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}}

.

~~~

— 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

Ken Chan 時光機 2.2

聖誕假後的常規課程中,Ken Chan 預計應該不會有時間,教光學和熱學。但是,他認為那兩個課題很淺易,所以,期望我們即使自修,也沒什麼難度。

但是,他後來改變主意,還是決定於,農歷新年的年初一、二、三,為我們補課,教回光學和熱學。

今次,那三天的課,他不單不收我們的學費,他還要自己,花時間找課室、花金錢付租金。

後來,他改為每天收 20 元,即是總共 60 元。他解釋道,補課社的職員因為他的補課,要犧牲那三天的農曆新年假期。那每人 60 元的學費是,全數慰勞他們的。

那三天的補課,好處是光學和熱學;壞處是,由於每天也補多個小時,我少了大量,各科的溫習時間。但是,那個農曆新年長假,因為我自己時間管理不善,而損失的時間,遠多於那三天的補課時間。

感謝 Ken Chan 的額外付出。我亦慶幸,生於那個時空,還是有真人授課的年代。出生遲十年,就已經只能透過錄影帶,來獲得 Ken Chan 的教導。

— Me@2019-01-06 02:18:47 PM

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2019.01.06 Sunday (c) All rights reserved by ACHK

Clasp

Overview

Clasp is a new Common Lisp implementation that seamlessly interoperates with C++ libraries and programs using LLVM for compilation to native code. This allows Clasp to take advantage of a vast array of preexisting libraries and programs, such as out of the scientific computing ecosystem. Embedding them in a Common Lisp environment allows you to make use of rapid prototyping, incremental development, and other capabilities that make it a powerful language.

— Clasp README.md

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I followed the official instructions to build Clasp:

d_2019_01_04__12_06_48_pm_

The building process had been going on for about an hour; and then I got this error:

d_2019_01_04__17_53_17_pm_

d_2019_01_04__23_07_39_pm_

— Me@2019-01-04 10:11:43 PM

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2019.01.04 Friday (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

Photon dynamics in the double-slit experiment, 5

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What is the relationship between a Maxwell photon and a quantum photon?

— Me@2012-04-09 7:38:06 PM

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The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence

Maxwell’s theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

That caught me by surprise, because I always thought, Maxwell’s equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967):

The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

Isn’t the view of Sakurai in contradiction to Gloge? Do Maxwell’s equation describe a single photon or an infinite number of photons? Or do Maxwell’s equations describe a single photon and also an infinite number of photons at the same time? But why do we need QED then at all?

— edited Nov 28 ’16 at 6:35
— tparker

— asked Nov 20 ’16 at 22:33
— asmaier

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Because photons do not interact, to very good approximation for frequencies lower than \displaystyle{m_e c^2 / h} (\displaystyle{m_e} = electron mass), the theory for one photon corresponds pretty well to the theory for an infinite number of them, modulo Bose-Einstein symmetry concerns. This is similar to most of the statistical theory of ideal gases being derivable from looking at the behavior of a single gas particle in kinetic theory.

Put another way, the single photon behavior \displaystyle{\leftrightarrow} Maxwell’s equations correspondence only holds if you look at the Fourier transform version of Maxwell’s equations. The real space-time version of Maxwell’s equations would require looking at a superposition of an infinite number of photons — one way to describe the taking [of] an inverse Fourier transform.

If you want to think of it in terms of Feynman diagrams, classical electromagnetism is described by a subset of the tree-level diagrams, while quantum field theory requires both tree level and diagrams that have closed loops in them. It is the fact that the lowest mass particle photons can produce a closed loop by interacting with, the electron, that keeps photons from scattering off of each other.

In sum: they’re both incorrect for not including frequency cutoff concerns (pair production), and they’re both right if you take the high frequency cutoff as a given, depending on how you look at things.

— edited Dec 3 ’16 at 6:28

— answered Nov 27 ’16 at 23:08

— Sean E. Lake

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Maxwells equations, which describe the wavefunction of a single noninteracting photon, don’t need Planck’s constant. I find that remarkable. – asmaier Dec 2 ’16 at 14:16

@asmaier : Maxwell’s equations predate the quantum nature of light, they weren’t enough to avoid the ultraviolet catastrophe. Note too that what people think of as Maxwell’s equations are in fact Heaviside’s equations, and IMHO some meaning has been lost. – John Duffield Dec 3 ’16 at 17:45

— Do Maxwell’s equations describe a single photon or an infinite number of photons?

— Physics StackExchange

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2019.01.03 Thursday ACHK

宇宙大戰 1.2

PhD, 2.4 | 故事連線 1.1.6 | 碩士 3.4

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(問:我也遇過類似的情境。

我和一位好朋友合作做小組習作時,雖然未至於反目,但總會有很多爭拗。和他合作前,明明和他感情要好。各自有什麼困難時,對方總會杖義相助。

為什麼人類會,那麼奇怪呢?)

.

簡單地說,即使是同一個人,其實也有不同方面,各樣性格。

做朋友時,你只需要接受小部分—你可以選擇,只接受他,最好的優點。但是,做工作伙伴時,你卻要接收大部分—你未必可以選擇,不接受他,最壞的缺點。

.

(問:那樣,如果要「複雜地說」呢?)

.

複雜地說,每個個體也透過自己,在這宇宙間的經歷,形成一個「主觀宇宙」,簡稱「世界觀」。

大部分人,也不自覺地,以為他的主觀宇宙,就是客觀宇宙的全部。這個不幸,源於每個人的主觀宇宙,是他唯一能夠觀察到的「客觀宇宙部分」;每個人當時的主觀宇宙,是他當時唯一能夠,觀察到的「客觀宇宙部分」。

只有一些「被選擇的心靈」,簡稱「半神人」,才會想像到,他的主觀世界,只是客觀世界的極小部分。所以,如果兩個人也不是「半神人」,而又要在工作上合作的話,其實就相當於,把兩個(主觀)宇宙的大部分,重疊在一起。

每個宇宙原本,都有各自的運行法則;貿然要求兩個宇宙,互相干涉對方內政,自然會十分危險。

六千五百萬年前,單單是一個小行星與地球相撞,就足以令大部分恐龍滅絕。試想想,兩個宇宙相撞,殺傷力會大多少倍。

— Me@2019-01-01 11:20:57 PM

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2019.01.01 Tuesday (c) All rights reserved by ACHK

Build cross-compiled ECL for Android, 3

EQL5-Android | Common Lisp for Android App Development 2018

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d_2018_12_19__22_56_00_PM_

After successfully running the command ./1-make-ecl-host.sh, when I tried to run the command ./2-make-ecl-android.sh, I got the following errors:

d_2018_12_29__23_30_22_PM_

— Me@2018-12-29 11:21:46 PM

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2018.12.30 Sunday (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}.

~~~

.

— This answer is my guess. —

.

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}}

.

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}}

.

— 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

The problem of induction 3.3

“Everything has no patterns” (or “there are no laws”) creates a paradox.

.

If “there are 100% no first order laws”, then it is itself a second order law (the law of no first-order laws), allowing you to use probability theory.

In this sense, probability theory is a second order law: the law of “there are 100% no first order laws”.

In this sense, probability theory is not for a single event, but statistical, for a meta-event: a collection of events.

Using meta-event patterns to predict the next single event, that is induction.

.

Induction is a kind of risk minimization.

— Me@2012-11-05 12:23:24 PM

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2018.12.28 Friday (c) All rights reserved by ACHK

Ken Chan 時光機 2.1

當年是 1996 年,我 9 月開始上他的課程。每星期上 1.5 小時,由星期六早上 8 時半,上到 10 時正。所以,每個星期六,我也要在還未睡夠時,就很早起床。去補習社的路程中,往往是一邊行,一邊流鼻水;唯有勉勵自己:「即使再辛苦,只要捱過『會考』(公開試),就可以有好日子過。」

記憶所及,大概在聖誕前,剛好完成了力學分部。

然後,他於聖誕假期中,有額外補課—歷時三天的全日課程,不再是每天 1.5 小時。就在那三天,他要完成波動分部。

他在公佈有這個聖誕波動補課時,我有兩點不開心。

第一點是,我要交額外的學費。不過,相對於第二點而言,那也只是小問題。

第二點是,那補課的學額有限,只有常規課程的兩成。我現在的記憶可能有誤,未必真的是兩成,但也必定是,遠少於他常規課程的學生數目。換句話說,如果供不應求,我就可能報讀不到。幸好,正正是因為這個擔憂,我及早報名,爭取到一個補課學位。

— Me@2018-12-27 03:50:13 PM

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2018.12.27 Thursday (c) All rights reserved by ACHK

Build cross-compiled ECL for Android, 2

EQL5-Android | Common Lisp for Android App Development 2018

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The step 2 is to “build cross-compiled ECL for Android”. It should have been straightforward to follow the instructions in the README page of the EQL5-Android project.

d_2018_12_19__22_56_00_PM_

However, when trying to run the command ./1-make-ecl-host.sh, I got the following error messages:

d_2018_12_19__23_11_14_PM_

The error was caused by the fact that my Ubuntu 18.04’s 64-bit gcc toolchain could not compile any source code to create 32-bit executables.

The solution is to run the following command to install the 32-bit gcc toolchain first:

sudo apt-get install g++-multilib libc6-dev-i386

— Me@2018-12-25 09:51:02 PM

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2018.12.25 Tuesday (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.

~~~

.

— This answer is my guess. —

.

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}}

.

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

.

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

Afshar experiment, 2

Double slit experiment, 8.2

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In the double slit experiment, the screen is used to detect interference pattern itself, causing the photon wavefunctions to “collapse”.

In the Afshar experiment, there is no classically definite position for a photon when the photon passes “through” the vertically wire slits. So there is no interference patterns “formed”, unless you put some kind of screen afterwards. [Me@2015-07-21 10:59 PM: i.e. making the observation, c.f. delayed choice experiment]

— Me@2012-04-09 12:19:52 AM

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Being massless, they cannot be localized without being destroyed…

— Photon dynamics in the double-slit experiment

— Wikipedia on Photon

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2018.12.23 Sunday (c) All rights reserved by ACHK

宇宙大戰 1.1

PhD, 2.3 | 故事連線 1.1.5 | 碩士 3.3

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(問:你好似講到,人類那麼危險?)

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因為事實上,人類的確是,那麼危險。

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剛才所講,有關選擇碩士或博士論文導師,所需的技巧,背後的精神,其實是通用的—同時適用於你將來選擇公司、上司、生意合作伙伴、配偶,等等。

選擇錯誤,同樣是有改變一生的後果。

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(問:人類真的那麼危險嗎?)

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你一日未試過,同一個人有工作關係,或者錢銀來往,你也不會知道,他的真面目。

時常會聽到一類故事:

甲和乙是幾十年的要好朋友。他們決定合作創業。不料,一同工作不出幾個月,就反目收場。

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(問:我也遇過類似的情境。

我和一位好朋友合作做小組習作時,雖然未至於反目,但總會有很多爭拗。和他合作前,明明和他感情要好。各自有什麼困難時,對方總會杖義相助。

為什麼人類會,那麼奇怪呢?)

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簡單地說,即使是同一個人,其實也有不同方面,各樣性格。

做朋友時,你只需要接受小部分—你可以選擇,只接受他,最好的優點。但是,做工作伙伴時,你卻要接收大部分—你未必可以選擇,不接受他,最壞的缺點。

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(問:那樣,如果要「複雜地說」呢?)

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— Me@2018-12-20 11:06:49 PM

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2018.12.20 Thursday (c) All rights reserved by ACHK