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  Mixed boundary condition of open string and spectrum counting

+ 2 like - 0 dislike

When an open string has mixed boundary conditions on both ends (for example, it stretches between \(Dp\) and \(Dq\) brane where \(p \neq q\)), then it might have both Dirichlet-Dirichlet, Neumann-Neumann, or Dirichlet-Neumann boundary condtion. Let us assume that the number of D-N boundary condition has total number of 4, so it becomes a standard \(Dp-D(p+4)\) system.

I am confused about matching the spectrum here. As noted in Polchinski, string theory, Vol II, under such conditions the NS sector and R sector are degenerate. This is because the zero point energy in both sectors are zero. As far as I understand, before GSO projection these two sectors have exactly same spectrum, with ground states four real fermionic degrees of freedom. But how could degenerate spectrum match? If they both have ground states to be fermions, where is the corresponding bosonic degrees of freedom?

asked Oct 27, 2014 in Theoretical Physics by Ke Ye (50 points) [ no revision ]

Are the ground states in both R and NS sectors fermionic only?


1 Answer

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Hi. I am going to lay out for you how to calculate the spectrum for mixed boundary conditions. I hope this helps because you might have some confusions on the zero point energies.

Let us take a specific example which you will be able to alter. Let us consider the D3-D7 brane system and consider open RNS superstrings. We can have four possible sets of boundary conditions which are NN, ND, DN, DD. N signifies Neumann and D signifies Dirichlet boundary conditions. It can be shown that the zero point energy in the R sector is zero, \(a_R=0\) . Also, it can be shown that for the NS sector we get \(a_{NS}=1/2\). Now, you must be familiar on what happens for NN and DD boundary conditions so I am not going to repeat it here. Instead, let's look at the mixed cases. As I said above there are two possibilities, ND and DN. It is possible to find the following contributions to the normal ordering constant \(a\)from half-integer or integer bosonic and fermionic modes. Note that all possibilities must be taken into account.

\(a_{bi}=\frac{1}{24}, \,\, a_{bh}=\frac{1}{48}, \,\, a_{fi}=-\frac{1}{24}, \,\, a_{fh}=\frac{1}{48}.\)

Where b,f stand for bosons and fermions respectively, while i and h stand for integer and half-integer respectively. Now, one requirement that must always be satisfied is that the zero-point energy of the R-sector is zero. Let us now consider that out of the d-2 polarization modes, ν of them are either ND or DN with half-integer bosonic modes. Then the contributions of the bosons to the zero-point energy is \(\frac{2(d-2)-3\nu}{48}.\)Therefore, the fermions in the R-sector should contribute \(\frac{3\nu-2(d-2)}{48}.\)The fermions which are polarized alonf d-2-ν NN or DD directions have integer modes and contribute \(-\frac{d-2-\nu}{24}.\)The fermions which are polarized along ν DN or ND directions therefore must have half-integer modes and contribute \(\frac{\nu}{48}.\)Then, in total, we get the required contribution \(\frac{2\nu-2(d-2)}{48}.\)This allows us to compute the zero point-energy for the NS sector which we find to be 

\(a_{NS} = \frac{1}{2}-\frac{\nu}{8}.\)

Remember that \(a_R\)is always zero and the ground state of a massless string is always in some fermionic representation. For \(\nu=4\) we get \(a_{NS}=0\) and therefore the mass of the lowest NS state is zero. This state is not projected out by the GSO so that we have an equal number of massless bosons and fermions building up the vector supermultiplet. 

Now after these details that are needed to convince yourself you are calculating stuff right we can move onto the spectrum for D3-D7 branes. The cases of NN and DD must be known to you. In the NS sector of D3's you get the \({\bf{8}}_V\)giving a gauge boson and 6 scalars in 4d. In the R sector you get the \({\bf{8}}_C\)getting four 2-spinors in 4d. For the NS sector of the D7's you get the \({\bf{8}}_V\)again (which has a gauge boson with six degrees of freedom and two scalars) and in the R sector you get the \({\bf{8}}_C\), a  \(SO(8)\) spinor.

Now, for the mixed part. Let us denote the coordinates as numbers. E.g. \(x_1 \equiv 1\)etc. In 0,...,3 we will have NN boundary conditions, in 4,...,7 we will have DN and in 8,9 we will have DD. Now, you would have to perform the oscillator expansions etc in order to be able to calculate the spectrum and I am not gonna do this here (but you can find help in Zweibach's book). Once you get the mass formulae you are able to compute the spectrum. For the NS sector we get three states which transform in the \({\bf{1}}, {\bf{4}}_C, {\bf{1}}\) under SO(4) respectively and give us 4 complex scalars. In the R sector we get two states transforming in the \({\bf{4}}_C\) under SO(4) and \({\bf{2}}_C\) under SO(2) giving an SO(2) spinor. 

Now, the D3-D7 intersection preserves 1/4 of the total susy of type IIB, i.e. 8 out of 32 supercharges and has an SO(4)xSO(2) isometry in the directions transverse to the D3-brane.  The SO(4) rotates in the 4,...,7 and the SO(2) rotates in the 8,9. Now, take a look at the moddings

\[\begin{array}{ l c c r } \text{Directions} & 0,1,2,3 & 4, 5, 6, 7 & 8,9 \\ \hline \text{Bosons} & \alpha_n & \alpha_{n+1/2}& \alpha_n \\ \text{NS}& b_{n+1/2} & b_n & b_{n+1/2} \\\text{R} & d_{n} & d_{n+1/2} & d_{n} \\ \hline \end{array}\]

Try to consider the states now. For the R-sector we get a SO(3,1) spinor with two degrees of freedom since Weyl spinors have two complex degrees of freedom and going on-shell cuts them in two, transforming under SO(2), in the \({\bf{2}}_C\) giving another 2 fermionic degrees of freedom and a 2d spinor transforming under SO(2) for the directions 8,9. The latter is only being seen as internal degrees of freedom for the R sector thus giving 4 degrees of freedom in total. By considering the 7-3 string we get another 4 degrees of freedom, thus in total we have 8 sermonic degrees of freedom. For the NS sector we get a scalar for 0,..., 3$and another scalar for 8,9. For 4,..., 7 we get a 4d spinor transforming under SO(4) with four degrees of freedom which will act as internal degrees of freedom. Finally we get another scalar from 8,9 so in the end multiplying out the total degrees of freedom we get 4. Now, we have to also take into account the 7-3 string just like we did for the fermions and not just the 3-7. Thus we get another 4 scalars and in the end the summation of total degrees of freedom yields 8. The previous result comes into agreement with the fact that the 3-7 string (together with the 7-3 in order the theory in order to be complete) saves 8 out of the 32 super symmetries and we see that the requirement \(N_B = N_F\) is satisfied indeed. Finally note that the states of the 3-7 string transform in the \({\bf{N}} \times {\bf{\bar{K}}}\) (and for the 7-3 \({\bf{K}} \times {\bf{\bar{N}}}\)) where the the D3-D3 system has the gauge group \(U(N)\) and the D7-D7 system has gauge group \(U(K)\). The bars above come from the fact that the Chan-Paton indices live on the fundamental and anti-fundamental of the gauge group.

Now, its almost straight forward to generalize to other systems as well.

I hope this helps a little bit. If you need more help ask here (maybe someone else can explain better than me) or try except for the classic textbooks the one by Ibanez and Uranga. 

answered Oct 28, 2014 by conformal_gk (3,625 points) [ revision history ]
edited Oct 30, 2014 by dimension10

I would like to ask some help from a moderator in order the table I have included to look properly. I cannot find how to make it work. Thanks.

@Dimension10 and @Polarkernel, do you have an idea how to fix the table here?

@conformal_gk First of all, thank you very much for your elaborated and patient explanation, I really appreciate your help.

I agree with most of your calculations, except some minor confusions. As I understand, in the spectrum of D3-D7 string, the ground states in NS and R sector are given by acting on \(b^{\mu}_0\), with \({\mu}=4,5,6,7\), and \(d^{\nu}_0\), with \({\nu}=0,1,2,3,8,9\), and light-cone gauge eliminate two out of six in the range of \({\nu}\). Therefore, both NS and R sector have four creation operators with subscript 0.

Now since these operator forms a Dirac algebra, the ground states in both sectors are fermionic. Each ground state carries a spinor indices and is transformed to one another by acting on \(b^{\mu}_0\) or \(d^{\nu}_0\). Higher excitations are obtained by acting on other creation operators. Since we start from a fermionic ground states, with same set of operators in NS and R sector, how could the excited states match?

If I understand correctly, you are refering your scalars as "ground state scalars" right? (Since higher excitation has mass, they actually correspond to different fields). But I do not see how scalars come out in ground states.

Thanks again for your explanation!


I cannot properly answer right now but first of all note we are always talking about the M=0 states while you get scalars from the reduction of the spinors that are in the 8_c. The way you do the reduction to lower dimensions is in Appendix B of Polchinski. I will try to augment my answer later, when i find some time.

@conformal_gk @Dilaton I tried playing with your latex code and it seems that Mathjax does not support the tabular environment. The best option you have is to change an array environment and manually do the formatting. I unfortunately cannot do it for you as I am  on a smartphone  with a malfunctioning keyboard at the moment (so I am afraid that I will break something). I will search for a Mathjax package for tabular and tell you if I find any. By the way, you should not be using dollar signs within the tabular environment. 

@dimension10 I got the table straight out of one of my Latex files as you see it (changed only a few details). I had it like this with dollar signs :s. By the way, since I just used it, is there any chance you introduce some sort of smiley faces to lighten the mood? Ok, maybe this is a naive suggestion.

@conformal_gk OK : ) Not all latex environments work with Mathjax, actually. @Dilaton it still doesn't seem to be working to me at least. Edit: OK I fixed it, I don't think it is possible to add the vertical column separators though. 

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