Daily Archives: December 18, 2018

Problem 14.5c6

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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. —

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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}}

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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

The problem of induction 3.1.2

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Square of opposition

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“everything has a pattern”?

“everything follows some pattern” –> no paradox

“everything follows no pattern” –> paradox

— Me@2012.11.05

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My above statements are meaningless, because they lack a precise meaning of the word “pattern”. In other words, whether each statement is correct or not, depends on the meaning of “pattern”.

In common usage, “pattern” has two possible meanings:

1. “X has a pattern” can mean that “X has repeated data“.

Since the data set X has repeated data, we can simplify X’s description.

For example, there is a die. You throw it a thousand times. The result is always 2. Then you do not have to record a thousand 2’s. Instead, you can just record “the result is always 2”.

2. “X has a pattern” can mean that “X’s are totally random, in the sense that individual result cannot be precisely predicted“.

Since the data set X is totally random, we can simplify the description using probabilistic terms.

For example, there is a die. You throw it a thousand times. The die lands on any of the 6 faces 1/6 of the times. Then you do not have to record those thousand results. Instead, you can just record “the result is random” or “the die is fair”.

— Me@2018-12-18 12:34:58 PM

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2018.12.18 Tuesday (c) All rights reserved by ACHK