζ₁,ζ₂,ζ₃}={1, 2 x/(1 + q), 2 y/(1 + q), (1 - q)/(1 + q)},

where I set x·x = x² + y² = q.

We act with the matrix  m on vector ζ to get:

ζ'={(1 + q + q t^2 + 2 t x)/(1 + q), (2 (q t + x))/(1 + q), (2 y)/(1 + q), (1 + q (-1 + t^2) + 2 t x)/(1 + q)}

From this we can read x' = ζ'₁/(ζ'₀+ζ^'₃) and y' = ζ'₂/(ζ'₀+ζ^'₃). The result is:

x' = (x + q t)/(1 + q t^2 + 2 t x), 

y' = y/(1 + q t^2 + 2 t x).

Now we can repeat the same procedure with

Y = (e² - ie¹)/2.

This time we get

x' = x/(1 + q t^2 + 2 t y), 

y' = (y +qt)/(1 + q t^2 + 2 t y).

We can then check the very important property: X and Y commute! We replace t by "a" for tX, and by "b" in tY. Then

g = exp(aX)exp(bY)=exp(bY)exp(aX)=exp(aX+bY).

The transformation implemented by such a g is then

x' =(x + qa)/(1+ q (a²+b²) + 2(xa+yb)),
y' =(y + qb)/(1+ q (a²+b²) + 2(xa+yb)).

This is known as a "special conformal transformation". Conformal transformations are known to preserve angles, and to transform circles into circles (straight lines are considered as infinite radius circles).

Exercise 1. Do the same with 3) and 4). This should be even easier. Why? You will see when you try.