duplicate

If (member o l) finds o in the list l, it also returns the cdr of l beginning with o. This return value can be used, for example, to test for duplication. If o is duplicated in l, then it will also be found in the cdr of the list returned by member. This idiom is embodied in the next utility, duplicate:

>(duplicate ’a ’(a b c a d))
(A D)

(defun duplicate (obj lst &key (test #’eql))
    (member obj (cdr (member obj lst :test test))
            :test test))

— p.51

— On Lisp

— Paul Graham

.

Exercise 4.4

Without using the existing function member, define duplicate as in

>(duplicate ’a ’(a b c a d))
(A D)

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

— This answer is my guess. —


(defun my-member (obj lst)
    (cond ((not lst) NIL)
          ((eq obj (car lst)) lst)
          (t (my-member obj (cdr lst)))))

— This answer is my guess. —

— Me@2019-01-21 06:34:46 AM

.

.

2019.01.21 Monday (c) All rights reserved by ACHK

Problem 14.5d2

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

.

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}.

~~~

.

— This answer is my guess. —

~~~

.

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:~~~~~&|{\bar \alpha}_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\  \end{aligned}}

.

\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}}

.

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}}

.

“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}}

.

\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}}

.

~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

.

.

2019.01.20 Sunday (c) All rights reserved by ACHK

Logical arrow of time, 6.4

The source of the macroscopic time asymmetry, aka the second law of thermodynamics, is the difference of prediction and retrodiction.

In a prediction, the deduction direction is the same as the physical/observer time direction.

In a retrodiction, the deduction direction is opposite to the physical/observer time direction.

.

— guess —

If a retrodiction is done by a time-opposite observer, he will see the entropy increasing. For him, he is really doing a prediction.

— guess —

.

— Me@2013-10-25 3:33 AM

.

The existence of the so-called “the paradox of the arrow of time” is fundamentally due to the fact that some people insist that physics is about an observer-independent objective truth of reality.

However, it is not the case. Physics is not about “objective” reality.  Instead, physics is always about what an observer would observe.

— Lubos Motl

— paraphrased

— Me@2019-01-19 10:25:15 PM

.

.

2019.01.19 Saturday (c) All rights reserved by ACHK

(反對)開夜車 2.1

Ken Chan 時光機 1.4.2.1

.

之前提到,Ken Chan 所讀的中學不正常:

在中四和中五的校內考試測驗中,不斷地考核高考課程。所以,他在中四時代開始,已經要鑽研,高考程度和大學程度的物理。

那樣,他何來有,那麼多的時間呢?

那時,他往往要讀書,至通宵達旦。

.

那樣,他的公開試成績,又如何呢?

根據他所講,他事後發現,會考物理科中的 MC(多項選擇題)部分,錯了兩題;他達不到他原本,「全部科目全部滿分」的理想。

.

那樣,他又做不做到「狀元」,全部科目「奪 A」呢?

他在另一天的課堂中,暗示自己當年,差一點才做到「狀元」:

(我暫時不記得,他以下說話的上半句是什麼。)

… 如何不是那樣,我就毋須於,放榜當天的晚上,在家裡哭。唉!還是不說了。

當時,我想知道詳情。可惜,他真的沒有說下去,我也沒有辦法。

雖然,主觀而言,從他自己的角度,成績不是理想,因為,那不是「全部科目全部滿分」;但是,如果不是對自己,那麼苛刻的話,客觀來說,他的成績是上乘的。

.

可能,因為他「溫習到凌晨」式的方法,對他來說有效,他亦鼓勵學生那樣做。

我現在的記憶,暫時未能確定,他在課堂中,有沒有明示推介過,用這個方法。但是,我記得在他派發的筆記中,其中一頁,有一個正在深夜讀書的漫畫。而在漫畫下面,有一句「study to 3 a.m.」(讀書至深夜三點鐘)。

.

長話短說,我是反對這個方法的,…

— Me@2019-01-18 03:47:50 PM

.

.

2019.01.19 Saturday (c) All rights reserved by ACHK

Equality Predicates

6.3. Equality Predicates

Common Lisp provides a spectrum of predicates for testing for equality of two objects: eq (the most specific), eql, equal, and equalp (the most general).

eq and equal have the meanings traditional in Lisp.

eql was added because it is frequently needed, and equalp was added primarily in order to have a version of equal that would ignore type differences when comparing numbers and case differences when comparing characters.

If two objects satisfy any one of these equality predicates, then they also satisfy all those that are more general.

.

[Function]
eq x y

(eq x y) is true if and only if x and y are the same identical object. (Implementationally, x and y are usually eq if and only if they address the same identical memory location.)

.

The predicate eql is the same as eq, except that if the arguments are characters or numbers of the same type then their values are compared. Thus eql tells whether two objects are conceptually the same, whereas eq tells whether two objects are implementationally identical. It is for this reason that eql, not eq, is the default comparison predicate for the sequence functions defined in chapter 14.

.

[Function]
eql x y

The eql predicate is true if its arguments are eq, or if they are numbers of the same type with the same value, or if they are character objects that represent the same character.

.

[Function]
equal x y

The equal predicate is true if its arguments are structurally similar (isomorphic) objects. A rough rule of thumb is that two objects are equal if and only if their printed representations are the same.

Numbers and characters are compared as for eql. Symbols are compared as for eq. This method of comparing symbols can violate the rule of thumb for equal and printed representations, but only in the infrequently occurring case of two distinct symbols with the same print name.

.

[Function]
equalp x y

Two objects are equalp if they are equal; if they are characters and satisfy char-equal, which ignores alphabetic case and certain other attributes of characters; if they are numbers and have the same numerical value, even if they are of different types; or if they have components that are all equalp.

— Common Lisp the Language, 2nd Edition

— Guy L. Steele Jr.

.

Conrad’s Rule of Thumb for Comparing Stuff:

1. Use eq to compare symbols

2. Use equal for everything else

— Land of Lisp, p.63

— Conrad Barski, M. D.

.

.

2019.01.16 Wednesday ACHK

Problem 14.5d1.1.2

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

.

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}.

~~~

.

— 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 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}} + ...}

.

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}}

.

\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}}

.

The left R’+ sector:

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

.

~~~

— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

.

.

2019.01.14 Monday (c) All rights reserved by ACHK

EPR paradox, 5.3

According to special relativity, in EPR, which of Alice and Bob collapses the wavefunction is not absolute. In other words, they do not have any causal relations.

— Me@2012-04-12 10:42:22 PM

.

.

2019.01.14 Monday (c) All rights reserved by ACHK

接觸永恆 2

Publish! 11

.

每個人,都需要有作品。

你的作品,就是你的人生意義。

每個人,都依靠自己的作品來生存。

作品,就是一些,不會隨自己的消失而消失的東西。

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

你,有沒有自己的作品?

— Me@2011.07.03

.

.

2019.01.13 Sunday (c) All rights reserved by ACHK

PhD, 3.1

故事連線 1.1.5.1 | 碩士 4.1 | On Keeping Your Soul, 2.2.1 | Release early. Release often, 3

這段改編自 2010 年 4 月 18 日的對話。

.

一般寫博士論文的方法過程,有如 closed source software(閉源軟件)的開發一樣,是不合理的。

那個方法是,花數年時間,編寫一個軟件程式,然後才正式出版。這個方法的危險之處是,如果軟件上市以後,才發現不受市場認同的話,那數年的投資,基本上是血本無歸的。

合理一點的開發模式是,open source software(開源軟件)開發時,一個經常採用的方法,叫做「release early, release often」(極速(而)頻繁(地)出版)。意思是,與其花數年時間,去編寫一個軟件程式的「完美」版,然後才正式出版,倒不如,先把那個軟件的「最粗疏但已經可用」版本,極速完成,然後立刻出版。

那樣,那公司就可以馬上得到,市場的初步反應;從而決定:是否繼續開發那個軟件:如果繼續的話,又應該改進哪些部分。

如果決定繼續開發,就立刻重覆這個「極速出版」流程。亦即是話,再把那個軟件改進部分的,「最粗疏但已經可用」版本,極速完成,然後立刻出版。

— Me@2019-01-13 06:22:43 PM

.

.

2019.01.13 Sunday (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

.

.

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

.

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}.

~~~

.

— 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

.

.

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

.

.

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

.

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

.

.

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

.

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. —

~~~

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

.

.

2019.01.03 Thursday (c) All rights reserved by ACHK

Photon dynamics in the double-slit experiment, 5

.

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

— Me@2012-04-09 7:38:06 PM

.

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

.

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

.

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

.

.

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

.

(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

.

.

2019.01.01 Tuesday (c) All rights reserved by ACHK