Problem 13.5b

A First Course in String Theory

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13.6 Unoriented closed strings

This problem is the closed string version of Problem 12.12. The closed string \displaystyle{X^{\mu} (\tau, \sigma)} with \displaystyle{\sigma \in [0, 2 \pi]} and fixed \displaystyle{\tau} is a parameterized closed curve in spacetime. The orientation of a string is the direction of the increasing \displaystyle{\sigma} on this curve.

Introduce an orientation reversing twist operator \displaystyle{\Omega} such that

\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)

Moreover, declare that

\displaystyle{\Omega x_0^- \Omega^{-1} = x_0^-}

\displaystyle{\Omega p^+ \Omega^{-1} = p^+}

(b) Used the closed string oscillator expansion (13.24) to calculate

\displaystyle{\Omega x_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_n^I \Omega^{-1}}

\displaystyle{\Omega \bar \alpha_n^I \Omega^{-1}}

~~~

Equation (13.24):

\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}

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\displaystyle{\begin{aligned}   X^{\mu} (\tau, \sigma) &= x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma}) \\   X^I (\tau, 2 \pi - \sigma)  &= x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} \left( \alpha_n^I e^{- in\sigma} + \bar \alpha_n^I e^{i n \sigma)} \right) \\   \end{aligned}}

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\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)

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By comparing \displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} with \displaystyle{X^I (\tau, 2 \pi - \sigma)}, we have:

\displaystyle{\begin{aligned}   \Omega x_0^I \Omega^{-1} &= x_0^I \\  \Omega \alpha_0^I \Omega^{-1} &= \alpha_0^I \\  \Omega \alpha_n^I \Omega^{-1} &= \bar \alpha_n^I \\  \Omega \bar \alpha_n^I \Omega^{-1} &= \alpha_n^I \\   \end{aligned}}

— Me@2019-11-24 04:33:23 PM

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2019.11.24 Sunday (c) All rights reserved by ACHK

Problem 13.6b

A First Course in String Theory | Topology, 2 | Manifold, 2

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13.6 Orientifold Op-planes

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In the mathematical disciplines of topology, geometry, and geometric group theory, an orbifold (for “orbit-manifold”) is a generalization of a manifold. It is a topological space (called the underlying space) with an orbifold structure.

The underlying space locally looks like the quotient space of a Euclidean space under the linear action of a finite group.

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In string theory, the word “orbifold” has a slightly new meaning. For mathematicians, an orbifold is a generalization of the notion of manifold that allows the presence of the points whose neighborhood is diffeomorphic to a quotient of \displaystyle{\mathbf{R}^n} by a finite group, i.e. \displaystyle{\mathbf{R}^n/\Gamma}. In physics, the notion of an orbifold usually describes an object that can be globally written as an orbit space \displaystyle{M/G} where \displaystyle{M} is a manifold (or a theory), and \displaystyle{G} is a group of its isometries (or symmetries) — not necessarily all of them. In string theory, these symmetries do not have to have a geometric interpretation.

— Wikipedia on Orbifold

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In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an \displaystyle{n}-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension \displaystyle{n}.

— Wikipedia on Manifold

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In topology and related branches of mathematics, a topological space may be defined as a set of points, along with a set of neighbourhoods for each point, satisfying a set of axioms relating points and neighbourhoods. The definition of a topological space relies only upon set theory and is the most general notion of a mathematical space that allows for the definition of concepts such as continuity, connectedness, and convergence. Other spaces, such as manifolds and metric spaces, are specializations of topological spaces with extra structures or constraints.

— Wikipedia on Topological space

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2019.09.26 Thursday ACHK

Problem 13.6

A First Course in String Theory

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13.6 Orientifold Op-planes

(a) For an O23-plane the two normal directions \displaystyle{x^{24}, x^{25}} can be represented by a plane. A closed string at a fixed \tau appears as a parameterized closed curve \displaystyle{X^a(\tau, \sigma)} in this plane. Draw such an oriented closed string that lies fully in the first quadrant of the \displaystyle{(x^{24}, x^{25})} plane. Draw also the string \displaystyle{\tilde{X}^a(\tau, \sigma) = -X^a(\tau, 2\pi - \sigma)}.

~~~

d_2019_08_26__22_32_32_PM_a

This one is wrong.

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d_2019_08_26__22_31_55_PM_b

— Me@2019-08-26 10:31:07 PM

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2019.08.26 Monday (c) All rights reserved by ACHK

Quick Calculation 15.1.2

A First Course in String Theory

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Recall that a group is a set which is closed under an associative multiplication; it contains an identity element, and each element has a multiplicative inverse. Verify that \displaystyle{U(1)} and \displaystyle{U(N)}, as described above, are groups.

~~~

Definition

A group is a set, G, together with an operation \displaystyle{\bullet} (called the group law of G) that combines any two elements a and b to form another element, denoted \displaystyle{a \bullet b} or \displaystyle{ab}. To qualify as a group, the set and operation, \displaystyle{(G, \bullet)}, must satisfy four requirements known as the group axioms:

Closure

For all a, b in G, the result of the operation, \displaystyle{a \bullet b}, is also in G.

Associativity

For all a, b and c in G, \displaystyle{(a \bullet b) \bullet c = a \bullet (b \bullet c)}.

Identity element

There exists an element e in G such that, for every element a in G, the equation \displaystyle{e \bullet a = a \bullet e = a} holds. Such an element is unique, and thus one speaks of the identity element.

Inverse element

For each a in G, there exists an element b in G, commonly denoted \displaystyle{a^{-1}} (or \displaystyle{-a}, if the operation is denoted “+”), such that \displaystyle{a \bullet b = b \bullet a = e}, where e is the identity element.

— Wikipedia on Group (mathematics)

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The axioms for a group are short and natural… Yet somehow hidden behind these axioms is the monster simple group, a huge and extraordinary mathematical object, which appears to rely on numerous bizarre coincidences to exist. The axioms for groups give no obvious hint that anything like this exists.

— Richard Borcherds in Mathematicians: An Outer View of the Inner World

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2019.07.28 Sunday ACHK

Quick Calculation 15.1

A First Course in String Theory

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Recall that a group is a set which is closed under an associative multiplication; it contains an identity element, and each element has a multiplicative inverse. Verify that \displaystyle{U(1)} and \displaystyle{U(N)}, as described above, are groups.

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What is \displaystyle{U(1)}?

— Me@2019-05-24 11:25:41 PM

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The set of all \displaystyle{1 \times 1} unitary matrices clearly coincides with the circle group; the unitary condition is equivalent to the condition that its element have absolute value 1. Therefore, the circle group is canonically isomorphic to \displaystyle{U(1)}, the first unitary group.

— Wikipedia on Circle group

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In mathematics, a complex square matrix \displaystyle{U} is unitary if its conjugate transpose \displaystyle{U^*} is also its inverse—that is, if

\displaystyle{U^{*}U=UU^{*}=I,}

where \displaystyle{I} is the identity matrix.

In physics, especially in quantum mechanics, the Hermitian conjugate of a matrix is denoted by a dagger (\displaystyle{\dagger}) and the equation above becomes

\displaystyle{U^{\dagger }U=UU^{\dagger }=I.}

The real analogue of a unitary matrix is an orthogonal matrix. Unitary matrices have significant importance in quantum mechanics because they preserve norms, and thus, probability amplitudes.

— Wikipedia on Unitary matrix

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2019.05.25 Saturday ACHK

Problem 14.5d3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

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— This answer is my guess. —

~~~

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

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\displaystyle{ \begin{aligned}   &f_{L, NS'+}(x) \\   &= a_{NS'+} (r) x^r \\  &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\   & \\  &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\   \end{aligned}}

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\displaystyle{\begin{aligned}  &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ & \\ &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5} + ...   \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_R^2: \end{aligned}}

\displaystyle{ \begin{aligned} f_{NS+}(x)  &= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ... \\ \end{aligned}}

\displaystyle{ \begin{aligned} f_{R-}(x)  &= 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ... \\ \end{aligned}}

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So the total number of states in heterotic string theory at \displaystyle{ \begin{aligned} \alpha' M^2 = 8 \end{aligned}} is

\displaystyle{ \begin{aligned}   &\left(1384320 + 1310720 \right) \times \left(1152 + 1152\right) \\  \end{aligned}}.

\displaystyle{ \begin{aligned}   &= 6209372160 \\ \end{aligned}}.

~~~

— This answer is my guess. —

— Me@2019-01-26 04:49:37 PM

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2019.01.27 Sunday (c) All rights reserved by ACHK

Problem 14.5d2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

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— This answer is my guess. —

~~~

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The left R’+ sector:

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\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

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\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_{\alpha} \rangle_L \\  \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&|{\bar \alpha}_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\  \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}  \left( 1 + \lambda_{-2} x^{2} \right)^{32}  ... \\  \end{aligned}}

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However, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha\rangle_L} and \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_R}:

\displaystyle{\begin{aligned}  (2^{15} + 2^{15})   \left[ \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}   ... \right] \\   \end{aligned}}

\displaystyle{\begin{aligned}  2^{16} \prod_{r=1}^\infty  \frac{1}{(1 - x^r)^8}  (1 + x^{r})^{32} \\  \end{aligned}}

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“Keep only states with \displaystyle{(-1)^{F_L} = +1}; this defines the left R’+ sector.”

\displaystyle{\begin{aligned}  \frac{2^{16}}{2} \prod_{r=1}^\infty  \frac{(1 + x^{r})^{32}}{(1 - x^r)^8}   \\  \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned}  &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\  & \\  &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5}+39739981824 \, x^{6} + ...   \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

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2019.01.20 Sunday (c) All rights reserved by ACHK

Problem 14.5d1.1.2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

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— This answer is my guess. —

~~~

p.322

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

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

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

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The left NS’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...   \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...   \right)^8  ...   \left( 1 + \lambda_{-\frac{1}{2}} x^{\frac{1}{2}} \right)^{32}   \left( 1 + \lambda_{-\frac{3}{2}} x^{\frac{3}{2}} \right)^{32}  ... \\  \end{aligned}}

\displaystyle{\begin{aligned}  \prod_{r=1}^\infty   \frac{1}{(1 - x^r)^8}  (1 + x^{r-\frac{1}{2}})^{32} \\   \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned}  &f_{L, NS'}(x) \\ &= a_{NS'} (r) x^r \\  &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   \end{aligned}}

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“The left NS’ sector is built with oscillators \displaystyle{\bar \alpha_{-n}^I} and \displaystyle{\lambda_{-r}^A} acting on the vacuum \displaystyle{|NS' \rangle_L}, declared to have \displaystyle{(-1)^{F_L} = + 1}:”

\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}

So all the states with integer \displaystyle{N^{\perp}} have \displaystyle{(-1)^F = +1}.

.

\displaystyle{ \begin{aligned} &f_{L, NS'}(x) \\ \end{aligned}}

\displaystyle{ = \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}

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Let

\displaystyle{ \begin{aligned}  &g (\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1+32 \, \sqrt{x}+504 \, x+5248 \, x^{\frac{3}{2}}+40996 \, x^{2}+258624 \, x^{\frac{5}{2}}+1384320 \, x^{3}+6512384 \, x^{\frac{7}{2}} + ... \\   \end{aligned}}

Then

\displaystyle{ \begin{aligned}  &g (-\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1 - 32 \sqrt{x} + 504 x - 5248 \, x^{\frac{3}{2}} + 40996 \, x^{2} - 258624 \, x^{\frac{5}{2}}+1384320 \, x^{3} - 6512384 x^{\frac{7}{2}} + ... \\   \\ \end{aligned}}

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\displaystyle{ \begin{aligned}  &f_{L, NS'+}(x) \\  &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\  &= \frac{1}{2x} \left[ g(\sqrt{x}) + g(-\sqrt{x}) \right] \\   &= \frac{1}{2x} \left[   \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   \right] \\   \end{aligned}}

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The left R’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

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2019.01.14 Monday (c) All rights reserved by ACHK

Problem 14.5d1.2 | SageMath

The generating function is an infinite product:

\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}

To evaluate the infinite product, you can use SageMath with the following commands:

typeset_mode(True)

(1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..oo))

a = (1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..200))

F = a.taylor(x,0,6)

g = "+".join(map(latex, sorted([f for f in F.operands()], key=lambda exp:exp.degree(x))))

g

\displaystyle{ \begin{aligned}  &f_{L, NS+}(x) \\  \end{aligned}}

\displaystyle{  \approx \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}

— Me@2019-01-11 11:52:33 AM

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2019.01.11 Friday (c) All rights reserved by ACHK

Problem 14.5d1

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

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— This answer is my guess. —

~~~

p.322

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

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

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

.

The left NS’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...   \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...   \right)^8  ...   \left( 1 + \lambda_{-\frac{1}{2}} x^{-\frac{1}{2}} \right)^{32}   \left( 1 + \lambda_{-\frac{3}{2}} x^{-\frac{3}{2}} \right)^{32}  ... \\  \end{aligned}}

\displaystyle{\begin{aligned}  \prod_{r=1}^\infty   \frac{1}{(1 - x^r)^8}  (1 + x^{r-\frac{1}{2}})^{32} \\   \end{aligned}}

.

\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned}  &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\  &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   \end{aligned}}

.

The left R’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-10 01:49:43 PM

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2019.01.10 Thursday (c) All rights reserved by ACHK

Problem 14.5c9

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) Calculate the total number of states in heterotic string theory (bosons plus fermions) at \displaystyle{\alpha' M^2 =4}.

~~~

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— This answer is my guess. —

~~~

spacetime bosons:

\displaystyle{NS'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-2}^I, \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

\displaystyle{\begin{aligned}  I, J, I' &= 2, 3, ..., 9 \\  A, B, C, D &= 1, 2, ..., 32 \\  \end{aligned}}

Number of states:

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

\displaystyle{\begin{aligned}  &\left( 8+ N(2,8) +8 \times {32 \choose 2} + 32^2 + {32 \choose 4} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 40996 \times 128  \end{aligned}}

.

\displaystyle{R'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

Following the same logic:

Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{2^{16}} degenerate Ramond ground states.

Therefore, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_L}.

— Me@2018-10-29 03:11:07 PM

\displaystyle{\begin{aligned}  I, J, K &= 2, 3, ..., 9 \\  \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 32768 \times 128  \end{aligned}}

~~~

spacetime fermions:

\displaystyle{NS'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 40996 \right) \times \left( 8^2 + 8^2 \right) \\  &= 40996 \times 128  \end{aligned}}

.

\displaystyle{R'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15}\right) \times \left( 8^2 + 8^2 \right) \\  &= 32768 \times 128  \end{aligned}}

~~~

— This answer is my guess. —

— Me@2019-01-03 05:26:59 PM

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2019.01.03 Thursday (c) All rights reserved by ACHK

Problem 14.5c8

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = 4k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Calculate the total number of states in heterotic string theory (bosons plus fermions) at \displaystyle{\alpha' M^2 =4}.

~~~

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— This answer is my guess. —

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When \displaystyle{\alpha' M^2 =4},

\displaystyle{\begin{aligned}  \alpha' M_L^2 &= 1 \\  \alpha' M_R^2 &= 1  \end{aligned}}

~~~

The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

The left R’+ sector:

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\  \end{aligned}}

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The right-moving NS+ states:

\displaystyle{\begin{aligned}  \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, \\  \end{aligned}}

The R- states (that used as right-moving states):

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\  \end{aligned}}

~~~

spacetime bosons:

\displaystyle{NS'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

\displaystyle{R'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

.

spacetime fermions:

\displaystyle{NS'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

\displaystyle{R'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

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— This answer is my guess. —

— Me@2018-12-28 11:12:59 PM

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2018.12.29 Saturday (c) All rights reserved by ACHK

Problem 14.5c7

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) … Write out the massless states of the theory (bosons and fermions) and describe the fields associated with the bosons.

~~~

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— This answer is my guess. —

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spacetime bosons:

NS'+ \otimes NS+

\displaystyle{\begin{aligned}  \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left(  b_{-1/2}^J~|NS \rangle \otimes |p^+, \vec p_T \rangle  \right)  \end{aligned}}

\displaystyle{\begin{aligned}  = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right)  \end{aligned}}

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What is the nature of each of the indices I, J, A, B?

The vector index J runs over eight values.

— c.f. p.323 A First Course in String Theory (Second Edition)

\displaystyle{I = 2,3,...,9}

\displaystyle{A, B = 1, 2, ..., 32}

— c.f. the blog post Problem 14.5a3

— Me@2018-12-24 10:04:52 PM

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For the states in the form

\displaystyle{  \bar \alpha_{-1}^I  b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle  \right)},

they

carry two independent vector indices I, J that run over eight values. There are therefore 64 bosonic states. Just like the massless states in bosonic closed string theory[,] they carry two vector indices. We therefore get a graviton, a Kalb-Ramond field, and a dilation:

(NS+, NS+) massless fields: g_{\mu \nu}, B_{\mu \nu}, \phi.

— p.323 A First Course in String Theory (Second Edition)

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Then how about the states in the form

\displaystyle{\begin{aligned}  \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right)  \end{aligned}}?

What kinds of fields do they represent?

— Me@2018-12-24 10:42:03 PM

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— This answer is my guess. —

— Me@2018-12-23 11:16:56 PM

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2018.12.24 Monday (c) All rights reserved by ACHK

Problem 14.5c6

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = 4k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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— This answer is my guess. —

.

The left NS’+ sector:

\displaystyle{\begin{aligned}  \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L  \end{aligned}}

The left R’+ sector has no massless states.

The right-moving NS+ states:

\displaystyle{\begin{aligned}  \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\  \end{aligned}}

The R- states (that used as right-moving states):

\displaystyle{\begin{aligned}  \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle \\  \end{aligned}}

~~~

Since R’+ has no massless states:

spacetime bosons:

NS'+ \otimes NS+

\displaystyle{\begin{aligned}  \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left(  b_{-1/2}^I~|NS \rangle \otimes |p^+, \vec p_T \rangle  \right)  \end{aligned}}

\displaystyle{\begin{aligned}  = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^I |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right)  \end{aligned}}

spacetime fermions:

NS'+ \otimes R-

\displaystyle{\begin{aligned}  \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \otimes |R_a \rangle \\  \end{aligned}}

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— This answer is my guess. —

— Me@2018-12-18 07:46:15 PM

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2018.12.18 Tuesday (c) All rights reserved by ACHK

Problem 14.5c5

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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We have the following well-known bosonic string mass formulae:

open string:

\displaystyle{\alpha' M^2 = N - a}

closed string:

\displaystyle{\frac{1}{4} \alpha' M^2 = N_L - a}
\displaystyle{\frac{1}{4} \alpha' M^2 = N_R - a}

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

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How come there is an extra \displaystyle{\frac{1}{4}} at the beginning of the closed string formula?

p.322

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

.

\displaystyle{\begin{aligned}  \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 \\  \end{aligned}}

— Me@2018-12-15 08:59:18 PM

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2018.12.15 Saturday (c) All rights reserved by ACHK

Slope parameter

Problem 14.5c4 | Counting states in heterotic SO(32) string theory

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c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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What is the meaning of \displaystyle{\alpha'}?

How come \displaystyle{\alpha' = \frac{1}{2}}?

— Me@2018-12-07 10:43:10 PM

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If we consider a rigidly rotating open string, \alpha' is the proportionality constant that relates the angular momentum J of the string, measured in units of \hbar, to the square of its energy E. More explicitly,

\displaystyle{  \frac{J}{\hbar} = \alpha' E^2  }

— p.68

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\displaystyle{  J = \frac{1}{ 2 \pi T_0 c} E^2  }

As anticipated, the angular momentum is proportional to the square of the energy of the string. Comparing with equation (8.69) we deduce that

\displaystyle{ \begin{aligned}   \alpha' &= \frac{1}{2 \pi T_0 \hbar c} \\   T_0 &= \frac{1}{2 \pi \alpha' \hbar c}  \end{aligned}}

These equations relate the slope parameter \alpha' to the string tension T_0.

— p.69

— A First Course in String Theory

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2018.12.11 Tuesday (c) All rights reserved by ACHK

Problem 14.5c3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

.

Open String:

\displaystyle{   \begin{aligned}  N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\   L_n^{\perp} &= \frac{1}{2} \sum_{-\infty}^{\infty} \alpha_{n-p}^I \alpha_{p}^I~~~(n \ne 0) \\   L_0^{\perp} &= \alpha' p^I p^I + N^\perp   \end{aligned}  }

.

Closed String:

\displaystyle{   \begin{aligned}  N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\   \bar N^\perp &= \sum_{p=1}^{\infty} \bar \alpha_{-p}^I \bar \alpha_p^I \\   L_0^\perp &= \frac{\alpha'}{4} p^I p^I + N^\perp \\   \bar L_0^\perp &= \frac{\alpha'}{4} p^I p^I + \bar N^\perp   \end{aligned}  }

— A First Course in String Theory

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We have the following well-known bosonic string mass formulae \displaystyle{\alpha' = \frac{1}{2}}:

open string:

\displaystyle{\frac{1}{2} M^2 = N - a}

closed string:

\displaystyle{\frac{1}{8} M^2 = N_L - a}
\displaystyle{\frac{1}{8} M^2 = N_R - a}

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

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What is the meaning of \displaystyle{\alpha'}?

How come \displaystyle{\alpha' = \frac{1}{2}}?

— Me@2018-12-07 10:43:10 PM

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2018.12.07 Friday (c) All rights reserved by ACHK

Problem 14.5c2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = 4k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

~~~

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— This answer is my guess. —

.

The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

.

The left R’+ sector:

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}

.

The right-moving NS+ states:

NS+ equations of (14.38):

\displaystyle{\begin{aligned}  \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 \end{aligned}}

.

The R- states (that used as right-moving states):

Mass levels of R- and R+ (Equations 14.54):

\displaystyle{\begin{aligned}  \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\  \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\  \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\  &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}

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There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with \displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}.

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— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

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2018.11.22 Thursday (c) All rights reserved by ACHK

Problem 14.5c

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

~~~

In heterotic (closed) string theory, there are left-moving part and right-moving part. Then, what is the meaning of “at any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string”?

— Me@2018-11-11 03:44:18 PM

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Type IIA/B 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

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

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

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

These are the reasons that any mass level of the heterotic string is always in the form \displaystyle{\alpha' M^2 = 4k}.

— Me@2018-11-12 03:09:11 PM

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Equation (14.77):

p.322

closed string sectors: (NS, NS), (NS, R), (R, NS), (R, R)

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

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

What is the difference between Type IIA/B closed superstrings and heterotic SO(32) strings?

— Me@2018-11-12 03:15:46 PM

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The five consistent superstring theories are:

d_2018_11_15__09_36_16_am_

  • The type I string has one supersymmetry in the ten-dimensional sense (16 supercharges). This theory is special in the sense that it is based on unoriented open and closed strings, while the rest are based on oriented closed strings.
  • The type II string theories have two supersymmetries in the ten-dimensional sense (32 supercharges). There are actually two kinds of type II strings called type IIA and type IIB. They differ mainly in the fact that the IIA theory is non-chiral (parity conserving) while the IIB theory is chiral (parity violating).
  • The heterotic string theories are based on a peculiar hybrid of a type I superstring and a bosonic string. There are two kinds of heterotic strings differing in their ten-dimensional gauge groups: the heterotic E8×E8 string and the heterotic SO(32) string. (The name heterotic SO(32) is slightly inaccurate since among the SO(32) Lie groups, string theory singles out a quotient Spin(32)/Z2 that is not equivalent to SO(32).)

— Wikipedia on Superstring theory

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2018.11.15 Thursday (c) All rights reserved by ACHK

Problem 14.5b2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) … Keep only states with \displaystyle{(-1)^{F_L}=+1}; this defines the left R’+ sector.

Write explicitly and count the states we keep for the two lowest mass levels, indicating the corresponding values of \displaystyle{\alpha' M_L^2}. [This is a shorter list.]

~~~

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

If we define N^\perp in a way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

.

\displaystyle{\begin{aligned}  (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\  (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

.

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\  \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\  \end{aligned}}

— This answer is my guess. —

— Me@2018-11-06 03:39:15 PM

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2018.11.06 Tuesday (c) All rights reserved by ACHK