Lecture 20
8.251 Spring 2007
Lecture 20 - Topics
• Closed Strings
Recall: For closed strings in light cone gauge:
σ ≈ σ + 2π
x + = α � p+ τ
P τ + constant
�
1 1
Ẋ ± X − = � + (Ẋ I ± X �I )2
α 2p
H = α� p+ p−
Pτµ =
1
Ẋ µ
2πα�
Open Strings:
[X I (τ, σ), P τ,J (τ, σ � )] = iη IJ δ(σ − σ � )
�
√
Ẋ I ± X �I = 2α�
αn e(−in(τ ±σ))
I
[αm
, αnI ] = mδm+n,0 δ IJ
Graviton States:
ξIJ aIJ+
p+ ,pτ |Ω�
ξIJ = ξJI = ξJI = 0
Solve and Find Mode Expansion of Closed Strings
µ
X µ (τ, σ) = XLµ (τ + σ) + XR
(τ − σ)
µ
Left and Right (XLµ and XR
) both solve wave equation, as goes their sum. Let:
u = τ + σ, v = τ − σ
1
Lecture 20
8.251 Spring 2007
xµ (τ, σ + 2π) = xµ (τ, σ)
True σ ≈ σ + 2π except when the wold has a compact dimension and x goes
aroudn a circle - even though back at same σ after 2π, at a different x coordinate.
XL (u) + XR (v) = XL (u + 2π) + XR (v − 2π)
XR (v) − XR (v − 2π) = XL (u + 2π) − XL (u)
This is the periodicity condition. XL and XR are independent variables. XL�
�
and XR
are periodic.
�
α� � µ (−inu)
αn e
2
n∈Z
�
α� � µ (−inv)
�µ
XR
(v) =
αn e
2
XL�µ (u)
=
n∈Z
Have 2 independent sets of oscillators. Not related to the open string oscillators.
�
α� µ
√
α0 u + i . . .
2
�
1 µ
α� µ
µ
XR (v) = XR0 +
α v + ...
2
2 0
XLµ (u)
1
= XLµ0 +
2
µ
All terms in XLµ and XR
have e−inu component except first two terms of each.
Periodicity Condition:
α0µ = αµ0 ∀µ
√
1 µ
µ
X (τ, σ) = (XL0
+ XR0
) + 2α� α0µ τ + i
2
µ
2
�
α� �
...
2
Lecture 20
8.251 Spring 2007
µ
µ
Let: xµ0 = 12 (XL0
+ XR0
)
Momentum of String:
µ
�
p =
2π
P τ µ dσ
0
� 2π
1
∂xµ
=
dσ
2πα� 0
∂τ
1 √ � µ
=
2α α0 (2π)
2πα�
�
2 µ
=
α
α� 0
µ�
Ẋ µ = XLµ � (τ + σ) + XR
(τ − σ)
µ�
X µ� = XLµ � (τ + σ) − XR
(τ − σ)
Ẋ µ + X µ� = 2XLµ � =
�
√
2α�
n ∈ Zαnµ e(−in(τ +σ))
µ�
Ẋ µ − X µ� = 2XR
=
�
√
2α�
n ∈ Zαnµ e(−in(τ −σ))
+
⊥
1� I I
α α
2 p p n−p
+
⊥
1� I I
α α
2 p
p n−p
(Ẋ I + X I � )2 = 4α�
�
Ln e(−in(τ +σ)) ⇒ Ln =
(Ẋ I − X I � )2 = 4α�
�
Ln e(−in(τ −σ)) ⇒ Ln =
�
1 1
(−in(τ +σ))
X˙ − + X −� = � + 4α�
L+
ne
α 2p
2 � + (−in(τ +σ))
= +
Ln
e
p
�
√
2 � ⊥ (−in(τ −σ))
(−in(τ +σ)) −
= 2α�
α−
Ẋ − X −� = +
Ln e
ne
p
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Lecture 20
8.251 Spring 2007
√
2 ⊥
2α� α−
n = + Ln
p
√
2
2α� αn− = + L⊥
p n
⊥
|L0 = L⊥
o | constraint on state space of theory.
Differences between closed and open string. Hilber space: can’t just double
everything for closed strings. X0 doesn’t double and momentum doesn’t double.
⊥
L0 = L⊥
0 constant.
1 I I
⊥
L⊥
← number operator =
0 = 2 α0 α0 + N
L⊥
0 =
L⊥
0 =
�
α I I
4 p p
α� I I
4 p p
�∞
p=1
I
α−p
αpI
+ N ⊥ (For open strings, did not have factor
�
⊥
⊥
+ N , N = p=1 αI−p αpI
⊥
⊥
L⊥
=N
0 = L0 ⇒ N
⊥
Recall:
√
2
⊥
2α� α−
n = + (Ln − 1)
p
√
2
2α� αn− = + (Ln⊥ − 1)
p
√
2α� α0− =
1
+
� −
(L⊥
0 + L0 − 2) = α p
+
p
⊥
H = α� p+ p− = L⊥
0 + L0 − 2
2
�
M =
Recall open string: M 2 =
1
α� (−1
2 ⊥
⊥
N +N −2
�
α
�
+ N ⊥)
⊥
I
[L⊥
0 + L0 , X (τ, σ)] = −i
4
∂xI
∂τ
1
4
Lecture 20
8.251 Spring 2007
⊥
I
[L⊥
0 − L0 , X (τ, σ)] = i
∂xI
∂τ
∂X I
∂σ
⊥
I
= X I (τ, σ) + [−i�(L⊥
0 − L0 ), X (τ, σ)]
X I (τ, σ+) = X I (τ, σ) +
∞ �
25
�
n=1 I=2
N
Ground state: N ⊥ = N
well understood.
⊥
∞ �
25
�
λn,I
(aI+
n )
λm,J
� + �
�p , pτ
m=1 J=2
⊥
⊥
=N
= 0, |p+ , pτ �, M 2 = − α4� . Close string tachyon. Not
J+ +
⊥
=N
Next state: M = RIJ aI+
1 a1 |p , pτ �, N
⊥
= 1, M 2 =
2
α� (1
+ 1 − 2) = 0.
We have a (D − 2)x(D − 2) matrix. Any matrix can be split into symmetric
and antisymmetric: RIJ = SIJ + AIJ
R + RT
2
R − RT
A=
2
S=
1
1
K
K
δIJ SK
)+(
δIJ SK
+ AIJ )
D−2
D−2
+ S � δIJ + AIJ
RIJ = (SIJ −
= ŜIJ
�
�
�
�
�
�
J+ � +
J+ � +
I+ � +
M = ŜIJ aI+
p , pτ + AIJ aI+
p , pτ + S � aI+
p , pτ
1 a1
1 a1
1 a1
SˆIJ ↔ ξIJ aIJ
p+ pτ |Ω� graviton states
AIJ : “Kalbra-mon” states
S � just one single state. No Lorentz index. Massless scalar. A real troublemaker.
Called dilation scalar. Tells us how strong strings interact.
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