Virasoro operators in bosonic String Theory

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SUMMARY

The discussion centers on the calculation of the zero mode Virasoro operator in bosonic String Theory, specifically addressing the divergent sum 1+2+3+... and its comparison to the Zeta function for regularization. Participants highlight that this divergent sum is equated to a finite negative constant, raising questions about the justification for this approach. The normal-ordered expression for the Virasoro generators is provided, illustrating that when acting on the vacuum state, all sums remain finite, thus avoiding regularization issues.

PREREQUISITES
  • Understanding of bosonic String Theory concepts
  • Familiarity with Virasoro operators and their properties
  • Knowledge of Zeta function regularization techniques
  • Basic grasp of normal ordering in quantum field theory
NEXT STEPS
  • Study the implications of Zeta function regularization in quantum field theory
  • Explore the derivation and properties of Virasoro operators in detail
  • Investigate the role of normal ordering in quantum mechanics
  • Review advanced topics in bosonic String Theory, focusing on central charge calculations
USEFUL FOR

The discussion is beneficial for theoretical physicists, string theorists, and graduate students specializing in quantum field theory and string theory, particularly those interested in operator formalism and regularization techniques.

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In a recent lecture on String Theory, we encountered the divergent sum 1+2+3+... when calculating the zero mode Virasoro operator in bosonic String Theory. This divergent sum is then set equal to a finite negative constant - the argument for doing so was a comparison with the definition of the Zeta function. However, is still have trouble with this argument, and my question is, wether there are any further justifications for taking this step. Thank you for your answers!
 
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A safe way to compute the central charge is to check how the Virasoro generators act on the vacuum. The normal-ordered expression

L_m = sum_{n=-infty}^infty : a_{m-n} a_n :

is equivalent to the following relations (perhaps modulo some signs that I don't have the energy to check) (and m > 0 in the first three lines):

L_m |0> = sum_{n=0}^m a_{m-n} a_n |0>

L_0 |0> = h |0>

L_-m |0> = 0

[L_m, a_n] = (m-n) a_{m+n}

The point is that in the second formulation, all sums are finite, so you don't have to worry about regularization.
 

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