Origin and Status of INSTANTONS

Transcription

1 Utrecht University Origin and Status of INSTANTONS Gerard t Hooft, Spinoza Institute. Erice 2013

2 The pre-qcd age (before 1971) d s u J PC = 0 + K o K + K* o K* + π η π o η π + ρ ω ρ o ϕ ρ + K K o K* J P = 1 K* o

3 Quark composition: the eta problem J P = 0 : π (u u d d ) J P = 1 + : ρ (u u d d ) η 1 6 η' 1 3 (u u + d d 2s s) (u u + d d + s s) ω 1 2 (u u + d d ) φ s s Compare: J /ψ c c ϒ b b

4 Spontaneous breakdown of chiral symmetry L = ψ γ ψ ψ i m ij ψ j The kinetic term has U ( N ) U ( N ) symmetry; the mass terms break that. This symmetry appears to be spontaneously broken towards U ( N ) diagonal. This could explain perfectly why m π 2 << m N 2 But, this would require m u,d 0, and this would keep U ( 2 ) U ( 2 ) unbroken. 4 parameters: 4 th pseudoscalar, η, should also be massless!

5 The eta problem: Explain the eta mass, and the eta mixing. Then came QCD It had even worse problems: Explain quark confinement!

6 Nielsen - Olesen: Magnetic Confinement In case of spontaneous "breakdown" of U(1) I L ( ) 1 A, ϕ = F F D ϕ D ϕ V( ϕ) 4 µν µν µ µ S N

7 Color Magnetic Super Conductivity N S Electric Super Conductor Magnetic Super Conductor + _

8 The Magnetic Monopole A.Polyakov, G. t Hooft (1974)

9 Hedgehogs: In 2 dimensions, we have a vortex (Niesen Olesen) You need a two-component (or complex U(1)) field In 3 dimensions, we have a particle (magnetic monopole) You need a 3-component ( or isospin 1) field In 4 dimensions?? A topological event?? Try a 4-component field, such as a (complex) isospin ½ field!

10 The Instanton Belavin, Polyakov, Schwartz, Tyupkin Group of Gauge Transformations SU (2) But it so happens that then you can discard the isospin ½ field!

11 The gauge transformation that transforms an isospin ½ hedgehog at infinity towards the form φ i F 0 creates a gauge potential field a A a 2 η x µν ν µ g with, by construction, x 2 a vanishing Now demand an extreme of the action F a µν S = 1 4 d 4 x F a µν F a µν without singularity at x = 0, gives the instanton solution:

12 A µ a (x) = 2 g a F µν = 4 g η a µν (x z) ν (x z) 2 + ρ 2 ρ 2 a η µν ((x z) 2 + ρ 2 ) 2 a η µν = ε aµν, µ,ν = 1, 2, 3 δ aν, µ = 4 δ aµ, ν = 4 0, µ = ν = 4 BPST observed: a F µν = F a µν = 1 a 2 ε µναβ F αβ ; 1 4 d 4 a a x F µν F µν = 1 4 d 4 a a x F µν F µν = S = 8π 2 g 2

13 Triangle diagram: µ J Aa µ = g2 a F 16π 2 µν F a µν µ J Aa µ d 4 x = ± 2 Apparently, two units of axial charge are created (or destroyed) by the instanton for every flavor! If we have just one flavor, this is a mass term: L R In case of many flavors: L L L R R R

14 u L κ d R m s π ο ο Inst π ο ο u R d L = + or d L κ u R u L u R s L κ s R d R d L d R u L + ( u d ) π 0 0 = 1 2 (uu + dd )

15 The case with massless fermions LEFT time Fermi level RIGHT Instanton

16 This topological transition is a tunnelling phenomenon, to be computed semiclassically in Euclidean spacetime ( t i x 4 ) ; the gauge fields take large values. One finds: F µν F µν d 4 x = 32π 2 g 2 If the original state and the gauge rotated state have i a relative phase e θ, then this can be represented in the functional integrals by means of the substitution e i Ld4 x i e Ld4 x+i θg 2 F 32π 2 µν F µν d 4 x L L+ θg2 32π 2 F µν F µν, or

17 (In the absence of massless fermions) L( A, ϕ) = 1 F F + 1 θg 2 F 4 µν µν 32π 2 µν F µν = = 1 2 ( E 2 B 2 ) + 1 8π 2 θg 2 E B = = 1 2 ( E + 1 8π 2 θg 2 B) ( π 4 θ 2 g 4 ) B 2 ( E + θ g 2 B 8π 2 ) = ρ ; D = E + θ g 2 B 8π 2 ; H = 1 2 ( E 2 + B 2 ) The electric and magnetic charge quanta are: Q e = g ; Q m = 4π g Conclude: magnetic monopoles B = Q m δ ( x) carry electric charge: θ 2π g

18 Condensation of Electric Charges Q magnetic 1 Dirac Unit θ = 0 O O O O O Q electric ΔQ Δ Q = 2π m e

19 Condensation of Magnetic Charges Q magnetic O O O O O O 1 Dirac Unit Q electric ΔQ Δ Q = 2π m θ = 0 e

20 Confinement when Q magnetic O O O O O O θ 0 O (No massless fermions present) Q electric

21 Oblique Confinement: θ ; π Q magnetic O O O (No massless fermions present) Q electric

22 String tension for oblique confinement ρ θ 0 π 2 π

23 Here, the true story about the instantons just begins! P. van Baal D. Diakonov

24 Put QCD in a box with periodic boundary conditions q q q

25 Twisted boundary conditions: instantons calorons Instanton number now may become fractional

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27 Yet another scalar meson problem: tetraquarks Take a baryon, and replace one quark (in a 3-representation of SU(3)-color) by an antisymmetric pair of antiquarks (also a color 3 state: [3.3] a = 3). We then have a 2quark+2antiquark state. Same tetraquarks can also be seen by replacing the quarks in a meson by antiquark bound states in the color-3 representation. They form a nonet:

28 σ [0] = [ud][ud ] κ = [su][ud ]; [sd][ud ] + f 0 [0] = 1 2 ([su][s u ]+ [sd][s d ]) a 0 = [su][sd ]; 1 2 ([su][s u ] [sd][s d ]); [sd][s u ] These states can be recognised by an inverted mass spectrum: the isotriplet has more strange quarks in it than the doublets. Why do they exist fairly prominently?

29 There s mixing between tetraquarks and diquarks, due to the instanton: Instanton Mixing is largest where the levels are closest together: ss [ud][u d ]

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