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Koide Formula for Neutrino Masses

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Koide Formula for Neutrino Masses Koide Formula for Neutrino Masses Carl Brannen Joint meeting of Pacific Region Particle Physics Communities 08:25AM, Waianae Room, November 1, 2006 Waikiki, Hawaii, United States www.DensityMatrix.com, Redmond Washington, November 1, 2006 1 Thanks to: Mark Mollo and Liquafaction Corporation Yoshio Koide Alejandro Rivero, Andre Gsponer, R. N. Mohapatra, Alexei Y. Smirnov, Hans Montanus Larry P. Horwitz Marni D. Sheppeard Jose Almeida www.PhysicsForums.com www.DensityMatrix.com, Redmond Washington, November 1, 2006 2 The ºR is not shown for clarity. 1.0 r r 6 "be¹L e¹R t0 " b " b b " u b 0.5 "r ¤R br e¹ bº¹R " R b b ¹ " u r r r b d¤L " ¤L r d¤L b " r u¤L 0.0 b"r º¹L ¹ r u¹¤R r d¤R r d -0.5 r e ¤R r º ¹ r r r bL " L d¤R b r " b u¹¤L " b " -1.0 b"r eR - ºL r ¡0:5 0:0 0:5 t3 www.DensityMatrix.com, Redmond Washington, November 1, 2006 3 Hermitian circulant matrix has ¹, ´, ± real: 0 1 1 ´e+i± ´e¡i± B C B ¡i± +i± C ¡(¹; ´; ±) = ¹ @ ´e 1 ´e A : ´e+i± ´e¡i± 1 The eigenvectors: 0 1 1 B C B +2in¼=3 C jni = @ e A n = 1; 2; 3 e¡2in¼=3 The eigenvalues: ¸n = ¹(1 + 2´ cos(± + 2n¼=3)) www.DensityMatrix.com, Redmond Washington, November 1, 2006 4 ¸1 + ¸2 + ¸3 = 3¹ 2 2 2 2 2 ¸1 + ¸2 + ¸3 = 3¹ (1 + 2´ ) so 2 2 2 2 3 ¸1+¸2+¸3 1 ´ = 2 ¡ 2 (¸1+¸2+¸3) 2 letting p p p ¸1 = me; ¸2 = m¹; ¸3 = m¿ gives 2 ´1 = 0:500003(23); ±1 = 0:2222220(19) www.DensityMatrix.com, Redmond Washington, November 1, 2006 5 Koide’s charged lepton mass formula: p p p 2 ( me+ m¹+ m¿ ) = 3 : me+m¹+m¿ 2 can be put to neutrinos if a square root is negative: mº1 = 0:000388(46) eV; mº2 = 0:00895(17) eV; mº3 = 0:0507(30) eV: p p p 2 (¡ mº1+ mº2+ mº3) = 3 mº1+mº2+mº3 2 Koide explains mass generational hierarchy for leptons. www.DensityMatrix.com, Redmond Washington, November 1, 2006 6 Hierarchy between electron and neutrino much bigger. Let ¹1; ´1, and ±1 apply to the charged leptons, and ¹0, ´0, and ±0 apply to the neutrinos. 2 Since ´1 = 0:5 within experimental error, we suppose that it is exact, and also that ´0 = ´1. Applying the restrictions to the measured neutrino mass differences we find: ¼ ±1 = ±0 ¡ 12 1:008(80); 11 ¹1=¹0 = (2:9999(71)) : www.DensityMatrix.com, Redmond Washington, November 1, 2006 7 If exact, can give precision estimates of neutrino masses: mº1 = 0:000383462480(38) eV; mº2 = 0:00891348724(79) eV; mº3 = 0:0507118044(45) eV; 2 2 ¡5 2 mº2 ¡ mº1 = 7:930321129(141) £ 10 eV ; 2 2 ¡3 2 mº3 ¡ mº2 = 2:49223685(44) £ 10 eV ; www.DensityMatrix.com, Redmond Washington, November 1, 2006 8 Various neutrino mass calculations, in eV: m1 :0000 :0000 :0000 :0004 :0030 :0041 :0049 :0170 m2 :0083 :0520 :0089 :0090 :0088 :0097 :0098 :0170 m3 :0500 :0530 :0508 :0508 :0500 :0510 :0500 :0530 hep-ph/0308097, hep-ph/0503159, hep-ph/0601098, this, hep-ph/0303256, hep-ph/0601104, hep-ph/0403077, hep-ph/0512009 www.DensityMatrix.com, Redmond Washington, November 1, 2006 9 Since ± is arbitrary modulo 2n¼=3, and since we can put ´0 = ¡´1, we can suppose (to first order): electron neutrino p p ´ + 0:5 ¡ 0:5 ± 2=9 2=9 ¡ ¼=4 ¹ 3¡1 3¡12 To see why these are “natural,” some theory is needed. www.DensityMatrix.com, Redmond Washington, November 1, 2006 10 From Feynman’s Nobel speech: “If every individual student follows the same current fashion in expressing and thinking about electrodynamics or field theory, then the variety of hypotheses being generated to understand strong interactions, say, is limited. Perhaps rightly so, for possibly the chance is high that the truth lies in the fashionable direction. But, on the off-chance that it is in another direction - a direction obvious from an unfashionable view of field theory - who will find it? Only someone who has sacrificed himself by teaching himself quantum electrodynamics from a peculiar and unusual point of view; one that he may have to invent for himself.” www.DensityMatrix.com, Redmond Washington, November 1, 2006 11 From the same speech: “I remember that when someone had started to teach me about creation and annihilation operators, that this operator creates an electron, I said, "how do you create an electron? It disagrees with the conservation of charge", and in that way, I blocked my mind from learning a very practical scheme of calculation.” We will use this as the unfashionable idea. www.DensityMatrix.com, Redmond Washington, November 1, 2006 12 Higgs model gives masses from arbitrary vacuum expectation values. String theory also has too many vacua, 10500. But Koide relation shows vevs are not arbitrary. So follow Feynman instinct and get rid of creation and annihilation operators and vacuum. www.DensityMatrix.com, Redmond Washington, November 1, 2006 13 Preons eliminate need for Higgs: “This problem does not exist in preon models for quark and lepton substructure with composite Z0 and W s, which, consequently, also avoid all other theoretical complications and paradoxes with the Higgs mechanism.” “Higgs pain? Take a preon!” hep-ph/9709227 J.-J. Dugne, S. Fredriksson, J. Hansson, E. Predazzi www.DensityMatrix.com, Redmond Washington, November 1, 2006 14 The square root in the mass formula “suggests that the charged lepton mass spectrum is not originated in the Yukawa coupling structure at the tree level, but it is given by a bilinear form” Y. Koide, hep-ph/0506247 Density matrices are bilinear. Spinors and density matrices have complementary linearity. Simple linear problems in one can be nonlinear in the other. www.DensityMatrix.com, Redmond Washington, November 1, 2006 15 Density operators allow unification of mathematics (in a Clifford algebra). Allows unification of physics concepts. Spinor Density Pure State: Vector Operator Mixed State: (can’t) Operator Operator: Operator Operator Potential Potential Operator www.DensityMatrix.com, Redmond Washington, November 1, 2006 16 Getting density from spinor is easy: ½A = jAihAj To get spinor from density, follow J. Schwinger, Quantum Kinematics and Dynamics, choose arbitrary constant 2 “vacuum” pure state (primitive idempotent) (½0) = ½0 and define: jAi = ½A ½0; hAj = ½0 ½A for any pure state ½A. www.DensityMatrix.com, Redmond Washington, November 1, 2006 17 Example. Choose arbitrary vacuum as projection for spin in +z: 0 1 1 0 ½0 = @ A 0 0 y The ket (a; b) becomes the density operator ½ab: 0 1 0 1 “ ” a a¤a b¤a ¤ ¤ ½ab = a b @ A = @ A ; b a¤b b¤b www.DensityMatrix.com, Redmond Washington, November 1, 2006 18 Converting back to spinor form we have 0 1 0 1 ¤ a a 0 ¤ a 0 jabi = ½ab½0 = @ A = a @ A ; a¤b 0 b 0 0 1 0 1 a¤a b¤a a¤ b¤ habj = ½0½ab = @ A = a @ A 0 0 0 0 See Clifford Algebras and Spinors by Pertti Lounesto. The factors of a and a¤ are just a rescaling, we leave them off. www.DensityMatrix.com, Redmond Washington, November 1, 2006 19 An amplitude habjMjcdi, in density formalism becomes: 0 10 10 1 ¤ ¤ a b m11 m12 c 0 ½0½abM½cd½0 = @ A@ A@ A ; 0 0 m m d 0 0 211 22 1 0 = habjMjcdi @ A = habjMjcdi ½0 0 0 In density formalism, amplitudes are operators. Complex numbers are no longer just complex numbers, they are operators, just like the states (a unification). www.DensityMatrix.com, Redmond Washington, November 1, 2006 20 Under rotation by 2¼, spinors negate themselves: jAi ! ¡jAi; hAj ! ¡hAj; however jAihAj ! +jAihAj = ½A; j0ih0j ! +j0ih0j = ½0; jAi ´ ½A½0 ! ½A½0: In density theory, spinors change sign because spinor theory forgets to rotate the vacuum and hAj0i changes sign. Naughty spinor behavior is a normalization issue, not a part of reality. www.DensityMatrix.com, Redmond Washington, November 1, 2006 21 0 1 i 0 σxσyσz = @ A ; 0 i so we can eliminate complex numbers from Pauli algebra by replacing i with σxσyσz. Get rid of the arbitrary choice of representation completely and write everything as sums over scalar (real) multiples of products of σx, σy, and σz. No arbitrariness in representation of operators. Everything written in geometric terms now (i.e. scalar, vector, pseudovector, pseudoscalar). No spinors. Same can be done to Dirac algebra. www.DensityMatrix.com, Redmond Washington, November 1, 2006 22 Find eigenstate of σy with eigenvalue +1: 2 σy (1 + σy) = (σy + σy) = +(1 + σy): Therefore eigenstate must be multiple of (1 + σy). We require ½2 = ½ so compute: 2 2 (1 + σy) = 1 + 2σy + σy = 2(1 + σy) therefore, ½+y = 0:5(1 + σy): www.DensityMatrix.com, Redmond Washington, November 1, 2006 23 Stern-Gerlach beam splitter: H H Stern-Gerlach H H H r H H Ãà ½+y = 0:5(1 + σy) ÃÃà ` σy ``` H `r ½ = 0:5(1 ¡ σ ) H ¡y y H HH 0:5(1 + σy) defines the state of a particle. It is an operator that projects out a portion of the beam. And it somehow describes the Stern-Gerlach apparatus itself (a field configuration). www.DensityMatrix.com, Redmond Washington, November 1, 2006 24 Chiral Dirac beam splitter, splits to eigenstates of helicity h^, and charge c^: H H H ^ H H H r eR = :25(1+h)(1+^c) H H H ^ r eL = :25(1¡h)(1+^c) ^ h c^ r ^ H r e¹L = :25(1+h)(1¡c^) H e¹ = :25(1¡h^)(1¡c^) H R HH Splitting pattern is square.
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