Some Properties of Classical Solutions in CSFT Isao Kishimoto (Univ. of Tokyo, Hongo) I.K.-K.Ohmori, hep-th/69 I.K.-T.Takahashi, hep-th/5nnn /5/9 seminar@komaba
Introduction Sen s conjecture There is a tachyon vacuum in bosonic open string theory. D5-brane vanishes on it and there is no open string excitation. open string (D5-brane) T 5 U Ψ Ψ Cubic String Field Theory (CSFT) [Witten(986)] Here we consider CSFT to investigate Sen s conjecture. S =, Q, g Ψ Ψ + Ψ Ψ Ψ 3 /5/9 seminar@komaba
Sen s conjecture can be rephrased in CSFT : There is a solution of equation of motion: Q Ψ+Ψ Ψ= Ψ The potential height equals to D5-brane tension: Ψ 6 5 The cohomology of new RST operator around it is trivial: Q' ψ = ψ = Q' φ, φ.. S / V = T. Q ' There are some numerical evidences, but an exact solution is necessary to prove them. /5/9 seminar@komaba 3
Another approach: Vacuum String Field Theory (VSFT) [(G)RSZ()] (Gaiotto-)Rastelli-Sen-Zwiebach proposed SFT around tachyon vacuum: S, V = κ Ψ QGRSZΨ + Ψ, Ψ Ψ, 3 ( ) Q = ci () c( i) = c + ( ) ( c + c ). n GRSZ n n i n Q GRSZ This has trivial cohomology. There are some solutions of equation of motion which are constructed by projectors with respect to. Can VSFT reconstruct CSFT? Recently, Okawa proved D5-brane tension can be reproduced. /5/9 seminar@komaba 4
Contents Introduction Some Solutions in CSFT Potential Height Cohomology of new RST operator Summary and Discussion /5/9 seminar@komaba 5
Some Solutions in CSFT Horowitz et al.(988) discussed rather formal solutions in the context of purely CSFT. Ψ = Q I + C( f ) I, f( π σ) = f( σ), f( π /) =. In particular it can be used to derive VSFT action from CSFT: f GRSZ ε ( ( ) ( )) ( σ) = lim δ σ ( π / ε) + δ ( π / + ε) σ. The corresponding solution is singular: /3 VSFT ( g κ ) iε iε iε iε Ψ = Q I + lim ( e c( ie ) e c( ie )) I ε 4i m /3 ( ) ( g κ ) = Q (m+ ) I + lim + cos kε c sin(k + ) ε ( c c ) I π m m ε + k k /5/9 seminar@komaba 6
There are some subtleties about identity string field, but here we treat I as the identity with respect to : A I = I A= A, Q I A= I Q A= Q A,... R Notation: - w i C C R - i w dw C( f) = f( w) c( w), C ( f) = f( w) c( w), πi C πi dw Q( f) = f( w) j ( w), Q ( f) = f( w) j ( w), RST πi C πi dw gh πi C πi gh RST q( f) = f( w) j ( w), q ( f) = f( w) j ( w),... dw dw dw i i RST 3 j ( w) = ct + : bc c : + = Q w, j ( w) = : bc : = q w,... X n n c n gh n n n /5/9 seminar@komaba 7
Takahashi and Tanimoto proposed a new regular solution of CSFT which might represent tachyon vacuum.() First, they found solutions of CSFT: ( h ) ( h ) ) Ψ = Q e I C ( h e I, where hw () is some function such that n n ( ) ( ) hw ( ) = h w + ( w ), h( ± i) =, h ± i =. n ut these solutions can be rewritten as pure gauge ones at least formally. Ψ = e Q e q( h) I q( h) I In fact, new RST operator is formally rewritten as Q = Q e C h e h h ' ( ) (( ) ) q( h) q( h) ' =. Q e Q e However they might become nontrivial solutions at some limits. /5/9 seminar@komaba 8
Example Noting ( ) log a ha w = + w+ w ( ) + + a + a qh = q Za + q h + q h ( ) ( ) ( a) log( ( )) ( a) ( a), Z( a) =, and ( + ) ( ) q ( ha), q ( ha) = log( Z( a) ), ( ± a ) = ( ) ( ) a ( ) ( + ) a a exp qh ( ) Za ( ) exp qlog( Za ( )) e e is well-defined for a > /. ± q ( h ) ± q ( h ) ecause Za= ( /) =, a = / case: ha /( w) log ( w w = = ) 4 might be nontrivial. /5/9 seminar@komaba 9
More Examples ( k ), h w ( k h w ) log( a) a Noting Za ( = /) =, Za ( = ) =, in the same sense, we have nontrivial limit at a respectively: l ( ) () l ( ) log ( l ( ) l h w = w w. 4 There is well-defined oscillator representation: m /5/9 seminar@komaba a ( ) ( ) l + l = = m ( ) 8l = ( Q (m+ ) + 4l c (m+ ) ) I (m+ )(m+ l)(m+ + l) l γ log ψ ψ + + + + ( c c ) I. π /, a. l () l ( ) ( ( )) l l l l l Ψ = Q + ( w + w ) I + C l w ( ) ( w + w ) I 4
Potential Height When we naively compute the value of the action at the solutions constructed on identity string field: ( π ) ( π ) n ( ) n I = b b exp α n α n + ( ) c nb n c c, 4i n n we encounter following divergence: ( ) 6/ + nm, nm, ( ) ( ) I I det δ δ =. nm, It is necessary to regularize I appropriately. /5/9 seminar@komaba
On the other hand, potential height is zero if we use e.o.m. naively in the action: Q QI QI =, C ( f) I C ( f) I =, Ψ +Ψ Ψ =, S Ψ VSFT = Ψ, Q Ψ = Ψ, Ψ Ψ =. 6g 6g Takahashi-Tanimoto solutions give also zero naively, because they are pure gauge formally. There is a possibility of = finite (!?) /5/9 seminar@komaba
Pacman regularization It is technically easy to use pacman regularization in the computation using PP+GGRT. [I.K.-K.Ohmori] regularize δ ut we could not get the definite value at δ =, for example, iε iε iε iε iε iε iε iε ( ( ) ( )) δ, ( ( ) ( )) e c ie e c ie I Q e c ie e c ie I i i sin ε tan tan 3 V6. ε δ ε δ = + + δ /5/9 seminar@komaba 3 δ
Regularization in oscillator language exp( π t ) If we multiply damping factor we get a formula for oscillator modes: on I, I q cmc m c m q I 3 = ( ) m+ m+ m3 4 i q f q V sinh( mπt)sinh( mπt)sinh( mπt) 6 3 m m m m ( q) q q ( q) det, m m m m ( q) q q ( q) m3 m3 m3 m 3 ( q) q q ( q) π t n /3 / /3 q: = e, f( q ) = ( q ) = ( π ) q ϑ it n= '(, ). Using some relations ( n ) j + Q I =, { Q, Q } =, { Q c } = c c,, n, n j n+ j j= we have the following expression: /5/9 seminar@komaba 4
Ψ () 6 = lim q ( + ) ( + ) t + 6 S / V = gv6 π / π / lim dx dy cos xcos y t + 6 g π / π / '(, ) ( ) ( ) I C w w w QC w w w I ( ϑ it ) ϑ4'( x / + π / iπt /4, it ) ϑ3 ' ϑ' ϑ' ( ) ( ) ( ) ϑ4( x / + π / iπt /4, it) ϑ3 ϑ ϑ det ϑ4 '( y /+ π / iπ t /4, it ) ϑ3 ' ϑ' ϑ'. ( ) ( ) ( ) ϑ4( y / + π / iπt /4, it) ϑ3 ϑ ϑ ϑ4'( y / + π / iπt /4, it) ϑ3 ' ϑ' ϑ' y y ( ) y ( ) y ( ) ϑ4( y / + π / iπt /4, it) ϑ3 ϑ ϑ We do not have definite (or exact ) value of the potential height for our solutions yet. Numerically, it tends to be divergent.(?) /5/9 seminar@komaba 5 iπ 4 q
Cohomology of new RST operator Kinetic term around a solution S Ψ, ',, + ψ = ψ Q ψ ψ ψ ψ S Ψ g + + 3 Q ' ψ = Q ψ +Ψ ψ ( ) ψ Ψ. ψ The new RST operator around our solution () Ψ Q ' = Q Q Q + c + c + c = R + R + R. 4 4 R± : = Q± + c±, R : = Q + c satisfy following relations: 4 R =, R R + R R =, R + R R + R R =. ± ± ± is /5/9 seminar@komaba 6
Q ' cohomology is trivial in ghost number states. Proof We consider the equation Q ' ψ = for the ghost number state ψ = ψ. N N h In other words, we solve the following equations: Rψ h k=, R ψ h k = R ψ h k, k =,, R = R R ψ, h k l ψ h k ( l ) ψ h k ( l ) l. /5/9 seminar@komaba 7
R is rewritten as R = Q + c = Q. 4 4 Q is given by replacing cb with c : = c, b : = b in =. n, n n n+ n n Q Q Solutions of ψ Qψ = are = ADDF, p + DDFc, p + Q [Kato-Ogawa,M.Henneaux, ] φ where ipx, p = e Ω, i p α ' p ± =±, =, DDF Ω : = c, c Ω =, n, b Ω =, n, DDF n i and A, are generated by DDF operators A n : i j i, j X i Am, A n = mδm+ n, δ, m, A n =. n /5/9 seminar@komaba 8
Similarly, solutions of Q ψ = are given by ψ = A, p + c, p + Q DDF DDF φ, p : = b b, p, c, p =, n, b, p =, n. n n Here h k is ghost number, R ψ ψ = ψ = R φ, φ. h k h k h k h k Then R ψ = R ψ = RR φ = RR φ, h k h k h k h k ψ = R φ + R φ, φ. h k h k h k 4 h k 4 /5/9 seminar@komaba 9
Suppose, ψ = R φ ( ) + R φ + R φ ( ),, then h k l h k l h k l h k l+ l = m m R ψ = R ψ R ψ h k ( m+ ) h k m h k ( m ) h k ( m+ ) ( ) φ h k m φ h k m φ h k m+ = R R + R + R ( ) ( ) ( φ ) h k m φ h k m φ h k m R R + R + R ( ) ( ) ( ) φ h k m φ h k m+ = R R + R ψ ( ) = R φ + R φ + R φ, φ h k m h k ( m+ ) h k ( m+ ) h k ( m+ ) y induction, we conclude ψ = R φ ( ) + R φ + R φ ( ), l, h k l h k l h k l h k l+ namely, ψ = Q φ, = φ. ' φ h k l k=,, l /5/9 seminar@komaba,
We have similar arguments for other solution by using R instead of R. l New RST charge around it is where Q = Q + l c ( ) ( R + R ), () l l l l ( l ) ± l = ± l + ± l = R Q c Q ± 4 4. Ψ () l () l is given by replacing with in. Q cn, b n c : = c, b : = b n n+ l n n l Q We can prove () l Q cohomology is trivial in ghost number states. /5/9 seminar@komaba
There are nontrivial solutions for of the form: ψ where ( l q f ) = exp ( ) () l () ( A b b b b p b b b p ), +,, DDF l l+ DDF l+ q f () l ( ) in other ghost number sector. This follows from the identity: ( ( ) ) ( ) ( ( ) q f Q q f ) l l l l exp ( ) exp ( ) = ( ) R. ( ) = n n n( + l) q ln Q ψ = () l () l l /5/9 seminar@komaba
Summary and Discussion We have investigated whether some solutions of CSFT using identity string field can be tachyon vacuum or not. Evaluation of potential height at the solutions expressed by identity string field is rather difficult by our two regularization methods. It is divergent at least naively. The new RST charge cohomology around Takahashi- Tanimoto limit solutions is trivial in ghost number states. This suggests they are nontrivial solutions of CSFT although we do not know yet that they are the tachyon vacuum in Sen s conjecture. /5/9 seminar@komaba 3
In the context of Takahashi-Tanimoto solutions, the solution which plugs CSFT into VSFT directly is very singular limit (?): ( h ) ( h) Ψ = Q e I C ( h) e I hw ( ) There is subtlety about QI Ψ = Q I + C( f) I. y CFT calculation, we have φ, QI QI = φ, QI =, but by oscillator calculation, we have : ( + ζ ) ζ () (). (?) Q I = Q c I = π /5/9 seminar@komaba 4
Is the identity string field well-defined? Mystery on [Rastelli-Zwiebach()] ( ) ( ) ( ) ( ) ( ci) A=, A. ca= c I A = ci A+ I ca = ci A+ ca, If we take ci A= I, ( ) ci = ci I = (??) Is there another exact solution which do not use identity string field in CSFT? /5/9 seminar@komaba 5
Solutions in CSFT CSFT Q Q () l VSFT QGRSZ pure gauge Ψ ( l ) singular VSFT Ψ Where is the tachyon vacuum solution? numerical solution in the Siegel gauge /5/9 seminar@komaba 6