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: . 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 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 directions. Then the projection operators are related by the equation:
where is the oriented area of the spherical triangle defined by the three unit vectors.
Some notes: The total area of the sphere is , 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 . 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 all have the same magnitude, . 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 , 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 so the total phase of this sequence of projection operators is , and the geometric phase is . 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 (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 . 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 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 that we need, two for the +z to +y to +x transitions: , and one for the overall transition which is +z to +x: . 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
, 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 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 . 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 , 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 . This gives a geometric phase factor of i, which matches the matrix calculation.
Let’s try another: . 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 .
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.