The role of double trace deformations in AdS/CMT Gary Horowitz UC Santa Barbara Based on Faulkner, Roberts, G.H. 1006.2387 and 1008.1581
Double trace deforma/ons can • Implement spontaneous symmetry breaking • Provide a new way to construct holographic superconductors • Provide a knob to tune the cri/cal temperature of holographic superconductors • Lead to new quantum cri/cal points with nontrivial cri/cal exponents
Consider the following ac/on: S = 1 d 4 x √− g � R − ( ∇ φ ) 2 − 2 V ( φ ) � � 2 where V ( φ ) = − 3 + 1 2 m 2 φ 2 + . . . Need m 2 > m BF 2 = ‐9/4 for a stable ground state. Start with asympto/cally (globally) AdS solu/ons ds 2 = r 2 ( − dt 2 + d Ω ) + dr 2 r 2
Asympto/cally, the scalar field is α β φ = r ∆ − + r ∆ + where � ∆ ± = 3 / 2 ± 9 / 4 + m 2 Usually normalizability requires α = 0, but if m 2 BF < m 2 < m 2 BF + 1 both modes are normalizable and one has a choice of boundary condi/ons.
If O is the operator dual to ϕ , α = 0 => O has dimension Δ + and <O> = β (standard quan/za/on) β = 0 => O has dimension Δ ‐ and <O> = α (alterna/ve quan/za/on) More generally, one can set β = W’(α) for any W(α)
Double trace deforma/ons In alterna/ve quan/za/on, Δ < 3/2, and one can modify the ac/on by � d 3 x κ O † O S → S − κ has dimension 3 ‐ 2Δ > 0 so this is a relevant coupling. On gravity side it corresponds to boundary condi/ons (Wiaen; Berkooz, Sever and Shomer) β = κα
Theory with κ < 0 can s/ll have a stable ground state. (Faulkner, Roberts, G.H., 2010) In the new ground state, <O> is nonzero. This is a classic example of spontaneous symmetry breaking. For real O, you break a Z 2 symmetry, but the argument can be extended to complex O. Then you break a U(1) symmetry.
We first review an earlier result and then extend it. (Assume m 2 = ‐2, so Δ ‐ =1, Δ + = 2.) Theorem (Amsel, Hertog, Hollands, Marolf, 2007) : If V( ϕ ) admits a suitable superpoten/al, and W(α) is bounded from below, then the total energy is bounded from below. Outline of proof: Let P( ϕ ) sa/sfy � 2 � dP − 3 P 2 V ( φ ) = 2 d φ Near ϕ = 0, a solu/on is P( ϕ ) = 1 + ϕ 2 /4 + O( ϕ 4 )
Following Wiaen and Townsend, let ∇ µ Ψ = ∇ µ Ψ + 1 ˆ 2 P ( φ ) Γ µ Ψ Given a spacelike surface Σ with boundary C, let Ψ Γ i ˆ be a solu/on to Wiaen's equa/on: ∇ i Ψ = 0 such that approaches − ¯ ∂ / ∂ t ΨΓ µ Ψ asympto/cally. B µ ν = ¯ ΨΓ [ µ Γ ν Γ ρ ] ˆ Let (Nester) ∇ ρ Ψ + h.c. � Then the spinor charge ∗ B Q = C sa/sfies Q ≥ 0.
In asympto/cally flat space/me, Q is the total energy E. But in AdS, with general boundary condi/ons: � � 1 � � 2 r α 2 − 2 r 3 ( P − 1) E = Q + [ W ( α ) + αβ ] + lim r →∞ Using the solu/on for P and the asympto/c form of ϕ � E = Q + W So E ≥ 4 π inf W
One can prove an even stronger posi/ve energy theorem (Faulkner, Roberts, G.H., 2010) : The equa/on for the superpoten/al: � 2 � dP − 3 P 2 V ( φ ) = 2 d φ admits a one parameter family of solu/ons for small ϕ (also no/ced by Papadimitriou, 2007): P s ( φ ) = 1 + 1 4 φ 2 − s 6 | φ | 3 + O ( φ 4 ) Repea/ng the above argument with this P( ϕ ) yields
� � W ( α ) + s 3 | α | 3 � E ≥ So the energy remains bounded from below even for, e.g., W = (κ/2)α 2 with κ < 0, corresponding to double trace deforma/ons with nega/ve coefficient! Of course, this assumes that solu/ons P s ( ϕ ) exist for all ϕ . This depends on V( ϕ ), but typically they do up to a cri/cal value s c . Thus � � W ( α ) + 1 3 s c | α | 3 E ≥ 4 π inf
Existence of superpoten/als The equa/on for P can be wriaen: � 3 P 2 + V ( φ ) P ′ ( φ ) = 2 2 Clearly, a solu/on fails to exist when the argument of the square root becomes nega/ve. Ini/ally, P ′ ( φ ) = 1 2[ φ − s φ 2 ]
V ′ = [4 P ′′ − 6 P ] P ′ Since one expects V’ = 0 if P’ = 0, but this is usually not the case. Instead, P ′ ∝ ± ( φ 1 − φ ) 1 / 2 The two branches of solu/ons meet at ϕ = ϕ 1 and P does not exist for ϕ > ϕ 1 . If V’ = 0 when P’ = 0, then P ′ ∝ ( φ 1 − φ ) and the solu/on exists for ϕ > ϕ 1 .
S c = 0 S c < .52 V �� 3 �Φ 2 1.0 S c = .52 This is the 0.5 solu/on we want. P � 0.0 S c > .52 � 0.5 � 1.0 0 1 2 3 4 5 Φ For a purely quadra/c poten/al with m 2 = ‐2 , s c = .52
V � 5 � 2 � 6 Cosh � Φ � 2 � � Cosh � 2 Φ �� 2 1.0 0.5 P � 0.0 � 0.5 � 1.0 0 1 2 3 4 Φ For the consistent trunca/on of supergravity used by Gauntlea, Sonner and Wiseman ( 2009) s c = .56
Candidate ground states Expect the ground state to be sta/c and spherically symmetric. Look for solu/ons of the form ds 2 = − f ( r ) dt 2 + dr 2 g ( r ) + r 2 d Ω , φ = φ ( r ) , The equa/ons of mo/on give three ODE’s for f(r), g (r), and ϕ (r). Solu/ons are solitons.
The general asympto/c solu/on is α /r + β /r 2 + . . . φ ( r ) = r 2 + (1 + α 2 / 2) − M 0 /r + . . . g ( r ) = r 2 + 1 − ( M 0 + 4 αβ / 3) /r + . . . f ( r ) = There are three undetermined parameters: α, β, M 0
Regularity at the origin requires φ 0 + V ′ ( φ 0 ) r 2 + . . . = φ 6 1 − V ( φ 0 ) r 2 + . . . = g 3 V ( φ 0 ) r 2 + . . . = f f 0 − f 0 3 But f 0 is fixed by requiring f = r 2 + … asympto/cally. So the only free parameter is ϕ 0 . This one parameter family of solitons define a curve β 0 (α).
For small α, one can determine the curve β 0 (α) by solving the linearized equa/on for ϕ in global AdS. The solu/on is ϕ = tan ‐1 (r), so for small α: β 0 ( α ) = − 2 π α For large α, one can show β 0 (α) = ‐s c α 2 These solu/ons are related to planar solu/ons in which there is a scaling symmetry r ‐> λr. Thus + β / λ 2 φ = α / λ r r 2
0 � 2 � 4 � 6 Β 0 � 8 � 10 � 12 0 1 2 3 4 5 Α The values of (α,β) realized by solitons. Blue line is for V = ‐3 – φ 2 Dashed line is for the poten/al in Gauntlea et al.: √ √ V = 5 / 2 − 6 cosh( φ / 2) + cosh( 2 φ ) / 2
Another poten/al coming from a consistent trunca/on of supergravity is (Duff and Liu, 1999) V = − cosh( a φ ) − 2 cosh( ab φ ) with a = [2/(1 + 2b 2 )] 1/2 . Below are soliton curves for b = 1, .5, .25, .1, 0 (dashed). Note: s c = 0 for b = 0. 0.0 � 0.5 � 1.0 Β 0 � 1.5 Found � 2.0 analy/cally by � 2.5 Papadimitriou � 3.0 0 1 2 3 4 5 6 Α
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