String Theory for Undergraduates- Lecture 26

0.12 MB·English

by  Alan Guth;Barton Zwiebach

Lecture 26 8.251 Spring 2007
8.251 Final Exam Review
Practice Problems
Problem 1.

Assume standard, space-filling Virasoro brane.

a. Calculate L⊥ |0� in terms of normal order oscillations.
−6
b. Calculate L⊥ αI |0�.
−6 −1
(Hint: αI |0� =0, � mom. vacuum)
0
Problem 2.
√
D2: z = 0, D1: y =0,z = π α�, D = 26.
a. Find the M2 formula in terms of N⊥
b. Ground states? Describe the lowest mass levels.

c. Generating f(x)= �a(r)xr where a(r) is the number of states of α�M2 = r.
Problem 3.

Consider the Ramond sector of a superstring in D = 18.

Naive α�M2 = 1 � αI αI + ndI dI.
2 n�=0 −n n −n n
a. Range of I? Write the precise (n.o) ∝ M 2
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Lecture 26 8.251 Spring 2007
b. Write the generating function f (x).
R
Solutions
Problem 1.
a.
1 �
L⊥ = α α
−6 2 −6−p p
p∈Z
p = 0:
1
(α ·α +α ·α +α ·α +...+α ·α +α α +α α +α α +α α +α α +α α )
2 −6 0 −7 1 −8 2 −5 −1 −4 −2 −3 3 −2 −4 −1 −5 0 −6 1 −7
Since n> 0, α’s are annihilation operators and kill the vacuum state.
1
= α α + α α + α α + α α +(α α + α α + ...)

2 −3 −3 −2 −4 −1 −5 0 −6 −7 1 −8 2
1

L⊥ |0� =( α α + α α + α α )|0�
−6 2 −3 −3 −2 −4 −1 −5
b.
1
L⊥ αI |0� = ( α α +α α +α α )αI |0�+α α αI |0� = L⊥ |0�+αI |0�
−6 −1 2 −3 −3 −2 −4 −1 −5 −1 −7 1 −1 −6 −7
Alternative method:
LI αI |0� =[L ,αJ ]|0� + αI L⊥ |0�
−6 −1 −6 1 −1 −6
� �� � � �� �
αI know
−7
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Lecture 26 8.251 Spring 2007
Problem 2.
a.
x+ ,x− y z xx4x25 . . .
D2 - - stretched · · localized
D1 - · · · localized
[D2, D1] NN ND DD DD DD
1 ND coordinate; all of the rest are DD
α�M2 = N⊥+ α�(TL)2
a
1 1 1
a = = − +
ND 48 24 16
1 15
−1+ = − = a
16 16
1
a = −1 + (p − q)
16
√
�π α� �2
α(TL)2 = α�
2πα�
15 4 11
α�M2 = N⊥ − + = N⊥ −
16 16 16
b.

Ground state: ��p+,p2,...,pq � in book, but here q = 1 ⇒ GS = |p+�. No mo­
mentum in ND (fractional oscillators) or all the DDs. The states live on the D1
brane.
c.
GS has α�M2 = − 11 . Suppose GS |0� and just oscillator α(3), then could build
16 −1
α(3) |0�, (α(3))2 |0�, ..., so:
−1 −1

� �
1
f(x)= x −1161 (1 + x + x2 + ...)= x− 1161
1 − x
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Lecture 26 8.251 Spring 2007
But we have more!
� 1 �23� 1 �23� 1 �23
f(x)= x− 1116 ...
1 − x 1 − x2 1 − x3
= x− 1116 �∞ 1
(1 − xn)23
n=1
Now have to account for the fractional oscillators.
f(x)= x− 1116 �∞ 1 1
(1 − xn)23 (1 − xn−1/2)
n=1
Plug into Mathematica to get specific answers.
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