String Theory I GEORGE SIOPSIS AND STUDENTS

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1 String Theory I GEORGE SIOPSIS AND STUDENTS Department of Physics and Astronomy The University of Tennessee Knoxville, TN U.S.A. siopsis@tennessee.edu Last update: 26

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3 Contents 4 Tree-level Amplitudes String Interactions A Short Course in Scattering Theory N-point open-string tree amplitudes Closed Strings Moduli BRST Invariance

4 UNIT 4 Tree-level Amplitudes 4. String Interactions In particle theory, we need to introduce a multi-particle space (Fock space) where creation and annhilation are possible. In string theory, the tools we have developed for one string are sufficient for the description of multi-string states and interactions! The entire quantum theory of strings is based on these tools! Example of particle interactions e e p 2 p 3 e e p 2 k p 3 γ k γ + e p p 4 + e e + p p 4 + e /k 2 : inverse of the Klein-Gordon operator φ =, /k 2 There is a pole at k 2, e.g., β-decay

5 6 UNIT 4: Tree-level Amplitudes p+ e W n Amplitude, pole at k 2 = m 2 k 2 m 2 W, resonance. W ν Strings The interaction consists of strings joining and splitting. Where do they join? This is a stupid question. It depends on the time slicing. Therefore this is a fuy interaction. Moreover, the shape (geometry) of the surface is not important, only the topology is important. There is one diagram for all tree diagrams.

6 4. String Interactions 6 Example + t B Outgoing string Interaction point (arbitrary) A Incoming string There is an arbitrary interaction point. The amplitude is constructed by joining two semi-infinite cylinders. Map the cylinders to a plane: B C C A cylinder C { } = S 2 (sphere). This is done through stereographic projection (sphere=fat cylinder).

7 62 UNIT 4: Tree-level Amplitudes N O P S P Amplitude: sphere with states (operator insertions) at North and South poles. Notice the equivalence of the two poles (clinder ). Open Strings Make a strip by cutting the cylinder in half along the axis. We then map the strip to the upper-half plane which can then be mapped to the unit circle via the mapping i +i. 8 Each string is a semi-infinite cylinder (or strip), which is mapped to a disk. When we put two on a sphere, they were simply represented by insertion of A() at =, =. Guess: For scattering of N strings we can do the same, i.e., on a sphere select points, 2,..., N and insert operators A i ( i ). Then the amplitude is A A ( )A 2 ( 2 )...A N ( N ).

8 4. String Interactions 63 d 2πi A(). Now A() is equivalent to C For conformal invariance, we require that all A i have dimension h i = so that da i () have ero dimension (conformally). Then we should define Amp d...d N A ( ).... In fact, the measure should read d 2...d 2 N, but we will not be writing the piece explicitly. The proper dimension of A i (, ) should be h i =, h i =. In general, A() : m X m2 X...e ik X :, where h = m + m α k 2 =. We shall work with the simplest case A() = e ik X, k 2 = α. The rest is similar. Complication: The amplitude is conformally invariant: + ɛv() where v() is analytic. v() should be analytic everywhere in C { }. We need to check that the transformation is analytic at infinity. So let =. δ = 2 δ = ɛ ( ) 2 v() = ɛ 2 v. therefore v() = a+b+c 2 so that 2 v ( ) is analytic. This is a six-parameter family of transformations. It includes SO(3) (rotation group). Special Cases: + ɛa generated by L. Recall [L m, A] = m+ A + h(m + ) m A where h = for BRST invariance. So [L, A] = A A i.e., L generates translations in. Finite transformation: + a, A() e al A()e al = A( + a). + ɛb = ( + ɛb) generated by L. Finite transformation: e b. + ɛc 2 generated by L. Finite transformations: [L, A] = A + A. A() e bl A()e bl = A(e b ). [L, A] = 2 A + A. c = A() e cl A()e cl = A ( ). c Combination of all three: a+b c+d, ad bc = defines the group SL(2, C) whose algebra is [L, L ] = 2L, [L, L ] = L, [L, L ] = L.

9 64 UNIT 4: Tree-level Amplitudes This is a closed algebra (no constant term) and is common in all conformal field theories. How come a matrix entered acting on a number? Answer: Consider the vector Let = / 2. Then ( a b (, 2 ) c d ) ( 2 ) = a + b 2 c + d 2 = a + b c + d. ( ) a + d 2. c + d 2 For open strings: the Real axis is a boundary, and the group of symmetries becomes SL(2, R), a, b, c, d R. Then under a+b c+d, is invariant. The upper-half plane maps to itself. Amplitude for open strings: Amp V ( )V ( 2 )...V ( N ), i R, where the product is time ordered and thus the i s are ordered. How do we integrate over i? Due to SL(2, R) symmetry, we have redundency, so naive integral would be proportional to the volume of SL(2, R) which is infinite! We need to fix the gauge by choosing three points. Easiest to fix them to (,, ). This is an arbitrary choice, but all choices are equivalent by the SL(2, R) symmetry. We will integrate over the rest of the parameters. Example : Three tachyons Consider three tachyons, V i () =: e iki X() : The amplitude is given by A V ( )V 2 ( 2 )V 3 ( 3 ), where X µ () = x µ i α α 2 pµ ln 2 + i 2 m ( m aµ m m + m). Since R, X µ reduces to X µ () = x µ iα p µ ln + i 2α m m aµ m m.

10 4. String Interactions 65 Since we can fix three points, let us choose ( =, 2 =, 3 = ), then The amplitude becomes V 3 ( 3 = ) = ; k 3, V ( = ) = ; k. A ; k V 2 ( 2 = ) ; k 3 = ; k ; k 2 + k 3 = δ D (k + k 2 + k 3 ). One can derive this for arbitrary, 2, 3, due to the SL(2,R) symmetry. Example 2: Two tachyons and one vector Consider two tachyons, V i () =: e iki X() : and a vector, V j () =: A µ X µ e ikj X() :, where kj 2 =. We may act the vertex operators on the vacuum states A ; k V 2 ()A µ X µ ; k 3 ; k e ik2 x e α 2 a k2 A µ α µ ; k 3 2α A k 2 δ D (k + k 2 + k 3 ). (4..) A is transverse to it s momentum (A k 3 = ) therefore, the amplitude is α A 2 A (k 2 k )δ D (k + k 2 + k 3 ), where the dot product represents the coupling of the electromagnetic potential to the charged scalar. We may check the gauge invariance of the amplitude. Using the gauge transformation A µ A µ + ωk µ 3, the amplitude becomes δ(a) k 3 (k 2 k ) = k2 2 k 2 =. Example 3: Four tachyons This is the first nontrivial amplitude. Due to the SL(2, R) symmetry, we may fix three operators. Now we have an extra operator we can not fix. We must integrate over its parameter. After we operate vertex operators on the vacuum states, the amplitude is given by A ; k : e ik2 X () :: e ik3 X () : ; k 3. This is a time-ordered product and we must integrate over from [, ]. The amplitude becomes A Using the mode expansion of X µ, A d ; k : e ik2 X () :: e ik3 X () : ; k 4. d ; k e ik2 x e 2α m> k2 αm/m e ik3 x e i α 2 k3 p ln e 2α n> k3 α n/n ; k 4.

11 66 UNIT 4: Tree-level Amplitudes Using the Hausdorff formula, e A e B = e [A,B] e B e A, A Define the Mandelstam variables d ; k 2α k 3 k 4 e 2α k 2 k 3 m /m ; k 3 + k 4 d 2α k 3 k 4 ( ) 2α k 2 k 3 δ D (k + k 2 + k 3 + k 4 ) k 3 k θ k 4 k 2 s = (k +k 2 ) 2 = (k 3 +k 4 ) 2 = 2k 3 k 4 2 α, t = (k 2 +k 3 ) 2, u = (k 2 +k 4 ) 2, and s + t + u = 4 α. The amplitude expressed in terms of Mandelstam variables becomes A d α s 2 ( ) α t 2 δ D (k + k 2 + k 3 + k 4 ) B( α s, α t ), where B is the Euler-beta function with the property B(x, y) = Γ(y)Γ(y) Γ(x + y). This is known as the Veneiano amplitude. Note, there are poles at α s = and α t =. Let us focus on the first pole ( α s = ). A Γ( α s ) = Γ( α s) α +... (Γ(x + ) = xγ(x)) s + α s The pole is due to an intermediate tachyon (s = /α ). Unitarity requires This checks, since The next pole is at α s =. Γ( α s ) = Γ( α s) α s + = Γ( α s + ) (α s + )(α s) = α s +... The amplitude becomes A Γ( α t ) α s Γ( α t 2) = Γ( α t 2) α s +... = u t 2s +...

12 4. String Interactions 67 where we used the condition s + t + u = 4/α. Check unitarity: The amplitude is gauge invariant. Summing over the polariations ɛ µ ɛ ν = η µν gives the amplitude A α (k k 2 )(k 3 k 4 ) 2k 2 = u t 2s. All the poles in α s : α s =,,, 2,... which are the masses of the open string states. (α s 2 = N from L = ). Curious Result: same structure of poles we obtain for α t, since the amplitude is symmetric in s and t. This would also be true of a field theory amplitude. Alternate derivation of the poles: It is instructive to find the poles without performing the integral for two reasons. (a) We can not always do the integral. (b) We can see what type of world-sheet contributes to the pole (physical picture for an effective field theory). Let A d α s 2 s= +... = α α +... = s α s + + analytic. Taylor expansion: A α s 2 ( ) α t 2 = d α s 2 ( + (α t + 2) +...), where the first and second terms in the expansion represent the α s = and α s = poles respectively. The other poles are acquired through higher order terms in the expansion. Poles in α t are obtained from. k k k 2 3 k 4 There is no reason to restrict d to d. We would like to extend the integral to d. The integral becomes + +. : ordering (k + k 2 + k 3 + k 4 ) which is with k 2 k. The effect is switching t and u. This can be seen through the transformation which maps (, ) (, ). Therefore, if = I(s, t) then = I(t, u). Similarly, = I(s, u). Therefore, the integral becomes 8 = I(s, t) + I(s, u) + I(t, u). Now the amplitude is completely symmetric in s, t, u.

13 68 UNIT 4: Tree-level Amplitudes BRST invariance If V () has weight h =, then dv () has weight h =. It is BRST invariant. Let us check this. [Q, V ()] = c n [L n, V ()] = c n ( n+ V () + (n + ) n V ()) = c() V () + c()v () = (c()v ()) Therefore [Q, V ] = (c()v ()) =. What happens with the three V s that we fixed? To turn them into h = operators, we multiply them by c(). Then c()v () has the weight h =. {Q, cv } = {Q, c}v c[q, V ] = c cv cc V c cv =. Now in the amplitude, we have three c()s, i =,,. The amplitude must be defined with respect to the SL(2, R) invariant vacuum. Recall: b ψ =, χ = c ψ. L bc m = n (2m n) : b n c m n : δ m, So, L bc = n n : b n c n :, L bc = n (2 n) : b n c n :, L bc = n ( 2 n) : b n c n :. The operators act on the states L bc ψ = ψ, Lbc ψ =, Lbc ψ = b χ, So psi is not invariant. Let = b ψ. Therefore, [L bc, b ] = b, [L bc, b ] = 2b, [L bc, b ] =. L bc = b ψ b ψ =, L bc = b b ψ =, L bc = b b χ =. So, is SL(2, R) invariant. The ghost contribution is c( )c()c(), c() = n c n n+. c() = c = ψ, c( ) = ψ, ψ c() ψ = ψ c ψ =.

14 4.2 A Short Course in Scattering Theory 69 High Energy k 3 k θ k 4 k 2 k µ = (E/2, p), kµ 2 = (E/2, p), kµ 3 = ( E/2, p ), k µ 4 = ( E/2, p ). where ( ) E 2 2 p 2 = m 2, p = p. The Mandelstam variables become s = (k +k 2 ) 2 = E 2, t = (k +k 3 ) 2 = (4m 2 E 2 ) sin 2 θ 2, u = (k +k 4 ) 2 = (4m 2 E 2 ) cos 2 θ 2. The high energy limit is equivalent to the small angle limit, where s and t is fixed. The gamma function is approximated by The amplitude is A Γ( α s )Γ( α t ) Γ( α s α t 2 Γ(x) x x e x 2π x. s α s s α s α t 2 eα t+ Γ( α t ) s α t+ Γ( α t ). This is the Regge behavior. At the poles α t n, the amplitude goes as A s n which is the exchange of a particle of spin n. For a fixed angle,θ=fixed: s, t, s/t = fixed. The amplitude becomes s t A s α α t (s + t) α s α t s α s t α t e α (s ln s+t ln t+u ln u) u α u e α (s ln(s/s)+t ln(t/s)+u ln(u/s)) e α s( t s ln t s + u s ln u s e α s( sin 2 θ 2 ln sin2 θ 2 cos2 θ 2 ln cos2 θ 2) e Cs, C >. unlike in field theory, where the amplitude goes as A s n. Therefore the underlying smooth extended object of sie α. 4.2 A Short Course in Scattering Theory We define the in state in the real infinite past (t ), and the out state in the infinite future (t ). These states are both described by free particles. There is an isomorphism in = S out, S = lim t e iht/..

15 7 UNIT 4: Tree-level Amplitudes To conserve probabilities, S must be unitary, S S = (c.f. unitarity of evolution operator, U = e iht/ ). The transition probability (S = I + it ) is i f 2 = i T f 2, where i > and f represent states in the same Hilbert space. We will discard the I because it represents i i (forward scattering i.e., along the beam: undetectable). Unitarity S S = I = I + i(t T ) + T T. Therefore Insert complete sets of physical states i T f i T f = i i T T f. i T f i T f = i n 2Im < i T f = n i T n n T f, < i T n f T n. (4.2.) Viewed as a function of s, i T f has poles in s. Away from the pole, i T f is real, so the left hand side vanishes. Near the pole, we obtain a behavior s+m (pole at s = m 2 ). to find the 2 imaginary part, first regulate the amplitude for small ɛ. Then s + m 2 s + m 2 + iɛ Im s + m 2 Im s + m 2 + iɛ = ɛ (s + m 2 ) 2 + ɛ 2 = πδ(s + m2 ). Therefore, for the imaginary part is This is in agreement with unitarity. INSERT FIGURE HERE INSERT FIGURE HERE 4.3 N-point open-string tree amplitudes Amp : e ik X ( ) :... : e k2 X ( n ) : = A(,..., n )

16 4.3 N-point open-string tree amplitudes 7 Consider A(,..., n ) : e ik X ( ) :... : e k2 X ( n ) : To evaluate this, consider the OPE ik X() : e ik X ( ) := α k 2 2( ) eik X() : + : e ik X ( ) : +... So, first replace : e ik X ( ) : by X() : e ik X ( ) : in Amp and define f µ () = X µ () : e ik X ( ) :... : e ikn X ( n ) : The singularity structure of f µ () can be deduced from OPEs C n X() : e ik X ( ) := iα k µ 2( ) eik X() : +... Therefore, n f µ () = iα 2 A() k µ +... i i= Behavior at : = which implies X µ = X µ = 2 X µ f µ () = 2 X µ Therefore, as, f µ () ( X µ :... : analytic at ) Therefore 2 the holomorphic piece vanishes and f µ () = iα n 2 A k µ i. i Now define a contour C surrounding all i s. There are two ways to evaluate the contour integral, d 2πi f(). Cauchy d 2πi f µ () = iα 2 A n k µ i i= i, or in the transformed coordinate =, C encircles = where f µ () is analytic. Therefore d n 2πi f µ () = k µ i = i= i=

17 72 UNIT 4: Tree-level Amplitudes The momentum is conserved. Now consider ik f and compare with the OPE ik x() : e ik X ( ) := α k 2 2( ) : eik X ( ) : + : e ik X ( ) : +... which implies Therefore Therefore ik f = α 2 A = α 2 Repeating for other points, k 2 A + α k k i A 2 i i k 2 A + α k k i A. 2 i i ln A = α 2 j ln A = α 2 i i j k k i i A. k j k i i j A. By integrating we obtain ln A = i<j ln i j ki kj + const where we added the piece. Therefore, A i j α k i k j. i<j For open strings, α 2α, so A i<j i j 2α k i k j. SL(2, R) Invariance = a + b c + d, c + d = a + b = d b, ad bc =. a c Therefore, i j = d i b a c i d j b a c j = i j (a c i )(a (4.3.) c j ). Therefore, A i j 2α k i k j i<j = i k i k j j 2α (a c i ) 2α k 2 i, i<j k 2 i = α. (4.3.2)

18 4.4 Closed Strings 73 d i = d i (a c i )2. If we let j i in (4.3.), we find that the amplitude is invariant under SL(2, R) transformations. The measure is given by di = d i (a c i ) 2, however, the last factor cancels with the overall factor in (4.3.2). 4.4 Closed Strings For open strings we found four tachyons, A open = d 2α k 3 k 4 ( ) 2α k 2 k 3 δ D (k + k 2 + k 3 + k 4 ) + + where = I(s, t) = d α s 2 ( ) a t 2 δ D (k + k 2 + k 3 + k 4 ) = I(t, u), = I(s, u), R. For closed strings, is the entire C and we need to multiply the holomorphic and anti-holomorphic pieces, so A closed d 2 α s/2 4 α t/2 4. Note: s s/4 is due to the different expansion of the X µ s. The tachyon mass is m 2 = 4 α, whereas for the open string it is, m 2 = α. To calculate the amplitude for the closed string, treat and as independent variables and deform the contour of integration until it coincides with the real axis. Then, R. We must take care with the branch cuts. There are three cases. (i) < : Contour for :

19 74 UNIT 4: Tree-level Amplitudes C has branch cuts on the same side and therefore contributes nothing. (ii) > : There is no contribution for the same reason as in (i). (iii) < < : C A closed d α s/4 2 ( ) α t/4 2 d α s/4 2 ( ) α t/4 2 Contribution from the upper side of C is The lower side gives dη η α s/4 2 e iπ(α t/4+2) η α t/4 2 I(s/4, t/4). dη η α s/4 2 e +iπ(α t/4+2) η α t/4 2 I(s/4, t/4). Therefore the amplitude for the closed string is A closed sin πα t I(t/4, u/4)i(s/4, t/4). 4 This can be cast in a symmmetric form by using the transformation properties of the Gamma function Γ(x)Γ( x) = π sin(πx), for α t 4 So, since s + t + u = 4m 2 = 6/α Γ( α t/4 )Γ(2 + α t/4) = Therefore the amplitude is given by π sin(α tπ/4). Γ( α s/4 )Γ( α t/4 )Γ( α u/4 ) A closed π Γ( α s/4 α t/4 )Γ( α t/4 α u/4 )Γ( α u/4 α s/4 ).

20 4.5 Moduli Moduli Build closed-string four-point amplitude as follows. In the -plane, drill holes. This will represent the diagram on the left with amputated legs. Now attach the legs by telescopically collapsing each semi-infinite tube to a disc: Next, patch the discs on the -plane. This produces a sphere with four punctures. There will be regions of overlap where = f(). of hole in plane of disk By conformal transformations, I can fix three points (due to SL(2, C) symmetry). The fourth point cannot be fixed. Call it. Punctured spheres with two different s, are not related by a conformal transformation. There are inequivalent surfaces. They are parametried by two parameters, and C. These parameters are called moduli and their space, moduli space (although it should be called modulus space) (c.f. vector space). They are also called Teichmuller parameters. They label conformally inequivalent surfaces. To calculate amplitudes, we need to integrate over the moduli.

21 76 UNIT 4: Tree-level Amplitudes E.g., the four-point amplitude, V ( )V 2 ()V 3 ( )V 4 () need to integrate over d 2. In general, N-point amplitudes integrate d 2...d 2 N 3 at =,, we specified V c c : e ik X :. We can do the same for the unfixed V s to put them all on equal footing. Thus, let V i = c c : e iki X :, i. Since we introduced an extra c, c, we need to compensate for it with a b, b insertion. To do this work as follows. Shift + δ. This is implemented in the -plane by a coordinate transformation + δ v (, ). where v is of course not conformal (depends on as well as ). Introduce the Beltrami differential. ψ = v There is a similar differential for the complex conjugate ψ = v If v represents a conformal transformation, then ψ, ψ =. Thus ψ encodes information about conformally inequivalent surfaces. We will insert 2π d 2 (pψ + b ψ) anti-holomorphic in the amplitude. We integrate over the patch that we will use so ( ) Amp d 2 V V 2 V 3 V 4 d 2 (pψ + 2π b ψ) (anti) Since b =, the integral is ) d 2 ( (bv ) + ( bv ). Therefore it can be written as (divergence theorem) B = 2πi where C is in the overlap region of and. C (d bv d bv ) C Explicitly, v =.

22 4.6 BRST Invariance 77 In the overlap region, = +, so v + =, therefore v =. Therefore = d bv = b, b() = b n n 2, c = c n n+. 2πi C d b V 3 = d b c c : e ik3 X ( ) := c : e ik3 X ( ) where b c = d b( )c( ) =. C d b c : e ik3 X ( ) :=: e ik3 X ( ) : so the b-insertions kill c c from V 3 and the amplitude is as before. 4.6 BRST Invariance Recall T ( )V 3 ( ) = {Q B, B } = d ( T v d T v ) 2πi c h ( ) 2 + V , h =! Therefore {Q B, B }V 3 d V 3 = unless the moduli space has (not true here, but argument is general and sometimes ).