How to show that $L_n^\dagger=L_{-n}$ for the Virasoro generators in CFT?
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Answered by
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Hey mate ^_^
Here the situation is opposite....
It seems to me that the integral around 00 vanishes in the z→∞z→∞ limit, since it grows as z3z3 and gets diminished by the factor z−4z−4....
On the other hand, the integral around ∞∞ seems to be non-vanishing as the integrand grows faster that in the preceding case, and ζ−4ζ−4 fall off of T(ζ)T(ζ) is not enough to make it regular at infinity.....
But I do not see how to bring this integral to the expected form ⟨A1(∞)L−2A2(0)⟩⟨A1(∞)L−2A2(0)⟩. Any help is appreciated....
#Be Brainly❤️
Here the situation is opposite....
It seems to me that the integral around 00 vanishes in the z→∞z→∞ limit, since it grows as z3z3 and gets diminished by the factor z−4z−4....
On the other hand, the integral around ∞∞ seems to be non-vanishing as the integrand grows faster that in the preceding case, and ζ−4ζ−4 fall off of T(ζ)T(ζ) is not enough to make it regular at infinity.....
But I do not see how to bring this integral to the expected form ⟨A1(∞)L−2A2(0)⟩⟨A1(∞)L−2A2(0)⟩. Any help is appreciated....
#Be Brainly❤️
Answered by
3
Hello mate here is your answer.
Well, I believe it is the usual prescription ⟨V1...⟩:=⟨0|V1...|0⟩⟨V1...⟩:=⟨0|V1...|0⟩ where |0⟩|0⟩ is the vacuum state. On the other hand, in CFT it usually suffices to consider the symmetry properties to fully determine the correlator. I expect this to be the case here.
Hope it helps you.
Well, I believe it is the usual prescription ⟨V1...⟩:=⟨0|V1...|0⟩⟨V1...⟩:=⟨0|V1...|0⟩ where |0⟩|0⟩ is the vacuum state. On the other hand, in CFT it usually suffices to consider the symmetry properties to fully determine the correlator. I expect this to be the case here.
Hope it helps you.
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