Lectures on String Theory

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– 59 – 4.2.1 Lorentz symmetry and critical dimension Studying the classical string in the light-cone gauge we realized that the generators J i− of the Lorentz symmetry become rather complicated functions of transversal oscillators J i− = x i p − − x − p i − i ∞∑ n=1 1 ( ∑ ∞ α i n −n αn − − α−nαn) − i 1 (ᾱi ) − i n −n ᾱn − − ᾱ−nᾱ − n i . n=1 We would like to ask a question whether we can use this expression in quantum theory regarding α n and ᾱ n as operators to define the quantum Lorentz generators? Due to the unusual canonical structure of the light-cone theory it is not obvious that consistent Lorentz generators should exist. In quantum theory we want to realize a unitary representation of the Poincaré group and therefore, we require the Lorentz generators to be hermitian, i.e. (J µν ) † = J µν , where we treat J µν as an operator acting in the Hilbert space. Also the Lorentz generators must be normal-ordered to have the well-defined action on the vacuum state. Consider the following ansatz for the Lorentz generators J i− , which are the most intricate generators to be defined in quantum theory, J i− = 1 ∞∑ 2 (xi p − + p − x i ) − x − p i −i } {{ } n=1 l i− 1 ( ∑ ∞ α i n −n αn − − α−nαn) − i − i n=1 1 (ᾱi ) n −n ᾱn − − ᾱ−nᾱ − n i . One can see that these generators are hermitian and normal-ordered so they can be considered as candidates to realize the Lorentz algebra symmetry. The latter requirement is equivalent to [J i− , J j− ] = 0 . This is an equation we would like to prove. First we discuss the orbital part. We have [l i− , l j− ] = [ 1 2 (xi p − + p − x i ) − x − p i , 1 2 (xj p − + p − x j ) − x − p j ] = 1 4 [xi p − , x j p − ] → i 4 (xj p i − x i p j ) p− p + + 1 4 [xi p − , p − x j ] → i 4 (pi x j − x i p j ) p− p + − 1 2 [xi p − , x − p j ] → − i 2 (x− p − δ ij − x i p j p− p + ) + 1 4 [p− x i , x j p − ] → − 4 i p − p + (pj x i − x j p i ) + 4 1 [p− x i , x − p j ] → − 4 i p − p + (pj x i − p i x j ) − − − 1 2 [p− x i , x − p j ] → i 2 ( p− p + pj x i − p − x − δ ij ) 1 2 [x− p i , x j p − ] → − i 2 (xj p i p− p + − x− p − δ ij ) 1 2 [x− p i , p − x j ] → − i 2 ( p− p + xj p i − x − p − δ ij ) .
– 60 – Here by arrow we indicated the explicit expression for the corresponding commutator. Reducing similar terms we get [l i− , l j− ] = 4 i [pi , x j ] p− p + + 4 i p − p + [pi , x j ] + 2 i [x− , p − ]δ ij = 0 . Thus, the generators of the orbital part of the total momentum have vanishing commutator. Next we compute [l i− , S j− ] + [S i− , l j− ] = −2 p− X ∞ 1 p + n=1 n α[i −n αj] n + (4.15) + 1 p + ∞ X 1 n=1 n h − (α j −n α− n − α− −n αj n )pi + (α i −n α− n − α− −n αi n )pj i . Here and below we use the concise notation α [i −n αj] n = α i −n αj n − αj −n αi n . Now we are in a position to study the most difficult commutator [S i− , S j− ]. We will do further analysis in several steps. 1. First we consider the following commutator Therefore, This further gives [S i− , α − ∞ m ] = −i X 1 n n [αi −n α− n − α− −n αi n , α− m ] (4.16) √ √ ∞X 1 4πT = −i − n=1 n p + nαi m−n α− n + αi −n [α− n , α− m ] −[α − 4πT −n , α− m ]αi n − p + nα− −n αi n+m . | {z } A [S i− , α − m ] = −i √ 4πT p + − i f(m) m αi m . X ∞ 1 − nα i m−n n=1 n α− n + αi −n (n − m)α− m+n + (n + m)α− m−n αi n − nα− −n αi n+m [S i− , α − m ] = −i √ 4πT + i p + √ 4πT p + X ∞ α i −n α− m+n − αi m−n α− n + α − m−n αi n −α − −n αi n+m n=1 | {z } | {z } A B X ∞ m n=1 n α i −n α− m+n − α− m−n αi n − i f(m) m αi m . The terms in the first line of the last equation can be partially cancelled upon changing the summation index and we find that [S i− , α − m ] = −i √ 4πT + i p + √ 4πT p + m X n=1 X ∞ m n=1 n − α i m−n α− n | {z } A + α − m−n αi n | {z } B α i −n α− m+n − α− m−n αi n − i f(m) m αi m . Since we have mX α − m−n αi n = X 0 α − m−1 k αi m−k ≡ X α − n αi m−n n=1 k=m−1 n=0 we see that mX − α i mX m−n α− n + α− m−n αi n = − α i m−1 m−n α− n + X α − n αi m−n n=1 n=1 n=0 = α − m−1 0 αi n − αi 0 α− m + X [α − n , αi m−n ] n=1 Using the fact that P m−1 n=1 (m − n) = 1 m(m − 1) we obtain 2 √ √ [S i− , α − 4πT 4πT m ] = i p + (αi 0 α− m − α− 0 αi m ) + i X ∞ m p + α i −n n=1 n α− m+n − α− m−n αi n 4πT m(m − 1) + i (p + ) 2 − f(m) ! α i m 2 m . (4.17) Thus, under the action of the Virasoro operators α − m the spin components Si− transform in a nontrivial manner. Note that the r.h.s. of eq.(4.17) is normal-ordered.
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Preprint typeset in JHEP style - HY
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- 2 - 6. Classical fermionic supers
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- 4 - bosons and Ramond’s fermion
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- 6 - ? Type I Type IIB E 8 x E 8 h
Page 9 and 10: - 8 - 2. Relativistic particle Cons
Page 11 and 12: - 10 - are called the first class c
Page 13 and 14: - 12 - which is in complete agreeme
Page 15 and 16: - 14 - The Nambu-Goto action ∫
Page 17 and 18: - 16 - Exercise Show that the Hessi
Page 19 and 20: - 18 - 3.2.3 Conformal gauge Consid
Page 21 and 22: - 20 - which result into We see tha
Page 23 and 24: - 22 - Analogously, {¯L m , X µ }
Page 25 and 26: - 24 - 3.3.1 Solutions of the equat
Page 27 and 28: - 26 - i.e. the total mass momentum
Page 29 and 30: - 28 - It follows from the Schwarz
Page 31 and 32: - 30 - 2. Varying w.r.t. γ τσ le
Page 33 and 34: - 32 - The physical Hamiltonian and
Page 35 and 36: - 34 - Orbits of the Virasoro algeb
Page 37 and 38: - 36 - Let us outline the computati
Page 39 and 40: - 38 - Thus, we arrive at {l i− ,
Page 41 and 42: - 40 - One can correctly anticipate
Page 43 and 44: - 42 - Using α µ q α µ m+n−q
Page 45 and 46: - 44 - Here N is the number operato
Page 47 and 48: - 46 - string spectrum and makes it
Page 49 and 50: - 48 - Thus, for τ > τ ′ one ob
Page 51 and 52: where the expression on the r.h.s.
Page 53 and 54: - 52 - Plugging everything together
Page 55 and 56: - 54 - where we assume that τ 1 >
Page 57 and 58: - 56 - It also follows from the sca
Page 59: - 58 - 4.2 Quantization in the phys
Page 63 and 64: - 62 - Thus, our commutator takes t
Page 65 and 66: - 64 - level α ′ mass 2 rep of S
Page 67 and 68: - 66 - representations of the trans
Page 69 and 70: - 68 - Let us illustrate this proce
Page 71 and 72: - 70 - Here c α is called a ghost
Page 73 and 74: - 72 - The ghosts and anti-ghosts a
Page 75 and 76: - 74 - The expression in the bracke
Page 77 and 78: - 76 - One can check that these obj
Page 79 and 80: ¡ ¡ £¢ ¢ - 78 - from the cylin
Page 81 and 82: - 80 - coordinate chart ¡ ¢¡¢¡
Page 83 and 84: - 82 - The last formula can be unde
Page 85 and 86: - 84 - i.e the conformal factor φ
Page 87 and 88: - 86 - The last term here reflects
Page 89 and 90: - 88 - i.e. it is not normalizable.
Page 91 and 92: - 90 - gluing the opposite sides to
Page 93 and 94: - 92 - Any point in the upper-half
Page 95 and 96: - 94 - d(d−1) 2 local Lorentz tra
Page 97 and 98: - 96 - The two-dimensional Dirac ma
Page 99 and 100: - 98 - The “flat” and “curved
Page 101 and 102: - 100 - Here D α ɛ = ∂ α ɛ
Page 103 and 104: - 102 - By using reparametrizations
Page 105 and 106: - 104 - Consider first the commutat
Page 107 and 108: - 106 - Now we note that in additio
Page 109 and 110: - 108 - One can also check that the
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- 110 - The oscillator ground state
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- 112 - and similar for ML 2 . For
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- 114 - real components. Thus, the
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- 116 - we get Since a general stat
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- 118 - that of Type IIB supergravi
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- 120 - In this form the Hamiltonia
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- 122 - group” was introduced by
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- 124 - where the operator Q k (x,
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- 126 - where we have substituted
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- 128 - D. Riemann normal coordinat
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- 130 - E. Exercises Exercise 1. Sh
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- 132 - Exercise 10. • Show that
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- 134 - Exercise 16. Prove that tha
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- 136 - where A(m) is a function of
Page 139 and 140:
- 138 - Exercise 38. Compute the th
Page 141 and 142:
- 140 - as Here Exercise 50. Under
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