Unparticles and Superconductivity

Unparticles and Superconductivity

Kiaran dave Charlie Kane Brandon LangleyBrandon Langley

Thanks to: NSF, EFRC (DOE)

J. A. HutasoitFriday, March 14, 14

Properties all particles share?

Friday, March 14, 14


fixed mass!charge



conserved(gauge

invariance)

fixed mass!charge



conserved(gauge

invariance)not conserved

fixed mass!charge


mass sets a scale


particles

E = "penergy


particles

E = "penergy

G(E) =1

E � "p


particles

E = "penergy

G(E) =1

E � "p

particle=infinity


particles

E = "penergy

Green function (propagator)

G(E) =1

E � "p

particle=infinity


particles

E = "penergy


G(E) =1

E � "p

particle=infinity

=poles in Green function


what would you see in a metal?



e�

e�e�

e�

e�

e�e�

e�

e�

e�

e�



e�

e�e�

e�

e�

e�e�

e�

e�

e�

e�



e�

e�e�

e�

e�

e�e�

e�

e�

e�

e�

two crossings:closed surface of excitations

p

"p


expect to see

px

py

closed surface


are there any exceptions?



where is this part?


where is this part?

Fermi arcs: (PDJ,JCC,ZXS)

Fermi Arcs


where are arcs seen?


ceramics

normalstate


what’s the explanation?

strange not strange

Two opposing Viewson Fermi arcs



strange not strange


this dispute has an answer!



strange not strange


this dispute has an answer!unparticles


Right Wingers: Fermi Arcs are not Strange

intensity too small to be seen

G(E) =Zp

! � "p

Zp ! 0pole (particle) exists but


Fermi Arcs are Strange

EF

seen not seen

k

no pole


poles

Re G<0

ReG>0

Problem for left-wingers (me)


poles

Re G<0

ReG>0



poles

Re G<0

ReG>0

How to account for the sign change without poles?



Fermi arcs

0

##

E � ✏p �1 = 0

pole no pole


do poles account for all the charged stuff?


if not?


if not?

charge stuff= particles + other stuff


if not?

charge stuff= particles + other stuff

unparticles!


counting particles


counting particles

1


counting particles

12


counting particles

12

3


counting particles

12

34

5 6

7 8


counting particles

12

34

5 6

7 8

is there a more efficient way?


Each particle has an energy

particles

E = "penergy


G(E) =1

E � "p

particle=infinity


Each particle has an energy

particles

E = "penergy


G(E) =1

E � "p

particle=infinity

particle count is deduciblefrom Green function


Luttinger’s Theorem

Green function (propagator) G(E) =1

E � "p

G(E)

"p E




E � "p

G(E)

"p E

E > "p




E � "p

G(E)

"p E

E > "p

E < "p




E � "p

G(E)

"p E

E > "p

E < "p

particle density= number of sign changes of G


Counting sign changes?



⇥(x)



⇥(x) ={ 0 x < 0



⇥(x) ={ 0 x < 01 x > 0



⇥(x) ={ 0 x < 01 x > 0

1/2 x = 0



⇥(x) ={ 0 x < 01 x > 0

1/2 x = 0

counts sign changes



⇥(x) ={ 0 x < 01 x > 0

1/2 x = 0

counts sign changes

Luttinger Theorem for electrons

n = 2X

k

⇥(<G(k,! = 0))


How do functions

change sign?

"p E

E > "p

E < "p


How do functions

change sign?

"p E

E > "p

E < "pdivergence


How do functions

change sign?

"p E

E > "p

E < "pdivergence

pole


Is there another way?


Yes


"p E

zero-crossing


"p E

zero-crossing

no divergence is necessary


closer look at Luttinger’s theorem

n = 2X

k

⇥(<G(k,! = 0))



divergences (poles)+ zeros

n = 2X

k

⇥(<G(k,! = 0))



divergences (poles)+ zeros

how can zeros affect the particle count?

n = 2X

k

⇥(<G(k,! = 0))


what are zeros?

are they (like poles) conserved?


what models have zeros?


Mott mechanism

t

NiO insulates?d8

Sir Neville


Mott mechanism

t

NiO insulates?d8

perhaps thiscosts energy

Sir Neville


Mott mechanism

t

U � t

NiO insulates?d8


Sir Neville


Mott mechanism

t

U � t

NiO insulates?d8


µ = 0

Sir Neville


Mott mechanism

t

U � t

NiO insulates?d8


µ = 0 no change insize of

Brillouin zoneSir Neville


Mott Problem


Mott Problem

Im G=0µ = 0

( ) d!

Z 1

�1!

ReG(0, p) = Kramers-Kronig


Mott Problem

= below gap+above gap

Im G=0µ = 0

( ) d!

Z 1

�1!



Mott Problem

= below gap+above gap = 0

Im G=0µ = 0

( ) d!

Z 1

�1!



Mott Problem


DetG(k,! = 0) = 0 (single band)

Im G=0µ = 0

( ) d!

Z 1

�1!



Mott Problem


DetG(k,! = 0) = 0 (single band)

Im G=0µ = 0

( ) d!

Z 1

�1!


DetReG(0,p)=0 MottnessFriday, March 14, 14


interactions dominate:Strong Coupling Physics

U/t = 10� 1


how do zeros show up?


poles

Re G<0

ReG>0




poles

0000000

00000

0 0 00 0

0

00

Re G<0

ReG>0




Only option: DetG=0! (zeros )

poles

0000000

00000

0 0 00 0

0

00

Re G<0

ReG>0




Fermi Arcs

zeros + poles

Luttinger, Dzyaloshinskii, Yang, Rice, Zhang,Tsvelik, Anderson (lots of smart people)...

n=zeros + poles


Fermi Liquids Mott Insulators


Is this famous theorem from 1960 correct?


A model with zerosbut Luttinger fails


e�t t t t t t t t



e�t t t t t t t t



e�t t t t t t t t


no hopping=> no propagation (zeros)


e�t t t t t t t t


no hopping=> no propagation (zeros)

N flavorsof e-


e� spin


e� spin

generalization

N flavors of spin


e� spin

generalization

N flavors of spinN = 5


e� spin

generalization

N flavors of spinN = 5


e� spin

generalization

N flavors of spin2E

1491625 N = 5


e� spin

generalization

N flavors of spin

H =U

2(n1 + · · ·nN )2

2E

1491625 N = 5


G↵�(! = 0) = #

✓2n�N

N

◆


G↵�(! = 0) = #

✓2n�N

N

◆


n = N⇥(2n�N)

Luttinger’s theorem


n = N⇥(2n�N)


{0, 1, 1/2


n = N⇥(2n�N)


{0, 1, 1/2n = 2

N = 3


n = N⇥(2n�N)


{0, 1, 1/2n = 2

N = 3

2 = 3


n = N⇥(2n�N)


{0, 1, 1/2n = 2

N = 3

2 = 3


n = N⇥(2n�N)


{0, 1, 1/2even

n = 2

N = 3

2 = 3


n = N⇥(2n�N)


{0, 1, 1/2even odd

n = 2

N = 3

2 = 3


n = N⇥(2n�N)


{0, 1, 1/2even odd

no solution

n = 2

N = 3

2 = 3


Problem

G=0


Problem

G=0

G =1

E � "p � ⌃

1


Problem

G=0

G =1

E � "p � ⌃

1

lifetime of a particle vanishes


Problem

G=0

G =1

E � "p � ⌃

1


=⌃ < ✏p


Problem

G=0

G =1

E � "p � ⌃

1


=⌃ < ✏p no particle


what went wrong?


what went wrong?

�I[G] =

Zd!⌃�G


what went wrong?

�I[G] =

Zd!⌃�G

if ⌃ ! 1


what went wrong?

�I[G] =

Zd!⌃�G

if ⌃ ! 1

integral does not exist


what went wrong?

�I[G] =

Zd!⌃�G

if ⌃ ! 1

integral does not exist

No Luttinger theorem!Friday, March 14, 14




experimental confirmation of violation?



`Luttinger’ countexperimentaldata (LSCO)

kF

1� xFS

Bi2212Zp ! 0



kF

1� xFS

Bi2212Zp ! 0

violation=> zeros are present



kF

1� xFS

each hole a single k-state6=

Bi2212Zp ! 0

violation=> zeros are present


strange



not strange


strange




how to count particles?

12

34

5 6

7 8


how to count particles?

12

34

5 6

7 8

some charged stuffhas no particle interpretation



what is the extra stuff?



scale invariance


new fixed point(scale invariance)

?Friday, March 14, 14




unparticles (IR)(H. Georgi)


what is scale invariance?

invariance on all length scales


f(x) = x

2


f(x) = x

2

f(x/�) = (x/�)2 scale change


f(x) = x

2


scale invariance


f(x) = x

2


f(x) = x

2�

�2g(�){1

scale invariance


f(x) = x

2


g(�) = �2

f(x) = x

2�

�2g(�){1

scale invariance


what’s the underlying theory?


L =1

2@µ�@µ�

x ! x/⇤

free field theory

�(x) ! �(x)

L ! ⇤2L


L =1

2@µ�@µ�

scale invariant

x ! x/⇤

free field theory

�(x) ! �(x)

L ! ⇤2L


L =1

2@µ�@µ�+m2�2

massive free theory

mass


L =1

2@µ�@µ�+m2�2

massive free theory

mass

x ! x/⇤�(x) ! �(x)


L =1

2@µ�@µ�+m2�2

massive free theory

mass

x ! x/⇤�(x) ! �(x)

m2�2


L =1

2@µ�@µ�+m2�2

massive free theory

mass

x ! x/⇤�(x) ! �(x)

m2�2⇤2

✓1

2@µ�@µ�

◆


L =1

2@µ�@µ�+m2�2

massive free theory

mass

x ! x/⇤�(x) ! �(x)

m2�2⇤2

✓1

2@µ�@µ�

◆


L =1

2@µ�@µ�+m2�2

massive free theory

mass

no scale invariance

x ! x/⇤�(x) ! �(x)

m2�2⇤2

✓1

2@µ�@µ�

◆



unparticlesfrom a massive theory?


L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�


L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2


theory with all possible mass!

L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2



L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2

� ! �(x,m2/⇤2)

x ! x/⇤

m2/⇤2 ! m2



L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2

scale invariance is restored!!

L ! ⇤4L

� ! �(x,m2/⇤2)

x ! x/⇤

m2/⇤2 ! m2



L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2

not particles


L ! ⇤4L

� ! �(x,m2/⇤2)

x ! x/⇤

m2/⇤2 ! m2


unparticles


L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0dm2

not particles


L ! ⇤4L

� ! �(x,m2/⇤2)

x ! x/⇤

m2/⇤2 ! m2


✓Z 1

0dm2m2� i

p2 �m2 + i✏

◆�1

/ p2|�|

propagator


✓Z 1

0dm2m2� i

p2 �m2 + i✏

◆�1

/ p2|�|

propagator

dU � 2


G / (E � "p)n�2

n massless particles


G / (E � "p)n�2


G / (E � "p)dU�2


G / (E � "p)n�2


unparticles=fractionalnumber of massless

particles

G / (E � "p)dU�2


G / (E � "p)n�2



particles

G / (E � "p)dU�2

zeros

dU > 2


G / (E � "p)n�2



particles

G / (E � "p)dU�2

zeros

dU > 2

no simple sign changeFriday, March 14, 14

dU?


what really is the summation

over mass?


mass=energy

what really is the summation

over mass?


high energy(UV)

low energy(IR)

related to sum over mass

QFT


high energy(UV)

low energy(IR)


QFT

dg(E)

dlnE= �(g(E))

locality in energyFriday, March 14, 14

high energy(UV)

low energy(IR)


implement E-scaling with an extra dimension

QFT

dg(E)

dlnE= �(g(E))




QFT

dg(E)

dlnE= �(g(E))




gauge-gravity duality(Maldacena, 1997)

UV

QFTIR gravityQFT

dg(E)

dlnE= �(g(E))





UV

QFTIR gravityQFT

dg(E)

dlnE= �(g(E))

locality in energy

no particles(conserved currents)





UV

QFTIR gravityQFT

dg(E)

dlnE= �(g(E))

locality in energy

no particles(conserved currents)

unparticles?


mass=extra dimension

fixed by metric


rewriting the action



L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0

m2�dm2



m = z�1

L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0

m2�dm2



m = z�1

L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0

m2�dm2

L =

Z 1

0dz

2R2

z5+2�

1

2

z2

R2⌘µ⌫(@µ�)(@⌫�) +

�2

2R2

�



m = z�1

L =�@

µ�(x,m)@µ�(x,m) +m

2�

2(x,m)�Z 1

0

m2�dm2

L =

Z 1

0dz

2R2

z5+2�

1

2

z2

R2⌘µ⌫(@µ�)(@⌫�) +

�2

2R2

�

can be absorbed with AdS metric


generating functional for unparticles

action on

S =1

2

Zd

4+2�x dz

p�g

✓@a�@

a�+�2

R

2

◆

AdS5+2�



action on

S =1

2

Zd

4+2�x dz

p�g

✓@a�@

a�+�2

R

2

◆

AdS5+2�

ds

2 =L

2

z

2

�⌘µ⌫dx

µdx

⌫ + dz

2� p

�g = (R/z)5+2�



action on

S =1

2

Zd

4+2�x dz

p�g

✓@a�@

a�+�2

R

2

◆

AdS5+2�

unparticle lives in

d = 4 + 2� � 0

ds

2 =L

2

z

2

�⌘µ⌫dx

µdx

⌫ + dz

2� p

�g = (R/z)5+2�


scaling dimension is fixed

m =1

z



m =1

z

m2AdS = dU (dU�d)

R2



m =1

z

m2AdS = dU (dU�d)

R21 = dU (dU � d)



m =1

z

dU =d

2+

pd2 + 4

2

m2AdS = dU (dU�d)

R21 = dU (dU � d)


GU (p) / p2(dU�d/2)

GU (0) = 0


GU (p) / p2(dU�d/2)

GU (0) = 0

unparticle (AdS) propagator has zeros!


interchanging unparticles

fractional (d_U) number of massless particles













ei⇡dU 6= �1, 0




ei⇡dU 6= �1, 0

fractional statistics in d=2+1


ei⇡dU 6= e�i⇡dU

dU =d

2+

pd2 + 4

2>

d

2

time-reversal symmetry breakingfrom unparticle (zeros=Fermi arcs) matter


g

unparticles BCS

Tc

d ln g

d ln�= 4dU � d > 0

tendency towards pairing (any instabilitywhich establishes a gap)

fermionswith unparticle propagator


High T_cunparticles

variable massUPt_3


emergentgravity

High T_cunparticles

variable massUPt_3


Documents

Unparticles and Superconductivity