# On the spectrum of a quantised open string

I'm learning string theory from the book by Zwiebach and others. I'm trying to understand the quantisation of the open string and its mass spectrum.

In light-cone gauge the mass-shell condition of an open string is given by:

$$M^2 = 2(N - 1)/l_s^2$$

where $$N = \sum_{i=1}^{D-2}\sum_{n=1}^\infty \alpha^i_{-n}\alpha^i_n$$ and $$l_s$$ is the string length scale.

Now to determine the mass spectrum of the string, we can look at the values of $$N$$:

• For $$N=0$$, there is a tachyon since $$M^2$$ is negative
• For $$N=1$$, there is a vector boson $$\alpha^i_-1 |0;k\rangle$$.
• For $$N=2$$, we have that $$M^2$$ is positive and the states are given by: $$\alpha^i_{-2}|0;k\rangle$$ and $$\alpha^i_{-1}\alpha^j_{-1}|0;k\rangle$$

Now this is what I don't understand:

• Why is the state with $$N=1$$ a vector, why is not a scalar? How does one determine if a state is a vector or scalar?

• In the material I've read, it is claimed that Lorentz invariance requires that the the state with $$N=1$$ is massless, but I don't understand why is this case.

• Finally, the number of states with $$N=2$$ is claimed to 324 because it is the number of independent components of a matrix representation of $$SO(25)$$, why is this? Also this state is said to have a single massive state with spin-2, why is this?

• Hint: how do the operators $\alpha^i$ transform? Second hint: Wigner's classification. May 14, 2021 at 13:54
• This issue is discussed in Tongs ST notes (ch-2,3). If you're not familiar with Wigner classification of unitary representation of poincare group you have to check that first. May 14, 2021 at 14:42

## Level 1 states

The core idea that we will use here and below is Wigner's classification. In a $$\mathbb{R}^{1,p}$$ spacetime, the $$\alpha_{n}^i, 0\le i\le p-2$$ transform as a vector of $$\mathrm{SO}(p-1)$$, which is itself a subgroup of $$\mathrm{SO}(1, p)$$. But massive states are classified by irreducible representations of $$\text{SO}(p)$$, the stabiliser subgroup of the momentum after moving to the rest frame. So there's no way that $$\alpha_{-1}^i|0, \vec k\rangle$$ is a massive state - the representations of $$\text{SO}(p)$$ are too large to hold it.

However, massless states are classified under $$\text{ISO}(p-1)\subset\mathrm{SO}(1, p)$$, which similarly makes sense when you go to a frame where the associated momentum only has a component in a single direction. This stabiliser subgroup is made even more manifest when this direction is chosen to be the non-time dimension that light-cone quantisation singles out.

The main problem with $$\text{ISO}(p-1)$$ is that it has infinite-dimensional unitary representations, which would label states by "continuous spin" parameters - something which we don't observe in nature. In fact, this is a mathematical requirement in string theory: one can rigorously show that continuous spin representations do not appear in the perturbative string spectrum, see: arxiv.org/abs/1302.4771 (this is an often overlooked "prediction" of string theory). However, as in Wigner's classification for single particle states in quantum field theory, we only consider the finite dimensional representations of $$\text{SO}(p-1)\subset \text{ISO}(p-1)$$, matching the transformation properties of the $$\alpha^i$$ and yielding discrete labels on the states corresponding to the $$D\!=\!4$$ notion of helicity.

The state $$\alpha_{-1}^i|0, \vec k\rangle$$ thus has the right number of degrees of freedom and the right transformation under the little group $$\mathrm{SO}(p-1)$$ to be considered a massless vector boson, conventionally called the photon. Masslessness of this state then demands that $$p=25$$ for consistency, though this can be derived in a host of other ways.

## Level 2 states

First, a quick sanity check on the number of states: at level-2, we have $$\alpha^i_{-1}\alpha^j_{-1}|0, k\rangle$$ having $$\frac{24\times 25}{2}$$ states due to the symmetry in $$i\leftrightarrow j$$, and $$\alpha^i_{-2}|0, k\rangle$$ having $$24$$ states, which sums to $$324$$. This is indeed equal to the DOF of the traceless symmetric representation of $$\mathrm{SO}(25)$$, as $$324=\frac{25\times 26}{2}-1$$.

These representations decompose as

States Representation Irreducible decomposition
$$\alpha^i_{-1}\alpha^j_{-1}|0, k\rangle$$ Symmetric $$\mathbf{300_s}$$ Traceless symmetric + Trivial $$\mathbf{299_t}\oplus\mathbf{1}$$
$$\alpha^i_{-2}|0, k\rangle$$ Vector $$\mathbf{24_v}$$ Vector $$\mathbf{24_v}$$

It is an artifact of the lightcone quantisation method that only the $$\mathrm{SO}(p-1)$$ subgroup of the true $$\mathrm{SO}(p)$$ spin symmetry group is manifest. The presence of the full symmetry group can be proven rigorously to all levels by a detour through the covariant method, but it is quite tedious. For the low-lying spectrum however, the analysis is easy. The goal is to package the level 2 states into a representation of $$\mathrm{SO}(25)$$ exactly.

On transforming under the $$\mathrm{SO}(24)$$ subgroup, the traceless symmetric representation $$\mathbf{324_t}$$ of $$\mathrm{SO}(25)$$ obviously breaks up into $$\mathbf{299_t}\oplus\mathbf{24}_v \oplus \mathbf 1$$:

$$\left[\begin{array}{cccc|c} 0&x_{1,0}&...&x_{23,0}&x_{24,0}\\ x_{1,0}&0&...&x_{23,1}&x_{24,1}\\ \small\vdots&\small\vdots&\small\ddots&\small\vdots&\small\vdots\\ x_{23,0}&x_{23,1}&...&0&x_{24,23}\\\hline x_{24,0}&x_{24,1}&...&x_{24,23}&x_{24,24} \end{array}\right]$$ $$\Downarrow$$ $$\left[\begin{array}{ccc|c} \\&\mathbf{299_t}&&\cong\!\mathbf{24_v}\\\\\hline &\mathbf{24_v}&&\mathbf{1} \end{array}\right]$$

The $$\mathbf{299_t}\oplus\mathbf 1$$ corresponds to the reducible $$\alpha^i_{-1}\alpha^j_{-1}|0, k\rangle$$ while the $$\mathbf{24}_v$$ corresponds to the irreducible $$\alpha^i_{-2}|0, k\rangle$$. Thus the level-2 states can be classified by representations of $$\mathrm{SO}(25)$$, unlike the level-1 states.

But according to Wigner's classification for massive particles mentioned above, this means that the spectrum at level 2 must correspond to a massive state, since it furnishes a representation of $$\mathrm{SO}(p)$$ - this same reasoning also applies to the higher level states as well. The fact that this particular representation is traceless symmetric means that the spin of this state is 2, as illustrated here.