§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Comments on the Takahashi-Tanimoto tachyonvacuum solution
N. Ishibashi
University of Tsukuba

July 29, 2014

1 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Classical solutions of the cubic SFT

QΨcl + Ψ2cl = 0Tachyon vacuum solution (Schnabl, Okawa, Erler,Erler-Schnabl, ...)
Marginal deformation (Kiermaier-Okawa-Rastelli-Zwiebach,Schnabl, Fuchs-Kroyter-Potting, ...)
Relevant deformation (Bonora-Maccaferri-Tolla, ...)···
“Any background” solution (Erler-Maccaferri)2 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Takahashi-Tanimoto (TT) solutions
tachyon vacuum solution∫))dξ (( hΨTT =e − 1 jB (ξ) − (∂h)2 eh c (ξ) I
CL 2πi3jB = cT m + bc∂c + ∂ 2 c2()211eh(ξ) = −ξ−4ξ

3 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Identity-based solutions

Ψcl = OI
I: identity string field

Impossible to calculate observables

4 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

SFT around the identity-based solutions
Ψ → Ψcl + ΨS′]
∫ [111= − 2
ΨQΨ + ΨΨcl Ψ + ΨΨΨg23]∫ [111= − 2ΨQ′ Ψ + ΨΨΨ
g23Q′ Ψ = QΨ + {Ψcl , Ψ}∗In the case of identity-based solutions, Q′ can be expressed byusing local fields on the worldsheet. For TT solution
Q′ =I)dξ ( he jB (ξ) − (∂h)2 eh c (ξ)
2πi
5 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Evidences
There are many evidences for the claim that ΨTT describes
tachyon vacuum:
No physical open string excitation around the backgroundΨTT (Kishimoto-Takahashi, Inatomi-Kishimoto-Takahashi)
Open string amplitudes vanish (Takahashi-Zeze)
Existence of an unstable solution around the background ΨTT(Takahashi, Kishimoto-Takahashi)

6 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

In this talkI would like to add one more to the list of these evidences.
I will consider the Erler-Schnabl solution in the SFT aroundthe TT solution.
I will calculate the observables of the solution and the results
indicate that the TT solution corresponds to the tachyon
vacuum.

I will study the SFT around the TT solution and discuss how
we should calculate various quantities.
c.f. Takahashi’s talk7 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Outline1Erler-Schnabl solution

2Observables

3SFT around the TT solution

4Conclusions and discussions

8 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Erler-Schnabl solution
Tachyon vacuum solution
1(c + Q (Bc))
1+K∫ i∞dz
B =b (z) I−i∞ 2πic = c (0) I

ΨES =K = QB
All the conditions in Sen’s conjectures are checked1
homotopy operator A = B 1+K, s.t. QA = 1VE [ΨES ] = − 2π2
9 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Erler-Schnabl solution around the TT solution
One can construct ES solution in the SFT around the TT solution
1S =− 2g′

∫ []11′
ΨQ Ψ + ΨΨΨ
23)
1 (′c+Q(Bc)
1 + K′= Q′ BΨ′ES =K′= K + {ΨTT , B}
1
With the homotopy operator A′ = B 1+K′ , this solution willcorrespond to the tachyon vacuum.
10 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Remark1
Here we assume that 1+K′ is well-defined with the definition

11 + K′

1
1 + K + {B, ΨTT }()n∞∑11n=(−1){B, ΨTT }
1+K
1+K

=n=0

11 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Observables
We will calculate the observables of the Ψ′ES[][]
E Ψ′ES = −S Ψ′ES
= EΨ′ES − ETT
TrV Ψ′ES = ⟨V c⟩Ψ′ − ⟨V c⟩TT
ESand show that they vanish.
Assuming Ψ′ES corrsponds to the tachyon vacuum, this implies
that the TT solution also coresponds to the tachyon vacuum.

12 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

RemarkRecently Maccaferri gives a way to construct a regular solution
out of an identity-based solution by a gauge transformation
ΨTT→ΨM = U QU −1 + U ΨTT U −1
U =1+B1
ΨTT
1+K

The observables become
E [ΨM ]=[]
E [ΨES ] − E Ψ′ES

TrV ΨM=TrV ΨES − TrV Ψ′ES

What we will show (E [Ψ′ES ] = TrV Ψ′ES = 0) implies that ΨM
is a tachyon vacuum solution.
c.f. Takahashi’s talk
13 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions§2 Observables
FromΨ′ES =)1 (c + Q′ (Bc)
′1+K

one can deriveE[Ψ′ES]

TrV Ψ′ES[]
1
11′= − TrcQc
6
1 + K′ 1 + K′[]
1c= TrV
1 + K′

We would like to show that the RHS vanish.

14 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Proof
One can show[]
1TrVc = 0
1 + K′[]
11′
Trc
Qc = 01 + K′ 1 + K′)(by using Q′ π12 b = 1, Q′ c = 0.
[]
[()]
1111′
TrVc = TrV √Qb √c
1 + K′π2
1 + K′
1 + K′= 015 / 34

§1 Erler-Schnabl solution

Q′§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions(1 )′π 2 b = 1, Q c = 0
Treating them more rigorously, these should be expressed as()1−ϵK ′eQb e−ϵK = e−2ϵKπ2e−ϵK Q′ ce−ϵK = 0
With the e−ϵK ’s, we have worldsheet with no operator insertions
and16 / 34

§1 Erler-Schnabl solution

e−ϵK Q′Q′(§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions( 1 ) −ϵK
= e−2ϵKπ2 b e)()
Idz
sin2 πz
14π 21b=− 2 jB (z) +b (0)c (z)4 πz2π22πicosπzcos
π0= 1e−ϵK (Q′ c) e−ϵK = 0 can be proven in the same way.
17 / 34

§1 Erler-Schnabl solution

TrV[11+K ′ c[TrV√

11 + K′
e−tK ′]§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions=0][)](1111′√√
c = TrVQbc
1 + K′π2
1 + K′
1 + K′
= 0

==

Γ1(1)

∫ ∞

2∞
∑

1

dtt− 2 e−t e−tK(−1)n∫ ∞0n=0×δ′0(n+1∑dt1 · · ·)∫ ∞dtn+10ti − t e−t1 K {B, ΨTT } e−t2 K · · · {B, ΨTT }e−tn+1 K

i=1Q′(1 )b = 1, Q′ (c) = 0 can be used safely.
π218 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

RemarkActually, since Q′ c = 0, Ψ′ES becomes identity-based
Ψ′ES =)
1 (c + Q′ (Bc) = c′1+K

One can avoid this by replacingc → cy = c (iy) I, (y ̸= 0)[TrV

11+K ′ cy][]
11′, Tr 1+K
are independent of y,′ cy 1+K ′ Q cy

and we get the same answers for the observables
19 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions§3 SFT around the TT solution
We have shown
E[Ψ′ES]

TrV Ψ′ES[]111′= − Trc
Qc =06
1 + K′ 1 + K′[]
1= TrVc =0
1 + K′

using the definition
e−tK′

≡

∞
∑

(−1)n∫ ∞0n=0×δ

(n+1∑dt1 · · ·)∫ ∞dtn+10ti − t e−t1 K {B, ΨTT } e−t2 K · · · {B, ΨTT } e−tn+1 K

i=120 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

SFT around the TT solution

1S =− 2g′

∫ []11′
ΨQ Ψ + ΨΨΨ23Since Q′ is given by a local operator, we should be able to show
[]
[ ′ ]
111′c
Qc =0E ΨES = − Tr
6
1 + K′ 1 + K′[]
1′c =0
TrV ΨES = TrV
1 + K′by dealing with the operator K ′ = Q′ B more directly.
We find that doing so is a little bit nontrivial.
21 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Similarity transformation
Since K ′ itself is still diﬃcult to deal with, we use the relation
discovered by Kishimoto-Takahshi
1Q′ = − U QU −14U =e−q(λ) U2q (λ) = 2∞∑
1n=1

jgh (ξ) =

∑

nq−2n

ξ −m−1 qmm
22 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

U2 :bc-shift operator

U2 cn U2−1 = cn+2
U2 bn U2−1 = bn−2U2 ϕm U2−1 = ϕmU2 |0⟩ = b−3 b−2 |0⟩⟨0| U2−1 = ⟨0| c−1 c0
U2 is of ghost number −223 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Useful relations

Q′=

1− U QU −1
4U b (ξ) U −1

=

U |0⟩

=

U −1 |0⟩

=

⟨0| U

=

( 2)2
ξ −1c (ξ)
ξ22
ξ
b (ξ)
2(ξ − 1)21
∂bb (1) ∂bb (−1) c0 c1 |0⟩161∂cc (1) ∂cc (−1) b−3 b−2 |0⟩16⟨0| b2 b3⟨0| U −1=⟨0| c−1 c0

U c (ξ) U−1=24 / 34§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Useful relationsQ′

1= − U QU −1
4U |I⟩

1∂bb (1) |I⟩32= 2∂cc (1) |I⟩

U −1 |I⟩=⟨I| U= 0⟨I| U −1

= 0

Using the similarity transformation and these relations, it should bepossible to calculate various quantities.
25 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Q′ cohomology
Kishimoto-Takahashi
1Q′ = − U QU −1
4the representative state of the cohomology of Q′
U cV (0) |0⟩ : gh# = −1
U ∂ccV (0) |0⟩ : gh# = 0
In conflict with the homotopy operator A = π12 b ?
{ ′}Q , b (1) = 1

26 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

U cV (0) |0⟩ , U ∂ccV (0) |0⟩
These are outside of the Fock space(Inatomi-Kishimoto-Takahashi)
U cV (0) |0⟩
1= ∂bb (1) ∂bb (−1) ∂ 2 c∂ccV (0) |0⟩32
AU cV (0) |0⟩ = b (1) U cV (0) |0⟩ = 0{Q′ , b (1)} = 1 is not correct with ∂bb (1).
Having some worldsheet with no operator insertions is crucial
for Q′ A = 1.
27 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Observables[][]
111′E Ψ′ES = − Trc
Qc=06
1 + K′ 1 + K′[]1′
TrV ΨES = TrVc =0
1 + K′()are derived from e−ϵK Q′ π12 b e−ϵK = e−2ϵK , e−ϵK Q′ ce−ϵK = 0.
Let us see if we can show)(1′−ϵK ′ ′b e−ϵKeQπ2′e−ϵK Q′ ce−ϵK′= e−2ϵK′= 0instead.
28 / 34

§1 Erler-Schnabl solution′e−ϵK Q′

§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions( 1 ) −ϵK ′′= e−2ϵKπ2 b e()()′′′e−ϵK |I⟩ ∗ Q′ (b (1) |I⟩) ∗ e−ϵK |I⟩ = e−2ϵK |I⟩

Inserting Q′ = − 41 U QU −1 , the left hand side becomes))(1(′′− e−ϵK |I⟩ ∗ U QU −1 (b (1) |I⟩) ∗ e−ϵK |I⟩4)
1 (′′= − U e−ϵK̃ |I⟩ ∗ Q (2c (1) |I⟩) ∗ e−ϵK̃ |I⟩4)
1 (′′→ − U e−ϵK̃ Q (2πc) e−ϵK̃4where′K̃ = −∫ i∞dzT (z) tan2 πz−i∞ 2πi29 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions∫ i∞ dzK̃ ′ = − −i∞ 2πi
T (z) tan2 πz
We can use the method of Kiermaier-Sen-Zwiebach to show thatK̃ ′ does not move the points on the boundary.

′We cannot prove e−ϵK Q′( 1 ) −ϵK ′′b e= e−2ϵK .
π230 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Regularization
The homotopy operator seems to be crucial for ΨTT to be atachyon vacuum solution.
The surface should be defined as a limit of regular surfaces
anyway.
We propose a regularization such that the homotopy operator
becomes well-defined.

31 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Regularization
Replace K̃ ′ byK̃δ′ =∫ i∞dz− sin2 πz + δT (z)cos2 πz−i∞ 2πiand take the limit δ → 032 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions

Homotopy operator
With this prescription,()
1′−ϵK ′ ′e
Qb e−ϵKπ2)
1 (′′− U e−ϵK̃ Q (2πc) e−ϵK̃
4()
1−ϵK̃δ′−ϵK̃δ′→ − U lim e
Q (2πc) e
δ→04()π′′
U lim e−ϵK̃δ ∂cce−ϵK̃δ=δ→0
4==′e−2ϵK′′

One also has e−ϵK Q′ ce−ϵK = 0 and we can derive
[TrV]
[]
111′c
=TrcQc=01 + K′
1 + K′ 1 + K′
33 / 34

§1 Erler-Schnabl solution§2 Observables§3 SFT around the TT solution§4 Conclusions and discussions§4 Conclusions and discussions
We have calculated the observables of the Erler-Schnabl
solution around the TT solution. The results imply that TT
solution corresponds to the tachyon vacuum.
We explain how to deal with kinetic operator of the SFTaround the TT solution.
We will be able to calculate various quantities from the SFT
around the TT solution. We may be able to see its relation tothe VSFT. (Drukker, Drukker-Okawa)

34 / 34