I'm currently reading this article, which is about Ghost D-branes in string theory. The authors define these objects as something that cancels the effects of an ordinary D-brane. However, there is something I do not understand.

On page 6, the authors ask us to consider N $Dp$-branes and $M$ ghost $Dp$-branes on top of each other. They claim the field content can be summarized by a hermitian supermatrix $$ \Phi = \begin{pmatrix} \phi_1 & \psi \\ \psi^{\dagger} & \phi_2 \end{pmatrix}. $$ Here $\phi_1$ and $\phi_2$ are boson hermitian matrices, while $\psi$ is a complex fermionic matrix. They say the following:

The diagonal part $\phi_1$ (or $\phi_2$) corresponds to the open strings between $D$-branes (or ghost $D$-branes). The off-diagonal part $\psi$ corresponds to the open strings between $D$-branes and ghost $D$-branes and thus they have the opposite statistics.

Now does anyone know why (1) the off-diagonal part would represent open strings stretched between $D$-branes and ghost $D$-branes? Why should I believe this? And also (2), how does this imply that the states of this open string have opposite statistics? I think this is far from trivial.

In the boundary state formalism, the amplitude for closed string propagation is then $$ \langle gD \mid \Delta \mid D \rangle = - \langle D \mid \Delta \mid D \rangle. $$ Here $\Delta$ is the closed string propagator. The minus sign is because, according to the authors, the boundary state of a ghost $D$-brane is minus the whole boundary state of an ordinary $D$-brane. In the open string channel, these amplitudes are interpreted as one-loop partition functions of open strings stretched between branes. Is it because I can attribute the minus sign to a fermionic path integral that this wrong statistics results? This is not clear to me.


An elementary way to motivate the "wrong statistics" of the off diagonal terms goes as follows:

The diagonal terms of the super-connection $\Phi$, denoted by $|i,j>_{NS}$, $|g(i),g(j)>_{NS}$ are just the usual Yang-Mills gluons arising from the NS sector of open superstrings with endpoints at brane-brane or ghost brane-ghost brane respectively.

Now, by definition, the wavefunction of an open superstring with an endpoint on a ghost brane changes the sign of the whole gluon wavefunction $|\psi,i,g(j)>$ as $$|\psi,i,g(j)>=|\psi,i,g(j)>_{NS} +|\psi,i,g(j)>_{R}=-|\psi,i,j>_{NS} -|\psi,i,j>_{R},$$

where $\psi$ denotes the gluon wave function (see Zwiebach, A first course in string theory, second edition, quick calculation 14.3 where the photon wavefunction is shown as the open superstring NS-state at level $N=\frac{1}{2}$).

Now notice that if $(-1)^{F}$ is the fermion number operator, then $(-1)^{F}|\psi,i,j>_{NS}=|\psi,i,j>_{NS}$ because the Yang-Mills gluon $|\psi,i,j>_{NS}$ is a bosonic state and arises from the NS sector of the open superstring.

The point is that the off-diagonal elements of the super-connection obey $$(-1)^{F}|\psi,i,g(j)>_{NS} = (-1)^{F}(-|\psi,i,j>_{NS}) = -|\psi,i,j>_{NS},$$

so they are fermions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.