By Headrick M

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**Additional resources for A solution manual for Polchinski's String Theory**

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The ground states |0; k of the closed string are invariant under ′ 2 ˜ orientation reversal. The Ωq L0 +L0 eigenvalue of a ground state is q α k /2 ; each raising operator α+ −m − m m multiplies that eigenvalue by q , while each raising operator α−m multiplies it by −q . Summing over all combinations of such operators, the partition function is ZX (t) = q −13/6 V26 ∞ d26 k α′ k2 /2 (1 − q m )−1 (1 + q m )−1 q (2π)26 m=1 = iV26 (4π 2 α′ t)−13 η(2it)−26 .

36), where v µ (y1 ) = −2iα′ k2µ kµ − 3 y12 y13 (22) 43 6 CHAPTER 6 (we leave out the contraction between X˙ µ (y1 ) and eik1 ·X(y1 ) because their product is already renormalized in the path integral). Momentum conservation, k1 + k2 + k3 = 0, and the mass shell conditions imply 2 + 2k2 · k3 , α′ 1 1 = k32 = (k1 + k2 )2 = ′ + 2k1 · k2 , ′ α α 1 1 = k22 = (k1 + k3 )2 = ′ + 2k1 · k3 , ′ α α 0 = k12 = (k2 + k3 )2 = (23) (24) (25) so that 2α′ k2 · k3 = −2, while 2α′ k1 · k2 = 2α′ k1 · k3 = 0. 11) therefore simplifies to ⋆ ⋆ X˙ µ eik1 ·X (y1 )⋆⋆ ⋆⋆ eik2 ·X (y2 )⋆⋆ ⋆⋆ eik3 ·X (y3 )⋆⋆ (26) D2 X = 2α′ CD (2π)26 δ26 (k1 + k2 + k3 ) 2 1 2 y23 kµ k2µ + 3 y12 y13 .

12) The L2 condition adds one more constraint: 0 = L2 |f, e; k √ = 2 2α′ kµ eµ + fµµ |0; k . (13) Using (12), this implies fii = 5f00 , (14) where fii is the trace on the spacelike part of f . There are D + 1 independent spurious states at this level: |g, γ; k = L−1 gµ αµ−1 + L−2 γ |0; k √ √ γ = 2α′ g(µ kν) + ηµν αµ−1 αν−1 |0; k + gµ + 2α′ γkµ αµ−2 |0; k . 2 (15) These states are physical and therefore null for g0 = γ = 0. Removing these D − 1 states from the spectrum leaves D(D − 1)/2 states, the extra one with respect to the light-cone quantization being the SO(D − 1) scalar, √ 2(D − 1) D−1 fij = f δij , f00 = f, e0 = f, (16) 5 5 with all other components zero.

### A solution manual for Polchinski's String Theory by Headrick M

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