Energy from the gauge invariant observables

Transcription

1 .... Energy from the gauge invariant observables Nobuyuki Ishibashi University of Tsukuba October / 31

2 Ÿ1 Introduction A great variety of analytic classical solutions of Witten type OSFT has been discovered, especially since the discovery of Schnabl's tachyon vacuum solution. Once a solution is found, there are two important gauge invariant quantities to be calculated. energy the gauge invariant observables 2 / 31

3 Energy For a static solution Ψ E S = 1 [ 1 g 2 2 Ψ Q Ψ + 1 ] 3 Ψ Ψ Ψ = E Ψ E 0 E Ψ : "energy" for the vacuum corresponding to Ψ E 0 : "energy" for the perturbative vacuum 3 / 31

4 the gauge invariant observables With an on-shell closed string vertex operator V = c cv m, one can construct I V (i) Ψ Hashimoto-Itzhaki, Gaiotto-Rastelli-Sen-Zwiebach I V (i) Ψ corresponds to the dierence of one point functions 4πi I V (i) Ψ = V c 0 B Ψ V c 0 B 0 Ellwood, Kiermaier-Okawa-Zwiebach 4 / 31

5 Energy and the gauge invariant observables Usually it is more dicult to calculate E = 1 [ 1 g 2 2 Ψ Q Ψ Ψ Ψ Ψ ] compared to the gauge invariant observable I V (i) Ψ. For V c c X 0 X 0, we expect that the gauge invariant observable is proportional to E. It will be useful to prove that the gauge invariant observable for such V yields E. 5 / 31

6 Energy from the gauge invariant observables We would like to show E = 1 I V (i) Ψ g2 for V = 2 πi c c X0 X 0 assuming that Ψ satises the equation of motion some regularity conditions Takayuki Baba and N. I. arxiv: / 31

7 Outline..1 Introduction.2. A proof of E = 1 I V (i) Ψ g 2.3. Solutions with K, B..4 Conclusions 7 / 31

8 Ÿ2 A proof of E = 1 g 2 I V (i) Ψ String eld Ψ = O Ψ (0) 0 Let us assume that O Ψ is expressed in terms of really local operators located away from the arc ξ = 1. 8 / 31

9 A proof of E = 1 g 2 I V (i) Ψ In order to prove E = 1 g 2 I V (i) Ψ, we consider for h 1 S h [ Ψ ] 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ + h I V (i) Ψ ] V = 2 πi c c X0 X 0 is a linear combination of graviton and dilaton vertex operators. c.f. Belopolsky-Zwiebach S h should describe the string eld theory in a constant metric and dilaton background. Zwiebach The constant metric can be gauged away and the eect of the constant dilaton is reduced to a rescaling of g. 9 / 31

10 Soft dilaton theorem. A "Soft dilaton theorem".. S h can be shown to be equivalent to the original SFT action with a. rescaling of g... S h [ Ψ ] = 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ + h I V (i) Ψ ] = 1 + h g 2 [ 1 Ψ Q Ψ + 1 Ψ Ψ Ψ ] + O ( h 2) 2 3 Ψ = Ψ + O (h) 10 / 31

11 E = 1 g 2 I V (i) Ψ from the soft dilaton theorem 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ + h I V (i) Ψ ] = 1 + h g 2 [ 1 Ψ Q Ψ + 1 Ψ Ψ Ψ ] + O ( h 2), 2 3 Substituting a classical solution Ψ cl into it E h g 2 I V (i) Ψ cl = (1 + h) E + O ( h 2) and comparing the O (h) terms E = 1 g 2 I V (i) Ψ cl 11 / 31

12 Soft dilaton theorem The soft dilaton theorem is proved in two steps. There exists χ such that V (i) = {Q, χ} Using this fact, we obtain S h [ Ψ ] = 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ + h I V (i) Ψ ] = 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ ] + O ( h 2) Ψ Ψ + hχ I Q Q h (χ χ ) 12 / 31

13 V (i) = {Q, χ} [ χ lim δ 0 P 1 dξ 2πi j ( ξ, ξ ) P 1 d ξ 2πi j ( ξ, ξ ) ( κ e iδ, e iδ)], j ( ξ, ξ ) 4 X 0 (ξ) c X 0 ( ξ), j ( ξ, ξ ) 4 X 0 ( ξ) c X 0 (ξ), κ ( ξ, ξ ) 1 ( X 0 ( ξ, πi ξ ) X 0 (i, i) ) ( c X 0 (ξ) c X 0 ( ξ)). 13 / 31

14 Soft dilaton theorem There exists G such that [Q, G] = χ χ GΨ 1 Ψ 2 + Ψ 1 GΨ 2 = Ψ 1 Ψ 2 GΨ 1 Ψ 2 Ψ 3 + Ψ 1 GΨ 2 Ψ 3 + Ψ 1 Ψ 2 GΨ 3 = Ψ 1 Ψ 2 Ψ 3 Using G, we eventually obtain S h [ Ψ ] 1 [ 1 Ψ ( ( g 2 Q h χ χ )) Ψ + 1 Ψ Ψ Ψ ] h g 2 [ 1 Ψ Q Ψ + 1 Ψ Ψ Ψ ] 2 3 Ψ (1 hg) Ψ Myers-Penati-Pernici-Strominger 14 / 31

15 [Q, G] = χ χ [ G lim δ 0 P 1 +P 2 dξ 2πi g ( d ξ ( ] ξ ξ, ξ) P1+ P2 2πi g ξ ξ, ξ) ( ( g ξ ξ, ξ) 2 X 0 ( ξ, ξ ) X 0 (i, i) ) X 0 (ξ) ( ( ξ, ξ) 2 X 0 ( ξ, ξ ) X 0 (i, i) ) 0 X ( ξ) g ξ 15 / 31

16 Remarks The proof is valid for O Ψ which does not aect the denition and the manipulations of χ, G. One can obtain the same relation for with h µ µ = 1. V = 1 πi c c Xµ X ν h µν For actual applications, it is desirable to nd a way to derive E = 1 I V (i) Ψ g 2 cl more directly using the properties of G, χ and the eom. 16 / 31

17 A more direct proof of E = 1 g 2 I V (i) Ψ G satises the following identities: GΨ Ψ Ψ = 1 [ GΨ Ψ Ψ + Ψ GΨ Ψ + Ψ Ψ GΨ ] 3 = 1 Ψ Ψ Ψ, 3 GΨ Q Ψ = 1 2 Ψ Q Ψ + 1 Ψ [Q, G] Ψ. 2 From these, we get E = 1 g 2 [ 1 2 Ψ Q Ψ Ψ Ψ Ψ ] = 1 g 2 [ GΨ (Q Ψ + Ψ Ψ ) 1 2 Ψ [Q, G] Ψ ] 17 / 31

18 A more direct proof eom implies E = 1 g 2 [ GΨ (Q Ψ + Ψ Ψ ) 1 2 Ψ [Q, G] Ψ ] E = 1 Ψ [Q, G] Ψ 2g2 = 1 ( 2g 2 Ψ χ χ ) Ψ = 1 I χ Ψ Ψ g2 = 1 I χq Ψ g2 = 1 I V (i) Ψ g2 18 / 31

19 Remarks If eom is not satised Q Ψ + Ψ Ψ = Γ 0, E = 1 g 2 [ GΨ (Q Ψ + Ψ Ψ ) 1 2 Ψ [Q, G] Ψ ] = 1 2g 2 Ψ [Q, G] Ψ + 1 g 2 GΨ Γ = 1 g 2 I χ Ψ Ψ + 1 g 2 GΨ Γ = 1 g 2 I V (i) Ψ 1 g 2 I χ Γ + 1 g 2 GΨ Γ 19 / 31

20 Ÿ3 Solutions with K, B Most of the solutions since Schnabl's one involve dξ π ( K 1 + ξ 2 ) T (ξ) 2πi 2 dξ π ( B 1 + ξ 2 ) b (ξ) 2πi 2 The denitions and the manipulations of operators G, χ are aected by the presence of K, B. 20 / 31

21 Okawa type solutions As an example of such solutions, let us consider Okawa type solutions (Okawa, Erler). K KB Ψ = F (K) c 2 cf (K) 1 F (K) B c 1 c (1) I π dξ π ( 1 + ξ 2 ) T (ξ) I 2πi 2 dξ π ( 1 + ξ 2 ) b (ξ) I 2πi 2 21 / 31

22 Okawa type solutions As an example of such solutions, let us consider Okawa type solutions (Okawa, Erler). KB Ψ = F (K) c 2 cf (K) 1 F (K) Ψ = F (K) = K 1 F 2 = 0 0 dle LK f (L) dle LK f (L) dl 1 dl 2 dl 3 e L 1K ce L 2K Bce L 3K f (L 1 ) f (L 2 ) f (L 3 ) 21 / 31

23 wedge state with insertions Ψ = dl 1 dl 2 dl 3 e L 1K ce L 2K Bce L 3K f (L 1 ) f (L 2 ) f (L 3 ) ψ (L) Ψ = 0 dle LK ψ (L) = dl 1 dl 2 dl 3 δ (L L 1 L 2 L 3 ) 0 dle LK L 1 {Ψ} (L) c (L 2 + L 3 ) Bc (L 3 ) f (L 1 ) f (L 2 ) f (L 3 ) 22 / 31

24 denition of G How should we dene G acting on wedge states? One way to do is We rather take 23 / 31

25 denition of G GΨ is dened so that for test state ϕ ϕ (0) 0 ϕ GΨ 0 dl e LK f ϕ (0) e LK G(L)ψ (L) C L+1 f (ξ) 2 π arctan ξ dz d z G (L) gz (z, z) g z (z, z) 2πi 2πi ψ (L) in the correlation function denotes the operator corresponding to the string eld ψ (L). 24 / 31

26 E = 1 g 2 I V (i) Ψ In order to obtain E = 1 g 2 I V (i) Ψ, we need GΨ 1 Ψ 2 + Ψ 1 GΨ 2 = Ψ 1 Ψ 2 GΨ 1 Ψ 2 Ψ 3 + Ψ 1 GΨ 2 Ψ 3 + Ψ 1 Ψ 2 GΨ 3 = Ψ 1 Ψ 2 Ψ 3 ( [Q, G] Ψ = χ χ ) Ψ If all these are satised, we get E = 1 g 2 I V (i) Ψ. It is straightforward to show the rst two for G dened in the previous slide. Showing the third one is not straightforward. 25 / 31

27 [Q, G] Ψ = ( χ χ ) Ψ [Q, G] Ψ = dle LK [Q, G (L)] ψ (L) + dle LK G (L) { QL 1 {Ψ} (L) L 1 {Qψ} (L) } ( ) QL 1 {Ψ} (L) L 1 {Qψ} (L) = e LK e LK α (L) δ (L) α (0) α (L) dl 1 dl 2 dl 3 δ (L L 1 L 2 L 3 ) c (L 2 + L 3) c (L 3) f (L 1) f (L 2) f (L 3) Assuming α ( ) = 0 and α (0) is nite, we are able to get [Q, G] Ψ = ( χ χ ) Ψ. 26 / 31

28 Solutions with K, B In summary, it is possible to show E = 1 I V (i) Ψ, even for the g 2 solutions with K, B provided α ( ) = 0 and α (0) is nite. α (L) dl 1 dl 2 dl 3 δ (L L 1 L 2 L 3 ) c (L 2 + L 3 ) c (L 3 ) f (L 1 ) f (L 2 ) f (L 3 ) These conditions for α (L) are concerned with the regularity of the function F (K) for K 0, K. If Q Ψ + Ψ Ψ = Γ = 0, we obtain E = 1 g 2 I V (i) Ψ 1 g 2 I χ Γ + 1 g 2 GΨ Γ 27 / 31

29 Ÿ5 Conclusions We have given a proof of E = 1 I V (i) Ψ g2 for a classical solution Ψ of Witten's SFT. The gauge invariant observables seem to have enough information to reproduce energy, boundary state, etc.. Kudrna-Maccaferri-Schnabl This identity can be applied to BMT, Murata-Schnabl solutions. It is straightforward to generalize the proof to modied cubic SSFT. T. Baba Masuda solution? Masuda-Noumi-Takahashi 28 / 31

30 Thank you 29 / 31

31 Murata-Schnabl solutions In order to calculate the gauge invariant observables Ψ Ψ ϵ F 2 K + ϵ (K + ϵ) cb 1 F 2 (K + ϵ) c The energy with this regularization can be calculated by our method: E = 1 ( ) N 1 g 2 2π 2 R N R N i N 2 8π 3 k=0 i N 1 8π 3 k=0 N! k!(k+2)!(n 2 k)! (1 N)! k!(k+2)!( N 1 k)! ( (2πi) k+2 ( 2πi) k+2), (N 1), ( (2πi) k+2 ( 2πi) k+2), (N 0). 30 / 31

32 Sliver frame, wedge states ξ = tan πz 2 dz K = 2πi T (z) 31 / 31