code 128 font vb.net Heterotic String Theory in Java Drawing QR-Code in Java Heterotic String Theory

CHAPTER 12 Heterotic String Theory
Scanning QR Code 2d Barcode In Java
Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications.
Printing Denso QR Bar Code In Java
Using Barcode generation for Java Control to generate, create QR Code JIS X 0510 image in Java applications.
( L0 a P ) = 0 ( L0 a A ) = 0 (P sector) (A sector) )
QR Code ISO/IEC18004 Scanner In Java
Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications.
Bar Code Creator In Java
Using Barcode generation for Java Control to generate, create barcode image in Java applications.
(12.23)
Barcode Recognizer In Java
Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications.
QR Code ISO/IEC18004 Generation In C#.NET
Using Barcode creator for .NET Control to generate, create QR Code image in .NET framework applications.
The L0 are not identical in these two equations, since we have the two number operators [Eq. (12.20)]. So we have p2 P + N L aP 8 p2 A + N L aA L0 = 8 L0 =
QR-Code Printer In Visual Studio .NET
Using Barcode printer for ASP.NET Control to generate, create Quick Response Code image in ASP.NET applications.
Denso QR Bar Code Creator In Visual Studio .NET
Using Barcode drawer for .NET framework Control to generate, create QR Code image in Visual Studio .NET applications.
(P sector) (12.24) (A sector) A
QR Code JIS X 0510 Maker In Visual Basic .NET
Using Barcode drawer for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in .NET framework applications.
Code 39 Extended Printer In Java
Using Barcode maker for Java Control to generate, create Code 3 of 9 image in Java applications.
The task now is to determine the values of a P and a A . It turns out that this can be done readily given what we know about superstrings. That is A periodic boson makes a contribution of 1/24 to the normal ordering constant. A periodic fermion makes a contribution of 1/24 to the normal ordering constant. The normal ordering constant is actually formed by the sum a = bosonic contribution + fermionic contribution i Going to the light-cone gauge, there are eight transverse bosonic components. We also have the 32 fermions A. So, the total normal ordering constant for the P sector is 1 1 aP = 8 + 32 = 1 24 24 Now for the A sector, we need to know that An antiperiodic fermion contributes 1/48 to the normal ordering constant. In the A sector, the bosonic contribution is the same. Hence 1 1 aP = 8 + 32 = 1 48 24 (12.26) (12.25)
Creating GTIN - 12 In Java
Using Barcode printer for Java Control to generate, create UPC A image in Java applications.
Data Matrix Printer In Java
Using Barcode creator for Java Control to generate, create Data Matrix image in Java applications.
String Theory Demysti ed
Generating Interleaved 2 Of 5 In Java
Using Barcode printer for Java Control to generate, create Uniform Symbology Specification ITF image in Java applications.
Drawing EAN / UCC - 13 In None
Using Barcode maker for Software Control to generate, create EAN128 image in Software applications.
You should be aware that the bosonic and fermionic contributions to the normal ordering constants are due to the zero point energy of the bosonic and fermionic modes. Note also that these zero energy contributions are nite. Using Eqs. (12.24), (12.25), and (12.26) and including the relation obtained for the right-moving modes, we obtain the mass formulas for the P and A sectors:
Code 39 Extended Recognizer In VB.NET
Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET applications.
UPC-A Supplement 5 Generation In None
Using Barcode creator for Font Control to generate, create UCC - 12 image in Font applications.
P m 2 = 8 N R = 8 ( N L + 1
Draw Matrix Barcode In VB.NET
Using Barcode drawer for .NET framework Control to generate, create Matrix 2D Barcode image in .NET applications.
Encode Code-39 In None
Using Barcode creator for Office Word Control to generate, create ANSI/AIM Code 39 image in Word applications.
m 2 = 8 N R
Read Bar Code In .NET
Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications.
Barcode Generation In Objective-C
Using Barcode creator for iPhone Control to generate, create barcode image in iPhone applications.
) = 8 ( N 1)
(P sector) (A sector) (12.27)
We have made the obvious leap of faith that the mass must be the same for a given string state whether looking at the right-moving or left-moving modes. We immediately notice that A state with zero mass has N R = 0 . Put another way, a state with zero mass in heterotic string theory has the rightmoving modes in the ground state. In addition, if m = 0, then
P For a state in the P sector, N L = 1. A For a state in the A sector, N L = +1.
Now, in ordinary quantum mechanics, you learned that a number operator satis es N 0. So, we must reject the rst possibility which translates into The P sector contains no massless states.
The Spectrum
To describe the spectrum of the theory, we follow the usual procedure of constructing states which are tensor products of left-moving and right-moving modes:
= left right
(12.28)
For the left movers, we just learned that the P sector contributes nothing to a A massless state. Since N L = +1 , this means that the state from the A sector is in the rst excited state. There are two possibilities. It can be a bosonic state: left = j1 0
(12.29)
CHAPTER 12 Heterotic String Theory
Or it can be a fermionic state, since we need to consider the A as well:
A B left = 1/ 2 1/ 2 0 L
(12.30)
For right movers, with N R = 0 , we can have a bosonic state: right = i Or a fermionic state a (12.32) (12.31)
Now let s consider the case when the left movers are in the bosonic state [Eq. (12.29)]. The bosonic sector is given by the tensor product with the bosonic states of the right movers [Eq. (12.31)]:
= j1 0 L i
(12.33)
The states [Eq. (12.33)] can be summarized as follows. The particle spectrum includes: A scalar, the dilaton An antisymmetric tensor state given by j1 0 L i The graviton which is the state
j 1 R i 1 0 L
0 L i
i 1
j L
Now let s take a look at the fermionic sector for the massless states. We can get this by pairing up the bosonic states from the left movers with the fermionic states from the right movers. This is going to give us the superpartners of Eq. (12.23). The states can be written as