Spin Chronicles Part 26: Imagine Clifford Algebra

 The song is in C major. Even if I like C minor better, I like the song.

"Imagine all the people

Livin' life in peace

You"

Can you imagine it? I can, even though it is not in my memory banks. Imagination is a powerful mechanism. I am reading "The Story of the Tower of Babel: Human Aspirations and Dreams".

There:

"Human imagination is a unique and powerful ability that allows us to conceive of things that do not exist, create new concepts, explore possibilities, and even transcend the current physical limitations in spirit. This ability is indeed rare in the natural world, and, at least based on our current knowledge, humans are unique in their use of complex symbolic thinking and creative imagination."

Imagination and creativity are close friends. I am imagining what I will write today. Then I start writing. The result will probably be not quite as I have imagined it.

"You may say I'm a dreamer

But I'm not the only one

I hope someday you'll join us

And the world will be as one"

You may say I'm a dreamer....

Yes, you may say I'm a dreamer. But I'm not the only one. Last night I dreamed about Hilbert Algebra. The dream had its roots in my past memory. My PhD thesis was partly about equilibrium states in quantum thermodynamics, the algebraic approach, with KMS (Kubo-Martin_Schwinger)  states. Tomita and Takesaki, the two Japanese mathematicians, developed the theory of Hilbert algebras and modular automorphisms. Much later, in 1994,  Alain Connes and Carlo Rovelli  applied it  to the "flow of time":

"We consider the cluster of problems raised by the relation between the notion of time, gravitational theory, quantum theory and thermodynamics; in particular, we address the problem of relating the “timelessness” of the hypothetical fundamental general covariant quantum field theory with the “evidence” of the flow of time. "

But now is now, and we are dealing with the Ur atom of space - the geometric Clifford algebra of 3D space Cl(V). We have an algebra. Buit where is "Hilbert"? Searching for "Hilbert Algebra" leads us to Wolfram MathWorld. There we find conditions 1, 2, 3, 4, 5 that need to be satisfied.

In our case only condition 3 counts. All other are automatically satisfied. So, we should have an algebra with involution (we have one), which is a Hilbert space (and Cl(V) is a Hilbert space, as we will see soon),  with a scalar product <a,b> satisfying

<ba,c> = <b,ca*>.                    (1)

We need to define Hilbert space scalar product on Cl(V). We have already done it in Part 11,

where we have defined:

"... But then we have also the second natural "bilinear" form:

B_τ(u,v) =t_n(τ(u)v),

In fact, this form is, when considered as a form on a complex vectors space,  Hermitian: it is complex linear in v, but complex anti-linear in u."

In Part 12 we further wrote:

B_τ(u,u) = |p₀|² + |p|².

This is our Hilbert space scalar product - the scalar part of τ(u)v: we set

<u,v> = the scalar part of the product τ(u)v. If u = (p₀,p), v = (q₀,q), then

<u,v> = p_0^*q₀ + p^*· q.

Then we need an involution in our algebra, we need a *-algebra. We take τ for our involution. We just set u* = τ(u). There may arise some confusion now. Because in my blog I often used the star to denote the complex conjugation. And now I am using the same symbol to denote the conjugation in the algebra! Well, the meaning will be defined by the context, and when confusion can arise, it will have to be explicitly addressed.

Let us recall the action of τ:

If u = (p₀,p), then τ(u) = (p₀*,p*), thus u* =  (p₀*,p*). In other words:

(p₀,p)* = (p₀*,p*).

Not too bad! I suppose we can live with that.

Now comes the crutial point:

Exercise 1. Prove (1).

Hint: there are at least two ways of proving it. Hard way, using definitions, and not-so-hard, if you are creative.

Spin Chronicles Part 25: QM ahead

 A few days ago, my neighbor—a passionate enthusiast of Clifford and other algebras—asked me: why do I use Cl(V) instead of Cl(3) like most others? This is a fair question, and my answer lies in a subtle but significant distinction. I am not entirely sure what people mean when they refer to Cl(3). Sometimes, they mean the Clifford algebra of R³, but our Cl(V) is not the same as that.

I equip V, our model of space, with a preferred orientation

So, what exactly distinguishes V from R³? In R³ there exists a canonical, distinguished basis consisting of the vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1). This basis is intrinsic to R³  and is the same for everyone. However, V, as we consider it, has no such distinguished basis. If we were to choose a basis for V, then V could indeed be identified with R³, but this identification depends entirely on the choice of basis. This dependency makes a world of difference, especially if we think of V as a model for physical space.

When we observe our physical space, do we see a distinguished basis? Not at all. One observer might adopt one basis, while another might choose a completely different one. In this sense, the basis is observer-dependent, and this observer dependence plays a crucial role in quantum theory. While in relativity—both special and general—we often strive to separate observer-dependent phenomena from observer-independent (or coordinate-independent, as in general relativity) quantities, quantum theory forces us to grapple directly with this observer dependence.

Therefore, I could use Cl(3) in my notation, and perhaps I should, provided we understand exactly what I mean by Cl(3). However, this might not align precisely with what others mean by the same term. Although the difference is often non-essential, it is critical to remember the freedom of choosing a basis. This subtle detail will soon become very important.

Introducing Cl(V)

Let’s delve into this intriguing construct, Cl(V). Since today is Sunday, we can afford to speak more about generalities and less about technicalities. Of course, technicalities matter—they give us the "real taste" of the tool we’re dealing with. After all, Cl(V) is a mathematical tool. It’s not a tangible object that burns your fingers when you touch it, yet it possesses a certain depth and elegance.

So far, our focus has been on uncovering the hidden structure within Cl(V). Through it, we’ve encountered rotations, translations, Lorentz boosts, space and time, and even electromagnetic field vectors E and B. All of these concepts lie embedded within Cl(V), waiting to be revealed by a proper lens. However, the ultimate goal of this series has been to understand spinors, and so far, they have remained elusive. Spinors, as you might know, originate from quantum theory—a topic we’ve yet to discuss in depth.

Interestingly, spinors are also valuable in classical relativity. They provide insights into certain reformulations of general relativity. Whether this utility is merely coincidental or indicative of something deeper remains an open question.

A Quantum Perspective on Cl(V)

Now, let’s start examining Cl(V) from a quantum perspective. The first thing to note is that Cl(V) is an algebra. This should immediately set off a mental alarm: quantum theory is fundamentally algebraic—at least in its operator formulation. (This stands in contrast to the wave mechanics picture championed by Schrödinger and de Broglie.) So, for now, let’s zero in on this algebraic aspect of Cl(V).

Observer Independence and Orientation

Before moving further, there’s another small but crucial detail worth highlighting. Earlier, I emphasized that our space has no preferred basis or direction. However, there is substantial evidence suggesting that it does have a preferred orientation—or “handedness,” if you prefer that term. This preference manifests in biological systems, in parity violation at the micro-level (and possibly at the scale of spiral galaxies, though proving this is more challenging). It also appears in the asymmetry between matter and antimatter in the observable universe.

This preference manifests in biological systems

For this reason, I equip V, our model of space, with a preferred orientation. This choice has significant algebraic consequences: Cl(V) acquires a natural complex structure when we identify e¹e²e³ with the imaginary unit "i". Importantly, this identification is basis-independent as long as we restrict ourselves to bases with the same orientation. Thus, while Cl(V) is fundamentally a real algebra, it gains a natural complex structure. Recognizing this, we also see that Cl(V) becomes isomorphic to the algebra of complex quaternions. A fascinating realization, indeed!

The Road Ahead

This interplay between algebra and complex structure unmistakably hints at quantum mechanics. In the posts to come, we will dig deeper into this connection, exploring how Cl(V) serves as a bridge between the abstract mathematical framework of Clifford algebras and the physical principles of quantum theory. 

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.