# New preprint on the weak quantum numbers

I’ve just submitted a paper, Density Matrices and the Weak Quantum Numbers to Foundations of Physics. There are things about the paper that I didn’t include, things that I didn’t think were appropriate to a journal submission and I thought I’d talk about them here, and explain what the paper is talking about to a more general (but still math/physics) audience.

The paper is on the subject of the weak quantum numbers of the left and right handed elementary fermions and anti-fermions. Ignoring color and generation, there are 16 of these quantum objects. I provide a method of defining these quantum numbers by an idempotency equation, that is, by solving an equation of the form $\rho^2 = \rho$. Since pure density matrices satisfy this equation, the calculation is a density matrix calculation based on the permutation group on 3 elements.

The usual method of elementary particles is to assume that a symmetry relates the quantum states. In this calculation, the quantum states themselves are assumed to be composed of group elements of the symmtry. This can be done in density matrix formalism because density matrices can operate on themselves. Also of interest are what happens when different density matrices operate on each other. Particularly when the density matrices are chosen from the basis states of a complete set of mutually unbiased bases. But that’s another paper (mostly written).

Elementary particle physics, and quantum mechanics more generally, has been wedded to symmetry techniques almost since the founding of the subject. The elementary particles are described with symmetries and Einstein’s relativity is also a symmetry (space = time, more or less). In the case of the weak quantum numbers, the appropriate symmetry is U(1) x SU(2). One normally sees this written in reverse order, as SU(2) x U(1), but in the above paper I’ve listed the weak quantum numbers in the order (U(1), SU(2)), so it seems more appropriate to reverse them here.

Brief Description of Weak Isospin
The quantum numbers of a representation or “rep” of SU(2) can be zero, positive or negative, but they are always multiples of 1/2. With the weak quantum numbers, SU(2) provides the symmetry of weak isospin which is designated as $t_3$. (By the way, the “3” refers to the 3rd Pauli spin matrix, $\sigma_3 = \sigma_z$, the one oriented in the z direction. This is because, in the state vector treatment of spin-1/2 spinors, it is customary to use the basis defined by spin up/down. In this basis, the vector (1,0) represents spin up while (0,1) represents spin down. And this choice is from the fact that the 3rd Pauli spin matrix is diagonal.)

Getting back to SU(2) weak isospin, only two representations are used. The first of these two representations is the SU(2) “singlet”. It has only one quantum state with quantum number 0 so for these particles, weak isospin $t_3 = 0$. Of the 16 left or right handed elementary fermions or anti fermions, 8 are weak isospin singlets. And of these 8, four are particles and four are antiparticles. The four particles are evenly split between leptons and quarks and the four antiparticles are also evenly split.

The other rep is the SU(2) “doublet”. This rep has two states with quantum number +1/2 and -1/2. The 8 left or right handed fermions or anti fermions that are SU(2) weak isospin doublets are grouped into four pairs of two each. Two of these pairs are particles and the other two are antiparticles. The two particle pairs include a pair of leptons and a pair of quarks. Similarly for the antiparticles. Still restricting ourselves to the first generation, the SU(2) doublet of leptons consists of a neutrino and an electron, while the SU(2) doublet of quarks has an up quark and a down quark.

The weak force is carried by the W and Z bosons. Of the handed elementary fermions, only the SU(2) doublets feel the weak force. That is, only they can absorb or emit a W or Z. The weak isospin singlets do not interact with the weak force at all. The W is a charged particle with charge +1 or -1. In order for a fermion to emit one of these, the charge on the fermion must change. When this happens, the fermion converts over to the other weak isospin doublet partner. Since the SU(2) partners are pairs of leptons or quarks, this means that a neutrino changes to an electron (or vice versa, or with the antiparticle pair), and similarly a down quark changes to an up quark or vice versa.

Brief Description of Weak Hypercharge
The U(1), or circle symmetry group has representations for all integers N, positive or negative. Unlike SU(2), all U(1) representations are singlets. These representations are homeomorphisms of the circle group: $e^{i\theta} \to e^{Ni\theta}$. The integer N is a “winding number”, it tells you how many times you wind around the circle in the map before you get back to the start. So one would think that the available quantum numbers for weak hypercharge would be all integers. Uh, well, that’s not how it turned out…

The left handed leptons have weak hypercharge quantum numbers of -1, so their transformation law is $\psi_L \to e^{-i\alpha}\psi_L$. Similarly, the right handed electron has quantum number -2 which gives $\psi_R \to e^{-2i\alpha}\psi_R$. (Or is it +2, I forget.) In these transformations, $\alpha$ is the “charge“, which means “the generator of a continuous symmetry.” In this phrase, the word “generator” means that you can make small (continuous) changes in the value while keeping your equations still satisfied.

Electrons and protons have the same (but opposite) electric charge and anything made from them has integer electric charge. Naturally people assumed that this would apply to any elementary particle and this assumption got frozen into the definitions. As a result, when it was found that it took three down quarks to equal the electron’s charge, the value of $\alpha$ for weak hypercharge ended up taking it on the chin. The quarks ended up with fractional weak hypercharge. If we multiplied all weak hypercharges by 3, they would be all integers, leptons and quarks. So if this bothers you, you can think of them this way, or perhaps write a physics paper to correct the injustice in a manner more complicated than to simply multiply all the weak hypercharge quantum numbers by 3. By the way, I should mention that there is already a factor of 2 between weak hypercharge and electric charge. This suggests that the true unit of electric charge is 1/6 the charge of the electron.

Symmetry
The course of elementary particle physics over the last 50 years has been to assume that the fundamental laws of physics are symmetry laws. One then makes the assumption that the symmetry laws are simple. After finding that a symmetry is only approximate, one uses the technique of broken symmetries. Basically, this idea is to suppose that the underlying differential equations are perfectly symmetric, but that their low temperature realization breaks the symmetry. Of course, as in epicycles of early astronomy, you can approximate any old thing by postulating complicated enough symmetry breaking.

The alternative technique I would prefer to use is to suppose that the universe is organized around differential equations and that these differential equations are simple. Symmetry principles are frequently useful in solving differential equations, but in my view, one should being so married to symmetry that one is unable to solve a problem any other way.

The Foundations of Physics
In the case of the weak quantum numbers, the foundational problem is that of explaining why they happen to be what they are. From the symmetry point of view, this is an unanswerable question. One might take the anthropic point of view and try to show that other universes would be incompatible with intelligent life, but this is a rather difficult task.

A few years ago, I independently realized that there was an alternative way of writing special relativity, a way so that it was no longer a principle of symmetry, but instead was a description of a perfectly realizable manifold with normal material properties (except that we cannot fashion things using an extra dimension). This form of special relativity is called Euclidean Relativity. My version amounts to assuming a single hidden, cyclic, very small, dimension. I assume that this is the origin of the U(1) symmetry. With this idea, all classical objects then move at speed c. If they are massive and are stationary, then their velocity in the hidden dimension is c.

The primary attraction of this idea for me was not that it eliminated symmetry, but instead that it would allow a definition of quantum mechanics where quantum waves would not be dispersive. A non dispersive wave travels at the same speed regardless of its frequency. In normal quantum mechanics, particles with different energies travel at different speeds and this requires dispersion. Dispersion is difficult from a mathematical point of view; having a non dispersive quantum wave theory means that calculations are easier. And in fact, the Wick rotation commonly used in quantum mechanics amounts to a transform to Euclidean relativistic coordinates.

It turns out that no one wants to hear about this sort of foundational ideas. It’s clear that one can move back and forth between standard relativity and Euclidean relativity so there is little motivation for physicists to look at ideas that come from Euclidean relativity. And most of the people working on Euclidean relativity ideas are amateurs. It smells of crankdom. However, if the underlying structure of the universe is, in fact, Euclidean instead of Einsteinian, then one is likely to find insights into the structure of the elementary particles by rewriting elementary partices from a Euclidean point of view.

If all you do is switch from Einstein’s relativity to Euclidean relativity you will accomplish nothing. To get anywhere you have to take the principle that you used to prefer Euclidean relativity (that of avoiding symmetry in favor of geometry), and apply it to all of elementary particles. Since elementary particles is built from symmetries, this is a tough roe to hoe. And I’ve got the calluses to prove it.

Density Matrices
The other part of the title of the paper, “density matrices” describes how the calculation is set up. Density matrices are an alternative formulation of quantum mechanics that is not as commonly use as the usual state vector formulation. The advantage of density matrices is that they do not possess the arbitrary U(1) symmetry (which is more or less related to weak hypercharge) of the state vector formulation. Since my objective has been to rewrite elementary particle physics without recourse to using symmetry principles in the foundation, it is natural for me to prefer the density matrix formulation.

So finding the equations that define the weak quantum numbers was not a difficult task. This was done after I had used a very similar method to rewrite Koide’s wave equation as an eigenvalue equation and extended it from the charged leptons to the neutrinos. Knowing that the elementary fermions are composed of three identical preons, I was sure that the appropriate group would be the symmetry group on 3 elements. From the permutation group on 3 elements, it takes only a few minutes to write down the 6 coupled quadratic equations. Actually solving them took me 2 or 3 days of hard work. When you’re as slow as I am at calculations it helps to be able to restrict the problems you work on to the ones that will give useful results.

The calculation as submitted is not at all how I thought of it when I first began working on it. To me, the six elements I, J, K, R, G, and B are analogs of Feynman diagrams. Each represents a quantum amplitude. The requirement of idempotency is equivalent to requiring that the particle be in a stable state. This is all more completely described in the paper I wrote up when I had first made the calculation Density Operators, Spinors and the Particle Generations. Like the just submitted paper, this was intended for Foundations of Physics. But the length basically grew to infinity. The new calculation is improved in that it uses the principle of Hermiticity to separate the extra solutions of the idempotency equation from the 16 that represent the 16 handed elementary fermions or anti fermions.

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### 11 responses to “New preprint on the weak quantum numbers”

1. carlbrannen

It’s been 2 days since I submitted the paper and already the editor has assigned reviewers for it. Current status with of the Editorial Manager Software is “Reviewers Assigned.”

I’m taking this as good news; I suppose the editor wouldn’t waste the time of reviewers if he didn’t think it worthwhile.

On the other hand, maybe it’s the standard way of dealing with papers from cranks. Make the reviewers explain why it’s wrong.

2. Kea

Yes, I suspect it is the standard way. Anyway, good luck with this and other papers. My move to Europe is becoming imminent. Maybe I will visit you in the US next year. Your state has some beautiful mountains, I believe, and your new president is not an idiot.

3. Wayne Jordan

I get in and out of studying fundamental particle physics and I must say that this synopsis seems like a thread between much of what I have read and that is good at helping me pull all I have read together. I will have to sit down and have a go at remapping from Einsteinian to Euclidean, but I am already picturing a hemisphere of points and I am connecting the lines back to an assumed point of reference.

4. Tim Black

We can write the identity I in two ways as a product of two group ele-
ments: I = II, and I = SS. To convert this into a quadratic equation, we
write I = I^2 + S^2 with I and S now thought of as complex numbers.

Similarly, we can write S in two ways as S = IS, and S = SI. This gives the quadratic equation S = 2IS

1. I don’t understand how the the quadratic equations I = I^2 + S^2 and S = 2IS emerge from the fact that you can write I as I = II, I = SS, S = IS = SI. Can you elaborate?

2. How are you adding group elements I^2 + S^2 – is this using the group algebra generated by I and S?

3. Why isn’t I^2 + S^2 just 2I, since I^2 = S^2 = I?
Similarly, why isn’t 2IS just 2S?

4. In the setup you define I and S as elements of a group (so they are not variables at that point). Later you use these symbols to stand for complex variables in a quadratic equation, and solve for numerical values of I and S. It appears to me that you’re trying to show that the properties of the original group lead to those solutions of the quadratic equation, but since I don’t understand how the equations emerge from the group properties, I can’t see how the group and the numerical values for I and S are connected.

5. carlbrannen

Tim, these are excellent questions. Clearly you’ve looked carefully at my paper and I appreciate that.

>>Can you elaborate?

On the one hand, you could just say that what I’ve described is just an arbitrary method of getting coupled quadratic equations from a finite group.

But what’s going on is that the group elements are being treated as if they represent Feynman diagrams. The “I” Feynman diagram leaves the particles unchanged, while the “S” Feynman diagram swaps them.

Now Feynman diagrams come about by looking at perturbation expansions of a Lagrangian / Hamiltonian and there is no such thing here. If we were to write it in perturbation theory, we’d have to make a guess for the Hamiltonian, and then the “I” would be the sum over all possible diagrams that leave the two particles unchanged, while “S” would be the sum over all possible diagrams that swap them.

But I’m not specifying a Lagrangian so there’s nothing to sum. Instead, I’m assuming somebody (or something) else did all this for me and all I’ve got is the result of the sum. You wouldn’t think that this sort of simplifying assumption would make any progress, eh, LOL.

A Feynman diagram eventually reduces to a complex number, and the same applies to a sum of them. So I and S are complex numbers. In other words, I abused the notation when I used “I” to represent both the permutation and the complex number. It gets worse. In the next section, I use “R” to represent (a) the elementy of the group being swapped, (b) the permutation group element that leaves that R unchanged, and (c) the complex number that represents the sum over those sorts of Feynman diagrams.

Sorry for the abuse… I’ve written the same thing up without the abuse and then the notation gets kind of long. It’s a simple idea really, but I’m not sure if it’s better with the abusive notation or the overly involved notation. And a problem for me is that I’ve been working on these things for years so it seems very natural to me.

> 3. Why isn’t I^2 + S^2 just 2I, since I^2 = S^2 = I? Similarly, why isn’t 2IS just 2S?

What’s going on with these Feynman diagrams is that they are representing activity that occurs to a bound state. “Activity” happens over time intervals. Suppose we look at the bound state at three times, t=1, 2, and 3. Then we have three intervals to deal with, (1,2), (2,3), and (1,3).

In general, the value of a Feynman diagram depends on the time interval between the incoming and outgoing states. But in the case of a bound state, the quantum state is stationary and so doesn’t change. In QM, the equivalent statement is that the wave function does not depend on time. So the complex numbers you get from the Feynman diagrams have to not depend on the interval. That simplicity makes them more confusing!

If we happen to know the Feynman diagrams (i.e. complex numbers) for the interval (1,2) and for (2,3), we can use these to get the complex numbers that correspond to (1,3). To do this, we sum over all the things that can happen at time 2.

For the swap group, there are two things that can happen at time 2. The group elements can be unchanged or they can be swapped.

Suppose I want to calculate the complex number for unchanged propagation from 1 to 3. I will call this I(1,3). Of course this is just a complex number, and by assumption, it does not depend on “1” or “3”, but leaving off the interval descriptions makes it confusing. So I’m going to write it like this so that it is more clear what is going on. Then, when we write the amplitude for unchanged propagation over the time interval (1,3) in terms of propagation (unchanged or swapped) over the time intervals (1,2) and (2,3) we have (a formula that works for any propagators, even if they depend on the time intervals):

I(1,3) = I(1,2)I(2,3) + S(1,2)S(2,3).

That is, to get the propagator I(1,3), I sum over all possible things that could happen at time 2. If time 2 is unchanged, the contribution is I(1,2)I(2,3) because I(1,2) is the propagator to get to 2 unchanged, and I(2,3) is the propagator to get from 2 to 3 unchanged. The other term, S(1,2) S(2,3) is the product of the propagators for when the state is swapped at 2, but returns to its unchanged state by time 3.

Getting rid of the dependency on time, you have
I = II + SS,

The next paper will explore these ideas more deeply. The title is “Density Matrices and Color Bound States.”

6. Tim Black

Thanks for that detailed reply Carl. It helps a bit but there’s still a lot I’m not clear on.

In general, the value of a Feynman diagram depends on the time interval between the incoming and outgoing states. But in the case of a bound state, the quantum state is stationary and so doesn’t change.
[…]
Suppose I want to calculate the complex number for unchanged propagation from 1 to 3. I will call this I(1,3). Of course this is just a complex number, and by assumption, it does not depend on “1″ or “3″

– In the paper you refer to I (or I(1,3)) as a transition amplitude right? I thought (at least in QM) that a bound state doesn’t change with time apart from an overall phase factor eg. if |P,t_0> is an eigenstate of the Hamiltonian with eigenvalue H then |P,t> = exp(-iH(t-t_0)/hbar)|P,t>. So my understanding was that for a bound state the transition amplitude does still have time-dependence even though the state doesn’t change with time.

– If you’re talking about bound states so the state at time 3 is required to be the same as at time 1, why is it allowed to (possibly) change into the swapped state at time 2?

7. Carl Brannen

Tim,

“I thought (at least in QM) that a bound state doesn’t change with time apart from an overall phase factor eg.”

You’re assuming the “Schroedinger picture” of QM. Look up “Heisenberg picture”.

“If you’re talking about bound states so the state at time 3 is required to be the same as at time 1, why is it allowed to (possibly) change into the swapped state at time 2?”

A stationary state is one that does not depend on time. It does not imply that all transition amplitudes are zero.

If that bothers you philosophically, another way of looking at this is to think of the matrices as “scattering matrices” instead of “density matrices”. My feeling is that what matters is the equations not necessarily the words that describe them. So I can be confusing.

8. Tim Black

Hi Carl, I hope you won’t mind indulging me with one more question.

If I understand what you’ve written above correctly, you are considering a system that can only be in one of two states: swapped, or not swapped. So say the system starts in the unswapped state at time 1, at time 2 it can either stay in the unswapped state with transition amplitude I(1,2), or go to the swapped state with transition amplitude S(1,2).

How come the corresponding probabilities don’t sum to unity? ie. why isn’t |I|^2 + |S|^2 = 1? In your table this is only true for the I = 1, S = 0 case.

9. carlbrannen

Tim, good question. To explain I need to back up to density matrix theory.

Let $\psi(x)$ be a wave function. It’s probability normalization is: $\int \psi^*(x)\psi(x)\;d^3x = 1$

The density matrix form is $\rho(x,x') = \psi^*(x)\psi(x')$. This guy’s normalization is more complicated, it’s given by: $\int \int \rho(x,x')\rho(x',x)\;d^3x\;d^3x' = 1$.

Writing this in matrix form, the probability normalization for a (pure of course) density matrix is: $\sum\sum a_{jk}a_{kj} = 1$ or $\sum \sum |a_{jk}|^2 = 1$
for an Hermitian density matrix.

In other words, to verify probabilities you have to sum over all the matrix elements squared magnitudes.

The I and S complex numbers appear in the density matrix in more than one place. For the problem you’re looking at, they appear twice each. The I is on the diagonal twice, while the S is in the two off diagonal positions. So when you check the total probability you expect: $2|I|^2 + 2|S|^2 = 1$.

So the (I,S) = (1/2,1/2) and (1/2,-1/2) are valid density matrices while the (1,0) and (0,0) states are not.

What the (1,0) gives when you compute the above sum is 2 instead of 1, and the (0,0) gives 0. These are also the values of the traces of the density matrices. You can think of these values, 2 and 0, as indicating that the actual state has a total of 2 and 0 particles. The concept of “0 particles” is that the number of particles and anti particles match and so the particle count is zero.

This method of treating the pure density matrices with trace non zero (i.e. idempotent Hermitian matrices) as representing states with differing number of particles is a slight modification of what Julian Schwinger did in his papers on the Measurement Algebra.

The modification is that Schwinger treated the states as representing beams of particle. For the Pauli spin matrix example, the state with trace 2 (which is just the unit matrix) would have two sorts of particles in it, spin up and spin down.

My modification is to treat the mathematical objects that Schwinger thinks of as representing a beam with two kinds of particles as instead representing a beam of bound states, each bound state having two particles in it, a spin up and a spin down.

10. Dave

“This suggests that the true unit of electric charge is 1/6 the charge of the electron.”

Indeed, if as I mentioned elsewhere, charge is a form of momentum (as it is in the current-force analogy between electrical and mechanical system dynamics modeling), then we have the following coincidence…

2pi*Yw/c = 5.596523E-028 kg

(Yw=e/6=weak hypercharge quantum)

Lepton “unity” mass constant = 5.594998E-028 kg