ngb and their parameters
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NGB and their parameters. Gradient expansion: parameters of the NGB’s Masses of the NGB’s The role of the chemical potential for scalar fields: BE condensation Dispersion relations for the gluons. Hierarchies of effective lagrangians. - PowerPoint PPT PresentationTRANSCRIPT
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NGB and their parametersNGB and their parameters
Gradient expansion: parameters of the NGB’s
Masses of the NGB’s
The role of the chemical potential for scalar fields: BE condensation
Dispersion relations for the gluons
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Hierarchies of effective Hierarchies of effective lagrangianslagrangians
Integrating out Integrating out heavy degrees of heavy degrees of freedom we have freedom we have
two scales. The gap two scales. The gap and a cutoff, and a cutoff, above which we above which we integrate out. integrate out.
Therefore: Therefore: two two different effective different effective theories, theories, LLHDETHDET and and
LLGoldsGolds
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Gradient Gradient expansion: NGB’s expansion: NGB’s
parametersparameters
AB ABA† BD *
AB AB
iV DdvL (L R)
iV D4
Recall from HDET that in the CFL Recall from HDET that in the CFL phasephase
9A
i A iA 1
1( )
2
and in the basis
AB A AB A
A 1, ,8A
A 9 29
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AA† AD
A
iVdvL (L R)
iV4
AABAB 2
A A
VS
V V V
Propagator
Coupling to the Coupling to the U(1)U(1) NGB: NGB:i / f i / fU e , V e
f , f
i( ) i( )L L R R
i i
e , e
U e U, V e V
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†2A† AA
D 2A
iV UdvL
4 U iV
Consider now the case of the U(1)B NGB. The invariant Lagrangian is:
At the lowest order in 2
2A† A
A 2
2
2i 20
f fdvL
4 2i 20
f f
generates 3-linear and 4-linear couplings
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Generating functional:†i A( )†Z[ ] D D e
21 A A
0 0 12
0 1
2i 2A( ) S
f f
0 1 0 1,
1 0 1 0
A10
A
VS
V
1Tr[log A( )]1/ 2 2Z[ ] (det[A( )]) e
eff
iS [ ] Tr[log A( )]
2
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21
0 12
n2n 1 21
0 12n 1
2i 2iTr[log A( )] iTr log S 1 S S
f f
( 1) 2i 2iTr logS i iS i iS i
n f f
At the lowest order:
eff 0 0
2
12
i iS(y,x)2i (x) iS(x, y)2i (y)S dxdyTr i i
4 f f
i iS(x,x)2 (x)dxTr i
2 f
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Feynman rules
For each fermionic internal line
AABAB AB 2
A A
ViiS i S(p)
V V V
For each vertex a term iLiLintint
For each internal momentum not For each internal momentum not constrained by momentum conservation:constrained by momentum conservation:
2 22
04 3
4d d d
(2 ) 4
Factor 2x(-1) from Fermi statistics and spin. A factor 1/2 from replica trick.
A statistical factor when needed.
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+2 2
Aeff I II 3 2
A
2 2 22 A
A A A
1 dviL iL (p) iL (p)
2 4 f
V ( p) V V ( p) V 2 2d
D ( p)D ( ) D ( )
2A AD ( ) V V i
I IIL (0) L (0) 0 Goldstone theorem:
Expanding in p/
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2
eff 2 2
9 1 dvL (x) (V ) (x)(V ) (x)
f 2 4
1 0 0 0
10 0 0
3dvV V 1
4 0 0 03
10 0 0
3
2 22 2
eff 02 2
22 2
2
1 9L (x) v
2 f
1 9v , f
3
CFL
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For the V NGB same result in CFL, whereas in 2SC
22 2
2
1 4v , f
3
2SC
With an analogous calculation:
2 8a 2 2 a 2
eff 02 2a 1
(21 8log 2) 1L ( ) v | |
36 F 2
22 2 2
T 2
1 (21 8log 2)v , F F
3 36
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Dispersion relation for the NGB’s
1E | p |
3
Different way of computing:
a b ab 0 10 | J | iF p , p p , p
3
Current conservation:
2 21p p E | p | 0
3
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Masses of the Masses of the NGB’sNGB’s
QCD mass term: L RM h.c.
2 LZM M
† T 4i T(Y X) e
1 † 2masses
† †
L c det[M]Tr[M h.c. c ' det( )Tr[(M ) ]
c" Tr[M ]Tr[M ]
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Calculation of the coefficients from QCD
Mass insertion in QCD
Effective 4-fermi
Contribution to the vacuum energy
2
2
3c , c ' 0, c" 0
2
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Consider:
a aQCD 0 L R R L
1L (iD ) M M G G
4
Solving for as in HDET,L ,L 0 ,L 0 ,R
1i D M
2
† † 2D ,L ,L ,L ,L
†
†
0 1L iV D ( D )
2
L R, M
g
M
iMM
2
like chemical potential
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Consider fermions at finite density:
0L gi i
as a gauge field A0
Invariant under: i ( t )e , (t)
Define: † †L R
1 1X MM , X M M
2 2
Invariance under:
,L ,L ,R ,R
† †L L 0
† †R R 0
L(t) , R(t) ,
X L(t)X L (t) iL(t) L (t),
X R(t)X R (t) iR(t) R (t)
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The same symmetry should hold at the level of the effective theory for the CFL phase (NGB’s), implying that
T T† †
0 0 0
MM M Mi i
2 2
The generic term in the derivative expansion of the NGB effective lagrangian has the form mn p† 2
2 2 q †r0NGB 2
iMM / 2 ML F
F
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mn p† 22 2 q †r0
NGB 2
iMM / 2 ML F
F
Compare the two contribution to quark masses:
kinetic term4 4
2 22 2 2 2
m 1 mF
F
mass insertion2 2 2
2 22 2 2
m 1 mF
F F F
Same order of magnitude for since
m F
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The role of the chemical The role of the chemical potential for scalar fields: potential for scalar fields:
Bose-Einstein condensationBose-Einstein condensation
A conserved current may be coupled to the a gauge field.
Chemical potential is coupled to a conserved charge.
The chemical potential must enter as the fourth component of a gauge field.
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Complex scalar field:
2† † 2 † †0 0
2† 2 † †0 0
2 † †
L i i m
im
negative mass term
breaks C
Mass spectrum:
2 2 20
2 2 2
p (m ) 2 Qp 0 (Q 1)
(E Q) m | p |
P,Pm m For < mm
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At = m = m, second order phase transition. Formation of a condensate obtained from:
22 2 † †V m
2 2
† mm
2
† 2 2V2 m
Charg
e densit
yGround state = Bose-Einstein condensate
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2 22v m
, v2
i (x ) / v1
(x) v h(x) e2
2 22 0
1 1L h h v h 2 h
2 2
Mass spectrum2 2
2
p 2 v 2i Edet 0
2i E p
At zero momentum
2 2 2 2M M 2 v 4 0
2
2 2 2
M 0
M 6 2m
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At small momentum
2 2
NGB 2 2
2 22 2 2
massive 2 2
mE | p |
3 m
9 mE 6 2m | p |
6 m
2 22NGB 2 2
m
m 33
1v
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Back to CFL. From the structure P,Pm m
0 0
2 2d u
u d s2
2 2s u
u s d2K
2 2s d
d s u2K ,K
m m 2cm (m m )m ,
2 F
m m 2cm (m m )m ,
2 F
m m 2cm (m m )m
2 F
First term from “chemical potential” like kinetic term, the second from mass insertions
†MM
2
2
3c
2
22
2
(21 8log 2)F
36
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For large values of ms:
0 0
u d s2
2 2s s
s d s u2 2K K ,K
2cm (m m )m ,
F
m 2c m 2cm m m , m m m
2 F 2 F
and the masses of K+ and K0 are pushed down. For the critical value
1/ 322 23 3
s u,d u,d2 2crit
s crit
12m m 3.03 m ,
F
m 40 110 MeV
masses vanish
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For larger values of ms these modes become unstable. Signal of condensation. Look for a kaon condensate of the type:
4i 24 4e 1 (cos 1) i sin
(In the CFL vacuum, = 1) and substitute inside the effective lagrangian
22 2s s
2
1 m 2cm mV( ) F sin (1 cos )
2 2 F
negative contribution from the “chemical potential”
positive contribution from mass insertion
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Defining
2
20s seff K 2
m 2cm m, m
2 F
2 2 2 0 2eff K
1V( ) F sin (m ) (1 cos )
2
with solution 20
K 0eff K2
eff
mcos , m
and hypercharge density
40K2
Y eff 4eff eff
mVn F 1
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Mass terms break original SU(3)c+L+R to
SU(2)IxU(1)Y. Kaon condensation breaks
this to U(1)
3 8
1 1Q , [Q, ] 0
2 3
I Y
0 0 0SU(2) U(1)
( , ) (K , K ) (K , K ) ( )
breaking through the doublet as in the SM
Only 2 NGB’s from K0, K+ instead of expected 3 (see Chada & Nielsen 1976)
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Chada and Nielsen theorem: The number of NGB’s depends on their dispersion relationI. If E is linear in k, one NGB for any
broken symmetry
II. If E is quadratic in k, one NGB for any two broken generators
In relativistic case always of type I, in the non-relativistic case both
possibilities arise, for instance in the ferromagnet there is one NGB of type II, whereas for the antiferromagnet there
are two NGB’s of type I
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Dispersion relations for the Dispersion relations for the gluonsgluons
The bare Meissner mass
The heavy field contribution comes from the term
2
† †h h h h
D DDP
2 iV D 2 iV D
1P g V V V V
2
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Notice that the first quantized hamiltonian is:
2
2 20 0
gH p gA eA | p | gA gv A | A | (v A)
2 | p |
Since the zero momentum propagator is the density one gets
3 22 2
f 3 2|p|
d p 1 (p A)g 2 N Tr A
(2 ) 2 | p | | p |
spin
2 2a a 2 a a
f BM2a a
2 22BM f 2
g 1 1N A A m A A ,
6 2 2
gm N
6
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Gluons self-energy
Vertices froma
aigA J
Consider first 2SC for the unbroken gluons:
00 2ab ab
kl kl,self klab ab ab
2 2 2 2 2
2
kl kl kl 20ab ab ab 02 2 2
0
2
2 2
2 2
2 2
k2 2
2a 20 k
b ab
(p) | p | ,
(p) (p) (p)
g p g1 p ,
3 6 3
(p)
g
18
g
18
p pg
18
2BMfrom m
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Bare Meissner mass cancels out the constant contribution from the s.e.
All the components of the vacuum polarization have the same wave function renormalization
a a b a a a a a aa ab i i i i i i
2 2
2 2
1 1 1 kL F F A A E E B B E E
4 2 2 2
gk
18
Dielectric constant = k+1, and magnetic permeability =1 1
vg
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Broken gluons
a (0) - ij(0)
1-3 0 0
4-7 3mg2/2 mg
2/2
8 3mg2 mg
2/3
2 22g 2
gm
3
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But physical masses depend on the wave function renormalization
2 2
2
g
Rest mass defined as the energy at zero momentum:
R
R
m 2 , a 4,5,6,7
gm , a 8
The expansion in p/ cannot be trusted, but numerically
Rm 0.9 , a 4,5,6,7
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In the CFL case one finds:2 2
2 2 2D 2
2 2 22 DM 2
gm (21 8log 2) g F
36
g 11 2 1 mm log 2
36 27 2 3
Recall that from the effective lagrangian we got:
2 2 2 2 2 2 2D T T M S Tm g F , m v g F
from bare Meissner mass
implying and fixing all the parameters.
S T 1
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We find:2 2
DR 1 2 2
1
R
m g 16m , 7 log 2
216 33
m 1.70
Numerically Rm 1.36
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Different quark massesDifferent quark masses
We have seen that for one massless flavors and a massive one (ms), the condensate may be disrupted for 2
sm
2
The radii of the Fermi spheres are:
1 2
22 2 s
F s F
mp m , p
2
As if the two quarks had different chemical potential (ms
2/2)
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Simulate the problem with two massless quarks with different chemical potentials:
u d
u d u d
,
,2 2
Can be described by an interaction hamiltonian
†I 3H
Lot of attention in normal SC.
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HI changes the inverse propagator
10 *
3
3
VS
V
and the gap equation (for spin up and down fermions):
2 2
2 2 2 20
dv dig
4 (2 ) ( )
This has two solutions:
2 20 0 0a) b, ): : 2
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Grand potential:
0
0
22
2 00
0( 0)
H dg| |
g g g
d
2
02
0
d 2 dg
g
Also:
20 0( ) (0)
2
Favored solution
0
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Also: 2 20 0( ) 2
4
First order transition to the normal state at
01
2
For constant Ginzburg-Landau expanding up to