NSR Vertex Operators and Picture Changing

The NSR formalism has a small but important extra layer that is absent in the bosonic string: a physical vertex operator is not specified only by its matter part. It also carries a superghost picture. Different pictures represent the same BRST cohomology class, but they are useful in different computations.

The key facts are these:

• NS vertices are often written in the $-1$ and $0$ pictures.
• Ramond vertices are often written in the $-1/2$ picture.
• At tree level on the sphere, the total left-moving picture number must be $-2$, and the total right-moving picture number must also be $-2$.
• At tree level on the disk, the total open-string picture number must be $-2$.

This page collects the standard NSR vertices and explains how picture-changing connects them.

Review: the superghost scalar and picture number

The commuting $\beta\gamma$ superghost system can be bosonized as

$$\beta = e^{-\phi}\partial \xi, \qquad \gamma = \eta e^{\phi},$$

where

$$\phi(z)\phi(w)\sim -\ln(z-w), \qquad \eta(z)\xi(w)\sim \frac{1}{z-w}.$$

The exponential $e^{q\phi}$ has conformal weight

$$h(e^{q\phi})=-\frac12 q(q+2).$$

The integer or half-integer $q$ is called the picture number. Thus

$$h(e^{-\phi})=\frac12, \qquad h(e^{-\phi/2})=\frac38.$$

These two numbers explain why the most economical massless vertices are $e^{-\phi}\psi^\mu e^{ik\cdot X}$ in the NS sector and $e^{-\phi/2}S_A e^{ik\cdot X}$ in the Ramond sector.

The NS sector is naturally written in pictures $-1$ and $0$, while the Ramond sector is naturally written in picture $-1/2$. Closed-string vertices are products of left- and right-moving open-string building blocks.

Open-string NS vertices

For an open-string gauge boson with momentum $k^\mu$ and polarization $\epsilon_\mu$, the $-1$ picture unintegrated vertex is

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

Here $y$ is a coordinate on the boundary of the disk or upper half-plane, and $\lambda$ is a Chan—Paton matrix. The physical-state conditions are

$$k^2=0, \qquad k\cdot \epsilon=0, \qquad \epsilon_\mu\sim \epsilon_\mu+\Lambda k_\mu.$$

The weight check is simple. On the boundary, the integrated vertex must have weight $1$. For the massless state,

$$h(e^{-\phi})=\frac12, \qquad h(\psi^\mu)=\frac12, \qquad h(e^{ik\cdot X})=0,$$

so the matter-plus-superghost part has $h=1$. Multiplication by $c$ then gives an unintegrated vertex of total weight zero.

The zero-picture gauge-boson vertex is obtained by picture-changing:

$$V_A^{(0)}(y) = g_o\,c(y)\,\lambda\,\epsilon_\mu \left( i\partial_y X^\mu +2\alpha' k_\nu\psi^\nu\psi^\mu \right)e^{ik\cdot X(y)},$$

up to a harmless overall normalization convention. The important structure is the sum

$$\epsilon\cdot \partial X \quad + \quad (k\cdot \psi)(\epsilon\cdot\psi).$$

The first term is the bosonic gauge-boson vertex. The second term is required by worldsheet supersymmetry.

The picture-changing operator turns $e^{-\phi}\epsilon\cdot\psi$ into the supersymmetric combination $\epsilon\cdot\partial X+(k\cdot\psi)(\epsilon\cdot\psi)$.

The picture-changing operator

The picture-changing operator is the BRST commutator

$$\mathcal X(z)=\{Q_{\mathrm{BRST}},\xi(z)\}.$$

Its leading matter term is

$$\mathcal X(z)=e^{\phi}(z)T_F^{\mathrm{m}}(z)+\text{ghost terms},$$

where the matter supercurrent is schematically

$$T_F^{\mathrm{m}}(z) \propto \psi_\mu\partial X^\mu.$$

If $V^{(q)}$ is a BRST-closed vertex operator in picture $q$, then a representative in picture $q+1$ is

$$V^{(q+1)}(w) = \lim_{z\to w}\mathcal X(z)V^{(q)}(w),$$

provided the limit is nonsingular after taking the appropriate OPE coefficient. The new vertex is BRST-equivalent to the old one.

Picture changing raises the picture number by one. Different pictures give different representatives of the same physical BRST cohomology class.

One should not think of picture-changing as an optional decoration. The path integral over the superghost zero modes forces a definite total picture number. The choice of which vertices carry which pictures is a gauge choice in the odd directions of supermoduli space.

A compact normalization checkpoint

Different books normalize the open-string zero-picture vertex in slightly different ways. The invariant statement is that picture-changing produces the supersymmetric combination

$$\epsilon\cdot \partial X + \text{constant}\times(k\cdot\psi)(\epsilon\cdot\psi).$$

The constant depends on whether one uses boundary or holomorphic-plane normalizations for $X$. All physical amplitudes are unchanged once the vertex normalization and coupling $g_o$ are chosen consistently.

Ramond vertices

A massless open-string Ramond vertex in the $-1/2$ picture is

$$V_u^{(-1/2)}(y) = g_o\,c(y)\,\lambda\,u_A S^A(y)e^{-\phi(y)/2}e^{ik\cdot X(y)}.$$

Here $S^A$ is a ten-dimensional spin field and $u_A$ is a spacetime spinor wavefunction. The conformal weights add to one:

$$h(S^A)=\frac58, \qquad h(e^{-\phi/2})=\frac38, \qquad h(e^{ik\cdot X})=0,$$

so the integrated Ramond vertex has weight $1$ when $k^2=0$.

The physical-state condition is the massless Dirac equation

$$k_\mu\Gamma^\mu u=0.$$

For a closed-string R—R state, the standard vertex is a product of two Ramond vertices:

$$V_{\mathrm{RR}}^{(-1/2,-1/2)}(z,\bar z) = g_c\,c\tilde c\, e^{-\phi/2}S_A(z) e^{-\tilde\phi/2}\tilde S_B(\bar z) \mathcal F^{AB}e^{ik\cdot X(z,\bar z)}.$$

The bispinor $\mathcal F^{AB}$ is equivalent to a sum of antisymmetric R—R field strengths,

$$\mathcal F^{AB} = \sum_p \frac{1}{p!}F_{\mu_1\cdots \mu_p} \left(C\Gamma^{\mu_1\cdots\mu_p}\right)^{AB},$$

with the allowed values of $p$ determined by the Type IIA or Type IIB chirality choice.

The open Ramond vertex contains one spin field and $e^{-\phi/2}$. The closed R—R vertex contains a left and right spin field and packages R—R field strengths into a spacetime bispinor.

Closed-string NS vertices

The massless NS—NS vertex in the $(-1,-1)$ picture is

$$V_{\zeta}^{(-1,-1)}(z,\bar z) = g_c\,c\tilde c\, \zeta_{\mu\nu} \psi^\mu(z)\tilde\psi^\nu(\bar z) e^{-\phi(z)}e^{-\tilde\phi(\bar z)} e^{ik\cdot X(z,\bar z)}.$$

The physical conditions are

$$k^2=0, \qquad k^\mu\zeta_{\mu\nu}=0, \qquad k^\nu\zeta_{\mu\nu}=0,$$

with gauge equivalences inherited from BRST-exact vertices. The polarization decomposes as

$$\zeta_{\mu\nu} = \zeta_{(\mu\nu)}^{\mathrm{traceless}} + \zeta_{[\mu\nu]} + \frac{1}{D-2}\eta_{\mu\nu}^{\perp}\zeta^\rho{}_{\rho},$$

corresponding to the graviton, $B$-field, and dilaton.

The $(0,0)$ picture vertex is obtained by applying picture-changing separately in the left and right sectors. Its matter part is schematically

$$\zeta_{\mu\nu} \left( \partial X^\mu + k\cdot\psi\,\psi^\mu \right) \left( \bar\partial X^\nu + k\cdot\tilde\psi\,\tilde\psi^\nu \right)e^{ik\cdot X}.$$

Again, the exact numerical coefficients depend on normalization conventions, but the tensor structure is fixed.

Picture bookkeeping on the sphere

For a closed-string sphere amplitude, the left and right picture numbers must each add to $-2$:

$$\sum_i q_i=-2, \qquad \sum_i \tilde q_i=-2.$$

A standard NS—NS tree-level choice is

$$V_1^{(-1,-1)}, \qquad V_2^{(-1,-1)}, \qquad V_3^{(0,0)}, \qquad \int d^2z_4\,V_4^{(0,0)}, \ldots.$$

The first three vertices also usually carry the $c\tilde c$ ghosts needed to fix the $SL(2,\mathbb C)$ conformal Killing group.

On the sphere, the total picture number is $-2$ separately in each chiral sector. Three unintegrated vertices supply the $c\tilde c$ ghost zero modes.

For amplitudes involving Ramond states, one distributes half-integer pictures so that the same total rule is satisfied. For example, a two-Ramond two-NS closed-string amplitude may use Ramond vertices in $(-1/2,-1/2)$ and NS vertices chosen to complete the $(-2,-2)$ total.

Picture bookkeeping on the disk

For an open-string disk amplitude, the total picture number is

$$\sum_i q_i=-2.$$

A common $n$-gluon choice is

$$V_1^{(-1)}, \qquad V_2^{(-1)}, \qquad V_3^{(0)}, \qquad \int dy_4\,V_4^{(0)}, \ldots, \int dy_n\,V_n^{(0)}.$$

The three unintegrated boundary vertices contain $c$ ghosts and fix the $PSL(2,\mathbb R)$ automorphism group of the disk.

Open-string disk amplitudes require total picture number $-2$. Three boundary positions are fixed by $PSL(2,\mathbb R)$ and are represented by unintegrated vertices carrying $c$ ghosts.

This is the practical rule used in almost all tree-level NSR amplitude computations.

Why amplitudes do not depend on PCO positions

If $\mathcal X(z)=\{Q_{\mathrm{BRST}},\xi(z)\}$, then moving a picture-changing operator changes the integrand by a BRST-exact term, at least away from singularities:

$$\partial_z\mathcal X(z) = \{Q_{\mathrm{BRST}},\partial_z\xi(z)\}.$$

In a BRST-invariant amplitude, BRST-exact insertions decouple. Thus the amplitude is independent of PCO positions, provided no spurious singularities are crossed. At higher genus this statement becomes subtle because of supermoduli space, but at tree level it is a reliable computational principle.

Exercises

Exercise 1

Verify that the open-string Ramond vertex

$$c S_A e^{-\phi/2}e^{ik\cdot X}$$

has total conformal weight zero when $k^2=0$.

Solution

The unintegrated vertex contains a $c$ ghost of weight $-1$. For the matter-superghost part,

$$h(S_A)=\frac58, \qquad h(e^{-\phi/2})=\frac38, \qquad h(e^{ik\cdot X})=0$$

for a massless state. Therefore

$$h_{\mathrm{matter+superghost}}=\frac58+\frac38=1.$$

Multiplication by $c$ gives total weight $1-1=0$.

Exercise 2

In an open-string disk amplitude with four gauge bosons, choose pictures and unintegrated/integrated vertices satisfying both the picture-number and $c$-ghost zero-mode rules.

Solution

The disk has three conformal Killing vectors, so three boundary vertices should be unintegrated and carry $c$ ghosts. The total picture number must be $-2$. A standard choice is

$$V_1^{(-1)}(y_1), \qquad V_2^{(-1)}(y_2), \qquad V_3^{(0)}(y_3), \qquad \int dy_4\,V_4^{(0)}(y_4).$$

The first three vertices are unintegrated. The picture numbers add to $-1-1+0+0=-2$.

Exercise 3

Show that the NS—NS vertex in the $(-1,-1)$ picture has the correct total conformal weights for an unintegrated closed-string vertex.

Solution

In each chiral sector,

$$h(\psi^\mu)=\frac12, \qquad h(e^{-\phi})=\frac12, \qquad h(e^{ik\cdot X})=0$$

when $k^2=0$. Thus the left matter-superghost factor has $h=1$, and the right factor has $\bar h=1$. Multiplication by $c\tilde c$ gives

$$(h,\bar h)=(1,1)+(-1,-1)=(0,0),$$

as required for an unintegrated closed-string insertion.

Exercise 4

Explain why applying one picture-changing operator to a $(-1,-1)$ NS—NS vertex produces a $(-1,0)$ vertex, not a $(0,0)$ vertex.

Solution

The left and right superghost systems are independent. A single holomorphic picture-changing operator raises only the left picture:

$$(-1,-1)\longrightarrow (0,-1).$$

A single antiholomorphic picture-changing operator raises only the right picture:

$$(-1,-1)\longrightarrow (-1,0).$$

To obtain a $(0,0)$ vertex one must apply picture-changing in both sectors.