The Exact Solution of the Schwarzian Theory
aa r X i v : . [ h e p - t h ] O c t Exact solution of the Schwarzian theory
Vladimir V. Belokurov ∗ Lomonosov Moscow State University, Leninskie gory 1, Moscow, 119991,Russia and Institute for Nuclear Research of the Russian Academy of Sciences,60th October Anniversary Prospect 7a, Moscow, 117312, Russia
Evgeniy T. Shavgulidze † Lomonosov Moscow State University, Leninskie gory 1, Moscow, 119991, Russia
The explicit evaluation of the partition function in the Schwarzian theory is presented.
The Schwarzian theory [1] is the basic element of var-ious physical models including the SYK model and thetwo-dimensional dilaton gravity (see, e.g., [2], [3], [4], [5],and references therein). The action of the theory is I = − g π Z (cid:20) S φ ( τ ) + 12 ( φ ′ ) ( τ ) (cid:21) dτ . (1)Here, S φ ( t ) ≡ (cid:18) φ ′′ ( t ) φ ′ ( t ) (cid:19) ′ − (cid:18) φ ′′ ( t ) φ ′ ( t ) (cid:19) (2)is the Schwarzian derivative, φ ∈ Dif f ([0 , π ]) , and φ ′ (0) = φ ′ (2 π ) . It is convenient to rewrite the action in the form I = − σ Z (cid:2) S ϕ ( t ) + 2 π ˙ ϕ ( t ) (cid:3) dt , (3)where σ = √ πg , t = 12 π τ , ϕ ( t ) = 12 π φ ( τ ) ,ϕ ∈ Dif f ([0 , , ˙ ϕ (0) = ˙ ϕ (1) . The functional integral for the partition function Z ( g ) = Z ˙ ϕ (0)= ˙ ϕ (1) exp {− I } dϕ (4)diverges [1]. However, as we will see later on (eq. (17)),the integral Z α ( g )= Z Diff ([0 , exp {− I } exp − (cid:2) π − α (cid:3) σ Z ˙ ϕ ( t ) dt dϕ = Z ˙ ϕ (0)= ˙ ϕ (1) exp σ Z (cid:2) S ϕ ( t ) + 2 α ˙ ϕ ( t ) (cid:3) dt dϕ (5)converges for 0 ≤ α < π . Therefore, let us evaluate theintegral (5) first.The measure µ σ ( X ) = Z X exp σ Z S ϕ ( t ) dt dϕ (6)is quasi-invariant, and the Radon - Nikodym derivativeof the measure is [6], [7] dµ fσ dµ σ ( ϕ ) = 1 q ˙ f (0) ˙ f (1) × exp σ " ¨ f (0)˙ f (0) ˙ ϕ (0) − ¨ f (1)˙ f (1) ˙ ϕ (1) + 1 σ Z S f ( ϕ ( t )) ˙ ϕ dt , (7)where µ fσ ( X ) = µ σ ( f ◦ X ) . Here, we have used the well known property of theSchwarzian derivative: S f ◦ ϕ ( t ) = S f ( ϕ ( t )) ˙ ϕ ( t )+ S ϕ ( t ) , ( f ◦ ϕ )( t ) = f ( ϕ ( t )) . Take the function f to be f ( t ) = 12 (cid:20) α tan (cid:18) α ( t −
12 ) (cid:19) + 1 (cid:21) . (8)In this case, S f ( t ) = 2 α , ˙ f (0) = ˙ f (1) = α sin α , − ¨ f (0)˙ f (0) = ¨ f (1)˙ f (1) = 2 α tan α . (9)Now we have the following functional integrals equality: α sin α Z ˙ ϕ (0)= ˙ ϕ (1) F ( ϕ ) µ σ ( dϕ )= Z ˙ ϕ (0)= ˙ ϕ (1) F ( f ( ϕ )) exp (cid:26) − ασ tan α ϕ (0) (cid:27) × exp σ Z (cid:2) S ϕ ( t ) + 2 α ˙ ϕ ( t ) (cid:3) dt dϕ . (10)The next step is the choice of the function F .