Problem 14.4b1.4

Closed string degeneracies | A First Course in String Theory

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What is the meaning of “With a = 1, ..., 8 and \bar b = \bar 1, ..., \bar 8, …”?

— Me@2015.09.14 12:11 PM

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p.315 “Explicitly, the eight states | R_a \rangle, a = 1, 2, ..., 8, with an even number of creation operators are … ”

p.316 “The eight states |R_{\bar{a}} \rangle, \bar a = \bar 1, \bar 2, ..., \bar 8, with an odd number of creation operators are … ”

— Me@2018-05-24 11:41:34 AM

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2018.05.24 Thursday (c) All rights reserved by ACHK

Problem 14.4b1.3

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

— This answer is my guess. —

Since for NS, the first 5 levels’ degeneracies are 8, 36, 128, 402, 1152, the degeneracies of (NS, NS) are 8^2, 36^2, 128^2, 402^2, 1152^2.

This is incorrect, for there are no (NS, NS) states. Instead, you should consider (NS+, NS+).

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Since for NS+, the first 5 levels’ degeneracies are 8, 128, 1152, 7680, 42112, the degeneracies of (NS+, NS+) are 8^2, 128^2, 1152^2, 7680^2, 42112^2.

p.317 Consider the relationship of the degeneracy of R+ and that of R-:

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How about the first 5 levels of R+?

The degeneracies are the same as those of R-.

p.317 Equation (14.54) “The appearance of an equal number of bosonic and fermionic states at every mass level is a signal of supersymmetry. This is, however, supersymmetry on the world-sheet.”

Equation (14.71):

f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...

p.321 “Indeed, the integer mass-squared levels in the NS generating function (14.67) have degeneracies that match those of (14.71) for the R- sector.”

— This answer is my guess. —

— Me@2018-05-14 02:51:55 PM

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2018.05.14 Monday (c) All rights reserved by ACHK

Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

NS+ equations of (14.38):

\alpha'M^2=0, ~~N^\perp = \frac{1}{2}: ~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle,
\alpha'M^2=1, ~~N^\perp = \frac{3}{2}: ~~~~\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle,
\alpha'M^2=2, ~~N^\perp = \frac{5}{2}: ~~~~\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}
\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \} |NS \rangle \otimes |p^+, \vec p_T \rangle,

For N^\perp = \frac{5}{2}, the number of states is

8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2
+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}
= 1152

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Since \alpha' M^2 = N^\perp - \frac{1}{2}, when N^\perp = \frac{5}{2}, \alpha' M^2 = 2.

Equation (14.67):

f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

Equation (14.66):

f_{NS} (x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8

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p.321

If we take f_{NS} (x) in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

\frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

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\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8 is the boson contribution.

\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8 is the fermion contribution.

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Turning Equation (14.67) into

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

is equivalent to turning all \sqrt{x} into - \sqrt{x}:

f_{NS?} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

Me@2015.08.29 12:49 PM: Somehow, \sqrt{x} represents “contribution from fermions”.

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Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all \sqrt{x} with y.

f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)

f_{NS} (x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

f_{NS+} (x)
= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)
= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...

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It is _not_ correct. Just consider it as \left(\sqrt{x} \to -\sqrt{x} \right) is not correct, since the beginning factor \frac{1}{\sqrt{x}} is not considered yet.

Instead, we should present in the following way:

f_{NS} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= \frac{1}{\sqrt{x}} g_{NS}(x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

g (\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

.

g (\sqrt{x}) - g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...

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f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]
= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]
= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]
= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...

— Me@2018-05-08 08:50:32 PM

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2018.05.08 Tuesday (c) All rights reserved by ACHK

Problem 14.4b1.1

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

Type IIA closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322 \alpha' M_L^2 = \alpha' M_R^2

 \frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2
 \alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right)
  = 4 \alpha' M_L^2

Equation (14.77):

\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)

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What is the difference of the meanings of R+ and R-?

R+ states are world-sheet bosonic states.

p.316

It thus follows that all eight | R_a \rangle states are fermionic and all | R_{\bar a} \rangle are bosonic.

Be careful:

Here, “fermionic”/”bosonic” refers to the world-sheet fermions/bosons, not the spacetime ones.

p.320

Identifying | R_a \rangle as spacetime fermions and | R_{\bar a} \rangle as spacetime bosons is not an alternative either, since spacetime bosons cannot carry a spinor index.

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How come the R+ cannot be the left-moving part?

p.320

A strategy then emerges. Since all states in the R sector have a spinor index, we will only attempt to get spacetime fermions from this sector. We also recognize that all fermions must arise from states with the same value of (-1)^F.

Me@2015.09.11 10:36 AM: In other words, it is a convention:

Following Gliozzi, Scherk, and Olive (GSO) we proceed to truncate the Ramond sector down to the set of states with (-1)^F = -1.

— Me@2018-05-01 05:59:53 PM

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2018.05.01 Tuesday (c) All rights reserved by ACHK

Problem 14.4a4

Closed string degeneracies | A First Course in String Theory

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(a) State the values of \alpha' M^2 and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

| ~\text{Number of states}
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right]^2
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)
{a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]
{a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~| ~(D-2)^2

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Can we create a formula for the number of states?

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\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2) + (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2
= ...
= (D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}
= 104976
= 324^2
= \left[ \frac{(D-2)(D-1)}{2} + (D-2) \right]^2

The result is the same as the square of the coefficients of x in Equation (14.63) on page 318.

\frac{1}{2} \alpha' M^2~| N~| ~\bar N~ |~\text{Number of states}
-2~| 0~| ~0~ |~1
0~| 1~| ~1~ |~(D-2)^2
2~| 2~| ~2~ |~(D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}
4~| 3~| ~3~ |~3200^2
8~| 4~| ~4~ |~25650^2

— Me@2018-04-25 05:13:04 PM

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2018.04.25 Wednesday (c) All rights reserved by ACHK

Problem 14.4a3

Closed string degeneracies | A First Course in String Theory

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(a) State the values of \alpha' M^2 and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

But for non-massless states, this probably is not true anymore:

| ~\text{Number of states}
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right]^2
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)
{a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]
{a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~| (D-2)^2

So the total number of states for \frac{1}{2} \alpha' M^2 = 2 (N = \bar N = 2) is

\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2)
+ (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2

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Should D be 10 or 26?

p.324 “Out of 26 left-moving bosonic coordinates of the bosonic factor only ten of them are matched by the right-moving bosonic coordinates of the superstring factor.”

D should be 26 for bosonic strings. So the total number of states is

\frac{1}{2} \alpha' M^2~| ~\text{Number of states}
2~| ~104976

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What does the difference of this part and Section 14.6 come from?

This part is for bosonic closed string, while Section 14.6 is for bosonic open string. There is no \bar N to consider in Section 14.6.

p.290 “A basis vector | \lambda, \bar \lambda \rangle belongs to the state space if and only if it satisfies the level-matching constraint”

N^\perp = \bar N^\perp.

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Can we create a formula for the number of states?

— Me@2018-04-23 11:31:16 AM

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2018.04.23 Monday (c) All rights reserved by ACHK

Quick Calculation 14.6b

A First Course in String Theory

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Construct explicitly all the states with \alpha' M^2=2 and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

~~~

Let N(n, k) = {n + k - 1 \choose k - 1}, the number of ways to put n indistinguishable balls into k boxes.

p.318 “For open bosonic strings \alpha' M^2 = N^\perp - 1, …”

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When \alpha' M^2 = 2, N^\perp = 3, the cases are:

1. three a_1^\dagger‘s:

N(3,24) = 2600

2. one a_2^\dagger and one a_1^\dagger:

24 \times 24 = 576

3. one a_3^\dagger:

24

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Total number of possible states for N^\perp = 3 is 3200.

— Me@2018-04-20 02:45:35 PM

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2018.04.20 Friday (c) All rights reserved by ACHK

Problem 14.4a2

Closed string degeneracies | A First Course in String Theory

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(a) State the values of \alpha' M^2 and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

p.288 Equation (13.48):

M^2 = \frac{2}{\alpha'} \left( N^\perp + \bar N^\perp - 2 \right)

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p.290 “A basis vector | \lambda, \bar \lambda \rangle belongs to the state space _if and only if_ it satisfies the level-matching constraint”

N^\perp = \bar N^\perp

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\frac{1}{2} \alpha' M^2~| ~\text{Number of states}
-2~| ~1
0~| ~(D-2)^2

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\frac{1}{2} \alpha' M^2~| ~\text{Number of states}
2~| ...

{a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle
{a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle

Assuming the signs of the wave functions do not matter:

 | |~\text{Number of states}
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger}~| \left[ \frac{(D-2)(D-1)}{2} \right]^2
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|
{a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|

p.291 How come the number of the components of the matrix represents the number of states?

p.292 For massless states, we have only one {a_1^{I}}^\dagger and \bar a_1^{J\dagger}, where a and \bar a cannot interchange. So I and J also cannot interchange. In other words, in Equation (13.69), there is no double count. All the states are independent.

But for non-massless states, this probably is not true anymore:

 | ~\text{Number of states}
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right]^2
{a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~| ~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)
{a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~| ~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]

— Me@2018-04-17 04:47:31 PM

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2018.04.17 Tuesday (c) All rights reserved by ACHK

Problem 14.4a1

Closed string degeneracies | A First Course in String Theory

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(a) State the values of \alpha' M^2 and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

What is the meaning of “closed bosonic string theory”?

Bosonic? Meaning commutating creation operators?

Closed?

String? Or superstring?

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p.308 “We describe the position of classical bosonic strings using the string coordinates X^\mu(\tau, \sigma). The X^\mu(\tau, \sigma) are classical commuting variables: …

In quantum theory the X^\mu‘s become operators that do not generally commute.”

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“To get fermions in string theory we introduce new dynamical world-sheet variables \psi_1^\mu(\tau, \sigma) and \psi_2^\mu(\tau, \sigma). The classical variables \psi_\alpha^\mu (\alpha = 1, 2) are not ordinary commuting variables, but rather anticommuting variables.”

p.324 “In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.”

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type IIA

(NS+, NS+) (bosons), (NS+, R+) (fermions), (R-, NS+) (fermions), (R-, R+) (bosons)

type IIB

(NS+, NS+) (bosons), (NS+, R-) (fermions), (R-, NS+) (fermions), (R-, R-) (bosons)

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p.323 “Just like the massless states in bosonic closed string theory they carry two vector indices.”

This line implies that the phrase “bosonic closed string theory” does not refer to superstrings.

— Me@2018-04-03 10:00:13 AM

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Bosonic string theory is the original version of string theory, developed in the late 1960s. It is so called because it only contains bosons in the spectrum.

In the 1980s, supersymmetry was discovered in the context of string theory, and a new version of string theory called superstring theory (supersymmetric string theory) became the real focus. Nevertheless, bosonic string theory remains a very useful model to understand many general features of perturbative string theory, and many theoretical difficulties of superstrings can actually already be found in the context of bosonic strings.

— Wikipedia on Bosonic string theory

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What is the difference between closed string and closed superstring?

— Me@2018-04-09 09:41:18 PM

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‘Superstring theory’ is a shorthand for supersymmetric string theory because unlike bosonic string theory, it is the version of string theory that accounts for both fermions and bosons and incorporates supersymmetry to model gravity.

— Wikipedia on Superstring theory

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2018.04.09 Monday (c) All rights reserved by ACHK

M-theory is _not_ a string theory

These five theories have all been known since the middle of the 1980s. Some relationships between them were found soon after their discovery, but a clearer picture emerged only in the late 1990s. The limit of type II theory as the string coupling is taken to infinity was shown to give a theory in eleven dimensions. This theory is called M-theory, with the meaning of M to be decided when the nature of the theory becomes clear. It is known, however, the M-theory is _not_ a string theory. M-theory contains membranes (2-branes) and 5-branes, and these branes are not D-branes. M-theory may end up playing a prominent role in understanding string theory. The discovery of many other relationships between the five string theories listed above and M-theory has made it clear that we really have just _one theory_. This is a fundamental result: there is a unique theory, and the five superstrings and M-theory are different limits of this unique theory.

— Second Edition p.325

— A First Course in String Theory

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2018.04.05 Thursday ACHK

Problem 14.4a

Closed string degeneracies.

For closed string states the left-moving and right-moving excitations are each described like states of open strings with identical values of \alpha' M^2. The value of \alpha' M^2 for the closed string state is four times that value.

How come?

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Equation (14.78):

\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2

Mass M is two times that value (M = 2 M_L or M = 2 M_R). So M^2 is four times.

— Me@2015.07.14 12:47 PM

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Not necessarily so.

— Me@2015.07.16 08:14 AM

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Instead, it is just due to this definition.

— Me@2015.07.30 09:19 AM

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p.322 “In the closed string theory the value of the mass-squared is given by…”

Equation (14.78) is actually a definition of the mass-squared of a closed string state.

Also, consider

p.322 “As befits closed strings there is also the level-matching condition \alpha_0^- = \bar \alpha_0^- on the states. This condition guarantees that the left and right sectors give identical contributions to the mass-squared: \alpha' M_L^2 = \alpha' M_R^2.”

Then

\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2

\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2

— Me@2018.03.02 11:05 AM

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2018.03.02 Friday (c) All rights reserved by ACHK

Problem 14.3b6

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring (\alpha' M^2 = 1 and \alpha' M^2 = 2). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

~~~

When \alpha' M^2 = 2, by Equation (14.54), the possible states are

\{ \alpha_{-2}^I, \alpha_{-1}^I \alpha_{-1}^J, d_{-1}^I d_{-1}^J \} | R_a \rangle, || ...

\{ \alpha_{-1}^I d_{-1}^J, d_{-2}^I \} | R_{\bar a} \rangle, || ...

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For \alpha_{-2}^I | R_a \rangle, the number of states is 8.

For \alpha_{-1}^I \alpha_{-1}^J | R_a \rangle, the number of states is \frac{8 \times 7}{2} + 8 = 36.

For d_{-1}^I d_{-1}^J | R_a \rangle, the number of states is \frac{8 \times 7}{2} = 28.

For \alpha_{-1}^I d_{-1}^J | R_{\bar a} \rangle, the number of states is 8 \times 8 = 64.

For d_{-2}^I | R_{\bar a} \rangle, the number of states is 8.

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However, since each of a and {\bar a} has 8 possible values, there is an additional multiple of 8.

The total number of states is 8 \left[ 8 + 36 + 28 + 64 + 8 \right].

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You can check this answer against Equation (14.67):

f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

— Me@2018.02.20 10:57 AM

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2018.02.20 Tuesday (c) All rights reserved by ACHK

Problem 14.3b5

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring (\alpha' M^2 = 1 and \alpha' M^2 = 2). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

~~~

When \alpha' M^2 = 2, by Equation (14.37),

M^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)

N^\perp = \frac{5}{2}

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\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M,

b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M,

b_{-5/2}^I,

\alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M,

\alpha_{-1}^I b_{-3/2}^J,

\alpha_{-2}^I b_{-1/2}^J,

\alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle

.

There are \frac{8 \times 7 \times 6 \times 5 \times 4}{5!} = {8 \choose 5} = 56 states for

\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M |NS \rangle \otimes |p^+, \vec p_T \rangle

There are 8 \times \frac{8 \times 7}{2} = 224 states for

\{ b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle

There are 8 states for

\{ b_{-5/2}^I \} |NS \rangle \otimes |p^+, \vec p_T \rangle

There are 8 \times {8 \choose 3} = 448 states for

\{ \alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle

There are 8 \times 8 = 64 states for

\{ \alpha_{-1}^I b_{-3/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle

There are 8 \times 8 = 64 states for

\{ \alpha_{-2}^I b_{-1/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle

There are \left( \frac{8 \times 7}{2!} + 8 \right) \times 8 = 288 states for

\{ \alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle

So the total number of states is 56 + 224 + 8 + 448 + 64 + 64 + 288 = 1152.

.

You can check this answer against Equation (14.67):

f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

— Me@2018-02-16 03:22:13 PM

.

.

2018.02.16 Friday (c) All rights reserved by ACHK

Ramond sector zero modes

Problem 14.3b4

A First Course in String Theory
 
 
What are \xi_1, \xi_2, \xi_3, \xi_4 in Equation (14.44)?

p.315 “Ramond fermions are more complicated than NS fermions because the eight fermionic zero mode d_0^I must be treated with care. It turns out that these eight operators can be organized by simple linear combinations into four creation operators and four annihilation operators. Let us call the four creation operators …”

Since there are 8 possible transverse directions, there are 8 possible d_0^I‘s, where I = 2,3, ..., 9.
 
What is the meaning of “… organized by simple linear combinations into four creation operators …”?

— Me@2015.11.01 03:53 AM
 
 
The d_0^I operators are similar to but different from other d_r^I operators.
 
d_0^I‘s and d_r^I‘s are similar in the sense that they all follow Equation (14.43):

\{ d_m^I, d_n^J \} = \delta_{m+n, 0} \delta^{IJ}

p.315 “Again, the negatively moded oscillators d_{-1}^I, d_{-2}^I, d_{-3}^I, ..., are creation operators, while the positively moded ones d_{1}^I, d_{2}^I, d_{3}^I, ... are annihilation operators.”
 
d_0^I‘s and d_r^I‘s are different in the sense that d_0^I‘s are neither creation nor annihilation operators.
 
 
(Based on the ideas from “Introduction to String Theory, A.N. Schellekens” and “A First Course in String Theory (Second Edition)” p.315:)
 
If we define d_0 | 0 \rangle = 0,

\{ d_0^I, d_0^J \} | 0 \rangle
= \left( d_0^I d_0^J + d_0^I d_0^J \right) | 0 \rangle
= 0

which does not match the requirement of

\{ d_0^I, d_0^J \} = \delta^{IJ}

So the definition d_0 | 0 \rangle = 0 does not work.
 
— Me@2015.11.12 11:30 AM
 
 
Instead, they are “organized by simple linear combinations into four creation operators”.
 
(Based on the idea from “Boundary Conformal Field Theory and the Worldsheet Approach to D-Branes, by Andreas Recknagel,Volker Schomerus” and “A First Course in String Theory (Second Edition)” p.315:)
 
Let

c_0^i = d_0^{i+1}
e_i = \frac{1}{\sqrt{2}} \left( c_0^{2i} - i c_0^{2i - 1} \right)
e_i^\dagger = \frac{1}{\sqrt{2}} \left( c_0^{2i} + i c_0^{2i - 1} \right).

Then

\left\{ e_i, e_j^\dagger \right\}
= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}
= \frac{1}{2} \delta^{ij} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right),  \left( c_0^{2i} + i c_0^{2i - 1} \right) \right\}

By p.315 Equation (14.43):

\{ d_0^I, d_0^J \} = \delta^{IJ}

In other words,

\{ c_0^{I-1}, c_0^{J-1} \} = \delta^{I-1,J-1}
\{ c_0^{I}, c_0^{J} \} = \delta^{IJ}

\left\{ e_i, e_j^\dagger \right\}
= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}
= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} - \left\{ i c_0^{2i - 1}, i c_0^{2i - 1} \right\} \right]
= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} + \left\{ c_0^{2i - 1}, c_0^{2i - 1} \right\} \right]
= \frac{1}{2} \delta^{ij} \left[1 + 1 \right]
= \delta^{ij}

This is compatible with the anti-commutator requirement for fermion creation and annihilation operators: 

\{a^{\,}_i, a^\dagger_j\} = \delta_{i j}

— Me@2015.11.13 11:14 PM
 
 
 
2015.12.12 Saturday (c) All rights reserved by ACHK

Ground states and Annihilation operators

1. The equation a | 0 \rangle = 0 means that the eigenvalue of a on | 0 \rangle is 0:

a | 0 \rangle = 0 | 0 \rangle

2. The length of the vector a | 0 \rangle is 0:

\langle 0 | a^\dagger a | 0 \rangle = 0

3. The physical meaning is that the probability of the system being at state a | 0 \rangle is 0.
 
In other words, there is no state with an eigen-energy lower than the ground state one.
 
 
4. For the equation a | 0 \rangle = 0 | 0 \rangle, the 0 at the right is a scalar.
 
 
5. For the equation a | 0 \rangle = 0, the 0 at the right is a zero vector – a state vector with length zero.
 
6. | 0 \rangle is a state vector. However, it is NOT the zero vector.

Instead, it is the state vector of the ground state. Its length is 1 unit.

— Me@2015-11-03 03:26:58 PM
 
 
 
2015.11.04 Wednesday (c) All rights reserved by ACHK

Problem 14.3b3

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.
 
~~~
 
What is a zero mode?
 
p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”
 
 
How come there are creation operators that do not contribute to the mass-squared of the states?
 
Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.
 
If an index can be zero, the corresponding term can be zero.
 
— Me@2015.09.26 08:44 PM
 
 
Consider the NS-sector:
 
Equation (14.37):
 
M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)
 
N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I
 
For the NS sector, the r values in b^I_{r} are half-integers, thus cannot be zero. So every creation operator b_{-r}^I contributes to the mass-squared.
 
(p.312 “… the negatively moded coefficients b_{-1/2}^I, b_{-3/2}^I, b_{-5/2}^I, …, are creation operators, …”)
 
 
Consider the R-sector:
 
Equation (14.53):

M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)
 
For the R sector, the n values in d^I_n are integers, thus can be zero. So some of the creation operators d^I_{-n} are zero modes.
 
p.315 “… the eight fermionic zero modes d_0^I…”
 
— Me@2015-10-11 11:01:44 AM
 
 
 
2015.10.11 Sunday (c) All rights reserved by ACHK

Problem 14.3b2

A First Course in String Theory
 
 
14.3 Massive level in the open superstring

~~~

How come R sector has a factor 16 while NS sector has not?

Equation (14.66):

f_{NS}(x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n - \frac{1}{2}}}{1-x^n} \right)^8

Equation (14.68):

f_R(x) = 16 \prod_{n=1}^\infty \left( \frac{1+x^n}{1-x^n} \right)^8

p.319 “The overall multiplicative factor appears because each combination of oscillators gives rise to 16 states by acting on each of the available ground states.”

p.319 “We note that the R coefficients are actually double the corresponding NS coefficients. This is not a coincidence, as we will see in the following section.”

p.320 “We have seen that the Ramond sector has world-sheet supersymmetry: there are equal numbers of fermionic and bosonic states at each mass level.”
 
 
With the factor 16, how come the R coefficients are only double, but not 16 times as big as the corresponding NS coefficients?
 
It is caused by the difference of x^n and x^{n-\frac{1}{2}}.

— Me@2015.10.06 08:23 AM
 
 
 
2015.10.06 Tuesday (c) All rights reserved by ACHK

Problem 14.3b1

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.

~~~

What is the difference between “open bosonic string” and “open superstring”?

p.320 “With our conventions, these are world-sheet fermionic states that are now recognized to be states of spacetime fermions. The resulting, truncated sector is called the R- sector.”

NS(+): Equation (14.38) \left(\alpha' M^2 = 1, N^\perp = \frac{3}{2}\right):

\left\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b^I_{-1/2} b^J_{-1/2} b^K_{-1/2} \right\} | NS \rangle \otimes | p^+, \vec p_R \rangle

There are 128 states (cf. Quick Calculation 14.4).

R: Equation (14.54) \left(\alpha' M^2 = 1\right):

\alpha^I_{-1}  | R_a \rangle, d_{-1}^I | R_{\bar a} \rangle~||~\alpha^I_{-1}  | R_{\bar a} \rangle, d_{-1}^I | R_{a} \rangle

Only the left hand side represents the R- states.

Each of index I and index a has 8 possible values. So there are totally 64 states in the form of \alpha^I_{-1} | R_a \rangle.

Similarly, there are totally 64 states in the form of d_{-1}^I | R_{\bar a} \rangle.

Totally, for \alpha' M^2 = 1, there are 128 R- states.

— Me@2015-10-02 02:16:23 PM
 
 
 
2015.10.02 Friday (c) All rights reserved by ACHK

Problem 14.3a

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.

~~~

b_i^2 = 0

b_i b_j = - b_j b_i

i_1 \ne i_2

Since the sign is ignored, the number of possible b^i b^j states is \frac{1}{2} (8) (7) = 28.

The number of possible b^i b^j b^k states is \frac{1}{6} (8) (7) (6) = 56.

The number of possible b^i b^j b^k b^l states is {8 \choose 4} = 70.

— Me@2015-09-27 10:40:30 AM
 
 
 
2015.09.27 Sunday (c) All rights reserved by ACHK

Problem 14.2.2

A First Course in String Theory
 
 
14.2 Generating function for the unoriented bosonic open string theory.

~~~

How to add “a term that implements the projection to unoriented states”?

What is the projector? \Omega?

f_{os} (x)

= \frac{1}{x} \prod_{n=1}^\infty \frac{1}{(1-x^n)^{24}} 

= \frac{1}{x} \left( 1 + 24 x + 324 x^2 + 3200 x^3 + ... \right)

To keep only the odd-power terms, we should construct an odd function:

f_{uos} (x)

= \frac{1}{2} \left( f_{os} (x) - f_{os} (-x) \right)

= \frac{1}{x} \prod_{n=1}^\infty \frac{1}{(1-x^n)^{24}} - \left( \frac{-1}{x} \prod_{n=1}^\infty \frac{1}{(1-(-x)^n)^{24}} \right)

— Me@2015-09-17 02:27:20 PM
 
 
 
2015.09.23 Wednesday (c) All rights reserved by ACHK