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  How exactly do superstrings reduce the number of dimensions in bosonic string theory from 26 to 10 and remove the tachyons?

+ 2 like - 0 dislike

In bosonic string theory, to obtain the photon as the first excited state, the ground state must have a negative mass (tachyon). By applying $1 + 2 + 3 + \cdots = -1/12$, it can be shown (in a simplified way...) that a total of 26 spacetime dimensions are needed to obtain such a tachyonic ground state.

Going from bosonic string theory to superstring theory, the number of spacetime dimensions is 10 and the mass of the ground state is 0, such that the tachyons are removed trom the spectrum of the theory.

Can somebody explain to me in some detail and step by step how the number of dimensions is reduced (or fixed) to 10 and how the tachyons are removed (or avoided) in superstring theory?

asked Dec 31, 2012 in Theoretical Physics by Dilaton (4,295 points) [ revision history ]
retagged May 7, 2014 by dimension10

2 Answers

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First, the critical dimension. There are many ways (seemingly inequivalent ways but ultimately bound to give the same result) to calculate $D=10$ for the superstring that mirror the methods to calculate $D=26$ for the bosonic string.

For the bosonic string, one may use a conformally invariant world sheet theory. Because of the residual conformal symmetry, it has to have $bc$ ghosts. The central charge of the $bc$ system is $c=1-3k^2$ where $k=2J-1$ where $J$ is the dimension of the $b$ antighost, in this case $J=2$. You see that my formulae imply $k=3$ and $c=1-27=-26$ so one has to add 26 bosons, i.e. 26 dimensions of spacetime, to get $c=0$ in total.

Now, for the superstring, the local symmetries on the world sheet are enhanced from the ordinary conformal group to the $N=1$ superconformal group. One needs to add the $\beta\gamma$ (bosonic) ghosts for the new (fermionic) generators. Their dimension is $J=3/2$, different from $J=2$ of $bc$ by $1/2$, as usual for the spin difference of things related by supersymmetry. You see that $k=2J-1=2$ and $3k^2-1=12-1=11$. Now, the central charge of $\beta\gamma$ is $3k^2-1$ and not $1-3k^2$, the sign is the opposite one, because they are bosons.

So the $bc$ and $\beta\gamma$ have $c=-26+11=-15$. This minus fifteen must be compensated by 10 bosonic fields and 10 fermionic fields (whose $c=1/2$ per dimension: note that a fermion is half a boson) and $10+10/2=15$ so that the total $c=0$. If some of the steps aren't understandable above, it's almost certainly because the reader isn't familiar with basics of conformal field theory and it is not possible to explain conformal field theory without conformal field theory. It's a whole subject, not something that should be written as one answer on this server.

In this formalism with the new world sheet fermions $\psi^\mu$ transforming as spacetime vectors, one has to protect the spin-statistics relationship. Vector-like fermions violate it so they are only allowed in pairs. This is achieved by the GSO projection – well, there are actually two GSO projections, one separate for left-movers and one for right-movers. Only 1/4 of the states are kept in the spectrum. The projection is a flip side of having four sectors – the left-moving and right-moving fermions may independently be periodic or antiperiodic. I wrote about the GSO projection a month ago:


Again, if anything is incomprehensible and incomplete, it's because it's not really one isolated insight that a layman may understand from one sentence. It's one of many technical results that follows from a large subject – string theory – that has to be systematically studied if one wants to understand it.

This post imported from StackExchange Physics at 2014-03-12 15:28 (UCT), posted by SE-user Luboš Motl
answered Dec 31, 2012 by Luboš Motl (10,248 points) [ no revision ]
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In my opinion the best way to see it is to use the worldsheet CFT as has already been explained. But, you can also use lightcone gauge to derive the critical dimension by requiring Lorentz invariance just as in the bosonic case.

You may know that states of the superstring are divided into an NS sector and an R sector depending on the boundary conditions of the worldsheet fields. The R sector has integer labeled fermionic modes and the NS sector has half integer labeled fermionic modes.

Now, just like in the lightcone quantization for the bosonic string we must enforce the vanishing of the zero mode of the virasoro algebra. However, here there is also a supercurrent whose zero mode will need to vanish on physical states. But, the supercurrent will only have a zero mode in the R sector because it has half integer modes in the NS sector.

In the NS sector we have a ground state, which will be a tachyon that is eliminated by the GSO projection. The first excited state is obtained by acting on the ground state with the operator $\psi_{-1/2}^i$. Just like in the bosonic case the first excited state transforms as an SO(D-2) vector, which means that this state must be massless in order for the theory to be Lorentz invariant. Also as in the bosonic case this fixes the normal ordering constant for to Virasoro zero mode, except now it is 1/2.

In the bosonic string we would do an explicit calculation of the normal ordering constant in order to related it to the spacetime dimension and use that to determine the critical dimension. However, here the vanishing of the supercurrent in the R sector can be used.

The supercurrent is given by $J=\psi^\mu \partial X^\mu$ so there can be no normal ordering constant. Also, the super-virasoro algebra says $$ \{J_0,J_0\} = 2\left(L_0-\frac{D-2}{16}\right) $$ So, compatibility with the vanishing of $L_0-1/2$ requires that $D=10$.

As for the tachyon, I mentioned that it is eliminated by the GSO projection, which eliminates all states in the NS sector with even fermion number.

This post imported from StackExchange Physics at 2014-03-12 15:28 (UCT), posted by SE-user ald5657
answered Dec 31, 2012 by ald5657 (40 points) [ no revision ]

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