D-Brane Interactions and the Annulus Amplitude

A D-brane is not merely a boundary condition for open strings. It is also a source for closed-string fields. The annulus is the cleanest place where these two meanings become the same calculation.

For two parallel D$p$-branes separated by a transverse distance $y$, the annulus may be read in two ways:

• in the open-string channel, it is a one-loop vacuum energy of strings stretched from one brane to the other;
• in the closed-string channel, it is tree-level exchange of closed strings between two boundary states.

The equality of these descriptions is much stronger than a qualitative slogan. The oscillator partition function knows the entire force law: NS—NS attraction, R—R repulsion, BPS cancellation, and the brane—antibrane tachyon.

Throughout this page, take two static flat D$p$-branes extended along $x^0,x^1,\ldots,x^p$ and separated in the transverse directions $x^i$, $i=p+1,\ldots,9$. Let

$$y^2=(y_1-y_2)^i(y_1-y_2)^i.$$

The regulated brane worldvolume is denoted by $V_{p+1}$.

The stretched-string spectrum

An open string stretched between two parallel branes has a classical piece in each transverse Dirichlet direction,

$$X^i_{\rm cl}(\tau,\sigma)=y_1^i+{y^i\over \pi}\sigma, \qquad 0\leq \sigma\leq \pi.$$

The classical length is $|y|$, so the string carries the stretching energy

$$E_{\rm stretch}=T_{\rm F}|y|={|y|\over 2\pi\alpha'}.$$

In the open-string mass formula this appears as

$$M_{\rm stretch}^2={y^2\over (2\pi\alpha')^2} ={y^2\over 4\pi^2\alpha'^2}.$$

Thus the NS and R sectors have

$$\boxed{ \alpha' M_{\rm NS}^2={y^2\over 4\pi^2\alpha'}+N_{\rm NS}-{1\over 2}, \qquad \alpha' M_{\rm R}^2={y^2\over 4\pi^2\alpha'}+N_{\rm R}. }$$

For two identical BPS D-branes, the GSO projection removes the NS ground-state tachyon. For a brane—antibrane pair, the GSO projection in the stretched sector is reversed, and the tachyon returns when $y$ is sufficiently small. We will see both statements directly in the annulus.

The annulus as an open-string trace

The open-channel annulus has modulus $t$. Its universal Schwinger form is

$$\mathcal A(y) =2V_{p+1}\int_0^\infty {dt\over 2t} \int {d^{p+1}k\over (2\pi)^{p+1}} \operatorname{Tr}_{\rm open} \exp \left[-2\pi t\left( \alpha'k^2+{y^2\over 4\pi^2\alpha'}+N-a \right)\right].$$

The factor $2$ counts the two orientations of a string stretched between the two branes. The factor $1/(2t)$ is the annulus modulus measure, including the residual conformal Killing symmetry of the cylinder.

The Gaussian integral over momentum tangent to the brane gives

$$\int {d^{p+1}k\over (2\pi)^{p+1}} \exp(-2\pi\alpha't k^2) =(8\pi^2\alpha't)^{-(p+1)/2}.$$

Therefore every parallel-brane annulus has the form

$$\boxed{ \mathcal A(y) =2V_{p+1}\int_0^\infty {dt\over 2t} (8\pi^2\alpha't)^{-(p+1)/2} \exp\left[-{y^2t\over 2\pi\alpha'}\right] Z_{\rm osc}(t). }$$

All the physics is in the oscillator partition function $Z_{\rm osc}(t)$.

The open-channel annulus is a one-loop trace over stretched open strings. For identical BPS branes, the NS and R oscillator sums combine into the Jacobi identity.

Warm-up: the bosonic string

For the bosonic string in $26$ dimensions, the transverse oscillator trace gives $24$ copies of a bosonic oscillator. With

$$q=e^{-\pi t}, \qquad f_1(q)=q^{1/12}\prod_{n=1}^{\infty}(1-q^{2n}),$$

the bosonic partition function is

$$Z_{\rm bos}(t)={1\over f_1(q)^{24}}.$$

Hence

$$\mathcal A_{\rm bos}(y) =2V_{p+1}\int_0^\infty {dt\over 2t} (8\pi^2\alpha't)^{-(p+1)/2} \exp\left[-{y^2t\over 2\pi\alpha'}\right] {1\over f_1(q)^{24}}.$$

The leading factor in $f_1(q)^{-24}$ is the open-string tachyon. This is not a small technical blemish: the bosonic D-brane vacuum is unstable because the underlying bosonic string vacuum is unstable.

The superstring changes the story in two ways. First, longitudinal matter and ghosts cancel, leaving only the eight physical transverse bosons. Second, the eight physical transverse fermions appear with different spin structures. Their delicate sum is the origin of the BPS no-force condition.

The four $f$-functions

In the RNS superstring it is convenient to use the standard functions

$$\begin{aligned} f_1(q)&=q^{1/12}\prod_{n=1}^{\infty}(1-q^{2n}),\\ f_2(q)&=\sqrt{2}\,q^{1/12}\prod_{n=1}^{\infty}(1+q^{2n}),\\ f_3(q)&=q^{-1/24}\prod_{n=1}^{\infty}(1+q^{2n-1}),\\ f_4(q)&=q^{-1/24}\prod_{n=1}^{\infty}(1-q^{2n-1}). \end{aligned}$$

The denominator $f_1(q)^8$ is the contribution of the eight transverse bosonic oscillators. The numerator distinguishes the fermionic spin structures:

Open-string trace	Oscillator factor	Interpretation
NS trace without $(-1)^F$	$f_3(q)^8$	all NS fermion oscillators
NS trace with $(-1)^F$	$f_4(q)^8$	insertion distinguishing worldsheet fermion number
R trace	$f_2(q)^8$	Ramond zero modes and oscillators

The Jacobi abstruse identity says

$$\boxed{ f_3(q)^8-f_4(q)^8-f_2(q)^8=0. }$$

This identity is much more than a theta-function curiosity. It is the worldsheet expression of spacetime supersymmetry for a pair of identical BPS branes.

Identical BPS D-branes: exact cancellation

For two identical parallel D$p$-branes, the GSO-projected open-string partition function is

$$\boxed{ Z_{\rm DD}(t) ={1\over 2} {f_3(q)^8-f_4(q)^8-f_2(q)^8\over f_1(q)^8}=0. }$$

The full annulus amplitude is therefore

$$\boxed{ \mathcal A_{\rm DD}(y) =2V_{p+1}\int_0^\infty {dt\over 2t} (8\pi^2\alpha't)^{-(p+1)/2} \exp\left[-{y^2t\over 2\pi\alpha'}\right] {f_3(q)^8-f_4(q)^8-f_2(q)^8\over 2f_1(q)^8} =0. }$$

This vanishing is exact for every value of the separation $y$. Since $y$ is a modulus, the static potential between parallel BPS D-branes is flat:

$$V_{\rm DD}(y)=0.$$

It is important not to misread this statement. The D-branes do couple to the graviton, dilaton, and Ramond—Ramond fields. The point is that the attractive NS—NS exchange and the repulsive R—R exchange cancel exactly.

One way to see the level-by-level cancellation is to expand the two pieces. The GSO-projected NS spectrum contributes

$${1\over2}{f_3(q)^8-f_4(q)^8\over f_1(q)^8} =8+128q^2+1152q^4+\cdots,$$

while the R spectrum contributes

$${1\over2}{f_2(q)^8\over f_1(q)^8} =8+128q^2+1152q^4+\cdots.$$

The same number of bosonic and fermionic open-string states appears at every mass level. The annulus vacuum energy vanishes because the trace includes the usual minus sign for spacetime fermions.

The abstruse identity is the oscillator form of the BPS no-force condition. In spacetime language, NS—NS attraction cancels R—R repulsion.

Closed-string channel and the force law

Set

$$\ell={1\over t}.$$

The same annulus is then a closed string propagating for proper time $\ell$ between two boundary states:

$$\mathcal A(y)=\langle B_p,y_1|\Delta|B_p,y_2\rangle.$$

The modular transformations of the $f$-functions are

$$\begin{aligned} f_1(e^{-\pi/\ell})&=\sqrt{\ell}\,f_1(e^{-\pi\ell}),\\ f_2(e^{-\pi/\ell})&=f_4(e^{-\pi\ell}),\\ f_3(e^{-\pi/\ell})&=f_3(e^{-\pi\ell}),\\ f_4(e^{-\pi/\ell})&=f_2(e^{-\pi\ell}). \end{aligned}$$

The region $t\to0$ is thus the region $\ell\to\infty$. What looks ultraviolet in the open-string loop is infrared in the closed-string channel. At large separation $y\gg\sqrt{\alpha'}$, the amplitude is dominated by massless closed strings.

The massless exchange in the $9-p$ transverse dimensions is governed by

$$G_{9-p}(y)=\int {d^{9-p}k\over (2\pi)^{9-p}}{e^{ik\cdot y}\over k^2} ={\Gamma\left({7-p\over2}\right)\over 4\pi^{(9-p)/2}}{1\over y^{7-p}}, \qquad p<7.$$

For two identical BPS D$p$-branes, the long-distance interaction has the schematic form

$$V_{\rm int}(y) \sim -2\kappa_0^2 \tau_p^2 G_{9-p}(y) +2\kappa_0^2 \mu_p^2 G_{9-p}(y).$$

The first term is NS—NS attraction, mainly graviton and dilaton exchange. The second term is R—R repulsion. Supersymmetry enforces

$$\mu_p=\tau_p,$$

so the two terms cancel.

Conversely, if one computes the massless exchange normalization in the closed channel and compares it with the DBI/Wess—Zumino couplings, one obtains

$$\boxed{ \tau_p^2={\pi\over \kappa_0^2}(4\pi^2\alpha')^{3-p}. }$$

Using

$$2\kappa_0^2=(2\pi)^7\alpha'^4,$$

this becomes

$$\boxed{ \tau_p=\mu_p={1\over (2\pi)^p(\alpha')^{(p+1)/2}}. }$$

This is the normalization used in the DBI and Wess—Zumino actions. The physical string-frame tension in a background with $g_s=e^{\Phi_\infty}$ is $T_p=\tau_p/g_s$.

Brane—antibrane pairs

Now replace one brane by an antibrane. The NS—NS tension is unchanged, but the R—R charge flips:

$$\tau_p\to \tau_p, \qquad \mu_p\to -\mu_p.$$

In the closed-string channel, this reverses the sign of the R—R exchange. The force is no longer zero:

$$V_{{\rm D}\overline{\rm D}}(y) \sim -2\kappa_0^2 \tau_p^2 G_{9-p}(y) -2\kappa_0^2 \mu_p^2 G_{9-p}(y),$$

so the two contributions are both attractive.

In the open-string channel, the same fact appears as a reversed GSO projection for strings stretched between the brane and the antibrane. The relevant oscillator sum is

$$\boxed{ Z_{{\rm D}\overline{\rm D}}(t) ={1\over 2} {f_3(q)^8+f_4(q)^8-f_2(q)^8\over f_1(q)^8}. }$$

The plus sign in front of $f_4(q)^8$ means that the NS ground state is no longer projected out. For the lightest stretched mode,

$$\boxed{ M_T^2(y)={y^2\over (2\pi\alpha')^2}-{1\over 2\alpha'}. }$$

It becomes tachyonic when

$$|y|<y_c, \qquad \boxed{y_c=\pi\sqrt{2\alpha'}}.$$

The long-distance attraction is the closed-string symptom of the same instability. At short distance, the open-string tachyon is the correct degree of freedom: tachyon condensation annihilates the brane—antibrane pair, or, for topologically nontrivial tachyon profiles, leaves lower-dimensional D-branes as defects.

A D$p$—$\overline{\mathrm{D}p}$ pair attracts at large separation. For $|y|<\pi\sqrt{2\alpha'}$, the lightest stretched open string becomes tachyonic.

What the annulus teaches us

The annulus is a compact lesson in string theory’s bookkeeping:

Open-channel statement	Closed-channel statement
stretched open-string vacuum energy	closed-string exchange between boundary states
$t\to\infty$	open-string infrared spectrum
$t\to0$	closed-string infrared spectrum
GSO projection	spacetime spin-statistics and R—R charge
Jacobi identity	BPS no-force cancellation
reversed GSO for D—$\overline{\rm D}$ strings	R—R force changes from repulsive to attractive

The most important point is conceptual. The vanishing force between parallel D-branes is not imposed by hand. It is simultaneously an identity of theta functions, a degeneracy of open-string bosons and fermions, and a cancellation between spacetime fields sourced by a BPS object. This is why D-branes became a central tool: a single worldsheet diagram knows about gauge theory, gravity, supersymmetry, and nonperturbative charge.

Exercises

Exercise 1. The stretched-string mass shift

Derive the contribution

$$\Delta M^2={y^2\over 4\pi^2\alpha'^2}$$

to the mass of an open string stretched between two parallel D-branes separated by $y$.

Solution

The Dirichlet zero mode is not constant. It is the linear classical solution

$$X^i_{\rm cl}(\sigma)=y_1^i+{y^i\over \pi}\sigma, \qquad y^i=y_2^i-y_1^i.$$

The string has physical length $|y|$. Since the fundamental-string tension is

$$T_{\rm F}={1\over 2\pi\alpha'},$$

the stretching energy is

$$E_{\rm stretch}=T_{\rm F}|y|={|y|\over 2\pi\alpha'}.$$

This contributes to the spacetime mass as

$$\Delta M^2=E_{\rm stretch}^2={y^2\over 4\pi^2\alpha'^2}.$$

Equivalently, in dimensionless form,

$$\alpha' \Delta M^2={y^2\over 4\pi^2\alpha'}.$$

Exercise 2. The Gaussian momentum integral

Show that

$$\int {d^nk\over (2\pi)^n} \exp(-2\pi\alpha't k^2) =(8\pi^2\alpha't)^{-n/2}.$$

Solution

Use

$$\int d^nk\,e^{-ak^2}=\left({\pi\over a}\right)^{n/2}.$$

Here $a=2\pi\alpha't$, so

$$\int {d^nk\over (2\pi)^n} \exp(-2\pi\alpha't k^2) ={1\over (2\pi)^n} \left({\pi\over 2\pi\alpha't}\right)^{n/2}.$$

Simplifying,

$${1\over (2\pi)^n} \left({1\over 2\alpha't}\right)^{n/2} = {1\over (8\pi^2\alpha't)^{n/2}}.$$

Setting $n=p+1$ gives the momentum factor in the annulus.

Exercise 3. Why the NS tachyon is removed for BPS branes

Use the leading behavior of $f_3$ and $f_4$ to explain why the combination

$${1\over2}{f_3(q)^8-f_4(q)^8\over f_1(q)^8}$$

has no $q^{-1}$ tachyon term.

Solution

For small $q$,

$$f_1(q)=q^{1/12}(1+\cdots),$$

while

$$f_3(q)=q^{-1/24}(1+q+\cdots), \qquad f_4(q)=q^{-1/24}(1-q+\cdots).$$

Therefore

$$f_3(q)^8=q^{-1/3}(1+8q+\cdots), \qquad f_4(q)^8=q^{-1/3}(1-8q+\cdots).$$

The leading terms cancel in the difference:

$$f_3(q)^8-f_4(q)^8=16q^{2/3}+\cdots.$$

Since

$$f_1(q)^8=q^{2/3}(1+\cdots),$$

the ratio begins at $q^0$, not at $q^{-1}$. Thus the GSO-projected NS sector has no tachyon. Its leading states are massless.

Exercise 4. Level-by-level Bose—Fermi cancellation

The Jacobi identity implies

$$f_3(q)^8-f_4(q)^8=f_2(q)^8.$$

Explain why this means that the number of bosonic and fermionic open-string states is equal at each mass level for strings stretched between identical parallel BPS D-branes.

Solution

The GSO-projected NS contribution to the open-string trace is

$$Z_{\rm NS,+}={1\over2}{f_3(q)^8-f_4(q)^8\over f_1(q)^8}.$$

The R contribution is

$$Z_{\rm R}={1\over2}{f_2(q)^8\over f_1(q)^8}.$$

The Jacobi identity says

$$Z_{\rm NS,+}=Z_{\rm R}.$$

Expanding both sides as power series in $q$ compares states at fixed open-string level. The equality of the coefficients means that, after the GSO projection, each mass level has the same number of bosonic and fermionic physical states. The annulus vacuum energy is the graded trace, so the bosonic and fermionic contributions cancel.

Exercise 5. Open UV as closed IR

Explain why the $t\to0$ region of the open-channel annulus is controlled by the lightest closed strings.

Solution

The open-channel modulus is $t$. The closed-channel modulus is

$$\ell={1\over t}.$$

Thus $t\to0$ means $\ell\to\infty$. In the closed-string channel, the amplitude is schematically

$$\langle B_1|e^{-\pi\ell(L_0+\widetilde L_0-a)}|B_2\rangle.$$

For large $\ell$, massive closed-string states are exponentially suppressed. The amplitude is therefore dominated by the lightest closed-string states. In a tachyon-free type II theory, these are the massless NS—NS and R—R fields. This is why the open-string ultraviolet region encodes long-range spacetime forces.

Exercise 6. Critical separation for a brane—antibrane tachyon

For a D$p$—$\overline{\rm D}p$ pair, the lightest stretched mode has

$$M_T^2(y)={y^2\over (2\pi\alpha')^2}-{1\over 2\alpha'}.$$

Find the critical separation at which it becomes massless.

Solution

Set $M_T^2(y_c)=0$:

$${y_c^2\over 4\pi^2\alpha'^2}={1\over 2\alpha'}.$$

Multiplying by $4\pi^2\alpha'^2$ gives

$$y_c^2=2\pi^2\alpha'.$$

Therefore

$$\boxed{y_c=\pi\sqrt{2\alpha'}}.$$

For $|y|<y_c$, the stretched mode is tachyonic and the brane—antibrane system is perturbatively unstable.

Exercise 7. Tension normalization from the annulus

Starting from

$$\tau_p^2={\pi\over \kappa_0^2}(4\pi^2\alpha')^{3-p}, \qquad 2\kappa_0^2=(2\pi)^7\alpha'^4,$$

show that

$$\tau_p={1\over (2\pi)^p(\alpha')^{(p+1)/2}}.$$

Solution

Since

$$\kappa_0^2={1\over2}(2\pi)^7\alpha'^4,$$

we have

$$\tau_p^2 ={\pi\over \kappa_0^2}(4\pi^2\alpha')^{3-p} ={2\pi\over (2\pi)^7\alpha'^4} \left((2\pi)^2\alpha'\right)^{3-p}.$$

Now simplify the powers:

$$\tau_p^2 ={(2\pi)^{1+6-2p}\over (2\pi)^7} \alpha'^{3-p-4} ={(2\pi)^{-2p}} \alpha'^{-(p+1)}.$$

Thus

$$\tau_p^2={1\over (2\pi)^{2p}\alpha'^{p+1}},$$

and hence

$$\boxed{ \tau_p={1\over (2\pi)^p(\alpha')^{(p+1)/2}}. }$$