Lecture 18
8.251 Spring 2007
Lecture 18 - Topics
• Open Strings
Still for open string:
τI
+
Heisenberg operators: X I (τ, σ), x−
0 , P (σ), p
[X I (σ), P τ J (τ, σ � )] = iη IJ δ(σ − σ � )
+
[x−
0 , p ] = −i
∂
∂
= 2α� p+ +
⇔
∂τ
∂X +
2α� p+ p−
� �� �
=H
Hamiltonian, H
−
�
p =
dσ(P −τ =
1 ∂X −
)
2πα� ∂τ
H = 2α� p+ p− = L+
0 from analysis of classical string
Are we sure H = 2α� p+ p− ? After all, p− is the product of lots of operators,
which can be ill-defined. Must be careful in our quantum case.
��
Ẍ I − X I = 0
X I (τ, σ) = xI0 +
�1
√
√
2α� α0I τ + i 2α�
αI cos(nσ)e−inτ
n n
n=0
�
PτJ =
(Ẋ I + X I � )(τ, σ) =
1 ∂xJ
2πα� ∂τ
�
√
2α�
αnI e(−in(τ +σ))
n∈Z
1
σ ∈ [0, π]
(1)
Lecture 18
8.251 Spring 2007
(X˙ I − X I � )(τ, σ) =
�
√
2α�
αnI e(−in(τ −σ))
σ ∈ [0, π]
(2)
n∈Z
This is an important computation. Later, we will do this for closed strings too,
and we’ll see very similar (though not same).
Best way to select Fourier modes is in [0, 2π] but σ ∈ [0, π]. σ → −σ
(Ẋ I − X I � )(τ, −σ) =
√
�
2α�
αnI e(−in(τ +σ))
(2’)
n∈Z
This makes sense when σ ∈ [−π, 0].
AI (τ, σ) =
√
2α�
�
αnI e(−in(τ +σ))
σ ∈ [−π, π]
n∈Z
�
=
(Ẋ I + X I � )(τ, σ)
(Ẋ I − X I � )
σ ∈ [0, π]
σ ∈ [−π, 0]
Now have σ defined over [−π, π].
[X I (τ, σ), Ẋ I (τ, σ � )] = 2πα� iη IJ δ(σ − σ � )
[Ẋ I (τ, σ), Ẋ J (τ, σ � )] = 0
[X I � (τ, σ), X J � (τ, σ � )] = 0
X’s commute at different σ’s so can then differentiate.
[(Ẋ I ± X I � )(τ, σ), (Ẋ J ± X J � )(τ, σ)] = [(Ẋ I ± X I � )(τ, σ), (Ẋ J ± X J � )(τ, σ)]
d
= ±4πα� iη IJ (σ − σ � )
dσ
2
Lecture 18
8.251 Spring 2007
[AI (τ, σ), AJ (τ, σ � )] = 2α�
�
�
�
�
I
J
e(−im (τ +σ)) e(−in (τ +σ )) [αm
� , αn� ]
m� ,n�
⎧
� IJ d
�
σ, σ � ∈ [0, π]
⎨ 4πα iη dσ δ(σ − σ )
� IJ d
�
4πα iη dσ δ(σ − σ ) = 0
σ ∈ [0, π], σ � ∈ [−π, 0]
=
⎩
d
� IJ
�
� IJ d
�
−4πα iη d(−σ) δ(σ − σ) = 4πα iη d(σ) δ(σ − σ ) σ, σ � ∈ [−π, 0]
�
�
�
�
I
J
IJ
e(−im (τ +σ)) e(−in (τ +σ )) [αm
� , αn� ] = 2πiη
m� ,n�
d
δ(σ − σ � )
dσ
Apply the following integral operations:
1
2π
�
π
1
·
2π
(imσ)
dσe
−π
�
σ
dσ � e(inσ)
−σ
Divide by e(−i(m+n)τ ) on both sides:
I
[αm
, αnI ] = −nη IJ δm+n,0 e(i(m+n)τ )
I
[αm
, αnI ] = mδm+n,0 η IJ
Commutation relation proved in book:
[xI0 , pJ ] = iη IJ
Note:
α0I =
√
2α� pI
I
[αm
, αnI ] = mη IJ δm,n
√
αnµ = anµ n
n>0
3
σ, σ � ∈ [−π, π]
Lecture 18
8.251 Spring 2007
√
µ
µ
α−n
= aµ+
n = (α+n
)+
n
n<0
Opposite signs for m and n
[aIm , aJn ] = 0
J+
[aI+
m , an ] = 0
m > 0,n > 0:
√
√
[aIm m, aJn n] = mη IJ δm,n
IJ
[aIm , aJ+
m ] = η δm,n
√
2α� αn− =
1 ⊥ n=0 + −
1
Ln → 2p p = � L⊥
+
p
α 0
L⊥
n =
1� I
αn−p αpI
2
p∈Z
Don’t have to worry if n �= 0. Might have to worry if n = 0.
⊥
� + −
But what we want is: H = L⊥
0 = 2α p p . L0 =
commute so don’t know if this is right.
M 2 = −p2 = 2p+ p− − pI pI =
1
2
�
p∈Z
I
α−p
αpI but α’s don’t
1 ⊥
L − pI pI
α� 0
∞
L⊥
0 =
1 I I 1� I I
I
α α +
(α α + αp α−p
)
2 0 0 2 p=1
−p p
= α� pI pI +
∞
�
�
1
α−p αpI + (D − 2)
p
2
p=1
p=1
Note αp>0 is destruction operation convention. αp<0 is creation operation con­
vention.
4
Lecture 18
8.251 Spring 2007
�∞
∞ �
�
1 � I+ I 1
pa
a
+
(D
−
2)
p
α� p=1 p p 2
p=1
M2 =
In classical theory, had
M2 =
�∞
∞ �
�
1 � I+ I
1
na
a
+
(D
−
2)
p
α� n=1 n n 2
p=1
Showed all states of string had mass ¿ 0. Couldn’t get anything intersting with­
out mass.
Would be great here if 12 (D − 2)
M2 =
�∞
p=1
p = −1. Then:
1 � I+ I
(
nan an − 1)
α�
Now want oscillation states without mass
∞
�
p = 1 + 2 + 3 + 4 + ... = −
p=1
1
12
Crazy, huh? Not true in general, of course, but almost true in one sense. Since
we want:
∞
�
1
(D − 2)
p = −1
2
p=1
�
�
1
1
(D − 2) −
= −1 ⇒ D = 26(dimension of string)
2
12
Now how is
�∞
p=1
1
p = − 12
?!
Recall Riemann Zeta Function:
5
Lecture 18
8.251 Spring 2007
ζ(s) =
ζ(s = −1) = −
∞
�
1
s
n
n=1
∞
∞
�
�
1
1
=
=
n
12 n=1 n−1
n=1
ζ(s) well-defined and convergent for s ≥ 2. Doesn’t converge for s = 1(pole). ζ
defined on complex plane.
The beauty of analytic functions: If you know it is defined in a very small finite
regin, you know it everywhere by the Cauchy-Riemann.
2p+ p− =
1 ⊥
(L + a)
α� 0
a = constant
Define for once and for all:
L⊥
0
∞
1 I I � I I
= α0 α0 +
α−p αp
2
p=1
[M −I (a, D), M −I (a, D)] = 0
Set standards of messy computation. All books omit at least some details.
+
J
≈ [Ln+ , L+
M −J ≈ αn− αm
m ] = (m − n)Lm+n + dim. of spacetime
So need to find algebra of Viroso operators.
6