String Theory Demystified

CHAPTER 13 D-Branes 229
fi rst excited state. We can act on the ground state with a i†
a
1 or with a1 using a i† fi rst. The state is
In this case
i† + i
a p , p
∞ p
∞ d
2 i† + i 1 ⎛
i † i
a †
ma p, p = naa ma a
1 ∑∑
+ n n ∑ ∑ m
′ ⎝
⎜
α n=
1 i=
2
m=
1 a=
p+
1
1 ⎛
=
′
a + i
Now, a p , p = 0, so
m
1
a
m
⎞
− 1
⎠
⎟ a p , p
1
i† + i
∞ p
∞ d
i † i i† + i
a † a i†
∑∑na a a p p +
n n 1 ∑ ∑ ma a a m m 1
α ⎝
⎜
n=
1 i=
2
m=
1 a= p+
1
† . Let’s consider
+ i i† + i
, p , p − a p , p
∞ p
∞ d
1 ⎛
i † i i†
+ i
a † i†
a + i
= ∑∑na a a p , p + ma a a p p −a
n n 1 ∑ ∑
,
m m
α ′ ⎝
⎜
n=
1 i=
2
m=
1 a= p+
1
∞ p
2 1 ⎛
i † i i† + i i† + i
m = ∑∑
nanana1p , p − a1p , p
α ′ ⎝
⎜
n=
1 i=
2
1 1
⎞
⎠
⎟
1
p , p
i† + i
i i †
i i†
i†
i
Now we use the commutator ⎡⎣ a , a ⎤⎦ = δ to write aa = δ + a a.
And so
m
n
∞ p
2 1 ⎛
i † i†
i + i
m = ∑∑
na δ + a a p , p a
n n1
1 n
α ′ ⎝
⎜
1 ⎛
=
α ′ ⎝
⎜
1
=
α ′
n=
1 i=
2
∞
p
∑∑
n=
1 i=
2
δ
mn
( ) −
−
i† + i i† + i
na p , p a p , p
n1n 1
i† + i i† + i
( a p , p − a p , p
1 1 )=
0
n
1 n1
1
i†
1
⎞
⎠
⎟
+
p , p
i† + i
Hence, the state a1p , p has mass m 2
= 0.
These states are characterized by an
index i which denotes coordinates on the brane. Since i= 2, ..., p,
there are a total
of ( p + 1) −2states.
Recall that a photon in a (3 + 1) dimensional theory has two
transverse states. So these states are photon states.
i
⎞
⎠
⎟
n
⎞
⎠
⎟
⎞
⎠
⎟