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Progress in String Field Theory

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String Field Theory

Syoji Zeze

Yokote Seiryo Gakuin High School

2 April, 2010 National Taiwan U.

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• 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]

(3)

(4)

(5)

Sen s conjecture

unstable D-brane no open string

1998

t V

tension of unstable

D-brane t0

V (t0) SFT can derive

this potential

(6)

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

Level truncation era

Ψ = t1 + t2 + · · ·

(7)

with no space time dependence

(8)

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

(9)

B

B

c = 2

π c(1) |I�

B = π

2 BL |I�

K = π

2 KL |I�

Okawa

(10)

0

−t

t

0

−t(1+K)

midpoint

π 2

π 2 t

(11)

dressed gauge

(12)

1 + K

f (K) = 1 + K

Erler-Schnabl

f (K) = t0 + t1K + t2K2 + · · ·

B

3

]

Dressed gauge condition

[B0{Ψ(1 + K)}] 1

1 + K = 0

(13)

3 The Tachyon Potential

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

2Tr[ΨQBΨ] + 1

3Tr[Ψ3] (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’ L0 operator with eigenvalue n − 1.1 Let us write the potential up to level N as

V = 1

2tmKmntn + 1

3Vmnptmtntp, (11) where the sum over m,n, p is taken from 0 to N. The coefficients Kmn and Vmnp can be obtained by plugging the level n field

ψn = cKnBc 1

1 + K (12)

= lim

s→0(s)n !

0 e−tdt cΩsBcΩt. (13) into the classical action. It is easily found that both Kmn and Vmnp can be evaluated in terms of the the well-known trace[2, 3]

g(r1,r2, r3,r4) ≡ Tr[BcΩr1cΩr2cΩr3cΩr4]. (14) More explicitly,

Kmn = lim

s1→0,s2→0(−s1)m(−s2)n !

0 dt1 !

0 dt2e−t1−t2h2(t1,t2, s1,s2), (15) where

h2(t1,t2, s1,s2) = − lim

u→0u"g(t2,s1 +t1,u,s2) − g(t2, s1,t1 + u,s2) + g(t2,s1,t1, u + s2) − g(t1,s2 + u,t2, s1)

+ g(u,t2,s1 +t1, s2) − g(u,t2,s1,t1 + s2)#

(16)

1The ‘dressed’ L0 is given by L Ψ = 12L0 {Ψ(1 + K)}, where L0 = L0 − L0.

3

and

Vmnp = lim

s1→0,s2→0,s3→0(−s1)m(−s2)n(−s3)p !

0 dt1 !

0 dt2 !

0 dt3

e−t1−t2−t3h3(t1,t2,t3, s1,s2,s3), (17) where

h3(t1,t2,t3,s1,s2,s3) = g(t3,s1 +t1, s2 +t2,s3) − g(t3, s1 +t1,s2,t2 + s3)

− g(t3, s1,t1 +t2 + s2,s3) + g(t3, s1,t1 + s2,t2 + s3). (18) In [1], the integrals in the kinetic term (which corresponds to Kmn in our paper) is performed with the help of the reparametrization

t1 = uv, t2 = u(1 − v) (19)

where u = t1 +t2 parametrize the width of the strip world sheet in the sliver frame.

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

!

0 dt1 !

0 dt2

!

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 Vmnp can be preformed with the replacement

t1 = uv1, t2 = uv2, t3 = u(1 − v1 − v2). (21) where u = t1 +t2 +t3 also represents the width of the world sheet. The integration mesure is also be rewritten into

!

0 dt1 !

0 dt2 !

0 dt3

!

0 du! 1

0 dv1 ! 1−v1

0 dv2 u2. (22) After integrating out v1, v2 and v3, we obtain the potentail as an integral with respetct to the width u as

V = !

0 due−uA(u,tn), (23)

where A(u,tn) 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[ψmQBψn]

Tr[ψmψnψp]

3 The Tachyon Potential

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

2Tr[ΨQBΨ] + 1

3Tr[Ψ3] (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’ L0 operator with eigenvalue n − 1.1 Let us write the potential up to level N as

V = 1

2tmKmntn + 1

3Vmnptmtntp, (11) where the sum over m,n, p is taken from 0 to N. The coefficients Kmn and Vmnp can be obtained by plugging the level n field

ψn = cKnBc 1

1 + K (12)

= lim

s→0(∂s)n !

0 e−tdt cΩsBcΩt. (13) into the classical action. It is easily found that both Kmn and Vmnp can be evaluated in terms of the the well-known trace[2, 3]

g(r1, r2, r3, r4) ≡ Tr[BcΩr1cΩr2cΩr3cΩr4]. (14) More explicitly,

Kmn = lim

s1→0,s2→0(−∂s1)m(−∂s2)n !

0 dt1 !

0 dt2e−t1−t2h2(t1,t2, s1, s2), (15) where

h2(t1,t2, s1, s2) = − lim

u→0u"

g(t2, s1 +t1, u,s2) − g(t2, s1,t1 + u,s2) + g(t2,s1,t1, u + s2) − g(t1, s2 + u,t2, s1)

+ g(u,t2, s1 +t1,s2) − g(u,t2, s1,t1 + s2)#

(16)

1The ‘dressed’ L0 is given by L Ψ = 12L0 {Ψ(1 + K)}, where L0 = L0 − L0.

3

(14)

s→0

s

i

i

2

i

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

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

(15)

(m,n, p) vm,n,p(u) (m,n, p) vm,n,p(u) (0,0,0) 3u54 (0,0,1) (−15+π2)u4

4

(0,1,1) (−15+π2)u3

4 (0,0,2) (−15+4π2)u3

4

(1,1,2) 2(−15+π2)u

π4 (0,2,2) 2(−15+4π2)u

π4

(1,2,2) 4(−15+π2)

π4 (0,0,3) 2(−6+π2)u2

2

(0,1,3) −2(45−9π24)u

4 (1,1,3) −4(−45+6π24)

4

(0,2,3) 4(45−15π2+2π4)

4

Table 1: A complete list of vm,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 π2

! 15

64π2t03

15t1

16π2t02 + 1

16t1t02 + 15

16π2t2t02 1

4t2t02 1

12π2t3t02 + 1

2t3t02 3

32t02 + 15

16π2t12t0 1

16t12t0 15

4π2t22t0 +t22t0 + 3

4t2t0 1

6 π2t1t3t0

15

2π2t1t3t0 + 3

2t1t3t0 + 4

3π2t2t3t0 + 30

π2t2t3t0 − 10t2t3t0 + π2t3t0 − 3t3t0 + 15

π2t1t22 −t1t22

15

4π2t12t2 + 1

4t12t2 1

3π2t12t3 + 15

π2t12t3 − 2t12t3

"

(28)

3.2 The instability of the classical potential

In last section, we have seen that the level expansion in the dressed B0 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 (t0,t1,t2,t3) = (1,1,0,0).

We have cheked that it is a stationary point of (28), and furthermore gives correct D-brane tension −1/2π2. To see whether the vacuum is stable or not, we evaluate

6

t0, t1, t2, t3

(16)

Is closed string vacuum stable ?

(t0, t1, t2, t3) = (1, 1, 0, 0)

Eigenvalues of Hessian

Hessian matrix at the closed string vacuum

Hi j = ∂ 2V

tit j . (29)

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

The Hessian matrix for this configuration is



180 + 6π2 −180 + 12π2 180 − 30π2 −180 + 48π2

−180 + 12π2 180 − 12π2 −180 + 12π2 180 − 12π2 − 12π4 180 − 30π2 −180 + 12π2 180 + 24π2 360 − 120π2 + 16π4

−180 + 48π2 180 − 12π2 − 12π4 360 − 120π2 + 16π4 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 B0 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 t0 as the tachyon mode, also such choice is ad-hoc, but seems to be natural. Therefore we have to solve

V

∂ ti = 0 (i = 0,1,2,3) (32)

for fields t1,t2,t3. In general, there appear many branches since the equations of motion is cubic in each ti. However, one can see that (28) is linear in t3. Therefore, t3 is an auxiliary field and can be eliminated by imposing

V

∂ t3 = 0. (33)

7

i j

2

i

j

i j

2

2

2

2

2

2

2

2

4

2

2

2

2

4

2

2

4

2

4

0

0

i

1

2

3

i

3

3

3

Unstable (classically)

Arroyo (2009)

(17)

�1.0 �0.5 0.0 0.5 1.0

�1.0

�0.5 0.0 0.5 1.0

t0 t1

V = −Tp

V = −0.71Tp

(18)

•

Integrate out t_1

•

Obtained analytically, but too long to present here

0.5 1.0 1.5 2.0 t0

�0.05 0.05 0.10

V1

(19)

fails?

(20)

s→0

s

s

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

Ψ = −cK (34)

= lim

s→0 scs. (35)

TrΨQBΨ = − lim

s→0 lim

u→0Tr[cΩscΩucΩs] (36) TrΨ3 = − lim

s→0 Tr[cΩsscs] (37)

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

s→0

s

s

B

s→0

u→0

s

u

s

3

s→0

s

s

s

10

Tr[cΩs1 cΩs2 cΩs3]

diﬀerent limits of

(21)

s→0

s

s

B

s→0

u→0

s

u

s

3

s→0

s

s

s

B

2

3

3

s→0

s

s

B

s→0

u→0

s

u

s

3

s→0

s

s

s

B

2

3

3

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

Tr[Ψ(QBΨ + Ψ2)] �= 0

(22)

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

(23)

Speculation

0

−u

A(u)

* u is length of the

boundaries of an open string

* small u divergence corresponds to

zero momentum closed string ?

another direction of world volume appears and resulting theory becomes (1+5)D Moreover, in this case, we can read the string coupling from the gauge field and this enables us to

A diamagnetic material placed in an external magnetic field B ext develops a magnetic dipole moment directed opposite B ext. If the field is nonuniform, the diamagnetic material

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In section29-8,we saw that if we put a closed conducting loop in a B and then send current through the loop, forces due to the magnetic field create a torque to turn the loopÆ

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The point should then be made that such a survey is inadequate to make general statements about the school (or even young people in Hong Kong) as the sample is not large enough

In addition to speed improvement, another advantage of using a function handle is that it provides access to subfunctions, which are normally not visible outside of their