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Kaplan

VIEWS: 16 PAGES: 111

  • pg 1
									                            Conformality Lost

                                              J.-W. Lee
                                              D. T. Son
                                              M. Stephanov
                                              D.B.K
                                          arXiv:0905.4752
                                          Phys.Rev.D80:125005,2009




David B. Kaplan                    INT               Feb. 22 , 2010
Monday, February 22, 2010
    Motivation: QCD at LARGE Nc and Nf
                                 lo rs      vo rs
                            Co           Fla




David B. Kaplan                    INT              Feb. 22 , 2010
Monday, February 22, 2010
    Motivation: QCD at LARGE Nc and Nf
                                               lo rs       vo rs
                                          Co            Fla

     Define x= Nf/Nc, treat as a continuous variable

                      asymptotic
                       freedom          conformal                  trivial
           0           ¯
                       ψψ = 0      xc                                           x
                                                       11/2




David B. Kaplan                                  INT                         Feb. 22 , 2010
Monday, February 22, 2010
    Motivation: QCD at LARGE Nc and Nf
                                                     lo rs       vo rs
                                                Co            Fla

     Define x= Nf/Nc, treat as a continuous variable

                      asymptotic
                       freedom              conformal                        trivial
           0           ¯
                       ψψ = 0          xc                                                     x
                                                             11/2
                                                                        ks
            gauge coupling: α✱              ?                        -Za nt
                                                                   ks oi               2
                                                                  n p
                                                                Ba ed           9,1
                                                                                    98
                                                                            :18
                                   0
                                                                  fix .B196
                                                                          .P hys
                                                                         l
                                                                    Nuc




David B. Kaplan                                        INT                                 Feb. 22 , 2010
Monday, February 22, 2010
    Motivation: QCD at LARGE Nc and Nf
                                                     lo rs       vo rs
                                                Co            Fla

     Define x= Nf/Nc, treat as a continuous variable

                      asymptotic
                       freedom              conformal                        trivial
           0           ¯
                       ψψ = 0          xc                                                     x
                                                             11/2
                                                                        ks
            gauge coupling: α✱              ?                        -Za nt
                                                                   ks oi               2
                                                                  n p
                                                                Ba ed           9,1
                                                                                    98
                                                                            :18
                                   0
                                                                  fix .B196
                                                                          .P hys
                                                                         l
                                                                    Nuc

     What is the nature of this transition?
     How does the IR scale appear as conformality is lost?
David B. Kaplan                                        INT                                 Feb. 22 , 2010
Monday, February 22, 2010
        Outline:
        I.         A mechanism for vanishing conformal invariance

        II. The Berezinskii-Kosterlitz-Thouless (BKT) transition

        III. A quantum mechanics model: the 1/r2 potential

        IV. AdS/CFT

        V. Relativistic model: defect Yang-Mills

        VI. QCD with many flavors? A partner theory QCD*
            with a nontrivial UV fixed point?

David B. Kaplan                         INT                Feb. 22 , 2010
Monday, February 22, 2010
   A theory with an infrared conformal fixed point at g=g has
   a zero in the beta function:

                               β(g)
              ∂g   ∂g
     β(g) = µ    =                            g
              ∂µ   ∂t                   g∗




David B. Kaplan               INT                 Feb. 22 , 2010
Monday, February 22, 2010
   A theory with an infrared conformal fixed point at g=g has
   a zero in the beta function:

                               β(g)
              ∂g   ∂g
     β(g) = µ    =                            g
              ∂µ   ∂t                   g∗




   Suppose the theory has another parameter κ such
   that the fixed point at g=g✱ vanishes for κ>κ✱




David B. Kaplan               INT                 Feb. 22 , 2010
Monday, February 22, 2010
   A theory with an infrared conformal fixed point at g=g has
   a zero in the beta function:

                                      β(g)
              ∂g   ∂g
     β(g) = µ    =                                     g
              ∂µ   ∂t                            g∗




   Suppose the theory has another parameter κ such
   that the fixed point at g=g✱ vanishes for κ>κ✱

   Example: supersymmetric QCD is conformal for 3/2 ≤ Nf/Nc ≤ 3




David B. Kaplan                     INT                    Feb. 22 , 2010
Monday, February 22, 2010
   A theory with an infrared conformal fixed point at g=g has
   a zero in the beta function:

                                      β(g)
              ∂g   ∂g
     β(g) = µ    =                                     g
              ∂µ   ∂t                            g∗




   Suppose the theory has another parameter κ such
   that the fixed point at g=g✱ vanishes for κ>κ✱

   Example: supersymmetric QCD is conformal for 3/2 ≤ Nf/Nc ≤ 3
            “κ” = Nf/Nc, “κ✱” = 3/2, 3



David B. Kaplan                     INT                    Feb. 22 , 2010
Monday, February 22, 2010
   A theory with an infrared conformal fixed point at g=g has
   a zero in the beta function:

                                      β(g)
              ∂g   ∂g
     β(g) = µ    =                                     g
              ∂µ   ∂t                            g∗




   Suppose the theory has another parameter κ such
   that the fixed point at g=g✱ vanishes for κ>κ✱

   Example: supersymmetric QCD is conformal for 3/2 ≤ Nf/Nc ≤ 3
            “κ” = Nf/Nc, “κ✱” = 3/2, 3

    How is conformality lost?
David B. Kaplan                     INT                    Feb. 22 , 2010
Monday, February 22, 2010
Three ways to lose an infrared fixed point:

  #1: Fixed point runs to zero:

            β(g; κ)                            β(g; κ)
                            ←   κ <κ ∗                   κ >κ ∗
                                         g                        g




David B. Kaplan                              INT          Feb. 22 , 2010
Monday, February 22, 2010
Three ways to lose an infrared fixed point:

  #1: Fixed point runs to zero:

            β(g; κ)                            β(g; κ)
                            ←   κ <κ ∗                   κ >κ ∗
                                         g                        g


  Example: Supersymmetric QCD at large Nc and Nf
  ➙ Increasing flavors, leave conformal window. κ=Nf/Nc, κ✱=3




David B. Kaplan                              INT          Feb. 22 , 2010
Monday, February 22, 2010
Three ways to lose an infrared fixed point:

  #1: Fixed point runs to zero:

            β(g; κ)                                 β(g; κ)
                            ←        κ <κ ∗                           κ >κ ∗
                                              g                                g


  Example: Supersymmetric QCD at large Nc and Nf
  ➙ Increasing flavors, leave conformal window. κ=Nf/Nc, κ✱=3
       Nf/Nc~< 3            weak coupling Banks-Zaks conformal fixed point

       Nf/Nc > 3            trivial QED-like “free electric” theory
                     ~



David B. Kaplan                                   INT                  Feb. 22 , 2010
Monday, February 22, 2010
Three ways to lose an infrared fixed point:

  #1: Fixed point runs to zero:

            β(g; κ)                                 β(g; κ)
                            ←        κ <κ ∗                           κ >κ ∗
                                              g                                g


  Example: Supersymmetric QCD at large Nc and Nf
  ➙ Increasing flavors, leave conformal window. κ=Nf/Nc, κ✱=3
       Nf/Nc~< 3            weak coupling Banks-Zaks conformal fixed point

       Nf/Nc > 3            trivial QED-like “free electric” theory
                     ~
                                           g2
                                  FE ∼ 2
                                      r ln (r ΛUV )
David B. Kaplan                                   INT                  Feb. 22 , 2010
Monday, February 22, 2010
  #2: Fixed point runs off to infinity:




David B. Kaplan                  INT     Feb. 22 , 2010
Monday, February 22, 2010
  #2: Fixed point runs off to infinity:

     β(g; α)                κ >κ ∗   β(g; α)   κ ≤ κ∗
                                                        ←




David B. Kaplan                         INT             Feb. 22 , 2010
Monday, February 22, 2010
  #2: Fixed point runs off to infinity:

     β(g; α)                κ >κ ∗   β(g; α)   κ ≤ κ∗
                                                            ←



     Possible example? SQCD again ➙ κ=Nf/Nc, κ✱=3/2

    For κ≤κ✱ get “free magnetic phase”          [Seiberg]




David B. Kaplan                         INT                 Feb. 22 , 2010
Monday, February 22, 2010
  #2: Fixed point runs off to infinity:

     β(g; α)                    κ >κ ∗   β(g; α)         κ ≤ κ∗
                                                                        ←



     Possible example? SQCD again ➙ κ=Nf/Nc, κ✱=3/2

    For κ≤κ✱ get “free magnetic phase”                      [Seiberg]

     ➤ electric theory dual to a QED-like
     magnetic theory:

             g ln (r ΛUV )  2                              2
                                                          gM
        FE ∼                             FM ∼                           gM ∼ 1/g
                  r2                               r2   ln (r ΛUV )
David B. Kaplan                             INT                         Feb. 22 , 2010
Monday, February 22, 2010
#3: UV and IR fixed points annihilate:


A toy model:
β(g; κ) = (κ − κ∗ ) − (g − g∗ )2


                                        √
   κ ≥ κ∗ :                 g± = g∗ ±       κ − κ∗
                                  UV, IR fixed points

   κ = κ∗                   fixed points merge

    κ <κ ∗                  conformality lost




David B. Kaplan                                      INT   Feb. 22 , 2010
Monday, February 22, 2010
#3: UV and IR fixed points annihilate:
                                                           β(g; κ)           κ >κ ∗
                                                                                      g
A toy model:                                                         g−     g+

β(g; κ) = (κ − κ∗ ) − (g − g∗ )2


                                        √
   κ ≥ κ∗ :                 g± = g∗ ±       κ − κ∗
                                  UV, IR fixed points

   κ = κ∗                   fixed points merge

    κ <κ ∗                  conformality lost




David B. Kaplan                                      INT                  Feb. 22 , 2010
Monday, February 22, 2010
#3: UV and IR fixed points annihilate:
                                                           β(g; κ)             κ >κ ∗
                                                                                        g
A toy model:                                                         g−        g+

β(g; κ) = (κ − κ∗ ) − (g − g∗ )2

                                                           β(g; κ)             κ = κ∗
                                        √
   κ ≥ κ∗ :                 g± = g∗ ±       κ − κ∗                        g∗            g

                                  UV, IR fixed points

   κ = κ∗                   fixed points merge

    κ <κ ∗                  conformality lost




David B. Kaplan                                      INT                  Feb. 22 , 2010
Monday, February 22, 2010
#3: UV and IR fixed points annihilate:
                                                           β(g; κ)             κ >κ ∗
                                                                                        g
A toy model:                                                         g−        g+

β(g; κ) = (κ − κ∗ ) − (g − g∗ )2

                                                           β(g; κ)             κ = κ∗
                                        √
   κ ≥ κ∗ :                 g± = g∗ ±       κ − κ∗                        g∗            g

                                  UV, IR fixed points

   κ = κ∗                   fixed points merge
                                                                                κ <κ ∗
                                                           β(g; κ)
    κ <κ ∗                  conformality lost
                                                                                        g




David B. Kaplan                                      INT                  Feb. 22 , 2010
Monday, February 22, 2010
What happens close to the transition on the nonconformal side?


     β(g; κ)
                       UV       g∗        IR


                            κ        κ∗




David B. Kaplan                                INT   Feb. 22 , 2010
Monday, February 22, 2010
What happens close to the transition on the nonconformal side?


     β(g; κ)
                       UV       g∗        IR


                            κ        κ∗

      i. Start: g = gUV < g✱ in the UV
      ii. g grows, stalling near g✱
      iii. g strong at scale ΛIR




David B. Kaplan                                INT   Feb. 22 , 2010
Monday, February 22, 2010
What happens close to the transition on the nonconformal side?


     β(g; κ)
                       UV       g∗        IR


                            κ        κ∗
                                                               = ln µ
      i. Start: g = gUV < g✱ in the UV                                  R    dg
                                                                   −
      ii. g grows, stalling near g✱                  ΛIR      ΛUV e         β(g)


      iii. g strong at scale ΛIR                                   −√         π
                                                           = ΛUV e          |κ−κ∗ |




David B. Kaplan                                INT              Feb. 22 , 2010
Monday, February 22, 2010
What happens close to the transition on the nonconformal side?


     β(g; κ)
                       UV       g∗        IR


                            κ        κ∗
                                                               = ln µ
      i. Start: g = gUV < g✱ in the UV                                   R    dg
                                                                     −
      ii. g grows, stalling near g✱                  ΛIR      ΛUV e          β(g)


      iii. g strong at scale ΛIR                                     −√        π
                                                           = ΛUV e           |κ−κ∗ |




    (Not like 2nd order phase transition:            ΛIR     ΛUV    |κ − κ∗ | )

David B. Kaplan                                INT                 Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                       ΛIR       ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)




David B. Kaplan                               INT                      Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                       ΛIR       ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)


 BKT: a classical phase transition in the 2-d XY-model




David B. Kaplan                               INT                      Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                       ΛIR       ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)


 BKT: a classical phase transition in the 2-d XY-model

 Vortices in XY model
 box size R, vortex core size a:




David B. Kaplan                               INT                      Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                        ΛIR      ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)


 BKT: a classical phase transition in the 2-d XY-model

 Vortices in XY model
 box size R, vortex core size a:
   E = E0 ln R/a ,            S = 2 ln R/a
  F = E − T S = (E0 − 2T ) ln R/a




David B. Kaplan                               INT                      Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                        ΛIR      ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)


 BKT: a classical phase transition in the 2-d XY-model

 Vortices in XY model
 box size R, vortex core size a:
   E = E0 ln R/a ,            S = 2 ln R/a
  F = E − T S = (E0 − 2T ) ln R/a

  Vortices condense for T>Tc = E0/2 ;                                 √
                                                                     b/ T −Tc
                                                           ξ     ae
  can show correlation length forms:
David B. Kaplan                               INT                         Feb. 22 , 2010
Monday, February 22, 2010
                                                             π
                                                      −√
                                        ΛIR      ΛUV e     |κ−κ∗ |




                            Scaling behavior of toy model is reminiscent of the
                            Berezinskii-Kosterlitz-Thouless (BKT) transition
                                    (an “infinite order” phase transition)


 BKT: a classical phase transition in the 2-d XY-model

 Vortices in XY model
 box size R, vortex core size a:
   E = E0 ln R/a ,            S = 2 ln R/a
  F = E − T S = (E0 − 2T ) ln R/a

  Vortices condense for T>Tc = E0/2 ;                                 √
                                                                     b/ T −Tc
                                                           ξ     ae
  can show correlation length forms:
David B. Kaplan                               INT                         Feb. 22 , 2010
Monday, February 22, 2010
    RG analysis of the BKT transition

 XY model = Coulomb gas
 (vortices = point-like charges with ln(r) Coulomb interaction):




David B. Kaplan                       INT                      Feb. 22 , 2010
Monday, February 22, 2010
    RG analysis of the BKT transition

 XY model = Coulomb gas
 (vortices = point-like charges with ln(r) Coulomb interaction):
                                        Sum over vortex
                            fugacity   positions/numbers
                             N+ N−     N+ N−
                            z z                                       R                     P
                                                                  −       d2 x T ( φ)2 +i
Z=N                                              d2 xi d2 yj   Dφ e            2                i,j (φ(xi )−φ(yj ))
                            N+ !N− !   i=1 j=1
              N+ ,N−
                                                         Coulomb field                 vortices
                                                                                            anti-vortices




David B. Kaplan                                          INT                                    Feb. 22 , 2010
Monday, February 22, 2010
    RG analysis of the BKT transition

 XY model = Coulomb gas
 (vortices = point-like charges with ln(r) Coulomb interaction):
                                             Sum over vortex
                            fugacity        positions/numbers
                             N+ N−           N+ N−
                            z z                                             R                     P
                                                                        −       d2 x T ( φ)2 +i
Z=N                                                    d2 xi d2 yj   Dφ e            2                i,j (φ(xi )−φ(yj ))
                            N+ !N− !         i=1 j=1
              N+ ,N−
                                                               Coulomb field                 vortices
                                                                                                  anti-vortices
                                   R
                               −       d2 x [ T ( φ)2 −2z cos φ]
    =N                Dφ e                    2



                                          temp.         fugacity




David B. Kaplan                                                INT                                    Feb. 22 , 2010
Monday, February 22, 2010
    RG analysis of the BKT transition

 XY model = Coulomb gas
 (vortices = point-like charges with ln(r) Coulomb interaction):
                                             Sum over vortex
                            fugacity        positions/numbers
                             N+ N−           N+ N−
                            z z                                             R                     P
                                                                        −       d2 x T ( φ)2 +i
Z=N                                                    d2 xi d2 yj   Dφ e            2                i,j (φ(xi )−φ(yj ))
                            N+ !N− !         i=1 j=1
              N+ ,N−
                                                               Coulomb field                 vortices
                                                                                                  anti-vortices
                                   R
                               −       d2 x [ T ( φ)2 −2z cos φ]
    =N                Dφ e                    2



                                          temp.         fugacity


                  The XY model is equivalent to the Sine-Gordon model

David B. Kaplan                                                INT                                    Feb. 22 , 2010
Monday, February 22, 2010
  Classical XY model BKT transition = zero temperature quantum transition in
  Sine-Gordon model:                     T
                                  L=        ( φ) − 2z cos φ
                                                2
                                        2




David B. Kaplan                        INT                       Feb. 22 , 2010
Monday, February 22, 2010
  Classical XY model BKT transition = zero temperature quantum transition in
  Sine-Gordon model:                     T
                                  L=        ( φ) − 2z cos φ
                                                2
                                        2
                                        1               2z
    New variables:                u=1−     ,        v=
                                       8πT             T Λ2

    Perturbative β-functions:     βu = −2v 2 ,      βv = −2uv




David B. Kaplan                        INT                       Feb. 22 , 2010
Monday, February 22, 2010
  Classical XY model BKT transition = zero temperature quantum transition in
  Sine-Gordon model:                     T
                                   L=        ( φ) − 2z cos φ
                                                2
                                         2
                                         1               2z
    New variables:                 u=1−     ,        v=
                                        8πT             T Λ2

    Perturbative β-functions:      βu = −2v 2 ,      βv = −2uv

    Λ = UV cutoff at vortex core
    Dimensionful quantities in
    units of XY model interaction strength




David B. Kaplan                         INT                      Feb. 22 , 2010
Monday, February 22, 2010
  Classical XY model BKT transition = zero temperature quantum transition in
  Sine-Gordon model:                     T
                                   L=        ( φ) − 2z cos φ
                                                2
                                         2
                                         1               2z
    New variables:                 u=1−     ,        v=
                                        8πT             T Λ2

    Perturbative β-functions:      βu = −2v 2 ,      βv = −2uv

    Λ = UV cutoff at vortex core
                                                               v
    Dimensionful quantities in
    units of XY model interaction strength




                            •T<Tc                                              u
                            •bound vortices
                            •trivially conformal

David B. Kaplan                         INT                        Feb. 22 , 2010
Monday, February 22, 2010
  Classical XY model BKT transition = zero temperature quantum transition in
  Sine-Gordon model:                     T
                                   L=        ( φ) − 2z cos φ
                                                2
                                         2
                                         1               2z
    New variables:                 u=1−     ,        v=
                                        8πT             T Λ2

    Perturbative β-functions:      βu = −2v 2 ,      βv = −2uv

    Λ = UV cutoff at vortex core
                                                               v
    Dimensionful quantities in
    units of XY model interaction strength




                            •T<Tc                                              u
                            •bound vortices                    •T>Tc
                            •trivially conformal               •Coulomb gas
                                                               •screening length
David B. Kaplan                         INT                        Feb. 22 , 2010
Monday, February 22, 2010
             1                  2z
       u=1−                 v=                                   v
            8πT
                ,
                               T Λ2                 κ/τ                       τ
       βu = −2v 2 ,         βv = −2uv

      Newer variables:                                                       u

      τ = (u + v) ,         κ = (u2 − v 2 )

       βτ = κ − τ 2 ,        βκ = 0


                                                          Nonperturbative
                                                          region


                                                              κ >κ ∗ (T<Tc)
                                                              κ = κ∗      κ∗ = 0
                                                              κ <κ ∗   (T>Tc)
David B. Kaplan                               INT                       Feb. 22 , 2010
Monday, February 22, 2010
 Correlation length in BKT transition:                  κ>0: Conformal
                                                       (bound vortices)


  For small negative κ, assume
  τ small & positive in UV
                                               T=Tc

                                                         κ<0 finite ξ
   τ blows up in RG time                              (unbound vortices)

                     dτ       π
    t=                    =− √
                    β(τ )   2 −κ




David B. Kaplan                          INT          Feb. 22 , 2010
Monday, February 22, 2010
 Correlation length in BKT transition:                                   κ>0: Conformal
                                                                        (bound vortices)


  For small negative κ, assume
  τ small & positive in UV
                                                  T=Tc

                                                                          κ<0 finite ξ
   τ blows up in RG time                                               (unbound vortices)

                     dτ       π
    t=                    =− √
                    β(τ )   2 −κ


      ...giving rise to an IR scale (like ΛQCD) which sets the scale for the finite
      correlation length for α<0:

                                           1 2√π
                                   ξBKT   ∼ e −α
                                           Λ

David B. Kaplan                            INT                          Feb. 22 , 2010
Monday, February 22, 2010
      So far:
      • BKT transition = loss of conformality via fixed point merger
      • Mechanism of fixed point merger in general gives rise to “BKT
          scaling”:

                                                π
                                         −√
                            ΛIR    ΛUV e      |κ−κ∗ |




David B. Kaplan                    INT                   Feb. 22 , 2010
Monday, February 22, 2010
      So far:
      • BKT transition = loss of conformality via fixed point merger
      • Mechanism of fixed point merger in general gives rise to “BKT
          scaling”:

                                                π
                                         −√
                            ΛIR    ΛUV e      |κ−κ∗ |




      Next: other examples:
      • QM with 1/r2 potential
      • AdS/CFT
      • Defect Yang-Mills
      • QCD with many flavors

David B. Kaplan                    INT                   Feb. 22 , 2010
Monday, February 22, 2010
 Example: QM in d-dimensions with 1/r2 potential
                                                                      Vr

                                                               κ
                −       2
                            + V (r) − k   2
                                              ψ=0,     V (r) = 2
                                                              r




David B. Kaplan                                      INT           Feb. 22 , 2010
Monday, February 22, 2010
 Example: QM in d-dimensions with 1/r2 potential
                                                                                  Vr

                                                                    κ
                −       2
                            + V (r) − k   2
                                              ψ=0,          V (r) = 2
                                                                   r


   k=0 solutions:                  ψ = c− r      ν−
                                                       + c+ r   ν+


                                                                           2
                             d−2          √                          d−2
              ν± −                   ±        κ − κ∗       κ∗ = −
                              2                                       2




David B. Kaplan                                          INT                   Feb. 22 , 2010
Monday, February 22, 2010
 Example: QM in d-dimensions with 1/r2 potential
                                                                                  Vr

                                                                    κ
                −       2
                            + V (r) − k   2
                                              ψ=0,          V (r) = 2
                                                                   r


   k=0 solutions:                  ψ = c− r      ν−
                                                       + c+ r   ν+


                                                                           2
                             d−2          √                          d−2
              ν± −                   ±        κ − κ∗       κ∗ = −
                              2                                       2

    •   valid for κ✱ < κ < (κ✱+1)
         •   κ < κ✱: ν± complex, no ground state
         •   κ = κ✱: ν+ = ν-
         •   κ > (κ✱+1): rν- too singular to normalize


David B. Kaplan                                          INT                   Feb. 22 , 2010
Monday, February 22, 2010
                                                                           κ
                  −         2
                                + V (r) − k   2
                                                   ψ=0,            V (r) = 2
                                                                          r
   k=0 solutions:                     ψ = c− r       ν−
                                                          + c+ r   ν+


                                                                                 2
                            d−2            √                               d−2
       ν± =                            ±          κ − κ∗ ,     κ∗ = −
                             2                                              2




David B. Kaplan                                              INT                     Feb. 22 , 2010
Monday, February 22, 2010
                                                                           κ
                  −         2
                                + V (r) − k   2
                                                   ψ=0,            V (r) = 2
                                                                          r
   k=0 solutions:                     ψ = c− r       ν−
                                                          + c+ r   ν+


                                                                                 2
                            d−2            √                               d−2
       ν± =                            ±          κ − κ∗ ,     κ∗ = −
                             2                                              2
   •   c+ =0 or c-=0 are scale invariant solutions
   •   If c+≠0, ψ → c+rν+ for large r (ν+ > ν-)
   •   to make sense of BC at r=0, introduce δ-function:




David B. Kaplan                                              INT                     Feb. 22 , 2010
Monday, February 22, 2010
                                                                           κ
                  −         2
                                + V (r) − k   2
                                                   ψ=0,            V (r) = 2
                                                                          r
   k=0 solutions:                     ψ = c− r       ν−
                                                          + c+ r   ν+


                                                                                 2
                            d−2            √                               d−2
       ν± =                            ±          κ − κ∗ ,     κ∗ = −
                             2                                              2
   •   c+ =0 or c-=0 are scale invariant solutions
   •   If c+≠0, ψ → c+rν+ for large r (ν+ > ν-)
   •   to make sense of BC at r=0, introduce δ-function:

                                          κ
                                  V (r) = 2 − gδ (d) (r)
                                         r




David B. Kaplan                                              INT                     Feb. 22 , 2010
Monday, February 22, 2010
                                                                           κ
                  −         2
                                + V (r) − k   2
                                                   ψ=0,            V (r) = 2
                                                                          r
   k=0 solutions:                     ψ = c− r       ν−
                                                          + c+ r   ν+


                                                                                 2
                            d−2            √                               d−2
       ν± =                            ±          κ − κ∗ ,     κ∗ = −
                             2                                              2
   •   c+ =0 or c-=0 are scale invariant solutions
   •   If c+≠0, ψ → c+rν+ for large r (ν+ > ν-)
   •   to make sense of BC at r=0, introduce δ-function:

                                          κ
                                  V (r) = 2 − gδ (d) (r)
                                         r

     • rν+      dominates at large r -- corresponds to IR fixed point of g
     • rν-      dominates at small r -- corresponds to UV fixed point of g
David B. Kaplan                                              INT                     Feb. 22 , 2010
Monday, February 22, 2010
  I. Non-perturbative RG treatment of 1/r2 potential:
                                                                        la tor
   regulate with square well:                                       egu
                                                                  Vr
                                                                 U r0
                            κ/r2     r > r0
     V (r) =                     2
                                                             2
                                                         −g/r0
                            −g/r0    r > r0                             κ/r2



   E=0 solution for r>r0:                ψ = c− r   ν−
                                                         + c+ r    ν+




David B. Kaplan                                     INT                          Feb. 22 , 2010
Monday, February 22, 2010
  I. Non-perturbative RG treatment of 1/r2 potential:
                                                                        la tor
   regulate with square well:                                       egu
                                                                  Vr
                                                                 U r0
                            κ/r2     r > r0
     V (r) =                     2
                                                             2
                                                         −g/r0
                            −g/r0    r > r0                             κ/r2



   E=0 solution for r>r0:                ψ = c− r   ν−
                                                         + c+ r    ν+


  Solve for c+/c- (a physical dimensionful quantity)
  and require invariance: d(c+/c-)/dr0 = 0:




David B. Kaplan                                     INT                          Feb. 22 , 2010
Monday, February 22, 2010
  I. Non-perturbative RG treatment of 1/r2 potential:
                                                                           la tor
   regulate with square well:                                       egu
                                                                  Vr
                                                                 U r0
                            κ/r2     r > r0
     V (r) =                     2
                                                             2
                                                         −g/r0
                            −g/r0    r > r0                                κ/r2



   E=0 solution for r>r0:                ψ = c− r   ν−
                                                         + c+ r    ν+


  Solve for c+/c- (a physical dimensionful quantity)
  and require invariance: d(c+/c-)/dr0 = 0:

  Find exact β-function for g. Eg, for d=3:                             Β g, Α

                       √      √    √          2√
                      2 g κ + g cot g − g cot g                                                        g
     β       =                 √   √      2 √g                                      1 g   2    3
                          − cot g + g csc                                                     Α 0
                                                                                              κ=0
                                                                                              Α 14
                                                                                              κ=-1/4
      κ✱ = -¼ , g✱ ≈ 1.36
                                                                                              Α 12
                                                                                              κ=-1/2
David B. Kaplan                                     INT                                   Feb. 22 , 2010
Monday, February 22, 2010
   Aside: Even better to define a modified coupling constant
                       √         √
                         g Jd/2 ( g)
                γ=              √
                        Jd/2−1 ( g)

   Condition d(c+/c-)/dr0 yields exact β-function in d-dimensions:
                                 ∂γ                                   d−2
                            βγ =    = (κ − κ∗ ) − (γ − γ∗ )2 ,   γ∗ =
                                 ∂t                                    2


   • Toy model is exact!
   • γ is a periodic function of g, γ=±∞
      equivalent
   • Aside: Limit cycle behavior for κ<κ✱:
      describes “Efimov states”


David B. Kaplan                                      INT                    Feb. 22 , 2010
Monday, February 22, 2010
   II. Perturbative RG treatment of κ/r2 potential:
   κ✱ ≡ -(d-2)2/4 so work in d=2+ε




David B. Kaplan                      INT              Feb. 22 , 2010
Monday, February 22, 2010
   II. Perturbative RG treatment of κ/r2 potential:
  κ✱ ≡ -(d-2)2/4 so work in d=2+ε
                                    2                 δ-function
                               | ψ|     gπ † †
  S =            d      †
             dt d x iψ ∂t ψ −         +    ψ ψ ψψ
                                2m       4
                                            κ
          − dt d x d y ψ (t, x)ψ (t, y)
                   d  d    †      †
                                                 ψ(t, y)ψ(t, x)
                                        |x − y|2
                  i
 propagator:
             ω − p2 /2m


 contact vertex:            iπgµ−
                   2πiκ 1
 “meson exchange”:
                       |q|




David B. Kaplan                      INT                     Feb. 22 , 2010
Monday, February 22, 2010
   II. Perturbative RG treatment of κ/r2 potential:
  κ✱ ≡ -(d-2)2/4 so work in d=2+ε
                                    2                 δ-function
                               | ψ|     gπ † †
  S =            d      †
             dt d x iψ ∂t ψ −         +    ψ ψ ψψ
                                2m       4
                                            κ
          − dt d x d y ψ (t, x)ψ (t, y)
                   d  d    †      †
                                                 ψ(t, y)ψ(t, x)
                                        |x − y|2
                  i
 propagator:
             ω − p2 /2m             Find g runs:            +
                                                            2
                                                ∂g
 contact vertex:            iπgµ−   β(g; κ) = µ    =   κ+        − (g − )2
                                                ∂µ          4
                   2πiκ 1
 “meson exchange”:
                       |q|




David B. Kaplan                       INT                       Feb. 22 , 2010
Monday, February 22, 2010
   II. Perturbative RG treatment of κ/r2 potential:
  κ✱ ≡ -(d-2)2/4 so work in d=2+ε
                                    2                 δ-function
                               | ψ|     gπ † †
  S =            d      †
             dt d x iψ ∂t ψ −         +    ψ ψ ψψ
                                2m       4
                                            κ
          − dt d x d y ψ (t, x)ψ (t, y)
                   d  d    †      †
                                                 ψ(t, y)ψ(t, x)
                                        |x − y|2
                  i
 propagator:
             ω − p2 /2m             Find g runs:            +
                                                             2
                                                ∂g
 contact vertex:            iπgµ−   β(g; κ) = µ    =   κ+         − (g − )2
                                                ∂µ          4
                   2πiκ 1           Same as toy model! κ✱ = -ε2/4, g✱ =ε
 “meson exchange”:
                       |q|
                                    Exact, ε=1: κ✱ = -1/4, g✱ =1.36




David B. Kaplan                       INT                        Feb. 22 , 2010
Monday, February 22, 2010
   II. Perturbative RG treatment of κ/r2 potential:
  κ✱ ≡ -(d-2)2/4 so work in d=2+ε
                                    2                 δ-function
                               | ψ|     gπ † †
  S =            d      †
             dt d x iψ ∂t ψ −         +    ψ ψ ψψ
                                2m       4
                                            κ
          − dt d x d y ψ (t, x)ψ (t, y)
                   d  d    †      †
                                                 ψ(t, y)ψ(t, x)
                                        |x − y|2
                  i
 propagator:
             ω − p2 /2m                  Find g runs:                  +
                                                                       2
                                                     ∂g
 contact vertex:            iπgµ−        β(g; κ) = µ    =         κ+        − (g − )2
                                                     ∂µ                4
                   2πiκ 1                Same as toy model! κ✱ = -ε2/4, g✱ =ε
 “meson exchange”:
                       |q|
                                         Exact, ε=1: κ✱ = -1/4, g✱ =1.36
κ>κ✱: conformal                           Λ2            Λ2        √
                                           IR            UV   −2π/ κ∗ −κ
κ=κ✱: critical                      B∼           ∼            e
                                          m              m
κ<κ✱: g blows up in IR                                                       BKT scaling
                                            bound state energy
David B. Kaplan                            INT                             Feb. 22 , 2010
Monday, February 22, 2010
    Conformal phases: measure correlations, not β-functions!
    Look at operator scaling dimensions:




David B. Kaplan                  INT                  Feb. 22 , 2010
Monday, February 22, 2010
    Conformal phases: measure correlations, not β-functions!
    Look at operator scaling dimensions:
     From Nishida & Son, 2007:
       • Replace V(r1-r2) ➞ V(r1-r2) + ½ ω2|r12+r22|
       • Compute 2-particle ground state energy E0
          • Operator dimension of ψψ is Δψψ =E0/ω




David B. Kaplan                            INT         Feb. 22 , 2010
Monday, February 22, 2010
    Conformal phases: measure correlations, not β-functions!
    Look at operator scaling dimensions:
     From Nishida & Son, 2007:
       • Replace V(r1-r2) ➞ V(r1-r2) + ½ ω2|r12+r22|
       • Compute 2-particle ground state energy E0
          • Operator dimension of ψψ is Δψψ =E0/ω
                                                  2-particle wave-
                                                 function at |r1-r2|=0




David B. Kaplan                            INT                           Feb. 22 , 2010
Monday, February 22, 2010
    Conformal phases: measure correlations, not β-functions!
    Look at operator scaling dimensions:
     From Nishida & Son, 2007:
       • Replace V(r1-r2) ➞ V(r1-r2) + ½ ω2|r12+r22|
       • Compute 2-particle ground state energy E0
          • Operator dimension of ψψ is Δψψ =E0/ω
                                                  2-particle wave-
                                                 function at |r1-r2|=0

      As the two conformal theories merge when κ➝ κ✱ , operator
      dimensions in the two CFTs merge




David B. Kaplan                            INT                           Feb. 22 , 2010
Monday, February 22, 2010
    Conformal phases: measure correlations, not β-functions!
    Look at operator scaling dimensions:
     From Nishida & Son, 2007:
       • Replace V(r1-r2) ➞ V(r1-r2) + ½ ω2|r12+r22|
       • Compute 2-particle ground state energy E0
          • Operator dimension of ψψ is Δψψ =E0/ω
                                                          2-particle wave-
                                                         function at |r1-r2|=0

      As the two conformal theories merge when κ➝ κ✱ , operator
      dimensions in the two CFTs merge


     For 1/r2 potential -- find for the two conformal theories:
                                               d+2       √            “+” = UV fixed point
[ψψ]:                       ∆± = (d + ν± ) =         ±       κ − κ∗
                                                2                     “-” = IR fixed point

   Note: (Δ++Δ-) = (d+2): scaling dimension of nonrelativistic spacetime.
David B. Kaplan                                  INT                             Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:




David B. Kaplan             INT   Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:
                                        d
                            1
 AdS:                  ds = 2
                            2
                                dz +
                                 2             2
                                             dxi
                           z           i=1




David B. Kaplan                                INT   Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:
                                                    d
                            1
 AdS:                  ds = 2
                            2
                                         dz +
                                          2                2
                                                         dxi
                           z                       i=1

Massive scalar in the bulk
two solutions to eq. of motion, corresponding to two different CFT’s:
     ϕ       =        c+ z ∆+ + c− z ∆−
                                                                                          r dim
                                                                                 op erato
 ∆±          =        d
                            ±   m2   +   d 2
                                               ≡   d
                                                        ±   m2   −   m2   Δ± =
                      2                  2         2                  ∗




David B. Kaplan                                             INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:
                                                     d
                            1
 AdS:                  ds = 2
                            2
                                          dz +
                                           2                2
                                                          dxi
                           z                        i=1

Massive scalar in the bulk
two solutions to eq. of motion, corresponding to two different CFT’s:
      ϕ      =        c+ z ∆+ + c− z ∆−
                                                                                               r dim
                                                                                      op erato
 ∆±          =        d
                            ±   m2    +   d 2
                                                ≡   d
                                                         ±   m2   −   m2       Δ± =
                      2                   2         2                  ∗

                                AdS                                              QM

  •   (Δ++Δ-)=d= spacetime dim of CFT                           •   (Δ+ψψ+Δ-ψψ)=(d+2)= conformal wt.
                                                                    of nonrelativistic d-space+time




David B. Kaplan                                              INT                         Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:
                                                     d
                            1
 AdS:                  ds = 2
                            2
                                          dz +
                                           2                2
                                                          dxi
                           z                        i=1

Massive scalar in the bulk
two solutions to eq. of motion, corresponding to two different CFT’s:
      ϕ      =        c+ z ∆+ + c− z ∆−
                                                                                               r dim
                                                                                      op erato
 ∆±          =        d
                            ±   m2    +   d 2
                                                ≡   d
                                                         ±   m2   −   m2       Δ± =
                      2                   2         2                  ∗

                                AdS                                              QM

  •   (Δ++Δ-)=d= spacetime dim of CFT                           •   (Δ+ψψ+Δ-ψψ)=(d+2)= conformal wt.
                                                                    of nonrelativistic d-space+time
  •   when m2 = m✱2 = -d2/4 , Δ±=d/2                            •   κ = κ✱ = -(d-2)2/4    Δ±=(d+2)/2




David B. Kaplan                                              INT                         Feb. 22 , 2010
Monday, February 22, 2010
  Analog in AdS/CFT:
                                                     d
                            1
 AdS:                  ds = 2
                            2
                                          dz +
                                           2                2
                                                          dxi
                           z                        i=1

Massive scalar in the bulk
two solutions to eq. of motion, corresponding to two different CFT’s:
      ϕ      =        c+ z ∆+ + c− z ∆−
                                                                                                r dim
                                                                                       op erato
 ∆±          =        d
                            ±   m2    +   d 2
                                                ≡   d
                                                         ±   m2   −   m2        Δ± =
                      2                   2         2                  ∗

                                AdS                                               QM

  •   (Δ++Δ-)=d= spacetime dim of CFT                           •   (Δ+ψψ+Δ-ψψ)=(d+2)= conformal wt.
                                                                    of nonrelativistic d-space+time
  •   when m2 = m✱2 = -d2/4 , Δ±=d/2                            •   κ = κ✱ = -(d-2)2/4    Δ±=(d+2)/2
  •   Instability (no AdS or CFT) for                           •   Conformality lost for κ < κ✱
      m2 < m✱2 (B-F bound)

David B. Kaplan                                              INT                          Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions   2 different CFTs




David B. Kaplan                INT                 Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions                         2 different CFTs

   ϕ = ϕ0 z ∆ + :           Zgrav.                    = ZCFT [ϕ0 ]
                                     ϕ − − ϕ 0 z ∆+
                                       −→
                                       z→0




David B. Kaplan                               INT                        Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions                         2 different CFTs

   ϕ = ϕ0 z ∆ + :           Zgrav.                    = ZCFT [ϕ0 ]
                                     ϕ − − ϕ 0 z ∆+
                                       −→
                                       z→0


                                                             S = SCFT +        d
                                                                              d x φ0 O




David B. Kaplan                               INT                        Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions                         2 different CFTs

   ϕ = ϕ0 z ∆ + :           Zgrav.                    = ZCFT [ϕ0 ]
                                     ϕ − − ϕ 0 z ∆+
                                       −→
                                       z→0


                                                             S = SCFT +             d
                                                                                d x φ0 O
   ϕ = J z ∆− :             Zgrav.                     = ZCFT [J]
                                     ϕ− − Jz ∆−
                                      −→
                                       z→0
                                                                                R
                                                                                    dd xJϕ
                                                       =      Dϕ ZCFT [ϕ]e  i




David B. Kaplan                               INT                        Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions                         2 different CFTs

   ϕ = ϕ0 z ∆ + :           Zgrav.                    = ZCFT [ϕ0 ]
                                     ϕ − − ϕ 0 z ∆+
                                       −→
                                       z→0


                                                             S = SCFT +             d
                                                                                d x φ0 O
   ϕ = J z ∆− :             Zgrav.                     = ZCFT [J]
                                     ϕ− − Jz ∆−
                                      −→
                                       z→0
                                                                                R
                                                                                    dd xJϕ
                                                       =      Dϕ ZCFT [ϕ]e  i




  UV fine-tuning: m2φ2...adds OO operator. Eg: O=ψψ, OO =ψψψψ


David B. Kaplan                               INT                        Feb. 22 , 2010
Monday, February 22, 2010
  AdS/CFT cont’d:

   As with QM example, 2 different solutions                         2 different CFTs

   ϕ = ϕ0 z ∆ + :           Zgrav.                    = ZCFT [ϕ0 ]
                                     ϕ − − ϕ 0 z ∆+
                                       −→
                                       z→0


                                                             S = SCFT +             d
                                                                                d x φ0 O
   ϕ = J z ∆− :             Zgrav.                     = ZCFT [J]
                                     ϕ− − Jz ∆−
                                      −→
                                       z→0
                                                                                R
                                                                                    dd xJϕ
                                                       =      Dϕ ZCFT [ϕ]e  i




  UV fine-tuning: m2φ2...adds OO operator. Eg: O=ψψ, OO =ψψψψ

                 analog of δ(r) in QM example tuned to unstable UV fixed pt.
David B. Kaplan                               INT                        Feb. 22 , 2010
Monday, February 22, 2010
  A relativistic example: defect Yang-Mills theory




                                  d
                                spatial
                              dimensions




David B. Kaplan                INT                   Feb. 22 , 2010
Monday, February 22, 2010
  A relativistic example: defect Yang-Mills theory




                                                       d
                                                     spatial
                                                  dimensions



 Charged relativistic fermions on a d-dimensional defect
 + 4D conformal gauge theory (eg, N=4 SYM)

                            S=   d   d+1      ¯ µ Dµ ψ − 1
                                           x iψγ                     a
                                                               d4 x Fµν F a,µν
                                                        4g 2




David B. Kaplan                                    INT                           Feb. 22 , 2010
Monday, February 22, 2010
  A relativistic example: defect Yang-Mills theory




                                                       d
                                                     spatial
                                                  dimensions



 Charged relativistic fermions on a d-dimensional defect
 + 4D conformal gauge theory (eg, N=4 SYM)

                            S=   d   d+1      ¯ µ Dµ ψ − 1
                                           x iψγ                     a
                                                               d4 x Fµν F a,µν
                                                        4g 2
                                                               g doesn’t run


David B. Kaplan                                    INT                           Feb. 22 , 2010
Monday, February 22, 2010
  g doesn’t run by construction


 Expect a phase transition as a function of g:
                            ¯      0      g < g∗
                            ψψ =
                                   ΛdIR   g > g∗




David B. Kaplan                               INT   Feb. 22 , 2010
Monday, February 22, 2010
  g doesn’t run by construction


 Expect a phase transition as a function of g:
                            ¯          0          g < g∗
                            ψψ =
                                       ΛdIR       g > g∗


 Add a contact interaction to the theory (as in QM & AdS/CFT
 examples!) and study its running:


                                                    c ¯
                            ∆S =   d   d+1
                                              x    − (ψγµ Ta ψ)2
                                                    2




David B. Kaplan                                       INT          Feb. 22 , 2010
Monday, February 22, 2010
  g doesn’t run by construction


 Expect a phase transition as a function of g:
                            ¯          0          g < g∗
                            ψψ =
                                       ΛdIR       g > g∗


 Add a contact interaction to the theory (as in QM & AdS/CFT
 examples!) and study its running:


                                                    c ¯
                            ∆S =   d   d+1
                                              x    − (ψγµ Ta ψ)2
                                                    2


   Phase transition is in perturbative regime for d=1+ε (spatial
   dimensions of “defect”): compute β-function
David B. Kaplan                                       INT          Feb. 22 , 2010
Monday, February 22, 2010
               =         cc (¯ µ ta ψ)2
                             ¯
            L = ······− (ψγ µta ψ)2
                       − ψγ
            L             2 1
           c ¯ µ a 22 1
µ ψAµ − c (¯ µ t ψ) ) −
 µ
             ψγ   a    2          d44 x Faa Faa + · · ·            (55)
  ψAµ − 2(ψγ t ψ) ) − 4 d x Fµν Fµν + · · ·
                                          µν µν                   (55)
          2                  4
      =
β(c)β(c):
 (c) = a 1/ factor from the gluon propagator (40) and
contains a 1/ factor from the gluon propagator (40) and
 ontains
   c:
r c:
                                                         1/ε pole for d=(1+ε)
            Nc 2 g 22  g                  1/! pole from photon
                                         1/! pole from photon
(c) = cc− Nc cc −
 c)  = −        2 −                     propagator in (1+!)+1 D
                                       propagator in (1+!)+1 D     (56)
                                                                  (56)
            2π
            2π        2π
                      2π
  g where β(c) has a double zero,
  g∗∗ where β(c) has a double zero,
       π22
g∗ = π
     =       π22 Same as gap eq, except 2CAA"N
                  Same as gap eq, except 2C "N
       Nc= π
 ∗      N
       g∗ c=
       g∗                                                      (57)
                                                              (57)
              N
             Ncc
 is substitution also works for mass gap at g>g
his substitution also works for mass gap at g>g**
  RG equation,
 RG equation,
         ∂c
        ∂c = β(c)                                              (58)
       ∂ln µ = β(c)
       ∂ ln µ
                                                              (58)
 David B. Kaplan                        INT                       Feb. 22 , 2010
  Monday, February 22, 2010
               =          cc (¯ µ ta ψ)2
                              ¯
             L = ······− (ψγ µta ψ)2
                       − ψγ
             L             2 1
            c ¯ µ a 22 1
µ ψAµ − c (¯ µ t ψ) ) −
 µ
              ψγ   a    2          d44 x Faa Faa + · · ·            (55)
  ψAµ − 2(ψγ t ψ) ) − 4 d x Fµν Fµν + · · ·µν µν                   (55)
           2                  4
       =
β(c)β(c):
 (c) = a 1/ factor from the gluon propagator (40) and
contains a 1/ factor from the gluon propagator (40) and
 ontains
   c:
r c:
                                                          1/ε pole for d=(1+ε)
             Nc 2g 2 g 22
                       g                   1/! pole from photon
                                          1/! pole from photon
             Nc − − − c − Nc c2 propagator in (1+!)+1 D
(c) = cc−= cc −
     = −
      β(c)      2                        propagator in (1+!)+1 D    (56)
 c)                                                                (56)
             2π 2π 2π
             2π        2π      2π
                                                       2
               1  π           2 2
                                N
                          2 zero, c               π
 g∗∗ where = 2πhas a double zero,
            β(c)
 g where β(c) has a double −
        2          Nc
                       −g
                                2π
                                              c−
                                                 Nc
      π2
      π
   =
g∗ =         π22 Same as gap eq, except 2CAA"N
                  Same as gap eq, except 2C "N
       Nc= π
 ∗     N
       g∗ c=
      g∗                                                        (57)
                                                               (57)
              N
             Ncc
 is substitution also works for mass gap at g>g
his substitution also works for mass gap at g>g**
  RG equation,
 RG equation,
         ∂c
        ∂c = β(c)                                               (58)
       ∂ln µ = β(c)
       ∂ ln µ
                                                               (58)
 David B. Kaplan                        INT                        Feb. 22 , 2010
  Monday, February 22, 2010
               =          cc (¯ µ ta ψ)2
                              ¯
             L = ······− (ψγ µta ψ)2
                       − ψγ
             L             2 1
            c ¯ µ a 22 1
µ ψAµ − c (¯ µ t ψ) ) −
 µ
              ψγ   a    2          d44 x Faa Faa + · · ·            (55)
  ψAµ − 2(ψγ t ψ) ) − 4 d x Fµν Fµν + · · ·µν µν                   (55)
           2                  4
       =
β(c)β(c):
 (c) = a 1/ factor from the gluon propagator (40) and
contains a 1/ factor from the gluon propagator (40) and
 ontains
   c:
r c:
                                                          1/ε pole for d=(1+ε)
             Nc 2g 2 g 22
                       g                   1/! pole from photon
                                          1/! pole from photon
             Nc − − − c − Nc c2 propagator in (1+!)+1 D
(c) = cc−= cc −
     = −
      β(c)      2                        propagator in (1+!)+1 D    (56)
 c)                                                                (56)
             2π 2π 2π
             2π        2π      2π
                                                       2
               1  π           2 2
                                N
                          2 zero, c               π
 g∗∗ where = 2πhas a double zero,
            β(c)
 g where β(c) has a double −
        2          Nc
                       −g
                                2π
                                              c−
                                                 Nc
      π2
      π
   =
g∗ =         π22 Same as gap eq, except 2CAA"N
                  Same as gap eq, except 2C "N
       Nc= π
 ∗      N
       g∗ c=
      g∗                                           (57)
                                                  (57)
    • Find BKT transition at g2 = g✱2 = (επ)2/Nc
              N
             Ncc
 is substitutionΛ exp[-π/√(g2-g gap at g>g
                   also works for mass 2
his substitution also works for mass gap at g>g**
         ΛIR ~
  RG equation, UV
       equation,                      ✱ )]
 RG
      • Schwinger-Dyson
        ∂c                 gap eq (rainbow approx) gives
        ∂c    = β(c)                                (58)
         qualitatively same results
       ∂ ln µ = β(c)                               (58)
       ∂ ln µ
 David B. Kaplan                        INT                        Feb. 22 , 2010
  Monday, February 22, 2010
 Back to QCD at LARGE Nc and Nf:
                                asymptotic
                                 freedom              conformal           trivial
                            0   ¯
                                ψψ = 0           xc                                     x
                                                                  11/2
                                                                             ks
                                                                           Za t
                        gauge coupling: α✱            ?                 ks- oin
                                                                      an d p
                                                                     B e
                                                                       fix
                                             0

      Transition at x=xc?




David B. Kaplan                                       INT                     Feb. 22 , 2010
Monday, February 22, 2010
 Back to QCD at LARGE Nc and Nf:
                                asymptotic
                                 freedom              conformal           trivial
                            0   ¯
                                ψψ = 0           xc                                     x
                                                                  11/2
                                                                             ks
                                                                           Za t
                        gauge coupling: α✱            ?                 ks- oin
                                                                      an d p
                                                                     B e
                                                                       fix
                                             0

      Transition at x=xc?
      Schwinger-Dyson (rainbow approximation):




David B. Kaplan                                       INT                     Feb. 22 , 2010
Monday, February 22, 2010
 Back to QCD at LARGE Nc and Nf:
                                asymptotic
                                 freedom              conformal           trivial
                            0   ¯
                                ψψ = 0           xc                                     x
                                                                  11/2
                                                                             ks
                                                                           Za t
                        gauge coupling: α✱            ?                 ks- oin
                                                                      an d p
                                                                     B e
                                                                       fix
                                             0

      Transition at x=xc?
      Schwinger-Dyson (rainbow approximation):




Found: BKT scaling for <ψψ>...not rigorous, but qualitatively correct?
David B. Kaplan                                       INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.




David B. Kaplan                    INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                  3
                                           Δ+
                                                      QCD

                            Δψψ   2                             Δ+ + Δ- = 4?

                                            -
                                                      QCD*
                                           Δ
                                  1                     xBZ      x
                                      x✱
                                                        =11/2




David B. Kaplan                                 INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                  3
                                           Δ+                    Free fermions
                                                      QCD

                            Δψψ   2                             Δ+ + Δ- = 4?

                                            -
                                                      QCD*
                                           Δ
                                  1                     xBZ      x
                                      x✱
                                                        =11/2




David B. Kaplan                                 INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                  3
                                           Δ+                    Free fermions
                                                      QCD

                            Δψψ   2                             Δ+ + Δ- = 4?

                                            -
                                                      QCD*
                                           Δ                     Free boson
                                  1                     xBZ      x
                                      x✱
                                                        =11/2




David B. Kaplan                                 INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                  3
                                           Δ+                    Free fermions
                                                      QCD

                            Δψψ   2                             Δ+ + Δ- = 4?

                                            -
                                                      QCD*
                                           Δ                     Free boson
                                  1                     xBZ      x
                                      x✱
                                                        =11/2




David B. Kaplan                                 INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                  3
                                           Δ+                    Free fermions
                                                      QCD

                            Δψψ   2                             Δ+ + Δ- = 4?

                                            -
                                                      QCD*
                                           Δ                     Free boson
                                  1                     xBZ      x
                                      x✱
                                                        =11/2


                             Near Banks-Zaks (IR) fixed point:




David B. Kaplan                                 INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                    3
                                             Δ+                    Free fermions
                                                        QCD

                              Δψψ   2                             Δ+ + Δ- = 4?

                                              -
                                                        QCD*
                                             Δ                     Free boson
                                    1                     xBZ      x
                                        x✱
                                                          =11/2


                               Near Banks-Zaks (IR) fixed point:
                       QCD:
         +
        Δ   ψψ = 3 - # g2Nc
         (almost free quarks)
David B. Kaplan                                   INT                     Feb. 22 , 2010
Monday, February 22, 2010
  Conjecture: loss of conformality for QCD at xc is of BKT type, due
  to fixed point merger.
                                    3
                                             Δ+                     Free fermions
                                                        QCD

                              Δψψ   2                              Δ+ + Δ- = 4?

                                              -
                                                        QCD*
                                             Δ                      Free boson
                                    1                        xBZ    x
                                        x✱
                                                          =11/2


                               Near Banks-Zaks (IR) fixed point:
                       QCD:                                   Partner theory QCD*:
         +                                               -              +
                                                        Δψψ = d-Δψψ = 1+ # g2Nc
        Δ   ψψ        =3-#     g2Nc
         (almost free quarks)                                 (almost free scalar?)
David B. Kaplan                                   INT                       Feb. 22 , 2010
Monday, February 22, 2010
                 WANTED
    ☛ Conformal theory
           defined at nontrivial
              UV fixed point
           to merge with QCD
                  at x=xc
            LAST SEEN WITH WEAKLY
                  COUPLED SCALAR




David B. Kaplan                     INT   Feb. 22 , 2010
Monday, February 22, 2010
                                    Haven’t found a Lorentz invariant
                 WANTED             perturbative example
                                    with:
    ☛ Conformal theory
                                    (i) weakly coupled scalar;
           defined at nontrivial
              UV fixed point        (ii) full SU(Nf)xSU(Nf) chiral symmetry
           to merge with QCD        (iii) Matching anomalies
                  at x=xc
            LAST SEEN WITH WEAKLY
                  COUPLED SCALAR




David B. Kaplan                              INT                        Feb. 22 , 2010
Monday, February 22, 2010
                                              Haven’t found a Lorentz invariant
                 WANTED                       perturbative example
                                              with:
    ☛ Conformal theory
                                              (i) weakly coupled scalar;
           defined at nontrivial
              UV fixed point                  (ii) full SU(Nf)xSU(Nf) chiral symmetry
           to merge with QCD                  (iii) Matching anomalies
                  at x=xc                     Look for nonperturbative QCD* on the lattice?
            LAST SEEN WITH WEAKLY             One place to start: strong/weak transition for QCD
                  COUPLED SCALAR              with Nf in conformal window?
                                              (A. Hasenfratz)

                            conformal phase            strong coupling phase

                             ∗                        ∗                  g
              0             g−                       g+
               QCD* possibly at g+* ?
David B. Kaplan                                        INT                        Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:




David B. Kaplan             INT   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality




David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality

        II. Leads to similar scaling as in the BKT transition:
            ΛIR ~ ΛUV e[-π/√(-κ-κ✱)]




David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality

        II. Leads to similar scaling as in the BKT transition:
            ΛIR ~ ΛUV e[-π/√(-κ-κ✱)]

        III. Both relativistic & non-relativistic examples




David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality

        II. Leads to similar scaling as in the BKT transition:
            ΛIR ~ ΛUV e[-π/√(-κ-κ✱)]

        III. Both relativistic & non-relativistic examples

        IV. Analog in AdS/CFT; implications for AdS below the
            Breitenlohner-Freedman bound?




David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality

        II. Leads to similar scaling as in the BKT transition:
            ΛIR ~ ΛUV e[-π/√(-κ-κ✱)]

        III. Both relativistic & non-relativistic examples

        IV. Analog in AdS/CFT; implications for AdS below the
            Breitenlohner-Freedman bound?

        V.        Implications for QCD with many flavors? Is there a pair
                  of conformal QCD theories? What is QCD*?




David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
       Conclusions:
        I.        Fixed point annihilation appears to be a generic
                  mechanism for the loss of conformality

        II. Leads to similar scaling as in the BKT transition:
            ΛIR ~ ΛUV e[-π/√(-κ-κ✱)]

        III. Both relativistic & non-relativistic examples

        IV. Analog in AdS/CFT; implications for AdS below the
            Breitenlohner-Freedman bound?

        V.        Implications for QCD with many flavors? Is there a pair
                  of conformal QCD theories? What is QCD*?

        VI. Finding QCD* should be on field theory / lattice QCD
            “to-do” list.

David B. Kaplan                            INT                   Feb. 22 , 2010
Monday, February 22, 2010
David B. Kaplan             INT   Feb. 22 , 2010
Monday, February 22, 2010

								
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