The Open Superstring Spectrum

We now build the open-string Hilbert space of the NSR theory. The two sectors have a wonderfully different character:

$$\text{NS sector} \longrightarrow \text{spacetime bosons}, \qquad \text{R sector} \longrightarrow \text{spacetime fermions}.$$

Before the GSO projection, the NS sector contains a tachyon. After the projection, the tachyon is removed and the massless NS vector combines with a massless R spinor into the ten-dimensional open-superstring vector multiplet. This page explains the spectrum before making the projection precise.

The open NSR string has two sectors. The NS sector has half-integer fermion modes and gives spacetime bosons; the R sector has integer fermion modes and gives spacetime fermions.

Open-string modes and mass formulas

For the open string, after doubling to a single holomorphic field, the standard mode expansions are

$$\partial X^\mu(z) = -i\sqrt{\frac{\alpha'}{2}} \sum_{n\in\mathbb Z}\alpha_n^\mu z^{-n-1},$$

and

$$\psi^\mu(z)=\sum_r \psi_r^\mu z^{-r-1/2}.$$

The oscillators obey

$$[\alpha_m^\mu,\alpha_n^\nu] =m\eta^{\mu\nu}\delta_{m+n,0}, \qquad \{\psi_r^\mu,\psi_s^\nu\} =\eta^{\mu\nu}\delta_{r+s,0}.$$

Creation operators have negative mode number. The matter number operator is

$$N = \sum_{n=1}^{\infty}\alpha_{-n}\cdot\alpha_n + \sum_{r>0}r\,\psi_{-r}\cdot\psi_r.$$

In the critical open superstring,

$$\boxed{\alpha'M^2_{\rm NS}=N-\frac12,} \qquad \boxed{\alpha'M^2_{\rm R}=N.}$$

The physical-state constraints are the quantum versions of $T=0$ and $G=0$:

$$(L_0-a)|\Psi\rangle=0, \qquad L_n|\Psi\rangle=0\quad(n>0),$$

and

$$G_r|\Psi\rangle=0\quad(r>0),$$

with the additional R-sector zero-mode condition

$$G_0|\Psi\rangle=0.$$

The supercurrent constraints are as important as the Virasoro constraints. In the R sector, $G_0$ becomes the spacetime Dirac operator on the massless ground state.

The NS sector: tachyon and massless vector

The NS vacuum satisfies

$$\alpha_n^\mu|0;k\rangle_{\rm NS}=0\quad(n>0), \qquad \psi_r^\mu|0;k\rangle_{\rm NS}=0\quad(r>0).$$

It has $N=0$, so

$$\alpha'M^2=-\frac12.$$

This is the open-string NS tachyon of the unprojected theory.

The first excited NS state is

$$|\zeta;k\rangle = \zeta_\mu\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}.$$

It has $N=1/2$, hence

$$M^2=0.$$

The state is therefore a candidate massless vector.

The unprojected NS sector contains a tachyonic ground state. The first fermionic excitation is a massless vector; this is the state that survives in the supersymmetric open string.

The physical-state condition $G_{1/2}|\zeta;k\rangle=0$ gives transversality. To see it, use the leading term

$$G_{1/2}\sim \sum_n \alpha_n\cdot\psi_{1/2-n}.$$

Only the zero-mode piece acts nontrivially on the one-fermion state, so

$$G_{1/2}\,\zeta_\mu\psi_{-1/2}^\mu|0;k\rangle \propto (k\cdot\zeta)|0;k\rangle.$$

Thus

$$k\cdot\zeta=0.$$

There is also a null state generated by $G_{-1/2}|0;k\rangle_{\rm NS}$, giving the equivalence

$$\zeta_\mu\sim \zeta_\mu+\lambda k_\mu.$$

This is exactly the linearized gauge redundancy of a massless vector.

The NS massless state is transverse, and longitudinal polarizations are null. In $D=10$ it has $D-2=8$ physical polarizations.

So the first non-tachyonic NS state is a spacetime gauge boson with $8$ transverse polarizations, transforming as the $8_v$ of the massless little group $SO(8)$.

Higher NS levels

Before the GSO projection, both even and odd worldsheet-fermion number states occur. The level immediately above the massless vector is $N=1$, with

$$\alpha'M^2=\frac12.$$

A simple representative is

$$\zeta_{\mu\nu}\psi_{-1/2}^\mu\psi_{-1/2}^\nu|0;k\rangle_{\rm NS},$$

where the two fermion oscillators make the polarization antisymmetric before constraints are imposed. The next odd-fermion level is $N=3/2$, with

$$\alpha'M^2=1.$$

At this level one encounters states of the schematic form

$$\psi_{-3/2}^\mu|0;k\rangle, \qquad \alpha_{-1}^\mu\psi_{-1/2}^\nu|0;k\rangle, \qquad \psi_{-1/2}^\mu\psi_{-1/2}^\nu\psi_{-1/2}^\rho|0;k\rangle.$$

The constraints remove time-like and longitudinal components, leaving massive representations of the little group $SO(9)$ in ten dimensions. The detailed decomposition is less important here than the pattern: the string has infinitely many massive higher-spin states, now arranged into NSR superconformal multiplets. In the conventional GSO projection, one keeps a definite fermion parity; this removes the NS tachyon and the even-fermion $N=1$ level while keeping the massless vector and the $N=3/2$ massive level.

The R sector: spacetime spinors

The Ramond vacuum is degenerate because the fermion zero modes satisfy

$$\{\psi_0^\mu,\psi_0^\nu\}=\eta^{\mu\nu}.$$

Equivalently, after defining $\Gamma^\mu=\sqrt2\psi_0^\mu$,

$$\{\Gamma^\mu,\Gamma^\nu\}=2\eta^{\mu\nu}.$$

So the ground state must carry a spinor index:

$$|0;k\rangle_{\rm R} \quad\longrightarrow\quad |u;k\rangle_{\rm R}.$$

The R ground state has $N=0$, and since $a_{\rm R}=0$,

$$M^2=0.$$

The R-sector zero-mode constraint becomes

$$G_0|u;k\rangle_{\rm R}=0.$$

On the ground state,

$$G_0\sim \sqrt{\frac{\alpha'}{2}}\,k_\mu\psi_0^\mu,$$

so the physical-state condition is the massless Dirac equation

$$k_\mu\Gamma^\mu u(k)=0.$$

The Ramond ground state is massless and spinorial. The constraint $G_0|u;k\rangle=0$ becomes the spacetime Dirac equation.

This is one of the most beautiful facts about the NSR formalism: spacetime fermions emerge from worldsheet fields $\psi^\mu$ that are spacetime vectors.

Preview: matching bosons and fermions

The physical massless NS vector has

$$D-2=8$$

polarizations in $D=10$. The R ground state, after the appropriate chirality projection, also has $8$ physical degrees of freedom. The GSO projection will remove the NS tachyon and choose one Ramond chirality, producing the open-superstring massless multiplet.

After the GSO projection, the massless open superstring contains an $SO(8)$ vector from the NS sector and one chiral $SO(8)$ spinor from the R sector. This is the ten-dimensional $\mathcal N=1$ vector multiplet.

The next two pages supply the missing ingredients: ten-dimensional spinors and the GSO projection.

Exercises

Exercise 1. The mass of the first NS excitation

Show that $\zeta_\mu\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}$ is massless.

Solution

The oscillator $\psi_{-1/2}$ raises the level by $1/2$, so

$$N=\frac12.$$

Using the NS mass formula

$$\alpha'M^2=N-\frac12,$$

we find

$$\alpha'M^2=\frac12-\frac12=0.$$

Thus the state is massless.

Exercise 2. Transversality from $G_{1/2}$

Derive $k\cdot\zeta=0$ for the NS massless vector.

Solution

At this level,

$$G_{1/2}\sim \alpha_0\cdot\psi_{1/2}+\cdots,$$

where the omitted terms annihilate the state. Since

$$\alpha_0^\mu=\sqrt{2\alpha'}\,k^\mu$$

for the open string, we get

$$G_{1/2}\zeta_\nu\psi_{-1/2}^\nu|0;k\rangle \propto k_\mu\zeta_\nu\{\psi_{1/2}^\mu,\psi_{-1/2}^\nu\}|0;k\rangle = (k\cdot\zeta)|0;k\rangle.$$

The physical-state condition sets this to zero, hence

$$k\cdot\zeta=0.$$

Exercise 3. Gauge equivalence from a null state

Show that $G_{-1/2}|0;k\rangle_{\rm NS}$ has the same form as a vector state with polarization proportional to $k_\mu$.

Solution

The relevant term in $G_{-1/2}$ is

$$G_{-1/2}\sim \alpha_0\cdot\psi_{-1/2}+\cdots.$$

Acting on the NS vacuum gives

$$G_{-1/2}|0;k\rangle_{\rm NS} \propto k_\mu\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}.$$

This is a vector state with polarization

$$\zeta_\mu\propto k_\mu.$$

It is null and is quotiented out of the physical Hilbert space. Therefore

$$\zeta_\mu\sim \zeta_\mu+\lambda k_\mu.$$

Exercise 4. The Ramond ground state is massless

Use the R-sector mass formula to show that the R ground state is massless.

Solution

The R ground state has no positive oscillator excitations, so

$$N=0.$$

The R-sector mass formula is

$$\alpha'M^2=N.$$

Therefore

$$M^2=0.$$

Exercise 5. The Dirac equation from $G_0$

Show that the R-sector condition $G_0|u;k\rangle=0$ gives $k_\mu\Gamma^\mu u=0$.

Solution

On the R ground state, oscillator terms with nonzero modes annihilate the state, so the relevant piece of $G_0$ is

$$G_0\sim \sqrt{\frac{\alpha'}{2}}\,k_\mu\psi_0^\mu.$$

The condition $G_0|u;k\rangle=0$ becomes

$$k_\mu\psi_0^\mu u=0.$$

Using $\Gamma^\mu=\sqrt2\psi_0^\mu$, this is equivalent to

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

This is the massless Dirac equation in momentum space.