Origin and Status o INSTANTONS · This topological transition is a tunnelling phenomenon F µν F...
Transcript of Origin and Status o INSTANTONS · This topological transition is a tunnelling phenomenon F µν F...
Origin and Status of INSTANTONS
Utrecht University
Gerard ’t Hooft, Spinoza Institute.
Erice 2013
The pre-QCD age (before 1971)
π+ π− πo η η’
Ko
K+
Ko
K−
J PC = 0−+
ρ+ ρ− ρo ω ϕ
K*o
K*+
K*o
K*−
J P = 1−−
d u
s
Quark composition: the eta problem
J P = 0− :π 0 ≈ 1
2 (u u − d d )
η ≈ 16 (u u + d d − 2s s )
η ' ≈ 13 (u u + d d + s s )
J P = 1+ :ρ0 ≈ 1
2 (u u − d d )
ω ≈ 12 (u u + d d )
φ ≈ s s
J /ψ ≈ c cϒ ≈ b b
Compare:
Spontaneous breakdown of chiral symmetry
L = −ψ γ ∂ψ − ψ imijψ 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 mπ
2 << mN2
But, this would require mu,d ≈ 0 , and this would keep U ( 2 ) × U ( 2 ) unbroken. 4 parameters: 4th pseudoscalar, η , should also be massless!
The eta problem: Explain the eta mass, and the eta mixing.
Then came QCD
It had even worse problems: Explain quark confinement !
Nielsen - Olesen: Magnetic Confinement
( ) †14, ( )A F F D D Vµν µν µ µϕ ϕ ϕ ϕ= − − −L
In case of spontaneous "breakdown" of (1)U I→
N
S
Color Magnetic Super Conductivity
N S
+ _
Electric Super Conductor
Magnetic Super Conductor
A.Polyakov, G. ’t Hooft (1974)
The Magnetic Monopole
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 !
Hedgehogs:
(2)SU
The Instanton Belavin, Polyakov, Schwartz, Tyupkin
Group of Gauge Transformations
But it so happens that then you can discard the isospin ½ field !
The gauge transformation that transforms an isospin ½ hedgehog at infinity towards the form
creates a gauge potential field
with, by construction, a vanishing
φi →F0
⎛⎝⎜
⎞⎠⎟
F µνa
Aµa → 2
gηµνa xνx2
Now demand an extreme of the action
without singularity at x = 0, gives the instanton solution:
S = − 14 d 4x∫ F µν
a F µνa
Aµa (x) = 2
gηµνa (x − z)ν
(x − z)2 + ρ2
F µνa = −4
gρ2 ηµν
a
(x − z)2 + ρ2( )2ηµνa =
ε aµν , µ,ν = 1,2, 3
−δ aν , µ = 4
δ aµ , ν = 4
0 , µ = ν = 4
⎧
⎨
⎪⎪
⎩
⎪⎪
BPST observed:
F µνa = F µν
a = 12 εµναβ F αβ
a ;14 d 4x∫ F µν
a F µνa = 1
4 d 4x∫ F µνa F µν
a = −S = 8π 2g2
Triangle diagram:
∂µ JµAa = g2
16π 2 F µνa F µν
a
∂µ JµAa d 4∫ x = ± 2
Apparently, two units of axial charge are created (or destroyed) by the instanton If we have just one flavor, this is a mass term:
L R
for every flavor !
In case of many flavors: R
R
R L
L
L
!"" !
""
Inst+= or
#uL
uR
dR
dL
#dL
dR
uR
uL
#
uL
uR
dR
dL
sR
sL
ms
+ ( u $ d )
π 00 = 1
2 (uu + dd )
Instanton
Fermi level
time LEFT
RIGHT
The case with massless fermions
This topological transition is a tunnelling phenomenon,
FµνFµν∫ d4x = 32π 2
g2
ei Ld4x∫ → e
i Ld4x+i θg2
32π 2 Fµν Fµν∫ d4x∫ , or
L→L+ θg2
32π 2 FµνFµν
If the original state and the gauge rotated state have a relative phase , then this can be represented in the functional integrals by means of the substitution
ie θ
to be computed semiclassically in Euclidean spacetime ( ) ; the gauge fields take large values. One finds: t→ i x4
(In the absence of massless fermions)
∂⋅(E + θg2 B
8π 2) = ρ ;
D =E + θg2 B
8π 2; H = 1
2 (E2 +
B2 )
L A,ϕ( ) = − 14 Fµν Fµν +
132π 2 θg2Fµν
Fµν =
= 12 (E 2 −
B 2 )+ 1
8π 2 θg2E ⋅B =
= 12 (E + 1
8π 2 θg2B)2 − 1
2 (1+ 164π 4 θ 2g4 )
B 2
The electric and magnetic charge quanta are:
Qe = g ; Qm = 4π
g
−θ2π
gConclude: magnetic monopoles
carry electric charge: ∂⋅B = Qmδ ( x)
• • • • • •• • • • • •• • • • • •
• • • • • •• • • •
O O O O O
• •• • • • • •
electricQ
1 Dirac Unit
2m eQ Q πΔ ⋅Δ =
magneticQCondensation of Electric Charges
0θ =
magneticQCondensation of Magnetic Charges
• • • • •• • • • •• • • • •• • • • •• • • • •• • •
O
O
O
O
O
• •• • • • •O
1 Dirac Unit
electricQ
2m eQ Q πΔ ⋅Δ =
0θ =
(No massless fermions present) magneticQ
electricQ
Confinement when
• • • • •• • • • •• • • • •• • • • •
• • • • •• • • • •• •
O
O
O
O
O
O
O • • •
0θ ≠
magneticQ Oblique Confinement:
• • • • •• • • • • • •• • • • •• • • • • •
• • • • •• • • • • ••
O
O
O
• • •
• • • • • •
θ π;
electricQ
(No massless fermions present)
String tension for oblique confinement
0 2 π π
ρ ↑
θ→
Here, the true story about the instantons just begins!
P. van Baal D. Diakonov
Put QCD in a box with periodic boundary conditions
+ !
a)
b)q q
c)q q
d)
q q
e)
q q q q
f )
q
Twisted boundary conditions: instantons calorons
Instanton number now may become fractional
Yet another scalar meson problem: tetraquarks
Take a baryon, and replace one quark (in a 3-repres-entation of SU(3)-color) by an antisymmetric pair of antiquarks (also a color 3 state: ). 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:
[3.3]a = 3
σ [0] = [ud][ud ]κ = [su][ud ]; [sd][ud ] +
f0[0] = 1
2 [su][s u ]+ [sd][s d ]( )a0 = [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?
There’s mixing between tetraquarks and diquarks, due to the instanton:
Instan- ton
Mixing is largest where the levels are closest together:
ss ↔ [ud][u d ]