# Fictitious Snuark Vacuum II

In the previous post we introduced a fictitious snuark vacuum with the intention of using it to simplify calculations. In the snuark algebra, complex amplitudes show up only as a geometric result of an interaction between two mass vertices. We’d like to convert this sort of computation, which uses Pauli spin matrices, into calculations where we associate an amplitude with each mass vertex by itself. There are 36 mass vertices, 12 of which are trivial. We found that we would know the remaining 24 by symmetry if we could solve for just four of them: $k_{+x,+y}, k_{+x,-y}, k_{-x,+y}, k_{-x,-y}$. In this post, we find these numbers, and then test them by comparison with explicit calculations using Pauli matrices.

A Beautiful Theorem: When visiting a strange land, it’s always nice to take in the pretty sights. For our utilitarian calculations, we will use a beautiful result for products of primitive idempotents (density matrix states) of the Pauli algebra:

Let $\vec{a}, \vec{b}, \vec{c}$ be three real unit vectors in 3 dimensions. They define the vertices of a spherical triangle. Let A, B, and C be the projection operators for Pauli spin in the $\vec{a}, \vec{b}, \vec{c}$ directions. Then the projection operators are related by the equation:

where $S_{ABC}$ is the oriented area of the spherical triangle defined by the three unit vectors.

Some notes: The total area of the sphere is $4\pi$, so the factor of 1/2 in the phase part of the above equation comes from the surface of the sphere being converted into an angle between 0 and $2\pi$. And the real factors inside square roots are related to the familiar rule for transition probabilities between two Pauli states with the angle theta between them:

The magnitude parts of the above theorem are just the familiar transition probabilites between quantum states. The phase part is less familiar. It can be generalized: the geometric phase arising in the Pauli algebra from a path on the surface of the sphere is given by half the oriented surface area bounded by the path. There are various proofs. There is a proof in sections 2.3 and 2.4 of my incomplete book on density operators.

From the theorem above, and the fact that (1+cos(pi/2))/2 = 0.5, it is clear that the $k_{\alpha,\beta}$ all have the same magnitude, $\sqrt{0.5}$. And the theorem gives us a way of computing the phase, all we have to do is to compute the oriented surface areas of the regions of the spherical surface given by the quadrilaterals, 0+X+Y0, 0+X-Y0, 0-X+Y0, and 0-X-Y0.

The six snuark directions, +x, -x, +y, -y, +z, and -z define six points on the sphere. Drawing the spherical triangles with these points as vertices cuts the surface of the sphere into 8 equal sized triangles. Each of these triangles has area $4\pi/8 = \pi/2$, and therefore contributes a phase of $\pi/4$. We can draw these areas on a flat surface and use the drawing to compute geometric phases. For example, if our path runs from +x to -z to -x to +z and back to +x, then the path encloses (in a positive direction) the four spherical triangles shaded brown:

There are four enclosed triangles, each contributing $\pi/4$ so the total phase of this sequence of projection operators is $\pi$, and the geometric phase is $\exp(i\pi)$. Note that there are only 7 triangles shown, the eighth is exterior to the -x, -y, -z triangle, or interior to the -z, -y, -x triangle. These sorts of calculations are particularly convenient when you are working with pen on paper; my notebooks are filled with them.

Our vacuum state is in the middle of the +x, +y, and +z unit vectors. If we add the vacuum point to the above drawing, we get a more complicated surface drawing. Instead of dividing the surface into 8 equal triangles, we now have 16 triangles of various sizes. In the following drawing we’ve color coded these 16 triangles (6 red, 6 green, and 4 blue) according to their surface areas:

With the above drawing, we can compute the phases of various vacuum calculations. Here are the four phase calculations we need, with the contributing triangles shown in color:

It’s convenient to write these results with the phases in multiples of $\pi/12$ (or 15 degrees). Each has the same magnitude (note that the magnitude contributions for things like 0X, 0Y, and X0, etc., are cancelled out):

The above are quite similar to each other. We will write $k_{\pm m} = \sqrt{0.5}\exp(\pm m i\pi/12)$. With this notation, we can write a table of the 36 geometric amplitudes:

The above is particularly convenient for calculation. So long as we avoid the 1s and 0s, the product of n factors of k will have a magnitude of 0.5 to the k/2 power, and the phase is obtained by summing the m indices in the $k_{\pm m}$ and multiplying by 15 degrees.

When we convert our calculations back to geometric form, we need to remember to divide by the k associated with the initial and final state. Our usual example may assist in explaining. We compute the geometric phase of the +z to +y to +x transition, which is represented by the matrix multiplication XYZ. There are three $k_{\alpha,\beta}$ that we need, two for the +z to +y to +x transitions: $k_{+x,+y} = k_{+1}, k_{+y,+z}=k_{+1}$, and one for the overall transition which is +z to +x: $k_{+x,+z} = k_{-1}$. The computation is as follows, which matches the Pauli matrix calculation:

Putting all this together, to compute the geometric amplitude for a product of m propagators, no two consecutive parallel or anti parallel, we obtain a magnitude of
$2^{(2-m)/2}$, and a phase given by the sum of the pairwise phase numbers minus the phase number for the initial and final states.

We will also need to make calculations where the initial and final states are the same. For these, the geometric amplitude between the initial and final states is unity so the magnitude will be $2^{(1-m)/2}$ while the phase is given just by the pairwise numbers.

This is very convenient for computer calculation, but first, we’d better try some explicit examples and see if it works. My experience in teaching is that examples are more convincing to the students than the proofs.

Sample Vacuum Snuark Calculations

Let’s compute the Pauli product $X\bar{Z}Y\bar{X}YZ$. Pulling together all the factors of 1/2 in the Pauli matrices, the Pauli calculation is as follows:

Multiplying this out and comparing it with XZ, we find that the geometric amplitude is i/4.

Using vacuum amplitudes: There are six projection operators so m=6. This gives the magnitude as $2^{(2-6)/2} = 1/4$, which matches the Pauli calculation. The phase is computed by looking at the consecutive elements. Using the table of 6×6 vacuum amplitudes, the phase contributions are: +x to -z gives -2; -z to +y gives +2; +y to -x gives +2; -x to +y gives -2; +y to +z gives +1; for a total of +5. The overall initial to final is a +x to +z transition which takes a phase of -1. We subtract -1 from +5 to get +6. Therefore the overall phase is $+6\pi/12 = \pi/2$. This gives a geometric phase factor of i, which matches the matrix calculation.

Let’s try another: $Z\bar{Y}\bar{X}Z$. The Pauli calculation is:
.
The use of Z as both the initial and final states makes the Pauli computation a little easier. The result is $(1-i)/4 = 2^{-1.5}\exp(-i\pi/4)$.

With vacuum amplitudes, we need the +Z to -Y, -Y to -X, and -X to +Z amplitudes on the numerator, and the +Z to +Z amplitude (i.e. unity), on the denominator. There are three factors of sqrt(0.5) which matches the Pauli result. For the complex phase, we have +z to -y gives +2, -y to -x gives -7, and -x to which is +2. The sum of the phases is +2-7+2 = -3, so the computed phase is -3 pi/12 = -pi/4, which matches the Pauli result.

Great, I’ve got the sign issues fixed. Next, on to the computer calculation.

Update. I’ve got the simulation running. The applet uses the above vacuum method to compute products of Pauli projection operators. Source code is available here.

Filed under physics

### 14 responses to “Fictitious Snuark Vacuum II”

1. Hi Carl,
If you remember, I’m the one who have complitely missed the gravitational calculations. But I think I have found something interesting this time.
I have found a structure of the Ether that follows the inverse square law and agrees with Newton gravitational law using “Working model” software for simulation.
This Ether definition is based on a springs structure
I would like to have your opinion on it
Think you

2. carlbrannen

Saad, I loved your graphics and added your new blog to my blogroll. To get general appreciation for a physical model of the ether, you need to get a LOT more out of it than the power law. The next thing that comes to mind is an approximate Lorentz symmetry.

Ironically, as the string theorists have demonstrated, if you begin with some horribly erudite mathematical structure, people are very impressed if you are able to get some sort of gravity out of it. They set the bar higher for people who start with simpler ideas because, well, they’re the ones who set the bar and they like string theory.

If you want to see a sort of idea of mine that is similar, look at the graphs on page 2 of this paper of mine.

3. Thank’s a lot for your comments. I’m very happy to see that I could transmit the visual image I have about matter. Actually my intellectual reasonning process is 100% visual and it’s hard to formalise a 4D process.
But it should be possible. It will certainly need a couple of weeks hard work!
what do you think about what I’ve xrote about Ether0 and blak holes?

4. carlbrannen

Saad, your beautiful work really shows how the visual element helps in explanation and intuitive understanding. I’d like to emulate that. But as for the foundations, I will stick to algebraic stuff.

Hi Carl
Could you please send me your Email. I’d have some files te share.

6. Mike R.

Carl

Your snuark graphs look almost identical to the Fano plane.

7. Why yes, Mike R, we are indeed interested in noting this comparison!

8. carlbrannen

I guess I see some similarity. However, the Fano plane has “Each pair of distinct points lies on a unique line. Each line contains three points”.

For the fictitious vacuum, this is true only for lines which go through the vacuum, or, the same thing, lines that connect antipodes. The other lines, the lines that don’t include the vacuum, all contain four points. Both these sorts of lines are “great circles” and carry a geometric phase of pi, that is, give a factor of -1.

What I find interesting about this way of doing SU(2) calculations (on a finite subalgebra) is that it models a non commutative algebra by converting products into complex numbers, which of course are commutative. The non commutativity shows up purely in a complex conjugacy for the definition of this mapping. The more familiar and more general method of modeling SU(2) uses matrices which are inherently non commutative, and involve a lot more numbers.

9. Mike R.

Any three nodes of the Fano plane define a quaternionic subalgebra of the octonions. It is well known that the units of such quaternionic subalgebras can be mapped to the generators of SU(2), i.e. the Pauli matrices.

So maybe the analogy is just more than pictorial.

10. Carl, if you forget the hidden octonionic ‘1’ at the back, you can delete the outside circle and you’ll basically get the Fano plane.

11. kneemo

Carl, are (1+x)/2, (1+y)/2, and (1+z)/2 supposed to be orthogonal? If so, which inner product are you using?

12. carlbrannen

Kneemo, as you know, (1+x)/2 is just the Pauli projection operator for spin in the x direction, which people usually write $(1+\sigma_x)/2$. As such, it and (1+y)/2 are not orthogonal in the usual sense.

But “the usual sense” applies to massive particles like electrons and neutrons. There are no massless fermions in the standard model (with massive neutrinos), so one cannot answer the question “are massless fermions travelling in perpendicular directions orthogonal?” in the context of current physics.

The usual “orthogonal” in the Pauli algebra, when written in projection operator / pure density operator form, is that two operators are orthogonal if they annihilate each other. Another way of saying the same things is that their transition probabilities are zero. This is too restrictive to arrange for three Pauli particles to be orthogonal.

Instead, I have to define “orthogonal” to be the traditional geometric meaning. Two vectors are orthogonal if they have a 90 degree angle between them. Two vectors that point in exactly opposite directions (such as spin up and spin down) are not orthogonal, but instead are antiparallel.

So from this I think you can see how you can write an inner product that will make (1+x)/2, (1+y)/2, and (1+z)/2 orthogonal. Let A and B be two primitive idempotents. They are orthogonal if (2A-1)(2B-1) = – (2B-1)(2A-1).

The subtraction of the 1/2 is completely a formal thing caused by our preferring a representation of “A” and “B” that is built around idempotency. Perhaps a better representation of them would be the A’ =2A-1 and B’ =2B-1 forms.

Then the inner product would be A’ B’ + B’ A’ , and the usual operator multiplication would be A’ x B’ = (AB)’ = 2 (A’ +1)/2 (B’ +1)/2 – 1 = (1+A’ )(1+B’ )/2 – 1, which makes A to A’ look like a Lie group to Lie algebra transformation.

13. kneemo

Thanks for the explanation, Carl. The Pauli matrices turn out to be orthogonal under the C*-algebra trace inner product $\langle A, B \rangle =tr(AB^{\ast})$. They are orthogonal in the usual sense that $tr(\sigma_i\sigma_j^{\ast})=0$. We get a 90 degree angle through the relation $cos(\theta)=\frac{\langle \sigma_i,\sigma_j \rangle}{||\sigma_i||||\sigma_j||}$.

I assumed this orthogonality would carry over to their corresponding projectors, but when I got around to checking I found they rather have 60 degree angles between them. I was if the 60 degree angles had any physical significance for you.