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code 128 generator vb.net Dbrane in Java
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Creating Code 128 Code Set B In Java Using Barcode creator for Java Control to generate, create Code 128B image in Java applications. EAN128 Generator In Java Using Barcode encoder for Java Control to generate, create UCC.EAN  128 image in Java applications. Once again we apply the quantization procedure, considering the bosonic string theory case. The rst step is to write down the modal expansions. These will be different depending on which coordinates we look at because now we have NN coordinates and DD coordinates. The rst step is to write out the modal expansions. Now let s recall the open string modal expansion, which can be written in the following way: Create Code 11 In Java Using Barcode creator for Java Control to generate, create Code 11 image in Java applications. Data Matrix Drawer In Java Using Barcode maker for BIRT Control to generate, create Data Matrix image in BIRT applications. X ( , ) = x0 + 2 p0 + i 2 n in e cos( n ) n 0 n
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Barcode Creation In ObjectiveC Using Barcode drawer for iPhone Control to generate, create barcode image in iPhone applications. Creating Code 128B In None Using Barcode printer for Software Control to generate, create Code 128 Code Set B image in Software applications. X ( , ) = i 2 n e in sin( n ) n 0 Clearly this expression satis es X =0
= 0 , which are the Neumann boundary conditions. So, we take the modal expansion for the X i to be
i i X i ( , ) = x0 + 2 p0 + i 2 i n in e cos( n ) n 0 n
(13.10) For the DD coordinates, we really have two requirements that have to be satis ed. We need the summation over the modes in the expansion to vanish at = 0, . This indicates that we should use sin( n ) instead of cos( n ) which is in the usual open string expansion. However, we also need X a ( 0, ) = X a ( , ) = x a. Looking at the usual modal expansion, this tells us that we must take a x0 x a a p0 0 n 0
String Theory Demysti ed
So, the presence of a momentum term in the modal expansion would mean that Dirichlet boundary conditions could not be satis ed. Putting everything together, the modal expansion for the DD coordinates is X a ( , ) = x a + i 2 i n in e sin( n ) n 0 n
(13.11) Quantization will involve imposing the usual commutators: X a ( , ), X b ( , ) = i ab ( ) a b m , n = m ab m +n ,0 (13.12) Now, for a moment we consider the general lightcone expansion of the string (so + for a moment we let i = 2, ..., 25). We gauge x by choosing n = 0 for all n 0 and so + X + ( , ) = x0 + 2 p +
The momentum p+ is de ned as p+ = 1 + 0 2 (13.13) The lightcone gauge condition is p+ = Now, X is an NN coordinate, so X ( , ) = x + 2 p + i 2 It follows that X X = 2 p + 2 n e in( ) n 0 n in e cos( n ) n 0 n
1 2 (13.14) (13.15) CHAPTER 13 DBranes
and i ( X X ) = 2 p i + 2 n e in ( ) n 0
i i 2 2
i Using the lightcone gauge condition, 2 p i = 1/ p +. By writing n in terms of n and looking at the n = 0 mode, we can use the expression for ( X i X i )2 to write 1 p pi pi = + ++ + p p p p+ p+
i n
i n
1 p p+
So, we nd that 1 i i 2 p+ p = pi pi + n n 1 = H n 0 (13.16) where we have introduced the normalordering constant a = 1 for bosonic string theory into the equations. To transition to the case of a Dbrane, all we have to do is have the modes split up into NN and DD coordinates. This means that 2 p+ p = 1 i i i i a a p p + n n + n n 1 = H n 0 This allows us to write down the mass: m 2 = p 2 = 2 p + p pi pi = 1 i i a a ( n n + n n ) 1 n 0 (13.17) We can de ne creation and annihilation operators: ai = aa =
1 m 1 m
i m
ai = ai =
a m a n
1 m 1 m
i m a m
a m i n
a , a = a , a = mn
String Theory Demysti ed
with all other commutators zero. So the mass can be written in the following way: m2 =
d 1 p i i a a nan an + mam am 1 n=1 i=2 m =1 a = p+1
Due to the presence of the Dbrane, the interpretation of the mass has changed. Lorentz invariance is restricted to the brane worldvolume, so we view this mass as a mass living in p + 1 dimensions. Now, recall that for a = p + 1, , d , the Dirichlet boundary conditions forced us to take the pa to vanish. This means that states will be of the form p + , pi (13.18) where i = 2, ..., p are the NN coordinates. Since the states only depend on pi , this means two things: Any elds we de ne are functions of the momenta pi . String states only have momentum in the NN directions along the brane. By writing the Fourier transform, we would see they are functions of the coordinates x i. So what does this mean The elds are de ned on coordinates that de ne the volume of the Dpbrane they have no coordinate dependence on the a = p + 1, , d and so are zero in the region outside the Dbrane. We summarize this by saying that the elds live on the Dpbrane. There are three states we can readily identify. The ground state p+ , pi is i a i a immediately annihilated by the terms an and am that is, an p + , p i = an p + , p i = 0 so the mass is

