Applied Conformal Field Theory by Paul Ginsparg

By Paul Ginsparg

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A zero eigenvector of this matrix gives a linear combination with zero norm, which must vanish in a positive definite Hilbert space. At level 2, for example, we work in the 2×2 basis L−2 |h , L2−1 |h , and calculate h|L2 L−2 |h h|L2 L2−1 |h h|L21 L−2 |h h|L21 L2−1 |h 4h + c/2 6h 6h 4h(1 + 2h) = . 4b) where h1,1 (c) = 0 and h1,2 , h2,1 = 1 16 (5 − c) ∓ 1 16 (1 − c)(25 − c). The h = 0 root is actually due to the null state at level 1, L−1 |0 = 0, which implies also the vanishing L−1 L−1 |0 = 0. This is a general feature: if a null state |h + n = 0 occurs at level n, then at level N there are P (N − n) null states L−n1 · · · L−nk |h + n = 0 (with i ni = N − n).

5. 16) created by a holomorphic field φ(z) of weight h. 17) 2πi so that Ln , φ(0) = 0, n > 0. The state |h thus satisfies L0 |h = h|h Ln |h = 0, n > 0 . 18a) with the replacements L → L, h → h. Since L0 ± L0 are the generators of dilatations and rotations, we identify h ± h as the scaling dimension and Euclidean spin of the state. 18a) is known as a highest weight state. States of the form L−n1 · · · L−nk |h (ni > 0) are known as descendant states. 10), evidently satisfies h|L0 = h h| h|Ln = 0, n < 0 .

As mentioned in the previous subsection, if the determinant is negative at any given level it means that there are negative norm states at that level and the representation is not unitary. If the determinant is greater than or equal to zero, further investigation can determine whether or not the representation at that level is unitary. 5) at any level. 5b) either have an imaginary part or (for p = q) are negative. For c ≥ 25 we can choose the branch −1 < m < 0 and find that all the hp,q ’s are negative.

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