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Generally, when you evaluate the 3 open string tachyon tree-level amplitude in CFT, you do a conformal transformation mapping the worldsheet to the upper half of the complex plane and the incoming and outgoing strings become points on the real axis.

However, in evaluating the string field theory tree-level amplitude, the problem involves a different procedure where the 3 string worldsheet are conformally transformed in to 3 different worldsheets. They are then glued together to form a circle, with the incoming and outgoing strings being the points on the circumference of the circle.

My question is: why the same amplitude is seen in two different ways?

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    $\begingroup$ "3 different worldsheets which are then glued together to form a circle" Could you please clarify this statement? How could 2d surfaces be glued to form a circle? $\endgroup$ Commented May 3, 2013 at 2:15
  • $\begingroup$ please see fig 4 and fig5 inthis reference arxiv.org/abs/hep-th/0311017 $\endgroup$ Commented May 5, 2013 at 4:06
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    $\begingroup$ @VaibhavWasnik could you not write a nice answer explaining it based on the two figures in the paper ;-)? $\endgroup$ Commented May 11, 2013 at 8:32

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The disk and upper half plane are related by an SL(2,R) transformation. Therefore correlators evaluated on the respective surfaces are identical.

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This disk is the interaction region on the S or T channel. One can have $2$ or more $e^{ikx}$ inputs on the boundary to compute vertices.

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