D-Brane Charges, BPS Bounds, and Dirac Quantization

D-branes entered the story as hyperplanes on which open strings can end. T-duality made them look geometrically inevitable. The next step is sharper and more physical: D-branes are charged, dynamical, BPS objects. They carry Ramond—Ramond charge, have a tension fixed exactly by supersymmetry, and obey a generalized Dirac quantization condition.

The core facts are these:

1. A D$p$-brane couples electrically to a Ramond—Ramond $(p+1)$-form potential $C_{p+1}$.
2. It is a half-BPS object: it preserves $16$ of the $32$ type II supercharges.
3. Two identical parallel BPS D-branes exert no static force on each other.
4. A D$p$-brane and a D$(6-p)$-brane are electric/magnetic duals in ten dimensions, and their charges obey Dirac quantization.

This page explains how these statements fit together. The punchline is the exact normalization

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

for the Ramond—Ramond charge and the string-frame DBI tension parameter of a single elementary D$p$-brane. In a constant dilaton background $e^{\Phi_\infty}=g_s$, the physical string-frame tension is

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

Many references denote the first quantity by $T_p$ rather than $\tau_p$. Here $\tau_p$ is used for the $g_s$-independent normalization appearing in the DBI/Wess—Zumino action, while $T_p=\tau_p/g_s$ is the actual energy per unit $p$-volume measured at infinity in string frame.

D-branes as sources for Ramond—Ramond fields

The low-energy action of a single abelian D$p$-brane has two universal pieces. The Dirac—Born—Infeld term gives the tension and coupling to NS—NS fields:

$$S_{\rm DBI} =-\tau_p\int_{\mathcal W_{p+1}} d^{p+1}\xi\, e^{-\Phi}\sqrt{-\det\left(P[G+B]_{ab}+2\pi\alpha'F_{ab}\right)}.$$

The Wess—Zumino term gives the Ramond—Ramond couplings:

$$S_{\rm WZ} =\mu_p\int_{\mathcal W_{p+1}} P\!\left[\sum_q C_q\right] e^{P[B]+2\pi\alpha'F}.$$

For the simplest flat brane with no worldvolume flux and no background $B$-field, this reduces to

$$S_{\rm Dp} =-\tau_p\int_{\mathcal W_{p+1}} e^{-\Phi}\sqrt{-\det P[G]} +\mu_p\int_{\mathcal W_{p+1}} C_{p+1}.$$

Thus a D$p$-brane is an electric source for $C_{p+1}$, just as a charged particle is an electric source for a one-form gauge potential $A_1$.

The corresponding field strength is

$$F_{p+2}=dC_{p+1},$$

up to the standard refinements involving $B$-fields and Chern—Simons terms. Ignoring those refinements for the moment, the electric equation of motion has the schematic form

$$d{*F_{p+2}}=2\kappa_0^2\mu_p\,\delta_{9-p}(\mathcal W_{p+1}),$$

where $\delta_{9-p}(\mathcal W_{p+1})$ is a delta-form localized in the $9-p$ transverse directions. Equivalently, on a sphere $S^{8-p}$ linking the D$p$-brane,

$$\int_{S^{8-p}} *F_{p+2}=2\kappa_0^2\mu_p.$$

Here $\kappa_0$ is the $g_s$-independent gravitational normalization convenient in the string-frame worldsheet conventions:

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

The physical ten-dimensional Newton constant includes the asymptotic string coupling,

$$2\kappa_{10}^2=(2\pi)^7g_s^2(\alpha')^4.$$

The magnetic source for $C_{p+1}$ is found by dualizing $F_{p+2}$. In ten dimensions,

$$*F_{p+2}\quad\text{has degree}\quad 8-p.$$

The dual potential has degree $7-p$, so the magnetic object has spatial dimension $6-p$:

$$\boxed{\text{magnetic dual of D}p\text{ is D}(6-p).}$$

For a magnetic D$(6-p)$-brane, the Bianchi identity is modified:

$$dF_{p+2}=2\kappa_0^2\mu_{6-p}\,\delta_{p+3}(\mathcal W_{7-p}),$$

or, on a sphere $S^{p+2}$ linking the magnetic brane,

$$\int_{S^{p+2}}F_{p+2}=2\kappa_0^2\mu_{6-p}.$$

A D$p$-brane is electrically charged under $C_{p+1}$, while the magnetic source for the same field strength is a D$(6-p)$-brane. Electric charge is measured by $\int *F_{p+2}$ on $S^{8-p}$; magnetic charge is measured by $\int F_{p+2}$ on $S^{p+2}$.

The most important dual pairs are

$$\begin{array}{c|c|c} \text{electric brane} & \text{magnetic dual} & \text{type II theory} \\ \hline \text{D}0 & \text{D}6 & \text{IIA} \\ \text{D}1 & \text{D}5 & \text{IIB} \\ \text{D}2 & \text{D}4 & \text{IIA} \\ \text{D}3 & \text{D}3 & \text{IIB} \end{array}$$

The D3-brane is special: it is self-dual under this electric/magnetic pairing. This is one of the reasons the D3-brane is central in type IIB string theory and in the gauge/gravity duality.

The BPS bound

A BPS object is not merely a stable soliton. It is a state or extended object whose energy is fixed by a charge appearing in the supersymmetry algebra. The simplest schematic form is

$$\{Q,Q\}\sim E-Z,$$

where $E$ is the energy and $Z$ is a central or tensorial charge. Since the left-hand side is a positive operator, one obtains

$$E\ge |Z|.$$

A BPS state saturates the inequality:

$$E=|Z|.$$

Saturation implies that some linear combination of supercharges annihilates the state. Those supercharges are preserved; the remaining ones are broken.

For an extended $p$-brane in ten dimensions, the relevant charge is not an ordinary scalar central charge. It is a $p$-form charge, or equivalently a charge density carried by a spatial $p$-volume. Schematically, the type II superalgebra contains terms of the form

$$\{Q,Q\} \supset (C\Gamma^M)P_M +\sum_p (C\Gamma_{M_1\cdots M_{p+1}})Z^{M_1\cdots M_{p+1}}.$$

For a static flat D$p$-brane extended along $x^1,\ldots,x^p$, the relevant component is

$$Z_{0\cdots p}.$$

Per unit spatial $p$-volume, the BPS inequality becomes

$$\mathcal M_p\ge |\mathcal Z_p|.$$

A supersymmetric D-brane saturates it:

$$\boxed{\mathcal M_p=|\mathcal Z_p|.}$$

In the DBI/WZ conventions above, this equality is the statement that the tension source strength and Ramond—Ramond charge source strength are equal after canonical normalization:

$$\boxed{\tau_p=|\mu_p|.}$$

The orientation of the brane determines the sign of the charge. A D$p$-brane has charge $+\mu_p$; an anti-D$p$-brane has charge $-\mu_p$. Both have positive tension.

Preserved supersymmetry

A flat D$p$-brane imposes a projection condition on the type II supersymmetry parameters. Up to signs and convention-dependent factors of $\Gamma_{11}$, the condition has the form

$$\epsilon_L=\eta\,\Gamma^{0\cdots p}\epsilon_R, \qquad \eta=+1\ \text{for a brane}, \qquad \eta=-1\ \text{for an antibrane}.$$

This equation relates the left- and right-moving supersymmetry parameters, so it cuts the number of independent supercharges in half. Therefore an isolated flat D$p$-brane is half-BPS:

$$32\quad\longrightarrow\quad 16\ \text{preserved}+16\ \text{broken}.$$

The broken supercharges generate fermionic zero modes on the brane. The preserved supercharges organize the massless open-string fields into a maximally supersymmetric vector multiplet on the D-brane worldvolume.

A brane and an antibrane impose opposite projections:

$$\epsilon_L=+\Gamma^{0\cdots p}\epsilon_R, \qquad \epsilon_L=-\Gamma^{0\cdots p}\epsilon_R.$$

The only simultaneous solution is $\epsilon_L=\epsilon_R=0$. Hence a brane—antibrane pair preserves no supersymmetry. This is the supersymmetry-algebra reason that the brane—antibrane force does not cancel and that a tachyon can appear at sufficiently small separation.

Why the static force cancels

Consider two parallel D$p$-branes separated by a transverse distance $y$. At large separation, the interaction is dominated by exchange of massless closed-string fields. The relevant transverse Green function is

$$G_d(y)=\int {d^d k\over (2\pi)^d}\,{e^{ik\cdot y}\over k^2}, \qquad d=9-p.$$

For $d>2$,

$$G_d(y)= {\Gamma\left({d-2\over2}\right)\over4\pi^{d/2}} {1\over y^{d-2}} = {\Gamma\left({7-p\over2}\right)\over4\pi^{(9-p)/2}} {1\over y^{7-p}}.$$

The NS—NS fields sourced by the DBI action are the graviton and the dilaton, and in more general backgrounds also the $B$-field. Between two identical static branes, the graviton/dilaton exchange gives an attractive contribution. The R—R potential $C_{p+1}$ gives a repulsive contribution between equal charges.

With a standard normalization of the exchanged massless fields, the long-distance amplitude has the schematic but very useful form

$$\mathcal A_{\rm tot}(y) =i\,2\kappa_0^2V_{p+1}G_{9-p}(y) \left(\tau_p^2-\mu_p^2\right).$$

The first term is the NS—NS contribution and the second is the R—R contribution. A BPS D-brane has

$$\tau_p=|\mu_p|,$$

so

$$\boxed{\mathcal A_{\rm tot}(y)=0.}$$

This is the spacetime version of the abstruse theta-function cancellation in the annulus amplitude. The open-string one-loop vacuum energy vanishes because the brane system is supersymmetric. In the closed-string channel, the same statement is that NS—NS attraction and R—R repulsion cancel exactly.

Two identical parallel BPS D-branes have equal tension and R—R charge. The NS—NS attraction is exactly cancelled by R—R repulsion, giving zero static force.

For a brane—antibrane pair, the tension is unchanged but the R—R charge reverses sign:

$$\mu_p\longrightarrow -\mu_p.$$

Then the R—R contribution changes sign and becomes attractive rather than repulsive. The force no longer cancels:

$$\mathcal A_{\rm brane\text{-}antibrane}(y) \propto \tau_p^2+\mu_p^2.$$

This explains why brane—antibrane systems are unstable at short distance and attractive at long distance.

The exact D-brane tension and charge

The elementary D-brane charge is fixed by several mutually consistent requirements:

• T-duality must map D$p$-brane tensions into D$(p\pm1)$-brane tensions.
• The annulus amplitude must factorize correctly onto massless closed-string exchange.
• The R—R charge must obey Dirac quantization with the magnetic dual brane.
• Supersymmetry requires the BPS equality between tension and charge.

The result is

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

Equivalently,

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

To check the equivalence, use

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

Then

$${\pi\over\kappa_0^2}\left(4\pi^2\alpha'\right)^{3-p} = {1\over (2\pi)^{2p}(\alpha')^{p+1}} =\tau_p^2.$$

A useful recursion relation follows immediately:

$$\tau_{p-1}=2\pi\sqrt{\alpha'}\,\tau_p.$$

This is precisely what T-duality requires. If a D$p$-brane wraps a circle of radius $R$, its mass per unit unwrapped volume is

$$2\pi R\,{\tau_p\over g_s}.$$

After T-duality along the circle,

$$R'={\alpha'\over R}, \qquad g_s'=g_s{\sqrt{\alpha'}\over R},$$

and the dual object is an unwrapped D$(p-1)$-brane with tension

$${\tau_{p-1}\over g_s'} = {2\pi\sqrt{\alpha'}\,\tau_p\over g_s\sqrt{\alpha'}/R} =2\pi R\,{\tau_p\over g_s}.$$

Thus the exact tension formula is not an isolated normalization convention; it is part of the T-duality structure of the full theory.

Dirac quantization for D-branes

Ordinary electric and magnetic charges obey the Dirac quantization condition

$$q_e q_m=2\pi n, \qquad n\in\mathbb Z,$$

in units where the gauge field kinetic term is canonically normalized. For higher-form gauge fields, the same logic applies, but the electric and magnetic objects are extended branes.

Let a D$p$-brane be the electric object for $C_{p+1}$ and a D$(6-p)$-brane be the magnetic object. Around the magnetic brane, choose a linking sphere $S^{p+2}$. The magnetic flux is

$$\int_{S^{p+2}}F_{p+2}=2\kappa_0^2\mu_{6-p}.$$

Now transport the electrically charged D$p$-brane around the magnetic brane in configuration space. The Wess—Zumino coupling gives an Aharonov—Bohm phase

$$\exp\left(i\mu_p\int C_{p+1}\right).$$

Changing the spanning surface shifts this phase by

$$\exp\left(i\mu_p\int_{S^{p+2}}F_{p+2}\right) = \exp\left(i\,2\kappa_0^2\mu_p\mu_{6-p}\right).$$

Quantum mechanics requires the phase to be single-valued, so

$$\boxed{ 2\kappa_0^2\mu_p\mu_{6-p}=2\pi n, \qquad n\in\mathbb Z. }$$

This is the generalized Dirac quantization condition for R—R charges.

Transporting an electrically charged D$p$-brane around its magnetic dual D$(6-p)$-brane gives an Aharonov—Bohm phase. Single-valuedness imposes $2\kappa_0^2\mu_p\mu_{6-p}=2\pi n$.

Using the elementary D-brane charge formula, the left-hand side becomes

$$\begin{aligned} 2\kappa_0^2\mu_p\mu_{6-p} &={(2\pi)^7(\alpha')^4\over (2\pi)^p(\alpha')^{(p+1)/2} (2\pi)^{6-p}(\alpha')^{(7-p)/2}} \\ &=2\pi. \end{aligned}$$

Therefore a single D$p$-brane and a single D$(6-p)$-brane carry minimal mutually allowed R—R charges:

$$\boxed{n=1.}$$

This is a deep consistency check. The same objects whose tension is inferred from open-string dynamics carry precisely the minimal R—R charge allowed by quantum mechanics.

Orientation, charge, and stacks

A stack of $N$ coincident D$p$-branes has total charge

$$Q_p=N\mu_p$$

and physical tension

$$T_p^{(N)}=N{\tau_p\over g_s}.$$

The open strings ending on the stack carry Chan—Paton indices, and the low-energy gauge symmetry is $U(N)$. The overall $U(1)$ couples to the center-of-mass brane position, while the nonabelian $SU(N)$ sector describes relative positions and strings stretched between different branes.

An anti-D$p$-brane has

$$Q_p=-\mu_p, \qquad T_p={\tau_p\over g_s}>0.$$

Therefore brane—antibrane annihilation can conserve total R—R charge while releasing positive energy. In the open-string description, the instability is represented by the brane—antibrane tachyon. In the spacetime description, it is the absence of a BPS bound protecting the pair.

What has been established

D-branes are not optional boundary conditions added to string theory by hand. They are the R—R charged BPS objects required by the consistency of type II theory. The equality

$$\tau_p=|\mu_p|$$

explains the no-force condition. The formula

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

makes D-branes the minimally charged objects in the R—R charge lattice. The generalized Dirac condition

$$2\kappa_0^2\mu_p\mu_{6-p}=2\pi n$$

then identifies the D$p$—D$(6-p)$ pair as the natural electric/magnetic pair of type II string theory.

These facts will be used repeatedly: in annulus force computations, intersecting brane systems, dissolved brane charges on worldvolumes, S-duality, supergravity brane solutions, and ultimately the D3-brane route to AdS/CFT.

Exercises

Exercise 1

Use

$$\mu_p={1\over(2\pi)^p(\alpha')^{(p+1)/2}}, \qquad 2\kappa_0^2=(2\pi)^7(\alpha')^4,$$

to show that a D$p$-brane and a D$(6-p)$-brane obey the minimal Dirac quantization condition.

Solution

The generalized Dirac product is

$$2\kappa_0^2\mu_p\mu_{6-p}.$$

Substituting the charge formula gives

$$\begin{aligned} 2\kappa_0^2\mu_p\mu_{6-p} &={(2\pi)^7(\alpha')^4\over (2\pi)^p(\alpha')^{(p+1)/2} (2\pi)^{6-p}(\alpha')^{(7-p)/2}}. \end{aligned}$$

The powers of $2\pi$ give

$${(2\pi)^7\over(2\pi)^p(2\pi)^{6-p}}=2\pi.$$

The powers of $\alpha'$ give

$${(\alpha')^4\over(\alpha')^{(p+1)/2}(\alpha')^{(7-p)/2}} ={ (\alpha')^4\over(\alpha')^4}=1.$$

Therefore

$$2\kappa_0^2\mu_p\mu_{6-p}=2\pi.$$

Thus the Dirac integer is $n=1$. Elementary D-branes carry the minimal mutually allowed R—R charge.

Exercise 2

Assume the long-distance interaction between two parallel D$p$-branes is proportional to

$$\mathcal A_{\rm tot}(y) \propto \left(\tau_p^{(1)}\tau_p^{(2)}-\mu_p^{(1)}\mu_p^{(2)}\right)G_{9-p}(y).$$

Show that identical BPS branes have no force, while a brane—antibrane pair is attractive.

Solution

For two identical BPS branes,

$$\tau_p^{(1)}=\tau_p^{(2)}=\tau_p, \qquad \mu_p^{(1)}=\mu_p^{(2)}=\mu_p, \qquad \tau_p=|\mu_p|.$$

Therefore

$$\tau_p^{(1)}\tau_p^{(2)}-\mu_p^{(1)}\mu_p^{(2)} =\tau_p^2-\mu_p^2=0.$$

The static force vanishes.

For a brane—antibrane pair, the tension is unchanged but the charge of the antibrane is reversed:

$$\tau_p^{(1)}=\tau_p^{(2)}=\tau_p, \qquad \mu_p^{(1)}=\mu_p, \qquad \mu_p^{(2)}=-\mu_p.$$

Then

$$\tau_p^{(1)}\tau_p^{(2)}-\mu_p^{(1)}\mu_p^{(2)} =\tau_p^2+\mu_p^2 =2\tau_p^2.$$

The NS—NS and R—R effects add instead of canceling. With the usual sign convention for the potential energy, this corresponds to an attractive interaction.

Exercise 3

In ten dimensions, determine the magnetic dual of a D$p$-brane by counting form degrees.

Solution

A D$p$-brane couples electrically to

$$C_{p+1}.$$

The field strength is

$$F_{p+2}=dC_{p+1}.$$

In ten dimensions, the Hodge dual has degree

$$10-(p+2)=8-p.$$

Thus

$$*F_{p+2}=F_{8-p}$$

is locally the field strength of a dual potential of degree $7-p$:

$$F_{8-p}=dC_{7-p}.$$

An object that couples electrically to $C_{7-p}$ has worldvolume dimension $7-p$, hence spatial dimension

$$(7-p)-1=6-p.$$

Therefore the magnetic dual of a D$p$-brane is a D$(6-p)$-brane.

Exercise 4

Show that the exact tension formula is consistent with T-duality along a circle. A D$p$-brane wraps a circle of radius $R$. After T-duality it becomes an unwrapped D$(p-1)$-brane. Use

$$R'={\alpha'\over R}, \qquad g_s'=g_s{\sqrt{\alpha'}\over R}, \qquad \tau_{p-1}=2\pi\sqrt{\alpha'}\,\tau_p.$$

Solution

The wrapped D$p$-brane has mass per unit unwrapped spatial volume

$$M_{\rm before}=2\pi R\,{\tau_p\over g_s}.$$

After T-duality, the object is an unwrapped D$(p-1)$-brane. Its physical tension is

$$M_{\rm after}={\tau_{p-1}\over g_s'}.$$

Substitute the T-duality rules:

$$M_{\rm after} ={2\pi\sqrt{\alpha'}\,\tau_p\over g_s\sqrt{\alpha'}/R} =2\pi R\,{\tau_p\over g_s}.$$

Thus

$$M_{\rm after}=M_{\rm before}.$$

The exact D-brane tension formula transforms correctly under T-duality.

Exercise 5

For a D$p$-brane with $p<7$, use the transverse Green function to determine the power-law falloff of the long-distance potential between two non-BPS objects with uncancelled source strength.

Solution

The number of transverse spatial dimensions is

$$d=9-p.$$

The massless Green function in $d>2$ dimensions behaves as

$$G_d(y)\sim {1\over y^{d-2}}.$$

Substituting $d=9-p$ gives

$$G_{9-p}(y)\sim {1\over y^{7-p}}.$$

Therefore any uncancelled long-distance force mediated by massless closed strings has potential-energy density scaling as

$$V(y)\sim {1\over y^{7-p}}, \qquad p<7.$$

For example, for D3-branes the transverse dimension is $6$, so the potential behaves as $1/y^4$ when the BPS cancellation is spoiled.

Exercise 6

Use the supersymmetry projection

$$\epsilon_L=\eta\,\Gamma^{0\cdots p}\epsilon_R, \qquad \eta=\pm1,$$

to explain why a brane—antibrane pair preserves no supersymmetry.

Solution

A D$p$-brane imposes

$$\epsilon_L=+\Gamma^{0\cdots p}\epsilon_R.$$

An anti-D$p$-brane imposes the opposite projection:

$$\epsilon_L=-\Gamma^{0\cdots p}\epsilon_R.$$

If both projections are to hold simultaneously, then

$$+\Gamma^{0\cdots p}\epsilon_R=-\Gamma^{0\cdots p}\epsilon_R.$$

Thus

$$\Gamma^{0\cdots p}\epsilon_R=0.$$

Since $\Gamma^{0\cdots p}$ is invertible, this implies

$$\epsilon_R=0,$$

and then also

$$\epsilon_L=0.$$

There is no nonzero preserved supersymmetry parameter. Hence a brane—antibrane pair is non-BPS, its long-distance force does not cancel, and its open-string spectrum may contain a tachyon.

Exercise 7

Suppose $N$ coincident D$p$-branes and $M$ coincident anti-D$p$-branes are placed together. What are the total R—R charge and the total physical tension before tachyon condensation?

Solution

Each D$p$-brane contributes charge $+\mu_p$ and physical tension $\tau_p/g_s$. Each anti-D$p$-brane contributes charge $-\mu_p$ and the same positive physical tension $\tau_p/g_s$.

Therefore the total R—R charge is

$$Q_{\rm total}=(N-M)\mu_p.$$

The total physical tension is

$$T_{\rm total}=(N+M){\tau_p\over g_s}.$$

The difference between these two expressions is the energetic reason brane—antibrane annihilation can release energy while preserving the conserved R—R charge. If $N=M$, the net R—R charge vanishes, and complete annihilation to the closed-string vacuum is allowed by charge conservation.