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The deformation of path theorem (see Complex Variables Demysti ed ) tells us that we can shrink or expand the contour used for the integration in Eq. (5.35) and leave the integral invariant. Since movement along the radial distance in the complex plane corresponds to time translation, this tells us that the Virasoro operators are invariant under a time translation, in other words the integrals [Eq. (5.35)] are related to conserved charges.
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To gain insight into the physics of the problem, we will begin by calculating propagators for the theory. In the text we will derive the closed string propagator, you can try the open string case in the chapter quiz. We begin by considering the propagator or Wick expansion for X ( , ) for the closed string. This is done by calculating X ( , ), X ( , ) = T ( X ( , ), X ( , )) : X ( , ) X ( , ) : This is also called the two-point function. There are two notations you should be aware of in this expression. The T indicates that the expression is time ordered. It is shorthand for
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T ( X ( , ), X ( , )) = X ( , ) X ( , ) ( ) + X ( , ) X ( , ) ( )
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To understand what we have here, note that the Heaviside function (t ) = 1 if t > 0, (t ) = 0 if t < 0 . This expression ensures that the term which occurs earlier in time is to the right. So if > , which means that is later than , then T ( X ( , ), X ( , )) = X ( , ) X ( , )
String Theory Demysti ed
and vice versa for the other possibility. The colons indicate normal ordering which puts all creating operators to the left of all annihilation operators. Now consider the fact that we can write the elds in terms of left movers and right movers. Then
0 X ( , ) X ( , ) 0 = 0 ( X L + X R )( X L + X R ) 0 = 0 XL XL + XL XR + XR XL + XR XR 0
Now let s move to the complex plane where X R = X R ( z ), X L = X L ( z ) so that 0 X ( , ) X ( , ) 0 0 X ( z, z ) X ( z , z ) 0 . Let s consider one term, 0 X R X R 0 using the expansion given in Eq. (5.30). Noting that p 0 = 0 and [ x , p ] = i , m 0 = 0 if m > 0 and 0 n = 0 for n < 0, we have
0 XR XR 0 =
2 2 1 0 x x 0 i s ln z 0 p x 0 s 4 4 2
m 0 n 0
mn
0 n m 0 z n z m
1 0 x x 0 4 2
2 ln z + s 2 2
m 0 n 0
mn
0 n m 0 z n z m
Moving from the rst to the second line, we used 0 p x 0 = 0 x p i 0 = 0 x p 0 i 0 0 = i Now the commutator of the creation and annihilation operators satis es
m , n = m m , n
And so this becomes
0 XR XR 0 =
1 0 x x 0 4 2
2 ln z 1 + s z n z n 2 2 n>0 n
CHAPTER 5 Conformal Field Theory Part I
Now, using the fact that
x /n = ln(1 x ), we can write this as
n n =1
0 XR XR 0 =
1 0 x x 0 4 4
1 0 x x 0 4 2
( z /z )n n n>0
2 s
ln z + ln(1 z /z ) 2 ln z
1 0 x x 0 + 4 4
2
ln( z z )
Similar calculations show that
0 XL XL 0 =
1 0 x x 0 + 4 4 4 4
ln z
2
ln( z z )
0 XR XL 0 =
ln z ln z
0 XL XR 0 =
Adding these terms up we nd that 0 X ( z , z ) X ( z , z ) 0 = 0 x x 0
2
ln[( z z )( z z )] = 0 X ( z , z ) X ( z , z ) 0
So the vacuum expectation value of the time ordered product is
0 T [ X ( z, z ) X ( z , z )] 0 = 0 x x 0 2
ln[( z z )( z z )]
Now let s return to the two-point function: X ( , ), X ( , ) = T ( X ( , ), X ( , )) : X ( , ) X ( , ) : Normal ordering puts all destruction operators to the right. So : p x : 0 = x p 0 = 0 . Also, normal ordering of terms like 1 /( mn ) 0 n m 0 z n z m
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