Introduction to String Theory and D–Branes - School of Natural ...

2.9.1 Zero Point Energy From the Exponential Map
After doing the transformation to the z–plane, it is interesting to note that the Fourier expansions we have
been working with to define the modes of the stress tensor become Laurent expansions on the complex plane,
e.g.:
Tzz(z) =
∞
m=−∞
Lm
.
zm+2 One of the most straightforward exercises is to compute the zero point energy of the cylinder or strip (for a
field of central charge c) by starting with the fact that the plane has no Casimir energy. One simply plugs
the exponential change of coordinates z = e w into the anomalous transformation for the energy momentum
tensor and compute the contribution to Tww starting with Tzz:
Tww = −z 2 Tzz − c
24 ,
which results in the Fourier expansion on the cylinder, in terms of the modes:
2.9.2 States and Operators
Tww(w) = −
∞
m=−∞
Lm − c
24 δm,0
e iσ−τ .
There is one thing which we might worry about. Have we lost any information about the state that the
string was in by performing this reduction of an entire string to a point? Should we not have some sort of
marker with which we label each point with the properties of the string it came from? The answer is in the
affirmative, and the object which should be inserted at these points is called a “vertex operator”. Let us see
where it comes from.
As we learned in the previous subsection, we can work on the complex plane with coordinate z. In these
coordinates, our mode expansions (36) and (37) become:
X µ (z, ¯z) = x µ − i
α ′
for the open string, and for the closed:
X µ (z, ¯z) = X µ
L
2
1/2
X µ 1
L (z) =
2 xµ − i
X µ 1
R (¯z) =
2 xµ − i
α µ
0 lnz¯z + i
(z) + Xµ
R (¯z)
′ α
2
α ′
2
1/2
1/2
′ 1/2
α 1
2 n αµ −n −n
n z + ¯z , (73)
α µ
0 lnz + i
˜α µ
0 ln ¯z + i
where we have used the zero mode relations (38). In fact, notice that:
n=0
∂zX µ ′ 1/2
α
(z) = −i α
2
n
µ nz −n−1
∂¯zX µ ′ 1/2
α
(¯z) = −i
2
20
n
′ 1/2
α 1
2 n αµ nz −n
n=0
′ 1/2
α 1
2 n ˜αµ n¯z −n ,
n=0
(74)
˜α µ n ¯z−n−1 , (75)