Bosonization, Spin Fields, and Superghosts

The Ramond sector forces us to describe operators that change the boundary condition of a worldsheet fermion. These operators are called spin fields. The most efficient way to construct them is bosonization: pairs of real fermions are rewritten as exponentials of chiral bosons.

The same idea is also used for the superconformal ghosts. The commuting $\beta\gamma$ ghost system is awkward in its original form, but it becomes transparent after writing it in terms of a scalar $\phi$ and an $\eta\xi$ system. This is the origin of picture number in the NSR formalism.

Bosonizing worldsheet fermions

For a holomorphic free boson normalized as

$$H_I(z)H_J(w)\sim -\delta_{IJ}\ln(z-w),$$

the exponential operator

$$:e^{i a H_I(z)}:$$

has conformal weight

$$h=\frac{a^2}{2}.$$

Now pair the real worldsheet fermions into complex fermions:

$$\psi^{\pm I} = \frac{1}{\sqrt2} \left(\psi^{2I-1}\pm i\psi^{2I}\right).$$

Bosonization writes

$$\psi^{+I}\sim e^{+iH_I}, \qquad \psi^{-I}\sim e^{-iH_I}.$$

The conformal weight is correct because $e^{\pm iH_I}$ has $h=1/2$, the weight of a chiral fermion.

Bosonization replaces a pair of real fermions by exponentials of a chiral boson. Cocycle factors are needed for exact anticommutation between different fermion pairs.

The symbol $\sim$ hides cocycle factors. These are important for global signs and gamma-matrix algebra, but for many local conformal-weight and OPE computations the exponential part carries the essential information.

Spin fields

The Ramond ground states are created by spin fields. For ten real fermions, we introduce five bosons $H_I$, $I=1,\ldots,5$, and define

$$S_{\vec s}(z) =: \exp\left(i\sum_{I=1}^{5}s_IH_I(z)\right):, \qquad s_I=\pm\frac12.$$

The conformal weight is

$$h(S_{\vec s}) = \frac12\sum_{I=1}^{5}s_I^2 = \frac12\cdot 5\cdot\frac14 =\frac58.$$

This is the correct weight of a ten-dimensional Ramond spin field. In light-cone gauge one often uses only the eight transverse fermions. Then there are four bosons $H_I$ and the transverse spin fields have

$$h=\frac12.$$

Spin fields have half-integer bosonized momenta $s_I=\pm 1/2$. Their conformal weight follows immediately from the free-boson exponential formula.

The chirality of a spin field is encoded in the pattern of signs in $\vec s$. A common convention is that the product of the signs determines the eigenvalue of $\Gamma_{11}$, but the precise labeling of $S_+$ and $S_-$ depends on the gamma-matrix convention. The physics is invariant under a simultaneous relabeling of the two chiralities.

Branch cuts and the Ramond sector

A spin field changes the boundary condition of a fermion. Locally, the OPE has a square-root singularity:

$$\psi^\mu(z)S_A(w) \sim \frac{1}{(z-w)^{1/2}} (\Gamma^\mu)_{AB}S^B(w)+\cdots.$$

This branch cut is not a bug; it is the point. Taking a fermion around a Ramond insertion changes its sign. That is the operator-state correspondence version of the Ramond boundary condition.

The OPE $\psi^\mu(z)S_A(w)$ contains a square-root branch cut. A Ramond spin field changes the fermion boundary condition.

In bosonized language this is immediate. Since

$$\psi^{+I}(z)S_{\vec s}(w) \sim (z-w)^{s_I} S_{\vec s+\vec e_I}(w),$$

the exponent is half-integer when $s_I=\pm 1/2$. The branch cut is the local signature of a spinor operator.

R-sector vertex operators before superghosts

Ignoring superghosts for a moment, the matter part of a massless Ramond vertex is

$$u_A S^A e^{ik\cdot X}.$$

For a massless state $k^2=0$, the plane-wave part contributes no conformal weight, while $S^A$ contributes $5/8$. A physical integrated open-string boundary vertex must have total weight $1$, so this is not enough. The missing $3/8$ comes from the superghost factor $e^{-\phi/2}$.

This is why superghosts are not cosmetic: they are required by worldsheet gauge fixing and by conformal weight.

The $\beta\gamma$ superghost system

Gauge fixing local worldsheet supersymmetry introduces a commuting ghost system

$$\beta(z),\gamma(z),$$

with conformal weights

$$h_\beta=\frac32, \qquad h_\gamma=-\frac12.$$

Their OPE is

$$\beta(z)\gamma(w)\sim \frac{1}{z-w},$$

and their central charge is

$$c_{\beta\gamma}=11.$$

This is exactly what is needed in the critical NSR string: the matter central charge is

$$c_{X,\psi}=D+\frac{D}{2}=\frac{3D}{2},$$

and the ghost central charge is

$$c_{bc}+c_{\beta\gamma}=-26+11=-15.$$

The total central charge vanishes for

$$\frac{3D}{2}-15=0, \qquad D=10.$$

The $\beta\gamma$ system is bosonized as

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

with

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

The scalar $\phi$ has a background charge. With the standard NSR convention,

$$h(e^{q\phi})=-\frac12q(q+2).$$

The superghost bosonization formula explains the characteristic factors $e^{-\phi}$ and $e^{-\phi/2}$ in NSR vertex operators.

Important examples are

$$h(e^{-\phi})=\frac12, \qquad h(e^{-\phi/2})=\frac38, \qquad h(e^{-3\phi/2})=\frac38.$$

Thus the standard massless vertices have the right conformal weights:

$$V_{\rm NS}^{(-1)} = \zeta_\mu e^{-\phi}\psi^\mu e^{ik\cdot X},$$

and

$$V_{\rm R}^{(-1/2)} = u_A e^{-\phi/2}S^A e^{ik\cdot X}.$$

For $k^2=0$, the NS vertex has weight

$$\frac12+\frac12=1,$$

and the R vertex has weight

$$\frac38+\frac58=1.$$

The factor $e^{-\phi/2}$ supplies the missing conformal weight needed by a massless Ramond vertex.

Picture number

The exponent of $e^{q\phi}$ is called the picture. For example,

$$V_{\rm NS}^{(-1)} \quad\text{has picture }-1,$$

while

$$V_{\rm R}^{(-1/2)} \quad\text{has picture }-\frac12.$$

Different pictures can describe the same physical state. The picture-changing operator is

$$X(z)=\{Q_{\rm BRST},\xi(z)\},$$

and its leading matter term is schematically

$$X(z)\sim e^{\phi}T_F(z)+\cdots.$$

Applying $X$ raises picture number by one. On the sphere, the total picture must be fixed:

$$\text{open string: total picture } -2,$$

and

$$\text{closed string: total picture } (-2,-2).$$

Picture-changing relates different representatives of the same BRST cohomology class. The total picture is fixed by the superghost zero modes.

The next page will use these ingredients to write the standard NSR vertex operators in the pictures most useful for amplitudes.

Exercises

Exercise 1. Weight of a bosonized fermion

Using $H(z)H(w)\sim-\ln(z-w)$, show that $e^{\pm iH}$ has conformal weight $1/2$.

Solution

For a free boson with this normalization,

$$h(e^{iaH})=\frac{a^2}{2}.$$

Setting $a=\pm1$ gives

$$h(e^{\pm iH})=\frac{1}{2}.$$

This matches the conformal weight of a chiral Majorana fermion.

Exercise 2. Weight of a ten-dimensional spin field

Compute the conformal weight of

$$S_{\vec s}=\exp\left(i\sum_{I=1}^{5}s_IH_I\right), \qquad s_I=\pm\frac12.$$

Solution

The weight is additive over the five independent bosons:

$$h=\frac12\sum_{I=1}^{5}s_I^2.$$

Since each $s_I^2=1/4$,

$$h=\frac12\cdot 5\cdot\frac14=\frac58.$$

Thus a ten-dimensional Ramond spin field has weight $5/8$.

Exercise 3. The branch cut from bosonization

For one bosonized pair, show that $\psi^{+}(z)S_s(w)$ has a square-root branch cut when $s=\pm 1/2$.

Solution

Using

$$\psi^+(z)=e^{iH(z)}, \qquad S_s(w)=e^{isH(w)},$$

and the exponential OPE,

$$e^{iaH(z)}e^{ibH(w)}\sim (z-w)^{ab}e^{i(a+b)H(w)},$$

we find

$$\psi^+(z)S_s(w) \sim (z-w)^s S_{s+1}(w).$$

For $s=\pm1/2$, the exponent is half-integer, so the OPE has a square-root branch cut.

Exercise 4. Superghost exponential dimensions

Use

$$h(e^{q\phi})=-\frac12q(q+2)$$

to compute $h(e^{-\phi})$, $h(e^{-\phi/2})$, and $h(e^{-3\phi/2})$.

Solution

For $q=-1$,

$$h(e^{-\phi})=-\frac12(-1)(1)=\frac12.$$

For $q=-1/2$,

$$h(e^{-\phi/2}) =-\frac12\left(-\frac12\right)\left(\frac32\right) =\frac38.$$

For $q=-3/2$,

$$h(e^{-3\phi/2}) =-\frac12\left(-\frac32\right)\left(\frac12\right) =\frac38.$$

Exercise 5. Dimension of the massless Ramond vertex

Show that

$$V_R^{(-1/2)}=u_A e^{-\phi/2}S^A e^{ik\cdot X}$$

has conformal weight one for $k^2=0$.

Solution

The three factors contribute

$$h(e^{-\phi/2})=\frac38,$$ $$h(S^A)=\frac58,$$

and, for a massless state,

$$h(e^{ik\cdot X})=0.$$

Therefore

$$h(V_R^{(-1/2)}) = \frac38+\frac58+0=1.$$

This is the required weight for an integrated open-string boundary vertex.