The Fermion Cube

CarlB · Feb 23, 2007

An elegant way of writing the standard model Lagrangian. The paper is titled "Standard Model Lagrangian" and is on this site:
http://federation.g3z.com/Physics/ [Broken]

This appears to fit well with my preon model of the fermions; and helps fill in how one connects up the gauge bosons in that sort of scheme. I don't know the name of the anonymous author.

john baez · Feb 23, 2007

CarlB said: ↑

I don't know the name of the anonymous author.

It's Mark Hopkins. Try his http://groups.google.com/group/sci.physics.research/msg/55f38fc6197d8bb7?dmode=source".

CarlB · Feb 24, 2007

john baez said: ↑

It's Mark Hopkins. Try his http://groups.google.com/group/sci.physics.research/msg/55f38fc6197d8bb7?dmode=source".

Thanks.

Now that I look at this more carefully, our use of cubes is quite different. I have every other row made of antiparticles while he uses particles throughout. And I am guessing the 1 3 3 1 pattern comes from the quarks and leptons being composite with three "snuarks" each:

(1) +++
(3) ++-, +-+, -++
(3) +--,-+-, --+
(1) ---

where position matters, that is, the three positions are RGB. And the 8 states described have to be used four times per generation. (2x from R/L, and 2x from particle / anti particle.)

So for me the cube is made by the three choices particle/antiparticle, left/right, and electron/neutrino, but these apply to the snuarks, not the particles.

And to get the right quantum numbers there has to be one more layer of condensation below the snuarks in that each of the +s and -s has to be made out of two "binons". Otherwise the quantum numbers can't come from the primitive idempotents of a Clifford algebra, which always take quantum numbers that come in +/- 1.

In the fundamental cube, the one defined by the Clifford algebra primitive idempotents, the binon quantum numbers are +++, +-+, ++-, and +-- on the top, and (-..)s on the bottom. In other words, the quantum numbers are all possible sets of three +-1s.

One combines two adjacent binons on either the top or bottom sides of the cube, to make a snuark that has quantum numbers given by the sum of the quantum numbers of those two corners. Writing +2/-2 as +/-, the four snuark quantum numbers on the top of the cube become +0+, +0-, ++0, and +-0. The bottom snuark quantum numbers are the same but begin with -. The effect is to make a new snuark cube that is rotated from the fundamental binon cube by 45 degrees.

This was the only way I could figure to put all the elementary fermions into binary form.

CarlB · Feb 24, 2007

john baez said: ↑

I'm fond of how this cube is http://groups.google.com/group/sci.physics.research/msg/4ae9e30b59031a43?dmode=source".

Half my problem with octonions is that I cannot come up with a physical justification for non associativity. In Schwinger's measurement algebra, multiplication means that two things follow one another in a particular order. Non commutativity means that order matters, and associativity is natural. (Addition means alternative ways of ending up with the same result, and so is naturally commutative.)

On the other hand, the octonions are not all that non associative in that a product of those seven roots of -1 will be either + or - depending on how one chooses the association, and that gets cleaned up when one computes probabilities. But I don't work with spinors, I work with density operators, and they don't have arbitrary phases to mess around with.

Everyone knows that the two slit experiment means that one cannot give a physical meaning to the intermediate states in a quantum experiment, but can only pay attention to the initial and final states. I take this one stage further and say that the initial and final states are not physical except when considered as a pair, hence the density matrix / operator formalism. If I multiply a bra by a phase, I also have to multiply an associated ket. Another way of saying this is that the phase is inherent in what happens to the particle (its path) rather than the particle itself.

If it turns out that octonions are useful in representing the fermions, this will translate into my version of quantum mechanics as an inadequacy of the spinor representations that will disappear when the problem is posed in density operator form.

The other half of my problem with octonions is that I don't know how to calculate with them efficiently. My computer does Clifford algebra calculations by using a table of signs to do the anticommutations of the basis vectors. To make it do octonionic multiplication is not difficult, I just change the number of degrees of freedom and choose the right signs. It's not like I'm representing the operator algebra as a matrix algebra which is inherently associative. Hmmmm....

kneemo · Feb 24, 2007

CarlB said: ↑

Half my problem with octonions is that I cannot come up with a physical justification for non associativity. In Schwinger's measurement algebra, multiplication means that two things follow one another in a particular order. Non commutativity means that order matters, and associativity is natural. (Addition means alternative ways of ending up with the same result, and so is naturally commutative.)

A good way to see nonassociativity in action is in trying to calculate the expectation value of a 3x3 Hermitian matrix A over the octonions [itex]\langle\alpha|A|\alpha\rangle[/itex], for a state [itex]|\alpha\rangle\in\mathbb{O}^3[/itex]. You'll see an ambiguity arise in this attempt, giving rise to two inequivalent cases: [itex](\langle\alpha|A)|\alpha\rangle[/itex] and [itex]\langle\alpha|(A|\alpha\rangle)[/itex]. In the Jordan-algebraic density matrix approach, there is no such ambiguity. So a quantum mechanics involving octonions must necessarily be a density matrix formulated QM.

The other half of my problem with octonions is that I don't know how to calculate with them efficiently. My computer does Clifford algebra calculations by using a table of signs to do the anticommutations of the basis vectors. To make it do octonionic multiplication is not difficult, I just change the number of degrees of freedom and choose the right signs. It's not like I'm representing the operator algebra as a matrix algebra which is inherently associative. Hmmmm....

If you enjoy a little C++ try using the octonion class http://www.boost.org/libs/math/octonion/octonion.html" [Broken].

Dcase · Feb 24, 2007

Curved cubes and possible relation to gravity waves

The most interesting cube may be the super-ellipsoid (<e,n>). As n -> oo, the structure becomes more of a cube.
http://mathworld.wolfram.com/Superellipsoid.html

As the catenoid becomes more important in curved space-time, the helicoid transformation becomes more important.
http://mathworld.wolfram.com/Catenoid.html

and Derek K Wise 'MacDowell-Mansouri gravity and Cartan geometry'
http://arxiv.org/PS_cache/gr-qc/pdf/0611/0611154.pdf [Broken]

The possible catenoid [and therefore helicoid] relationship to gravity waves might be apparent when the catenary is considered:
“The curve a hanging flexible wire or chain assumes when supported at its ends and acted upon by a uniform gravitational force.”
http://mathworld.wolfram.com/Catenary.html

CarlB · Feb 24, 2007

Dcase said: ↑

The most interesting cube may be the super-ellipsoid (<e,n>). As n -> oo, the structure becomes more of a cube.
http://mathworld.wolfram.com/Superellipsoid.html

Interesting. The superellipsoid formulas with e=n = 2N can be rewritten in a scale free (i.e. pure number) form as:
[tex]f_N(x,y,z) = (|x|^{1/N} + |y|^{1/N} + |z|^{1/N} )^N = 1[/tex]

This reminds me of the Koide relation for the masses, a, b, c, of the charged leptons. The experimental data show that if

[tex]f_2(a,b,c) = (a^{1/2} + b^{1/2} + c^{1/2})^2[/tex]

and [tex]f_1(a,b,c) = a+b+c[/tex]

then [tex]f_1/f_2 = 2/3[/tex].

The Koide relation puts a single restriction on the masses of the charged leptons. Ignoring the overall scale, three masses have two degrees of freedom. So if one could write another relation, for example comparing f_3(a,b,c) with f_4(a,b,c), then one could completely determine the ratios of the charged lepton masses.

It might be worthwhile to compute [tex]f_N(a,b,c)[/tex] for various values of N to see if there are any simple rational ratios like the one observed by Koide.

While I can see justification for the Koide relation from the principles of density operators, I can't see justification for other relations. That doesn't mean I'm not going to code up the Java to look for such relations, just that if I find one, I'm going to call it "numerology".

[edit] Well I didn't find one. A table of Koide type ratios up to powers of 15:
http://brannenworks.com/kfs.txt
In the above, the Koide formula is in the top left corner. N runs from 2 to 15 on the left side of each triangle. The columns run from M=1 to 15, and the numbers are [tex]f_M/f_N[/tex] for M<N. The first triangle gives best mass result. The second triangle gives the smallest the ratio can be if the three masses are within the PDG errors. The third column gives the maximum the ratios can be. As you can see, the Koide number is 2/3 and is fairly close to the center of minimum and maximum over the range of measured masses. The java source code follows.[/edit]

arivero · Feb 26, 2007

CarlB said: ↑

(1) +++
(3) ++-, +-+, -++
(3) +--,-+-, --+
(1) ---

This is also related to the use of spinors to build representations of SO(2n).

I thought a little about it, the passing from a - to a + being equivalent to the application of a fermionic creation operator. If this operator has some charge, then we see how the charge progresses. For instance assume we use two fermionic operators of electric charge +1 and we start from a particle of charge q. We have

-- : one particle of charge q
+-, -+: two particles of charge q+1
++: one particle of charge q+2

and if we ask the final particle to be opposite of the starting one, then -q=q+2, and q=-1. Obviusly, with three operators, -q=q+3, and q=-3/2 and so on.

If we ask the final particle to have the same charge than the starting one, then there is no way to solve q=q+2; we need more quanta of charge in the SU(n) way: colours. So for 2 creation operators

--: neutral
-+: North, +-: South
++: NorthSouth, ie neutral again

For 3:

--- neutral
+-- Red -+- Blue --+ Green
++- RB, +-+ RG, -++BG
+++ RGB

and so on. But then what is funny is to consider both

----- neutral
+--- R, -+--- G, --+-- B, ---+- N, ----+ S
.
.
.
+++++ RGBNS

For the last one to be neutral, you may to consider that NS cancel as well as RGB. Got 32 particles

12 coloured: (RGB)0, (RGB)N, (RGB)S, (RGB)NS
12 anticoloured: [RGB]0, [RGB]N, [RGB]S, [RGB]NS
8 uncoloured: 0, N, S, NS, RGB, RGBN, RGBS, RGBNS

No surprise as we are building a kind of SU(5) or SO(10), the GUT groups.

Looking to the uncoloured sector, aiming to get leptons, we can first try to set the electric charge of N to be +1, the one of S to be -1, and then... the quarks do not fit.

We need to relax the N,S sector. For instance, take N to be of electric charge +1, S to be 0, and RGB of charge -1/3. Then RGBNS is still neutral (but neither RGB or NS are). You can check that with this assignation we get the right electric charge:

(RGB)0 = -1/3
(RGB)N = +2/3
(RGB)S= -1/3
(RGB)NS= +2/3
etc.

I think this is one of the traditional explanations of how to fit the standard model into SO(10).

A worry happening here is that if we consider that we are jumping from ----- to +---- and then to +--+- and so on by using fermionic creation operators, then half of the fermions have a sort of wrong grading. The answer could be to use a sixth generator, neutral, only to duplicate the number of elements. Then for each combination of charges we have one element with even grade and another one with odd grade.

Superpartners, if you wish. But with a minor objection: if every application of the fermionic creation is expected to increase the spin by +1/2, we need to start with a neutral particle of spin -3/2 to get 15 particles of spin -1/2, another 15 of +1/2 and another one of +3/2. Note here "a particle" means "a degree of freedom".

kneemo · Feb 27, 2007

arivero said: ↑

But then what is funny is to consider both

----- neutral
+--- R, -+--- G, --+-- B, ---+- N, ----+ S
.
.
.
+++++ RGBNS

For the last one to be neutral, you may to consider that NS cancel as well as RGB. Got 32 particles

12 coloured: (RGB)0, (RGB)N, (RGB)S, (RGB)NS
12 anticoloured: [RGB]0, [RGB]N, [RGB]S, [RGB]NS
8 uncoloured: 0, N, S, NS, RGB, RGBN, RGBS, RGBNS

No surprise as we are building a kind of SU(5) or SO(10), the GUT groups.

Very nice post Alejandro. When counting the RGBNS possibilities I recovered:
[itex]\sum_{p=0}^5 \left(\begin{array}{c} 5 \\ p \end{array}\right) = \textbf{1}\oplus\textbf{5}\oplus\textbf{10}\oplus\overline{\textbf{10}}\oplus\overline{\textbf{5}}\oplus\textbf{1} = \textbf{32}[/itex].

Indeed, in the SU(5) theory the [itex]\overline{\textbf{5}}[/itex] 5D spinor breaks down into reps of SU(3)xSU(2) as:
[itex]\overline{\textbf{5}}=(\overline{\textbf{3}},\textbf{1})\oplus(\textbf{1},\overline{\textbf{2}})[/itex]

The [itex]\textbf{10}[/itex] breaks up as:
[itex]\textbf{10}=(\textbf{3},\textbf{2})\oplus(\overline{\textbf{3}},\textbf{1})\oplus(\textbf{1},\textbf{1})[/itex].

The gauge mesons transform in the adjoint rep of SU(5), giving the [itex]\textbf{24}[/itex], that we later find in the SO(10) adjoint rep as:
[itex]\textbf{45}=\textbf{24}\oplus\textbf{1}\oplus\textbf{10}\oplus\overline{\textbf{10}}[/itex].

Of course, the SO(10) nicely fits inside E6, which gets us back to 3x3 Hermitian matrices once again. The E6 is the automorphism group of a structure called the exceptional Jordan C*-algebra [itex]J(3,\mathbb{O}_{\mathbb{C}})[/itex]. Carl's lepton model is likely related to this algebra.

arivero · Feb 27, 2007

Thanks very much kneemo!

(Er, got a typo there: 1+5+10+10+5+1 -> 32.)

I should stress that the idea of building SO(2n) as powers of spinorial representations comes from Zee and Wilczek. In his last papers, and in the book, Zee is still fond of using the +++--- notation CarlB has used above.

And yet we are back to octonions and all that if we wish. But I find another two approaches are interesting. On one hand, to take seriously the fermionic generators and tell that we are in N=6 supersymmetry (IE: that the usual, non susy, standard model is a N=6 SUSY theory !?!?). On other hand, that the discrete groups that CarlB, Ma and Zee himself are using to account for the generations are the breaking of SU(2) or O(3) or something peculiar enough to justify another two fermionic generators in the game... and then N=8.

Dcase · Feb 27, 2007

Koide relationship - also in game theory?

I was unaware of the Koide relationship.
I reviewed FEB and APR 2006 PF posts and other sources on this subject.

From CarlB edit, 02-24-2007 06:08 PM:

“... the Koide number is 2/3 and is fairly close to the center of minimum and maximum over the range of measured masses ...”

I begin to suspect some type of relation to the mini-max theorem of von Neumann in his work with game theory. Since von Neumann also contributed algebraic work to physics, I strongly suspect a relationship, but lack the skill to be rigorous.

1 - However, in Basar [engineer] and Olsder [mathematician], ‘Dynamic Noncooperative Game Theory’ are equations surprisingly similar to the 1/3, 2/3 but with 7/9 rather than 2/9.

a - [p 293] within 6.3.1 A three-person dynamic game illustrating informational nonuniqueness; “... side conditions on the equilibrium strategies ... which depend on the equilibrium strategy [small case... gamma-sup3] of P3 {player 3]. Besides these side conditions, the Nash equilibrium strategies of P1 and P2 have no other natural constraints imposed on tem ...”:
(6.33a)....small case... gamma-sup1 * (x-bar-sub2,x-sub-1 = -(2/3)*(x-bar-sub2)
(6.33b)....small case... gamma-sup2 * (x-bar-sub2,x-sub-1 = +(1/3)*(x-bar-sub2)

b - [p 294] Proposition 6.4 proof for a Nash equilibrium for the nonzero-sum dynamic game [7/9 v 2/9]:
(6.35) ... ((2/3)+p)^2 + 2pq - q^2 + (7/9) > 0

c - [p 330] Example 6.3 of scalar linear-quadratic two-person zero-sum differential game “... the differential game under consideration admits the unique feedback saddle-point solution ...”, t in [0,2]:
small case... gamma-sup(1*) (t,x) = -[2/(3-t)] * x * t
small case... gamma-sup(2*) (t,x) = -(1/(3-t)] * x * t

2 - An Ellipsoid algorithm with figure 2 are discussed within lecture 7 [Avner Magen?, U-Toronto], notes by Xuming He. “CSC2411 - Linear Programming and Combinatorial Optimization - von Neumann minimax theorem”.
http://www.cs.toronto.edu/~avner/teaching/S5-2411/ln/lecture7.pdf

3 - Terry Gannon in ‘Monster Beyond the Moonshine’ comments on Braid and Ribbon Categories, especially braid group B3; in particular a possible relation of the trefoil to the Dedekind eta function. [p 164-167 in 2.4.3 Braided #2: from the trefoil to the Dedekind]

Dcase · Feb 27, 2007

Koide relation as a 45 degree structure?

The 45 degree angle associated with the Koide relation suggests the possibility of conic, hourglass or catenoid structure.

Look at the symmetry of the superellipse which should be present in the superellipsoid.
http://mathworld.wolfram.com/Superellipse.html

W Tucker, H Tananbaum, A Fabian, 'Black Hole Blowback', SCIAM, March 2007, p 42-49 in The Anatomy of a Cluster diagram illustrates an hourglass figure [p43].

The Eight Surface tends to mimic the houglass, conic surface or even catenoid [referenced my post 02-24-2007 05:36 AM]. The 2D catenary curve is a gravity curve between two fixed points and related to a rolling parabola.
http://mathworld.wolfram.com/EightSurface.html
http://mathworld.wolfram.com/Cone.html

See also catenary involute and evolute.
http://mathworld.wolfram.com/CatenaryInvolute.html
http://mathworld.wolfram.com/CatenaryEvolute.html

Dcase · Mar 2, 2007

Cantor Time Fractions 1/3 2/3 2/9 or 7/9 in Koide Relation

Someone referred me to this paper:
M Bohner, A Peterson [U-MO-Rolla]
'Dynamic Equations on Time Scales'

Chapter 1: The Time Scales Calculus
Figure 1.5, page 20 deals with a Cantor Time Set
Only 70 of 353 pages with 253 references
http://www.math.unl.edu/~apeterson1/sample_book.pdf

The fractions 1/3 and 2/3 perhaps even 2/9 and 7/9 may in some manner deal with the oscillation of fermion phase transitions?

arivero · Jul 11, 2007

The fermion cube, also in the context of composites, is drawn by Adler in the second page of Phys Rev D vol 21, p2903 (1980). He attributes it to S. L. Glashow in Proceedings of the 1979 Cargese Summer Institute (unpublished at that time).
The cube is drawn, from top to bottom of the figure, in the order: e+, u, \bar d, \nu_e

Glashow preprint, under the title "http://www.slac.stanford.edu/spires/find/hep/www?j=NASBD,59,687 [Broken]", is scanned in KEK. The cube is drawn in page 29

CarlB · Jul 11, 2007

Interesting. Another reference I will have to add.

I split the fermions into left and right, and into particle and antiparticle, so I've got 4x as many states in my cube. With my cube, the cube is oriented vertically along the electron / positron axis. Th electron and positron correspond to a single point in their cube.

To put their cube into mine, you have to split it into four copies, and then ignore the color dimension in each copy, and then the four copies make up four parallel edges of my cube.

Another way of describing this is that their cube uses charge as the vertical axis, and the two dimensions of color as the perpendicular axes. This is a true 3-dimensional cube. My version of the cube is only 2-dimensional; the axes are weak hypercharge and weak isospin. In expanding this to a 3-dimensional cube, one has to add another quantum number that is not observed.

By the way, that paper I wrote on the neutrino masses now has four journal citations (but still isn't even published on arXiv). The most recent papers to get published are Koide's replies using S_3 symmetry. I wrote them up here:
https://www.physicsforums.com/showpost.php?p=1376238&postcount=40

arivero · Jul 12, 2007

CarlB said: ↑

Interesting. Another reference I will have to add.

I split the fermions into left and right, and into particle and antiparticle,

Adler was writing in the age of left neutrinos, so later he was forced to drawn a double cube sharing the neutrino corner

About left and right, I am bit of worried, because I am not sure about how to allocate L and R in each row.

jal · Jul 12, 2007

There might be some information in this paper that might help with what you are doing.
http://www.slac.stanford.edu/pubs/slacpubs/12000/slac-pub-12632.html [Broken]
SLAC-PUB-12632
Novel QCD Phenomena
Stanley J. Brodsky∗†
Stanford Linear Accelerator Center, Stanford University, Stanford, CA, 94309
june 2007
-------------
jal

CarlB · Jul 22, 2007

jal, I didn't see anything useful to me there.

The most useful paper I've seen recently is one that does QFT with Feynman diagrams the same way that I do, on the qubit space:
http://www.arxiv.org/abs/0705.2121

Since the fermions are point particles, there's no reason I can see, in searching for preons that describe them, to do anything more spatially complicated than qubits. When you restrict to qubits, QFT becomes a lot easier to compute.

arivero · Jul 29, 2008

It seems that Glashow's cube reappears in an article of the Scientific American by H. Georgi in 1981. Full splash in figure 1, page 2 of A Unified Theory of Elementary Particles and Forces (by. Howard Georgi, Scientific American, April 1981)

arivero said: ↑

The fermion cube, also in the context of composites, is drawn by Adler in the second page of Phys Rev D vol 21, p2903 (1980). He attributes it to S. L. Glashow in Proceedings of the 1979 Cargese Summer Institute (unpublished at that time).
The cube is drawn, from top to bottom of the figure, in the order: e+, u, \bar d, \nu_e

Glashow preprint, under the title "http://www.slac.stanford.edu/spires/find/hep/www?j=NASBD,59,687 [Broken]", is scanned in KEK. The cube is drawn in page 29

Federation 2005 · Oct 13, 2010

The original article that Baez quoted was from 1999 and did not have both the particles and anti-particles together. But the pictures you were referring to from the (defunct) website had both particles and anti-particles together and matches your preon model. There were numerical representations for the 3 bits in the fermion cube. But note: there was also the flavor square, which gave 2 more bits for a total of 5.

What underlies the regularities cited are the idea that SU(3)_color x U(1)_{Baryon - Lepton} form U(3) -- hence the cube. The cubic spectrum is characteristic of SO(6) = SU(4).

The flavor square is related to the idea that SU(2)_isospin x U(1)_{Hypercharge - (Baryon - Leoton)/2} forms U(2) -- hence the square. This seems to fit naturally into a SU(2)_L x SU(2)_R, the latter incorporating the U(1) part with the right-handed analogue of isospin being Hypercharge - (Baryon - Lepton)/2.

Together, this seems to lead to U(3) x U(2) which fits nicely into SO(6) x SO(4). In turn, this leads naturally to an underlying SO(10) model. The 5 bits are then naturally associated with the 5 pairs of dimensions involved in SO(10).

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The Fermion Cube

CarlB

john baez

CarlB

CarlB

kneemo

Dcase

CarlB

arivero

kneemo

arivero

Dcase

Dcase

Dcase

arivero

CarlB

arivero

jal

CarlB

arivero

Federation 2005

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