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CHAPTER 4 String Quantization
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To obtain a formula for the mass operator in string theory, which is denoted by 2 M , we use the condition on physical states. Taking L0 = 1/ 2 0 + n n , using n =1 L0 a with a = 1 , we arrive at the condition
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1 2 ( L0 a) = 0 0 + n n 1 = 0 2 n =1 1 2 0 + n n 1 = 0 2 n =1 The rst term in this expression is nothing other than the mass squared: 2 (1 / 2 ) 0 = M 2 where = 1 / (2 T ). So, in bosonic string theory the mass shell condition becomes M2 = 1 ( N 1) (4.22)
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where N is the total number operator. The term 2 T sets the energy scale of the theory, it is taken to be on the order of the Planck mass. This is the origin of the high energy scale of string theory. It can be shown that M2 = 1 D 2 N 24
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Notice that setting a = 1forces us to take D = 26. The number operator acts on the ground state as N 0 =0 Hence the mass of the ground state is M2 0 = 1 D 2 1 D 2 1 0 = 0 N 0 = 24 24
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So the ground state of bosonic string theory in the open string case has negative mass. This means that the ground state is a Tachyon. This is an unphysical state which travels faster than the speed of light. Consistency of bosonic string theory requires that we choose a = 1, so the Tachyon cannot be removed from the theory.
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String Theory Demysti ed
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It will turn out that the introduction of supersymmetry (that is the introduction of fermionic states) into the theory will get rid of the Tachyon, giving us a realistic string theory. We will see that this also changes the number of space-time dimensions. Now let s consider the mass of the rst excited state. The rst excited state is i i = 1 0 where i is a spatial index. Here, i = 1, ..., D 2, and the state transforms as a vector in space-time. You will recall from your studies of quantum eld theory that a vector is a spin-1 particle that in general has D 1 components, the fact that this state has D 2 components implies that it is a massless state. An example of a massless vector is the photon, it only has transverse components of spin. This explains why there are D 2 components rather than D 1. The mass of the state is
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i M 2 1 0 =
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1 D 2 i 1 26 D i 1 1 0 = 1 0 24 24
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In order for the state to be massless, 26 D / 24 must vanish, once again setting the number of space-time dimensions D to 26. Physicists refer to the bosonic string theory with a = 1, D = 26 as critical and call D = 26 the critical dimension.
CLOSED STRING SPECTRUM
In the case of the closed string, things are a little more complicated than what you re used to from the harmonic oscillator in ordinary quantum theory due to the fact that we have a second commutation relation in addition to m , n = m m+n ,0 that must be satis ed, namely, m , n = m m+n ,0 . What this is going to mean is that we need to de ne two number operators. These are de ned by in nite sums over the modes: N R = m m
m =1
N L = m m
m =1
(4.23)
Together with the momentum operator p , the number operators N R and N L serve to characterize the state of a closed string. Let us denote a state by n, k . As in the open string case, the momentum operator will act according to p 0, k = k 0, k (4.24)
Therefore, the state 0, k of the string carries momentum k . Turning our attention to the number operators, rst let s specify the action of the raising and
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