### Progress in

### String Field Theory

Syoji Zeze

Yokote Seiryo Gakuin High School

2 April, 2010 National Taiwan U.

### • A very simple solution for tachyon condensation in string ﬁeld theory is found

[Erler-Schnabl, arXiv: 0906.0979]

### • We apply their solution to various problems of tachyon condensation

[S.Z., JPS in Okayama, 2010, to appear in arXiv]

### • Classical potential

### • Eﬀective potential

### • Interpretation from world sheet geometry

### They are exact result

## Claim

## Tachyon

## Condensation

## in SFT

## Sen s conjecture

unstable D-brane no open string

1998

*t*
*V*

tension of unstable

D-brane
t_{0}

V (t0) SFT can derive

this potential

## 10 yeas history

year brane tension

(ratio) method ^{authors}

1999 0.9864 level 4 Sen-Zwiebach, (Kostelecky-

Samuel)

2000 0.9912 level 10 Moeller-Taylor

2002 1.00049 level 18 Gaiotto-Rastelli

2005 1 exact Schnabl

2009 1 exact Erler-Schnabl

### Exact era

### Level truncation era

_{Ψ = t}

_{1}

_{+ t}

_{2}

_{+ · · ·}

### Erler-Shnabl solution

### • Analytic solution of the classical SFT

### • Lorentz invariant, i.e. scalar ﬁeld

### with no space time dependence

Schnabl Erler Schnabl form of the

solution

Inﬁnite

sum Integral

`Phantom

term exists none

D-brane

tension OK OK

Homotopy

operator exists exists

Gauge equivalent

## K B c subalgebra

### B, c, K: string ﬁelds (identity based)

### {B, c} = 1 ^{[K, B] = 0}

### Q _{B} c = cKc

### Q

_{B}

### K = 0

### Q

_{B}

### B = K

_{c =}

^{2}

π c(1) |I�

B = π

2 B_{L} |I�

K = π

2 K_{L} |I�

Okawa

### = �

_{∞}

0

### dt e

^{−t}

### c(1 + K)BcΩ

^{t}

### = �

_{∞}

0

### dt c(1 + K)Bce

^{−t(1+K)}

### Ψ = c(1 + K)Bc 1

### 1 + K

### Ω

midpoint

π 2

### Ω ^{t}

π 2 t

## Level expansion in

## dressed gauge

### Level expansion

### Ψ = cf(K)Bc 1

### 1 + K

f (K) = 1 + K

Erler-Schnabl

f (K) = t_{0} + t_{1}K + t_{2}K^{2} + · · ·

### S[Ψ] = T r[ 1

### 2 ΨQ

_{B}

### Ψ + 1

### 3 Ψ

^{3}

### ]

Dressed gauge condition

[B_{0}^{−}{Ψ(1 + K)}] 1

1 + K = 0

### 3 The Tachyon Potential

In this section, we evaluate the tachyon potential V = 1

2Tr[ΨQ_{B}Ψ] + 1

3Tr[Ψ^{3}] (10)

for the string field (8). All of the calculation can be preformed by employing a
procedure developed in Ref. [1]. We evaluate the tachyon potential order by order
with respect to the expansion (9). We assign ‘level’ n to the nth order term in f (K),
since the corresponding string field is an eigenstate of ‘dressed’ L_{0} operator with
eigenvalue n − 1.^{1} Let us write the potential up to level N as

V = 1

2t_{m}K_{mn}t_{n} + 1

3V_{mnp}t_{m}t_{n}t_{p}, (11)
where the sum over m,n, p is taken from 0 to N. The coefficients K_{mn} and V_{mnp}
can be obtained by plugging the level n field

ψ_{n} ^{= cK}^{n}^{Bc} ^{1}

1 + K (12)

= lim

s→0(∂_{s}^{)}^{n} ^{!} ^{∞}

0 e^{−t}dt cΩ^{s}BcΩ^{t}. (13)
into the classical action. It is easily found that both K_{mn} and V_{mnp} can be evaluated
in terms of the the well-known trace[2, 3]

g(r_{1},r_{2}, r_{3},r_{4}) ≡ Tr[BcΩ^{r}^{1}cΩ^{r}^{2}cΩ^{r}^{3}cΩ^{r}^{4}]. (14)
More explicitly,

K_{mn} = lim

s_{1}→0,s2→0(−∂_{s}_{1}^{)}^{m}_{(−}∂_{s}_{2}^{)}^{n} ^{!} ^{∞}

0 dt_{1} ^{!} ^{∞}

0 dt_{2}e^{−t}^{1}^{−t}^{2}h_{2}(t_{1},t_{2}, s_{1},s_{2}), (15)
where

h_{2}(t_{1},t_{2}, s_{1},s_{2}) = − lim

u→0∂_{u}^{"}^{g(t}_{2}^{,}^{s}_{1} ^{+t}_{1}^{,}^{u,s}_{2}_{) − g(t}_{2}^{,} ^{s}_{1}^{,}^{t}_{1} ^{+ u,s}_{2}^{)}
+ g(t_{2},s_{1},t_{1}, u + s_{2}) − g(t1,s_{2} + u,t_{2}, s_{1})

+ g(u,t_{2},s_{1} +t_{1}, s_{2}) − g(u,t^{2},s_{1},t_{1} + s_{2})#

(16)

1The ‘dressed’ L0 is given by L Ψ = ^{1}_{2}L_{0}^{−} {Ψ(1 + K)}, where L_{0}^{−} = L_{0} − L_{0}^{†}.

3

and

V_{mnp} = lim

s_{1}→0,s2→0,s3→0(−∂_{s}_{1}^{)}^{m}_{(−}∂_{s}_{2}^{)}^{n}_{(−}∂_{s}_{3}^{)}^{p} ^{!} ^{∞}

0 dt_{1} ^{!} ^{∞}

0 dt_{2} ^{!} ^{∞}

0 dt_{3}

e^{−t}^{1}^{−t}^{2}^{−t}^{3}h_{3}(t_{1},t_{2},t_{3}, s_{1},s_{2},s_{3}), (17)
where

h_{3}(t_{1},t_{2},t_{3},s_{1},s_{2},s_{3}) = g(t_{3},s_{1} +t_{1}, s_{2} +t_{2},s_{3}) − g(t3, s_{1} +t_{1},s_{2},t_{2} + s_{3})

− g(t3, s_{1},t_{1} +t_{2} + s_{2},s_{3}) + g(t_{3}, s_{1},t_{1} + s_{2},t_{2} + s_{3}). (18)
In [1], the integrals in the kinetic term (which corresponds to K_{mn} in our paper) is
performed with the help of the reparametrization

t_{1} = uv, t_{2} = u(1 − v) (19)

where u = t_{1} +t_{2} parametrize the width of the strip world sheet in the sliver frame.

Under this reparametrization, the masure of the integral is transformed as

! _{∞}

0 dt_{1} ^{!} ^{∞}

0 dt_{2} →

! _{∞}

0 du ^{!} ^{1}

0 dvu. (20)

While the qubic term is not calculated in [1], we find that similar reparametrization
also works. Thus the integrals in V_{mnp} can be preformed with the replacement

t_{1} = uv_{1}, t_{2} = uv_{2}, t_{3} = u(1 − v1 − v2). (21)
where u = t_{1} +t_{2} +t_{3} also represents the width of the world sheet. The integration
mesure is also be rewritten into

! _{∞}

0 dt_{1} ^{!} ^{∞}

0 dt_{2} ^{!} ^{∞}

0 dt_{3} →

! _{∞}

0 du^{!} ^{1}

0 dv_{1} ^{!} ^{1−v}^{1}

0 dv_{2} u^{2}. (22)
After integrating out v_{1}, v_{2} and v_{3}, we obtain the potentail as an integral with
respetct to the width u as

V = ^{!} ^{∞}

0 due^{−u}A(u,t_{n}), (23)

where A(u,t_{n}) is finite order in u, possibly includes negative powers of u. This
prescription in terms of world sheet width u helps us to understand the relation
between world sheet fluctuation and the tachyon potential.

4

Tr[ψ_{m}Q_{B}ψ_{n}]

Tr[ψ_{m}ψ_{n}ψ_{p}]

### 3 The Tachyon Potential

In this section, we evaluate the tachyon potential V = 1

2Tr[ΨQ_{B}Ψ] + 1

3Tr[Ψ^{3}] (10)

for the string field (8). All of the calculation can be preformed by employing a
procedure developed in Ref. [1]. We evaluate the tachyon potential order by order
with respect to the expansion (9). We assign ‘level’ n to the nth order term in f (K),
since the corresponding string field is an eigenstate of ‘dressed’ L_{0} operator with
eigenvalue n − 1.^{1} Let us write the potential up to level N as

V = 1

2t_{m}K_{mn}t_{n} + 1

3V_{mnp}t_{m}t_{n}t_{p}, (11)
where the sum over m,n, p is taken from 0 to N. The coefficients K_{mn} and V_{mnp}
can be obtained by plugging the level n field

ψ_{n} = cK^{n}Bc 1

1 + K (12)

= lim

s→0(∂_{s})^{n} ^{!} ^{∞}

0 e^{−t}dt cΩ^{s}BcΩ^{t}. (13)
into the classical action. It is easily found that both K_{mn} and V_{mnp} can be evaluated
in terms of the the well-known trace[2, 3]

g(r_{1}, r_{2}, r_{3}, r_{4}) ≡ Tr[BcΩ^{r}^{1}cΩ^{r}^{2}cΩ^{r}^{3}cΩ^{r}^{4}]. (14)
More explicitly,

K_{mn} = lim

s_{1}→0,s2→0(−∂^{s}_{1})^{m}(−∂^{s}_{2})^{n} ^{!} ^{∞}

0 dt_{1} ^{!} ^{∞}

0 dt_{2}e^{−t}^{1}^{−t}^{2}h_{2}(t_{1},t_{2}, s_{1}, s_{2}), (15)
where

h_{2}(t_{1},t_{2}, s_{1}, s_{2}) = − lim

u→0 ∂_{u}"

g(t_{2}, s_{1} +t_{1}, u,s_{2}) − g(t2, s_{1},t_{1} + u,s_{2})
+ g(t_{2},s_{1},t_{1}, u + s_{2}) − g(t^{1}, s_{2} + u,t_{2}, s_{1})

+ g(u,t_{2}, s_{1} +t_{1},s_{2}) − g(u,t^{2}, s_{1},t_{1} + s_{2})#

(16)

1The ‘dressed’ L_{0} is given by L Ψ = ^{1}_{2}L_{0}^{−} {Ψ(1 + K)}, where L_{0}^{−} = L_{0} − L_{0}^{†}.

3

### Level expansion terminates at level 3 because

### tr[K · · · ] → lim

s→0

### ∂

_{s}

### sin t

_{i}

### u + s ∼ t

^{i}

### u

^{2}

### cos t

_{i}

### u

### higher order of K higher rank derivative integrand contains sin(s/u) factor

(Erler-Shnabl: solution of EOM terminates at level 1)

(m,n, p) v_{m,n,p}(u) (m,n, p) v_{m,n,p}(u)
(0,0,0) _{4π}^{3u}^{5}_{4} (0,0,1) (−15+π^{2})^{u}^{4}

6π^{4}

(0,1,1) −(−15+π^{2})^{u}^{3}

3π^{4} (0,0,2) −(−15+4π^{2})^{u}^{3}

3π^{4}

(1,1,2) ^{2}(−15+π^{2})^{u}

π^{4} (0,2,2) ^{2}(−15+4π^{2})^{u}

π^{4}

(1,2,2) −^{4}(−15+π^{2})

π^{4} (0,0,3) −^{2}(−6+π^{2})^{u}^{2}

3π^{2}

(0,1,3) −^{2}(45−9π^{2}+π^{4})^{u}

3π^{4} (1,1,3) −^{4}(−45+6π^{2}+π^{4})

3π^{4}

(0,2,3) ^{4}(45−15π^{2}+2π^{4})

3π^{4}

Table 1: A complete list of v_{m,n,p}(u) up to level 3. Other coefficients are zero.

swer for the tachyon potential is obtained by performing integration with respect
to u, together with the e^{−u} factor. We present the final expression below.

V = 1
π^{2}

! 15

64π^{2}^{t}^{0}^{3} ^{−}

15t_{1}

16π^{2}^{t}^{0}^{2} ^{+}
1

16t_{1}t_{0}^{2} + 15

16π^{2}^{t}^{2}^{t}^{0}^{2} ^{−}
1

4t_{2}t_{0}^{2} − 1

12π^{2}^{t}_{3}^{t}_{0}^{2}
+ 1

2t_{3}t_{0}^{2} − 3

32t_{0}^{2} + 15

16π^{2}^{t}^{1}^{2}^{t}^{0} ^{−}
1

16t_{1}^{2}t_{0} − 15

4π^{2}^{t}^{2}^{2}^{t}^{0} ^{+t}^{2}^{2}^{t}^{0} ^{+}
3

4t_{2}t_{0} − 1

6 π^{2}t_{1}t_{3}t_{0}

− 15

2π^{2}^{t}^{1}^{t}^{3}^{t}^{0} ^{+}
3

2t_{1}t_{3}t_{0} + 4

3π^{2}t_{2}t_{3}t_{0} + 30

π^{2}^{t}^{2}^{t}^{3}^{t}^{0} ^{− 10t}^{2}^{t}^{3}^{t}^{0} ^{+} π^{2}t_{3}t_{0} − 3t^{3}t_{0}
+ 15

π^{2}^{t}^{1}^{t}^{2}^{2} ^{−t}^{1}^{t}^{2}^{2} ^{−}

15

4π^{2}^{t}^{1}^{2}^{t}^{2} ^{+}
1

4t_{1}^{2}t_{2} − 1

3π^{2}t_{1}^{2}t_{3} + 15

π^{2}^{t}^{1}^{2}^{t}^{3} ^{− 2t}^{1}^{2}^{t}^{3}

"

(28)

3.2 The instability of the classical potential

In last section, we have seen that the level expansion in the dressed B_{0} gauge
termnates at level 3 within the KBc subalgegra. This implies that any result ob-
tained from the potential (28) is exact within our setup, although the KBc subalge-
bra covers very limited part of the whole spcae of the string fields. Therefore we
can expect some exact result for the tachyon potential from our analysis. Since the
potential is completely known, we only need to follow the proceder as in Ref. [5].

The closed string vacuum is given by the configuration (t_{0},t_{1},t_{2},t_{3}) = (1,1,0,0).

We have cheked that it is a stationary point of (28), and furthermore gives correct
D-brane tension −1/2π^{2}. To see whether the vacuum is stable or not, we evaluate

6

## Level 3 potential

### exact !

### ﬁelds:

^{t}

^{0}

^{, t}

^{1}

^{, t}

^{2}

^{, t}

^{3}

## Is closed string vacuum stable ?

(t_{0}, t_{1}, t_{2}, t_{3}) = (1, 1, 0, 0)

### Eigenvalues of Hessian

Hessian matrix at the closed string vacuum

H_{i j} = ∂ ^{2}V

∂^{t}_{i}∂^{t} _{j} ^{.} ^{(29)}

The closed string vacuum is stable(unstable) if all eigenvalues of H_{i j} is posi-
tive(negative). Otherwise, it is a saddle point.

The Hessian matrix for this configuration is

180 + 6π^{2} _{−180 + 12}π^{2} _{180 − 30}π^{2} _{−180 + 48}π^{2}

−180 + 12π^{2} _{180 − 12}π^{2} _{−180 + 12}π^{2} _{180 − 12}π^{2} _{− 12}π^{4}
180 − 30π^{2} _{−180 + 12}π^{2} 180 + 24π^{2} _{360 − 120}π^{2} + 16π^{4}

−180 + 48π^{2} _{180 − 12}π^{2} _{− 12}π^{4} _{360 − 120}π^{2} + 16π^{4} 0

, and eigenvalues is found to be (30)

1487.14, −1261.59, 412.919, 79.1831. (31)
Hessian matrix has a negative eigenvalue while others are positive. This con-
cludes that the closed string vacuum of Erler and Schnabl is a suddle point. It
should be stressed that the four eigenvalues obtained here do not change even if
we include other field outside KBc subalgebra or outside the dressed B_{0} gauge,
since such fields are set to zero on the closed string vacuum. Therefore our result
is completely exact and does not depend on choice of basis for string fields.

### 3.3 Effective Potential

The effective potential for tachyon can be obtained by eliminating fields other than
tahyon field from the classical potential by solving equations of motion. In this
paper, we identified t_{0} as the tachyon mode, also such choice is ad-hoc, but seems
to be natural. Therefore we have to solve

∂^{V}

∂ t_{i} = 0 (i = 0,1,2,3) (32)

for fields t_{1},t_{2},t_{3}. In general, there appear many branches since the equations of
motion is cubic in each t_{i}. However, one can see that (28) is linear in t_{3}. Therefore,
t_{3} is an auxiliary field and can be eliminated by imposing

∂^{V}

∂ t_{3} = 0. (33)

7

### Hessian matrix at the closed string vacuum

### H

_{i j}

### = ∂

^{2}

^{V}

### ∂ ^{t}

_{i}

### ∂ ^{t}

_{j}

^{.} ^{(29)}

### The closed string vacuum is stable(unstable) if all eigenvalues of H

_{i j}

### is posi- tive(negative). Otherwise, it is a saddle point.

### The Hessian matrix for this configuration is

###

###

###

### 180 + 6 π

^{2}

_{−180 + 12} π

^{2}

_{180 − 30} π

^{2}

_{−180 + 48} π

^{2}

### −180 + 12 π

^{2}

_{180 − 12} π

^{2}

_{−180 + 12} π

^{2}

_{180 − 12} π

^{2}

_{− 12} π

^{4}

### 180 − 30 π

^{2}

_{−180 + 12} π

^{2}

### 180 + 24 π

^{2}

_{360 − 120} π

^{2}

### + 16 π

^{4}

### −180 + 48 π

^{2}

_{180 − 12} π

^{2}

_{− 12} π

^{4}

_{360 − 120} π

^{2}

### + 16 π

^{4}

### 0

###

###

### , (30)

### and eigenvalues is found to be

### 1487.14, −1261.59, 412.919, 79.1831. (31) Hessian matrix has a negative eigenvalue while others are positive. This con- cludes that the closed string vacuum of Erler and Schnabl is a suddle point. It should be stressed that the four eigenvalues obtained here do not change even if we include other field outside KBc subalgebra or outside the dressed B

_{0}

### gauge, since such fields are set to zero on the closed string vacuum. Therefore our result is completely exact and does not depend on choice of basis for string fields.

### 3.3 Effective Potential

### The effective potential for tachyon can be obtained by eliminating fields other than tahyon field from the classical potential by solving equations of motion. In this paper, we identified t

_{0}

### as the tachyon mode, also such choice is ad-hoc, but seems to be natural. Therefore we have to solve

### ∂ ^{V}

### ∂ ^{t}

_{i}

^{= 0} (i = 0,1,2,3) (32)

### for fields t

_{1}

### , t

_{2}

### , t

_{3}

### . In general, there appear many branches since the equations of motion is cubic in each t

_{i}

### . However, one can see that (28) is linear in t

_{3}

### . Therefore, t

_{3}

### is an auxiliary field and can be eliminated by imposing

### ∂ ^{V}

### ∂ ^{t}

_{3}

^{= 0.} ^{(33)}

### 7

### Unstable (classically)

Arroyo (2009)

�1.0 �0.5 0.0 0.5 1.0

�1.0

�0.5 0.0 0.5 1.0

*t*_{0}
*t*1

### Eﬀective Potential

### • t_3 and t_2 can be integrated out without specifying branch

### closed string vacuum

V = −T^{p}

### Fake

### vacuum ?

V = −0.71T^{p}

### Eﬀective potential

### •

Integrate out t_1### •

Obtained analytically, but too long to present here0.5 1.0 1.5 2.0 *t*_{0}

�0.05 0.05 0.10

*V*_{1}

### • Closed string vacuum only

### • Stable, no fall-oﬀ

### • deﬁned for positive

### t_0

## How identity

## based solution

## fails?

### 4 Analysys in terms of width of the strip 5 Identity based solution

### Ψ = −cK (34)

### = lim

s→0

### ∂

_{s}

### cΩ

^{s}

### . (35)

### asd f (36)

### A Formlas

### B Summary and Discussion References

### [1] T. Erler and M. Schnabl, JHEP 0910, 066 (2009) [arXiv:0906.0979 [hep- th]].

### [2] Y. Okawa, JHEP 0604, 055 (2006) [arXiv:hep-th/0603159].

### [3] T. Erler, JHEP 0705, 083 (2007) [arXiv:hep-th/0611200].

### [4] T. Erler, JHEP 0705, 084 (2007) [arXiv:hep-th/0612050].

### [5] E. Aldo Arroyo, JHEP 0910, 056 (2009) [arXiv:0907.4939 [hep-th]].

### 10

### Equation of motion contracted with itself

### 4 Analysys in terms of width of the strip 5 Identity based solution

Ψ = −cK (34)

= lim

s→0 ∂_{s}cΩ^{s}. (35)

TrΨQ_{B}Ψ = − lim

s→0 lim

u→0Tr[cΩ^{s}cΩ^{u}cΩ^{s}] (36)
TrΨ^{3} = − lim

s→0 Tr[cΩ^{s}Ω^{s}cΩ^{s}] (37)

### A Formlas

### B Summary and Discussion References

[1] T. Erler and M. Schnabl, JHEP 0910, 066 (2009) [arXiv:0906.0979 [hep- th]].

[2] Y. Okawa, JHEP 0604, 055 (2006) [arXiv:hep-th/0603159].

[3] T. Erler, JHEP 0705, 083 (2007) [arXiv:hep-th/0611200].

[4] T. Erler, JHEP 0705, 084 (2007) [arXiv:hep-th/0612050].

[5] E. Aldo Arroyo, JHEP 0910, 056 (2009) [arXiv:0907.4939 [hep-th]].

10

take this ﬁrst

They diﬀer at ﬁnite s because width are diﬀerent

### 4 Analysys in terms of width of the strip 5 Identity based solution

### Ψ = −cK (34)

### = lim

s→0

### ∂

_{s}

### cΩ

^{s}

### . (35)

### TrΨQ

_{B}

### Ψ = − lim

s→0

### lim

u→0

### Tr[cΩ

^{s}

### cΩ

^{u}

### cΩ

^{s}

### ] (36) TrΨ

^{3}

### = − lim

s→0

### Tr[cΩ

^{s}

### cΩ

^{s}

### cΩ

^{s}

### ] (37)

### A Formlas

### B Summary and Discussion References

### [1] T. Erler and M. Schnabl, JHEP 0910, 066 (2009) [arXiv:0906.0979 [hep- th]].

### [2] Y. Okawa, JHEP 0604, 055 (2006) [arXiv:hep-th/0603159].

### [3] T. Erler, JHEP 0705, 083 (2007) [arXiv:hep-th/0611200].

### [4] T. Erler, JHEP 0705, 084 (2007) [arXiv:hep-th/0612050].

### [5] E. Aldo Arroyo, JHEP 0910, 056 (2009) [arXiv:0907.4939 [hep-th]].

### 10

Tr[cΩ^{s}^{1} cΩ^{s}^{2} cΩ^{s}^{3}]

diﬀerent limits of

### 4 Analysys in terms of width of the strip 5 Identity based solution

### Ψ = −cK (34)

### = lim

s→0

### ∂

_{s}

### cΩ

^{s}

### . (35)

### TrΨQ

_{B}

### Ψ = − lim

s→0

### lim

u→0

### Tr[cΩ

^{s}

### cΩ

^{u}

### cΩ

^{s}

### ] (36) Tr Ψ

^{3}

### = − lim

s→0

### Tr[c Ω

^{s}

### c Ω

^{s}

### c Ω

^{s}

### ] (37) TrΨQ

_{B}

### Ψ = 1

### 2 + 2

### π

^{2}

^{(38)}

### TrΨ

^{3}

### = 2

### 9 + 9 √ 3

### 4 π

^{3}

^{+}

### √ 3

### 2 π ^{(39)}

### A Formlas

### B Summary and Discussion References

### [1] T. Erler and M. Schnabl, JHEP 0910, 066 (2009) [arXiv:0906.0979 [hep- th]].

### [2] Y. Okawa, JHEP 0604, 055 (2006) [arXiv:hep-th/0603159].

### [3] T. Erler, JHEP 0705, 083 (2007) [arXiv:hep-th/0611200].

### [4] T. Erler, JHEP 0705, 084 (2007) [arXiv:hep-th/0612050].

### [5] E. Aldo Arroyo, JHEP 0910, 056 (2009) [arXiv:0907.4939 [hep-th]].

### 10

### 4 Analysys in terms of width of the strip 5 Identity based solution

### Ψ = −cK (34)

### = lim

s→0

### ∂

_{s}

### cΩ

^{s}

### . (35)

### TrΨQ

_{B}

### Ψ = − lim

s→0

### lim

u→0

### Tr[cΩ

^{s}

### cΩ

^{u}

### cΩ

^{s}

### ] (36) TrΨ

^{3}

### = − lim

s→0

### Tr[cΩ

^{s}

### cΩ

^{s}

### cΩ

^{s}

### ] (37) TrΨQ

_{B}

### Ψ = 1

### 2 + 2

### π

^{2}

^{(38)}

### TrΨ

^{3}

### = 2

### 9 + 9 √ 3

### 4 π

^{3}

^{+}

### √ 3

### 2 π ^{(39)}

### A Formlas

### B Summary and Discussion References

### [1] T. Erler and M. Schnabl, JHEP 0910, 066 (2009) [arXiv:0906.0979 [hep- th]].

### [2] Y. Okawa, JHEP 0604, 055 (2006) [arXiv:hep-th/0603159].

### [3] T. Erler, JHEP 0705, 083 (2007) [arXiv:hep-th/0611200].

### [4] T. Erler, JHEP 0705, 084 (2007) [arXiv:hep-th/0612050].

### [5] E. Aldo Arroyo, JHEP 0910, 056 (2009) [arXiv:0907.4939 [hep-th]].

### 10

### Not a solution !

### Finite value of action (D-brane tension) But they doesn t depend of s !

Tr[Ψ(Q_{B}Ψ + Ψ^{2})] �= 0

## Summary

### •

Tachyon potential in dressed Schnabl gauge is obtained### •

Potential terminates at level 3### •

Closed string vacuum: saddle point (classically)stable in the eﬀective potential

### •

Explanation for identity based solution## Speculation

### V =

### �

_{∞}

0

### du e

^{−u}

### A(u)

* u is length of the

boundaries of an open string

* small u divergence corresponds to

zero momentum closed string ?