Hagedorn Growth and String Thermodynamics

In light-cone gauge the physical Hilbert space of the critical bosonic string is completely transparent: the only creation operators are transverse oscillators. For the open string in $D=26$,

$$\alpha_{-n}^i, \qquad n=1,2,\ldots, \qquad i=1,\ldots,24.$$

The level operator is

$$N=\sum_{n=1}^\infty n\,N_n, \qquad N_n=\sum_{i=1}^{24}N_{n,i},$$

and the mass formula is

$$M^2=\frac{N-1}{\alpha'}.$$

The first few levels looked innocent:

$$d_0=1, \qquad d_1=24, \qquad d_2=324, \qquad d_3=3200, \qquad d_4=25650.$$

But the sequence grows much faster than any power of $N$. This rapid growth is not a minor combinatorial curiosity. It is one of the first signs that string theory has a characteristic high-energy thermodynamics, controlled by a limiting scale called the Hagedorn temperature.

What is being counted?

Here $d_N$ denotes the number of physical transverse oscillator states at fixed oscillator level $N$ for the open critical bosonic string. We are temporarily ignoring the center-of-mass momentum and the continuous phase-space degeneracy. Those affect the full thermodynamic partition function, but not the essential exponential growth in the internal string oscillator spectrum.

Counting open-string states

At level $N$, a state is built by applying oscillators whose mode numbers add up to $N$:

$$\alpha_{-n_1}^{i_1}\alpha_{-n_2}^{i_2}\cdots\alpha_{-n_k}^{i_k}|0;p\rangle, \qquad n_1+\cdots+n_k=N.$$

Because the oscillators are bosonic, repeated oscillators are symmetrized. Thus the problem is a partition problem, but with 24 colors, one for each transverse direction.

For example, at level $N=2$ there are two possibilities:

$$\alpha_{-2}^i|0;p\rangle, \qquad \alpha_{-1}^i\alpha_{-1}^j|0;p\rangle.$$

The first gives $24$ states. The second is symmetric in $i,j$, so it gives

$$\dim \operatorname{Sym}^2(\mathbb R^{24}) = \binom{24+2-1}{2} = \binom{25}{2} = 300.$$

Therefore

$$d_2=24+300=324.$$

Similarly, at level $N=3$,

$$3, \qquad 2+1, \qquad 1+1+1$$

give

$$d_3 = 24+24^2+\binom{24+3-1}{3} = 24+576+2600 = 3200.$$

This combinatorics is best packaged in a generating function.

Each oscillator mode $n$ can be occupied any nonnegative number of times in each of the $24$ transverse directions. A state at level $N$ is therefore a $24$-colored partition of $N$.

For each fixed pair $(n,i)$, the occupation number can be $0,1,2,\ldots$, so its contribution to the generating function is

$$1+w^n+w^{2n}+\cdots = \frac{1}{1-w^n}.$$

Multiplying over all oscillator modes and transverse directions gives

$$G(w) = \operatorname{Tr} w^N = \sum_{N=0}^\infty d_N w^N = \prod_{n=1}^\infty \frac{1}{(1-w^n)^{24}}.$$

Equivalently,

$$d_N=[w^N]\,G(w).$$

The notation $[w^N]$ means “coefficient of $w^N$.”

Coefficients from a contour integral

The coefficient $d_N$ may be extracted by Cauchy’s theorem:

$$d_N = \frac{1}{2\pi i} \oint_C \frac{dw}{w^{N+1}}G(w),$$

where $C$ is a small contour around the origin. For fixed, small $N$, this is just a formal way of extracting a Taylor coefficient. For large $N$, it becomes a saddle-point problem.

The dominant large-$N$ contribution comes from the singularity of $G(w)$ closest to the origin. Since

$$G(w)=\prod_{n=1}^\infty (1-w^n)^{-24},$$

the most important singularity is at

$$w=1.$$

Physically, $w\to1$ weights high oscillator levels less and less, so the trace over the infinitely many string states becomes singular.

The degeneracy $d_N$ is a contour integral in the complex $w$-plane. For large $N$, the contour can be analyzed by deforming it toward the dominant singularity near $w=1$.

To analyze this singularity, set

$$w=e^{-\beta}, \qquad \beta>0,$$

and take $\beta\to0^+$. Then

$$G(e^{-\beta}) = \prod_{n=1}^\infty (1-e^{-\beta n})^{-24}.$$

This product is closely related to the Dedekind eta function. With

$$q=e^{2\pi i\tau}, \qquad \eta(\tau)=q^{1/24}\prod_{n=1}^\infty(1-q^n),$$

we use

$$q=e^{-\beta}, \qquad \tau=\frac{i\beta}{2\pi}.$$

The modular transformation

$$\eta(-1/\tau)=(-i\tau)^{1/2}\eta(\tau)$$

implies, for $\beta\to0^+$,

$$\prod_{n=1}^\infty(1-e^{-\beta n}) \sim \left(\frac{2\pi}{\beta}\right)^{1/2} \exp\left(-\frac{\pi^2}{6\beta}\right).$$

Therefore

$$G(e^{-\beta}) \sim \left(\frac{\beta}{2\pi}\right)^{12} \exp\left(\frac{4\pi^2}{\beta}\right).$$

The exponential factor is the key. The power $\beta^{12}$ affects only the power-law prefactor in $d_N$, not the leading exponential growth.

The modular transformation of the eta function turns the $w\to1$ singularity into an explicit exponential. A saddle-point estimate then gives $d_N\sim e^{4\pi\sqrt N}$.

Using

$$w^{-N-1}\,dw \simeq -e^{N\beta}\,d\beta$$

near $w=e^{-\beta}$, the exponential part of the coefficient integral is

$$d_N \sim \int d\beta\, \exp\left( N\beta+\frac{4\pi^2}{\beta} \right).$$

The saddle point satisfies

$$0 = \frac{d}{d\beta} \left( N\beta+\frac{4\pi^2}{\beta} \right) = N-\frac{4\pi^2}{\beta^2},$$

so

$$\beta_*=\frac{2\pi}{\sqrt N}.$$

At the saddle,

$$N\beta_*+\frac{4\pi^2}{\beta_*} = 2\pi\sqrt N+2\pi\sqrt N = 4\pi\sqrt N.$$

Thus the open-string level degeneracy has the leading asymptotic behavior

$$d_N \sim \exp(4\pi\sqrt N), \qquad N\gg1.$$

More precisely,

$$d_N = \text{const}\times N^{-27/4}\exp(4\pi\sqrt N) \left(1+O(N^{-1/2})\right),$$

but the exponential is what controls the Hagedorn phenomenon.

Cardy formula viewpoint

This result is the same as the Cardy formula for a two-dimensional CFT with effective central charge

$$c_{\mathrm{eff}}=D-2=24.$$

The Cardy estimate gives

$$d_N\sim \exp\left(2\pi\sqrt{\frac{c_{\mathrm{eff}}N}{6}}\right) = \exp(4\pi\sqrt N).$$

This is an early preview of why two-dimensional CFT becomes the natural language for perturbative string theory.

From level degeneracy to density of massive string states

The open-string mass formula is

$$M^2=\frac{N-1}{\alpha'}.$$

At large level, the intercept is negligible:

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

Therefore

$$d_N \sim \exp(4\pi\sqrt N) \sim \exp(4\pi\sqrt{\alpha'}\,M).$$

This means that the density of string states grows exponentially with the mass:

$$\rho(M) \sim \exp(\beta_H M),$$

with

$$\boxed{ \beta_H=4\pi\sqrt{\alpha'}, \qquad T_H=\frac{1}{4\pi\sqrt{\alpha'}}. }$$

Here we use units $k_B=\hbar=c=1$.

For the closed string, the same Hagedorn temperature appears. The critical closed bosonic string has independent left- and right-moving transverse oscillators, with

$$N=\widetilde N$$

by level matching and

$$M^2=\frac{4}{\alpha'}(N-1).$$

The degeneracy at fixed matched level is roughly

$$d_N^{\mathrm{closed}} \sim d_N\,\widetilde d_N \sim \exp(8\pi\sqrt N).$$

But

$$N\simeq \frac{\alpha' M^2}{4},$$

so

$$\exp(8\pi\sqrt N) \simeq \exp(8\pi\cdot \frac{\sqrt{\alpha'}M}{2}) = \exp(4\pi\sqrt{\alpha'}M).$$

Thus both open and closed critical bosonic strings have the same leading Hagedorn inverse temperature:

$$\beta_H=4\pi\sqrt{\alpha'}.$$

Open strings have one tower of oscillators, while closed strings have left- and right-moving towers. Level matching changes the mass-level relation in just the right way for the leading Hagedorn temperature to agree.

Why the canonical partition function diverges

Consider the single-string canonical partition function, suppressing momentum-space powers:

$$Z_1(\beta) \sim \int^\infty dM\,\rho(M)e^{-\beta M}.$$

Using

$$\rho(M)\sim e^{\beta_H M},$$

we get

$$Z_1(\beta) \sim \int^\infty dM\, e^{-(\beta-\beta_H)M}.$$

This integral converges only when

$$\beta>\beta_H,$$

or equivalently

$$T<T_H.$$

For

$$\beta\leq \beta_H,$$

the exponential growth of the number of string states overwhelms the Boltzmann suppression.

The density of states grows as $\rho(M)\sim e^{\beta_H M}$. The canonical Boltzmann factor $e^{-\beta M}$ suppresses high masses only when $\beta>\beta_H$.

This is the string-theoretic analog of the Hagedorn behavior originally discovered in hadronic physics: adding energy does not simply populate a fixed number of particle species with higher momenta. Instead, it opens up exponentially many internal string oscillator states.

Bosonic tachyon caveat

The critical bosonic string has a tachyon, so its thermal ensemble is not a stable vacuum ensemble in the same sense as a tachyon-free superstring. The Hagedorn calculation above isolates the high-level oscillator growth. That growth, and the corresponding Hagedorn singularity, survive in tachyon-free string theories, although the numerical value of $T_H$ changes when the worldsheet degrees of freedom and GSO projection change.

Physical interpretation

The Hagedorn scale has several complementary meanings.

First, it is a statement about the spectrum. Strings have infinitely many oscillator modes, and the number of ways to distribute a large level $N$ among them grows exponentially in $\sqrt N$. Since $M\sim\sqrt N$, this becomes exponential growth in $M$.

Second, it is a statement about thermal equilibrium. In a canonical ensemble, high-mass string states are Boltzmann-suppressed by $e^{-\beta M}$, but the number of such states is enhanced by $e^{\beta_H M}$. When $\beta$ approaches $\beta_H$ from above, the competition becomes marginal.

Third, it is a statement about the extended nature of strings. A point particle with finitely many species has a density of internal states that is finite or polynomial. A string has an infinite tower of vibrational modes, and the high-energy thermodynamics remembers this tower.

At this stage in the course, the Hagedorn calculation also serves a structural purpose. We have seen that even the free string knows about modular functions, contour integrals, and asymptotic CFT state counting. These tools will reappear immediately when we study perturbative string interactions as sums over worldsheet geometries.

Exercises

Exercise 1. Low-level degeneracies

Compute $d_2$, $d_3$, and $d_4$ for the open critical bosonic string using only transverse oscillators.

Solution

At level $N=2$, the partitions are $2$ and $1+1$:

$$d_2 = 24+\binom{25}{2} = 24+300 = 324.$$

At level $N=3$, the partitions are $3$, $2+1$, and $1+1+1$:

$$d_3 = 24+24^2+\binom{26}{3} = 24+576+2600 = 3200.$$

At level $N=4$, the partitions are

$$4,\qquad 3+1,\qquad 2+2,\qquad 2+1+1,\qquad 1+1+1+1.$$

Thus

$$d_4 = 24 + 24^2 + \binom{25}{2} + 24\binom{25}{2} + \binom{27}{4}.$$

Using

$$\binom{25}{2}=300, \qquad \binom{27}{4}=17550,$$

we find

$$d_4 = 24+576+300+7200+17550 = 25650.$$

Exercise 2. Derive the generating function

Show that

$$G(w)=\operatorname{Tr}w^N = \prod_{n=1}^{\infty}(1-w^n)^{-24}.$$

Solution

For each oscillator $\alpha_{-n}^i$, the occupation number $N_{n,i}$ can be any nonnegative integer. Its contribution to the trace is

$$\sum_{k=0}^{\infty}w^{nk} = \frac{1}{1-w^n}.$$

There are $24$ transverse directions for each $n$, so fixed $n$ contributes

$$\prod_{i=1}^{24}\frac{1}{1-w^n} = (1-w^n)^{-24}.$$

Multiplying over all positive modes gives

$$G(w) = \prod_{n=1}^{\infty}(1-w^n)^{-24}.$$

Exercise 3. Saddle-point estimate

Assume

$$G(e^{-\beta}) \sim \left(\frac{\beta}{2\pi}\right)^{12} e^{4\pi^2/\beta}$$

as $\beta\to0^+$. Ignoring the power of $\beta$, use a saddle-point approximation to show that

$$d_N\sim e^{4\pi\sqrt N}.$$

Solution

Near $w=e^{-\beta}$, the coefficient integral has exponential part

$$d_N \sim \int d\beta\, \exp\left( N\beta+\frac{4\pi^2}{\beta} \right).$$

Define

$$S(\beta)=N\beta+\frac{4\pi^2}{\beta}.$$

The saddle satisfies

$$S'(\beta)=N-\frac{4\pi^2}{\beta^2}=0,$$

so

$$\beta_*=\frac{2\pi}{\sqrt N}.$$

Then

$$S(\beta_*) = N\frac{2\pi}{\sqrt N} + \frac{4\pi^2}{2\pi/\sqrt N} = 2\pi\sqrt N+2\pi\sqrt N = 4\pi\sqrt N.$$

Therefore

$$d_N\sim e^{4\pi\sqrt N},$$

up to a power-law prefactor.

Exercise 4. Same Hagedorn temperature for closed strings

Using

$$d_N^{\mathrm{open}}\sim e^{4\pi\sqrt N},$$

show that the closed bosonic string has the same leading $\beta_H$ as the open bosonic string.

Solution

For a closed string, level matching gives

$$N=\widetilde N,$$

and the large-level degeneracy is approximately

$$d_N^{\mathrm{closed}} \sim d_N\widetilde d_N \sim e^{4\pi\sqrt N}e^{4\pi\sqrt N} = e^{8\pi\sqrt N}.$$

The closed-string mass formula is

$$M^2=\frac{4}{\alpha'}(N-1).$$

At large $N$,

$$N\simeq \frac{\alpha'M^2}{4}, \qquad \sqrt N\simeq \frac{\sqrt{\alpha'}M}{2}.$$

Therefore

$$d_N^{\mathrm{closed}} \sim \exp\left(8\pi\cdot\frac{\sqrt{\alpha'}M}{2}\right) = \exp(4\pi\sqrt{\alpha'}M).$$

Thus

$$\beta_H=4\pi\sqrt{\alpha'}, \qquad T_H=\frac{1}{4\pi\sqrt{\alpha'}}.$$

Exercise 5. General number of transverse bosons

Suppose a light-cone string had $d$ transverse bosonic oscillators, so that

$$G_d(w)=\prod_{n=1}^\infty (1-w^n)^{-d}.$$

Show that the leading open-string growth is

$$d_N\sim \exp\left(2\pi\sqrt{\frac{dN}{6}}\right),$$

and find the corresponding open-string Hagedorn inverse temperature.

Solution

For $d$ transverse bosons, the Cardy estimate gives

$$d_N \sim \exp\left(2\pi\sqrt{\frac{dN}{6}}\right).$$

For an open string with

$$M^2\simeq \frac{N}{\alpha'},$$

we have

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

Therefore

$$d_N \sim \exp\left( 2\pi\sqrt{\frac{d\alpha'}{6}}\,M \right).$$

Thus the Hagedorn inverse temperature is

$$\beta_H = 2\pi\sqrt{\frac{d\alpha'}{6}}.$$

For the critical bosonic string, $d=24$, so

$$\beta_H = 2\pi\sqrt{4\alpha'} = 4\pi\sqrt{\alpha'}.$$