Tree Amplitudes, Chan–Paton Factors, and Gauge Theory from Open Strings

The previous page introduced physical string states as BRST cohomology classes of vertex operators. We now begin using those vertices to compute scattering amplitudes. The first genuinely stringy answer is already visible at tree level: the four-point open-string amplitude is not a sum of a few Feynman diagrams, but an integral over the moduli space of a punctured disk. That integral produces the Euler beta function, whose poles describe the whole infinite tower of open-string states.

The second key idea is that open strings may carry labels at their endpoints. These labels, introduced by Chan and Paton, turn the massless open-string vector into a nonabelian gauge boson. In modern language an oriented open string carries a fundamental index at one end and an antifundamental index at the other. The disk boundary remembers the cyclic order of external states, and the product of endpoint matrices around the boundary gives the familiar color trace.

The slogan for this page is

$$\boxed{ \text{open-string disk amplitudes} =\text{color-ordered gauge amplitudes dressed by an infinite string form factor}. }$$

At energies much smaller than the string scale, the string form factor tends to one, and the amplitudes reduce to ten-dimensional Yang–Mills theory. At finite $\alpha'$, the same amplitudes contain all massive Regge excitations and all higher-derivative corrections.

Disk amplitudes from boundary vertex operators

For open strings, vertex operators are inserted on the boundary of the disk. By a conformal transformation the disk may be mapped to the upper half-plane, so the boundary is the real line and the insertion points are real numbers $x_i$.

At tree level the disk has a conformal Killing group

$$PSL(2,\mathbb R),$$

which acts on the boundary coordinate by fractional linear transformations. Therefore three boundary insertions may be fixed. For a four-point amplitude one usually chooses

$$x_1=0,\qquad x_3=1,\qquad x_4=\infty,$$

and integrates over the single remaining modulus

$$x_2=x,\qquad 0<x<1,$$

for the cyclic ordering $(1,2,3,4)$.

A color-ordered four-point disk amplitude is an integral over one boundary modulus. The limits $x\to0$ and $x\to1$ are the $s$- and $t$-channel factorization regions of the same worldsheet.

The basic free-field correlator of boundary plane waves is

$$\left\langle \prod_{i=1}^n e^{i k_i\cdot X(x_i)}\right\rangle =(2\pi)^D\delta^{(D)}\!\left(\sum_i k_i\right) \prod_{i<j}|x_i-x_j|^{2\alpha' k_i\cdot k_j},$$

up to the conventional normalization of $X$. This formula is the open-string analogue of the Koba–Nielsen factor. All oscillator, fermion, ghost, and superghost contractions multiply this universal exponential correlator by a kinematic numerator.

For the bosonic open-string tachyon, the unintegrated boundary vertex is schematically

$$\mathcal V_T=c\,e^{ik\cdot X}\,\lambda,$$

where $\lambda$ is a Chan–Paton matrix, to be discussed below. The ordered four-tachyon amplitude has the characteristic integral

$$A(1,2,3,4) \propto \int_0^1 dx\,x^{-\alpha' s-2}(1-x)^{-\alpha' t-2},$$

where, in mostly-plus signature,

$$s=-(k_1+k_2)^2, \qquad t=-(k_2+k_3)^2, \qquad u=-(k_1+k_3)^2.$$

The integral is Euler’s beta function,

$$\boxed{ A(1,2,3,4) \propto B(-1-\alpha's,-1-\alpha't) ={\Gamma(-1-\alpha's)\Gamma(-1-\alpha't) \over \Gamma(-2-\alpha's-\alpha't)}. }$$

This is the Veneziano amplitude. The precise shift by $1$ is the open bosonic intercept: the bosonic open-string spectrum has

$$\alpha' M_N^2=N-1, \qquad N=0,1,2,\ldots.$$

Thus the poles of the amplitude occur at

$$\alpha's=N-1,$$

which are exactly the masses of the intermediate open-string states.

One worldsheet, many channels

The Veneziano integral is over the interval $0<x<1$. The endpoints of this interval are not ultraviolet singularities of a point-particle diagram; they are degeneration limits of the string worldsheet.

When $x\to0$, the vertex operators $1$ and $2$ approach each other. The disk develops a long strip carrying momentum $k_1+k_2$. This is the $s$-channel factorization region. When $x\to1$, operators $2$ and $3$ approach each other, and the same integral factorizes in the $t$-channel.

This is the original duality of the Veneziano amplitude: the amplitude can be expanded in $s$-channel poles or in $t$-channel poles, but these are not separate diagrams to be added. They are different boundary regions of the same moduli-space integral.

Near an $s$-channel pole, the beta function has the form

$$B(-1-\alpha's,-1-\alpha't) \sim {\mathcal R_N(t)\over \alpha's-(N-1)},$$

where the residue $\mathcal R_N(t)$ is a polynomial in $t$ whose degree grows with $N$. This polynomial residue encodes the finite number of spin states at oscillator level $N$. At high mass the highest spins lie on a Regge trajectory,

$$J\sim \alpha'M^2+\text{constant},$$

which is why the open string naturally produces Regge behavior.

For the superstring, the NS tachyon is removed by the GSO projection. The four-gluon amplitude is better written as

$$\boxed{ A_4^{\rm open}(1,2,3,4) = A_4^{\rm YM}(1,2,3,4)\,F_{\rm str}(s,t), }$$

where $A_4^{\rm YM}$ is the color-ordered Yang–Mills partial amplitude and

$$\boxed{ F_{\rm str}(s,t) = {\Gamma(1-\alpha's)\Gamma(1-\alpha't) \over \Gamma(1-\alpha's-\alpha't)} }$$

in a standard open-string convention. The Yang–Mills poles are contained in $A_4^{\rm YM}$, while $F_{\rm str}$ packages the massive string poles and contact corrections. Its low-energy expansion is

$$F_{\rm str}(s,t) =1-\zeta(2)\alpha'^2 st -\zeta(3)\alpha'^3 st(s+t)+O(\alpha'^4).$$

The absence of an $O(\alpha')$ correction is a useful consequence of supersymmetry. The first genuine superstring correction to the four-gluon effective action is of the schematic form $\alpha'^2\operatorname{Tr}F^4$.

Chan–Paton endpoint labels

Now add internal labels to the two endpoints of an oriented open string. Let the left endpoint carry an index $i=1,\ldots,N$, and the right endpoint carry an index $j=1,\ldots,N$. A general open-string state is then

$$|\Psi;\lambda\rangle =\sum_{i,j=1}^N \lambda^i{}_{j}\,|\Psi;i,j\rangle,$$

where $\lambda$ is an $N\times N$ matrix.

Under a global unitary rotation of the endpoint labels,

$$|i\rangle\mapsto U^i{}_{k}|k\rangle, \qquad |j\rangle\mapsto |\ell\rangle (U^{-1})^\ell{}_{j},$$

the Chan–Paton wavefunction transforms as

$$\lambda\mapsto U\lambda U^{-1}.$$

Therefore the massless open-string vector transforms in the adjoint of $U(N)$, plus the overall $U(1)$ if it is not projected out. The corresponding vertex operator is simply the usual open-string vector vertex multiplied by a matrix:

$$\mathcal V_A^{(-1)} = c\,e^{-\phi}\,\epsilon_\mu\psi^\mu e^{ik\cdot X}\,\lambda.$$

In the old endpoint language, this is why an open string looks like a meson: one end carries a fundamental index and the other carries an antifundamental index. In the modern D-brane language, the same labels count which brane an endpoint lies on; that interpretation will become central later.

Endpoint labels make open-string states matrix-valued. On a disk, the cyclic order of boundary insertions multiplies these matrices in order and produces a color trace.

For an oriented disk amplitude, the Chan–Paton factor is obtained by multiplying the matrices in their cyclic order around the boundary:

$$\boxed{ \lambda_1\lambda_2\cdots\lambda_n \quad\longrightarrow\quad \operatorname{Tr}(\lambda_1\lambda_2\cdots\lambda_n). }$$

Thus the full tree amplitude is a sum over inequivalent cyclic orderings,

$$\boxed{ \mathcal A_n^{\rm disk} = g_o^{\,n-2} \sum_{\sigma\in S_n/\mathbb Z_n} \operatorname{Tr}\!\big(\lambda_{\sigma(1)}\lambda_{\sigma(2)}\cdots\lambda_{\sigma(n)}\big) A_n\big(\sigma(1),\sigma(2),\ldots,\sigma(n)\big). }$$

Here $A_n(1,2,\ldots,n)$ is the color-ordered partial amplitude computed with the boundary ordering $x_1<x_2<\cdots<x_n$. The power $g_o^{n-2}$ comes from assigning one open-string coupling to each external vertex and one disk normalization proportional to $g_o^{-2}$.

Cyclicity of the trace reflects the fact that the disk boundary has no preferred starting point. Reversing the orientation of the boundary sends

$$\operatorname{Tr}(\lambda_1\lambda_2\cdots\lambda_n) \mapsto \operatorname{Tr}(\lambda_n\cdots\lambda_2\lambda_1).$$

For an oriented open-string theory, both orderings are included as different cyclic orderings. For an unoriented projection, orientation reversal becomes a gauge identification and imposes constraints on the allowed Chan–Paton matrices.

The three-gluon amplitude and the Yang–Mills cubic vertex

The massless open-string vector is the candidate gauge boson. A sharp test is the three-point amplitude. Choose two gauge-boson vertices in the $-1$ picture and one in the $0$ picture:

$$V_A^{(-1)}=c e^{-\phi}\epsilon\cdot\psi\,e^{ikX}\lambda,$$ $$V_A^{(0)}= \epsilon_\mu\left(\partial X^\mu+i\alpha' k\cdot\psi\,\psi^\mu\right)e^{ikX}\lambda,$$

where overall factors depend on convention. The disk correlator gives the kinematic structure

$$K_3(1,2,3) = (\epsilon_1\cdot\epsilon_2)(k_1-k_2)\cdot\epsilon_3 +(\epsilon_2\cdot\epsilon_3)(k_2-k_3)\cdot\epsilon_1 +(\epsilon_3\cdot\epsilon_1)(k_3-k_1)\cdot\epsilon_2.$$

The color-dressed amplitude has the schematic form

$$\mathcal A_3 \propto K_3(1,2,3) \left[ \operatorname{Tr}(\lambda_1\lambda_2\lambda_3) - \operatorname{Tr}(\lambda_2\lambda_1\lambda_3) \right].$$

With generators $T^a$ normalized by

$$\operatorname{Tr}(T^aT^b)={1\over2}\delta^{ab}, \qquad [T^a,T^b]=i f^{abc}T^c,$$

we have

$$\operatorname{Tr}([T^{a_1},T^{a_2}]T^{a_3}) ={i\over2}f^{a_1a_2a_3}.$$

Therefore the string amplitude reproduces the Yang–Mills cubic vertex:

$$\boxed{ \mathcal A_3 \propto g_{\rm YM}\,f^{a_1a_2a_3} K_3(1,2,3). }$$

Gauge invariance is built in. Replacing $\epsilon_\mu$ by $k_\mu$ makes the corresponding vertex BRST exact, so its insertion decouples from physical amplitudes. In the low-energy field theory this is the Ward identity.

The same logic applies to the massless Ramond states. Their vertex operator is matrix-valued,

$$V_\chi^{(-1/2)} =c e^{-\phi/2}u^\alpha S_\alpha e^{ikX}\lambda,$$

and the disk amplitude with two fermions and one vector gives the ten-dimensional gaugino coupling

$$\bar u_1\Gamma^\mu u_2\,\epsilon_\mu^3 \operatorname{Tr}(\lambda_1\lambda_2\lambda_3)+\cdots.$$

Together, the vector and the GSO-projected Ramond fermion form the ten-dimensional super Yang–Mills multiplet.

Charge conjugation and unoriented projections

The orientation-reversal operation exchanges the two endpoints of an open string. On Chan–Paton wavefunctions it acts as transposition, possibly conjugated by a fixed matrix $\gamma_\Omega$:

$$\Omega:\lambda\mapsto \gamma_\Omega\lambda^T\gamma_\Omega^{-1}.$$

The worldsheet oscillator part of a state also has an intrinsic sign. Therefore an unoriented projection keeps states obeying

$$\boxed{ \lambda=\eta_\Psi\,\gamma_\Omega\lambda^T\gamma_\Omega^{-1}, }$$

where $\eta_\Psi=\pm1$ depends on the open-string state. For the massless gauge boson, the standard orthogonal choice $\gamma_\Omega=1$ keeps antisymmetric matrices,

$$\lambda=-\lambda^T,$$

which form the Lie algebra $\mathfrak{so}(N)$. With a symplectic choice of $\gamma_\Omega$, one obtains a symplectic gauge algebra instead. In the type I superstring, consistency at one loop selects the famous $SO(32)$ gauge group; the mechanism behind that statement will appear when we discuss unoriented one-loop worldsheets.

This charge-conjugation projection is conceptually simple but physically important: it shows that gauge groups are not inserted by hand into the low-energy theory. They arise from the allowed endpoint wavefunctions of open strings, subject to worldsheet consistency conditions.

Large-$N$ interpretation

Chan–Paton factors also explain why open-string perturbation theory naturally organizes itself like a large-$N$ gauge theory. Draw each matrix propagator as a double line: one line carries the fundamental index and one carries the antifundamental index. A disk boundary with ordered insertions is a single color trace. Adding open-string loops adds additional boundaries, just as adding index loops changes the topology of a large-$N$ Feynman diagram.

Chan–Paton indices make open-string diagrams look like large-$N$ double-line diagrams. Planar color orderings correspond to disk boundaries; nonplanar reorderings are associated with higher worldsheet topology or additional boundaries.

The disk amplitude is the leading open-string tree contribution. An annulus is an open-string one-loop diagram, but after a modular transformation it can also be interpreted as closed-string exchange between two boundaries. This open/closed duality will later become the key to D-brane physics.

At the level of color factors, the leading large-$N$ contributions are single-trace planar amplitudes,

$$\operatorname{Tr}(T^{a_1}T^{a_2}\cdots T^{a_n})A_n(1,2,\ldots,n).$$

Multi-trace structures arise from worldsheets with more than one boundary or from nonplanar contractions. This is the same topology expansion that appears in the ‘t Hooft expansion of matrix-valued gauge fields.

The field-theory limit

The field-theory limit of open strings is

$$\alpha'\to0$$

with the external momenta fixed and much smaller than the string scale. The massive string poles move to infinite mass because

$$M_N^2\sim {N\over\alpha'}, \qquad N\ge1,$$

while the massless vector and its superpartner remain in the spectrum.

For the four-gluon amplitude,

$$F_{\rm str}(s,t)=1+O(\alpha'^2),$$

so the disk amplitude reduces to the color decomposition of Yang–Mills theory:

$$\boxed{ \mathcal A_n^{\rm open} \xrightarrow[\alpha'\to0]{} \mathcal A_n^{\rm Yang\text{--}Mills}. }$$

In this sense ten-dimensional super Yang–Mills theory is not an independent ingredient. It is the low-energy theory of massless open superstrings. The string calculation also predicts how Yang–Mills theory is completed at high energy: by an infinite tower of massive higher-spin states and an infinite sequence of higher-derivative interactions.

A useful way to summarize the hierarchy is

$$S_{\rm eff} = -{1\over4g_{\rm YM}^2}\int d^{10}x\,\operatorname{Tr}F_{\mu\nu}F^{\mu\nu} +\alpha'^2\int d^{10}x\,\operatorname{Tr}F^4 +O(\alpha'^3),$$

where the detailed tensor contractions in $\operatorname{Tr}F^4$ are fixed by the string amplitude. Fermions complete this to ten-dimensional super Yang–Mills plus its stringy corrections.

Exercises

Exercise 1: deriving the Veneziano integral

For four open bosonic tachyons, fix the boundary insertion points by

$$x_1=0, \qquad x_2=x, \qquad x_3=1, \qquad x_4=\infty.$$

Using the boundary Koba–Nielsen correlator, show that the ordered amplitude is proportional to

$$\int_0^1 dx\,x^{-\alpha's-2}(1-x)^{-\alpha't-2}.$$

Solution

The boundary plane-wave correlator gives

$$\prod_{i<j}|x_i-x_j|^{2\alpha' k_i\cdot k_j}.$$

After fixing three positions, all $x$-dependence comes from the factors involving $x_2=x$:

$$|x_{12}|^{2\alpha' k_1\cdot k_2}|x_{23}|^{2\alpha' k_2\cdot k_3} = x^{2\alpha' k_1\cdot k_2}(1-x)^{2\alpha' k_2\cdot k_3}.$$

For open-string tachyons, $k_i^2=1/\alpha'$ in mostly-plus signature, since $M^2=-1/\alpha'$ and $k_i^2=-M^2$. Therefore

$$s=-(k_1+k_2)^2 =-k_1^2-k_2^2-2k_1\cdot k_2 =-{2\over\alpha'}-2k_1\cdot k_2,$$

so

$$2\alpha' k_1\cdot k_2=-\alpha's-2.$$

Similarly,

$$2\alpha' k_2\cdot k_3=-\alpha't-2.$$

The remaining factors are independent of $x$ after the usual treatment of the vertex at infinity. Thus

$$A(1,2,3,4) \propto \int_0^1 dx\,x^{-\alpha's-2}(1-x)^{-\alpha't-2}.$$

Exercise 2: locating the open-string poles

Show that

$$B(-1-\alpha's,-1-\alpha't)$$

has $s$-channel poles at

$$\alpha's=N-1, \qquad N=0,1,2,\ldots.$$

Explain why this agrees with the open bosonic string spectrum.

Solution

The gamma function $\Gamma(z)$ has simple poles at

$$z=0,-1,-2,\ldots.$$

The factor $\Gamma(-1-\alpha's)$ therefore has poles when

$$-1-\alpha's=-N, \qquad N=0,1,2,\ldots.$$

Equivalently,

$$\alpha's=N-1.$$

The open bosonic string mass formula is

$$\alpha'M_N^2=N-1.$$

Since $s$ is the invariant mass squared carried by the intermediate state, the beta-function poles occur precisely when the intermediate string goes on shell.

Exercise 3: color traces from Chan–Paton matrices

Consider three open-string gauge bosons with Chan–Paton matrices $T^{a_1}$, $T^{a_2}$, $T^{a_3}$. Show that the difference between the two boundary orderings $(1,2,3)$ and $(2,1,3)$ is proportional to the structure constant $f^{a_1a_2a_3}$.

Solution

The two color factors are

$$\operatorname{Tr}(T^{a_1}T^{a_2}T^{a_3}) \quad\text{and}\quad \operatorname{Tr}(T^{a_2}T^{a_1}T^{a_3}).$$

Their difference is

$$\operatorname{Tr}\big((T^{a_1}T^{a_2}-T^{a_2}T^{a_1})T^{a_3}\big) = \operatorname{Tr}([T^{a_1},T^{a_2}]T^{a_3}).$$

Using

$$[T^a,T^b]=i f^{abc}T^c, \qquad \operatorname{Tr}(T^aT^b)={1\over2}\delta^{ab},$$

we find

$$\operatorname{Tr}([T^{a_1},T^{a_2}]T^{a_3}) ={i\over2}f^{a_1a_2a_3}.$$

Thus the antisymmetric part of the Chan–Paton trace is exactly the Yang–Mills color factor.

Exercise 4: the low-energy expansion of the string form factor

Use

$$\log\Gamma(1-z)=\gamma z+\sum_{n=2}^{\infty}{\zeta(n)\over n}z^n$$

for small $z$ to show that

$${\Gamma(1-x)\Gamma(1-y)\over\Gamma(1-x-y)} =1-\zeta(2)xy+O(z^3),$$

where $x$ and $y$ are small.

Solution

Let

$$F(x,y)={\Gamma(1-x)\Gamma(1-y)\over\Gamma(1-x-y)}.$$

Taking the logarithm gives

$$\log F =\log\Gamma(1-x)+\log\Gamma(1-y)-\log\Gamma(1-x-y).$$

Using the expansion,

$$\log F =\gamma x+\gamma y-\gamma(x+y) +{\zeta(2)\over2}\left[x^2+y^2-(x+y)^2\right]+O(z^3).$$

The linear terms cancel. The quadratic term is

$${\zeta(2)\over2}[-2xy] =-\zeta(2)xy.$$

Since $\log F=-\zeta(2)xy+O(z^3)$, exponentiating gives

$$F(x,y)=1-\zeta(2)xy+O(z^3).$$

With $x=\alpha's$ and $y=\alpha't$, this gives the leading correction quoted in the text.

Exercise 5: the orthogonal Chan–Paton projection

Suppose an unoriented projection keeps gauge-boson Chan–Paton matrices satisfying

$$\lambda=-\lambda^T.$$

Show that these matrices form the Lie algebra $\mathfrak{so}(N)$.

Solution

The Lie algebra $\mathfrak{so}(N)$ may be represented by real antisymmetric $N\times N$ matrices. If

$$A^T=-A, \qquad B^T=-B,$$

then their commutator is also antisymmetric:

$$[A,B]^T=(AB-BA)^T=B^T A^T-A^T B^T =(-B)(-A)-(-A)(-B) =BA-AB=-[A,B].$$

Thus antisymmetric matrices close under the Lie bracket. Their number is

$${N(N-1)\over2},$$

which is the dimension of $\mathfrak{so}(N)$. Therefore the projected massless vectors are gauge bosons of an $SO(N)$ gauge theory.

Exercise 6: why three boundary points are fixed

Explain why a disk amplitude with $n$ open-string insertions has $n-3$ real integration variables after gauge fixing.

Solution

Each open-string vertex insertion lies on the one-dimensional boundary of the disk, so before gauge fixing there are $n$ real insertion coordinates.

The disk conformal Killing group preserving the boundary is

$$PSL(2,\mathbb R),$$

which is three-dimensional. It acts on the real boundary coordinate by

$$x\mapsto {ax+b\over cx+d}, \qquad ad-bc=1,$$

with three independent real parameters. Therefore three insertion points can be fixed, leaving

$$n-3$$

real moduli. For $n=4$ there is one modulus, the variable $x$ in the Veneziano integral.