Ken Chan 時光機 1.4.1

迷宮直昇機 5.1

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至於第三個技巧,則實在是最宏觀,亦可能是最重要。

Ken Chan 說:「你們現在會盤算,在會考中,各科太概會有什麼目標,奪取什麼等級的成績。但是,我當年全部科目,只有一個目標,就是要『攞 full』」。

那就即是要,全部科目中的每一科,不只要甲等,而是要滿分;因為在當年,如果可以在全部科目奪得滿分,考生會獲頒一張特別證書。他那時就為了,那張證書而努力。

但是,那是如何辦到的呢?

考生在中五時參加「會考」,中七時參加「高考」。所以,正常的學校,會在中四和中五的校內考試測驗中,考核會考課程。

但是,他的學校不正常——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。

結果,到真正會考時,由於「只會」考核中五程度的東西,他會突然覺得十分容易。

— Me@2018-11-18 10:06:43 PM

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2018.11.18 Sunday (c) All rights reserved by ACHK

defmacro, 2

Defining the defmacro function using only LISP primitives?

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McCarthy’s Elementary S-functions and predicates were

atom, eq, car, cdr, cons

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He then went on to add to his basic notation, to enable writing what he called S-functions:

quote, cond, lambda, label

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On that basis, we’ll call these “the LISP primitives”…

How would you define the defmacro function using only these primitives in the LISP of your choice?

edited Aug 21 ’10 at 2:47
Isaac

asked Aug 21 ’10 at 2:02
hawkeye

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Every macro in Lisp is just a symbol bound to a lambda with a little flag set somewhere, somehow, that eval checks and that, if set, causes eval to call the lambda at macro expansion time and substitute the form with its return value. If you look at the defmacro macro itself, you can see that all it’s doing is rearranging things so you get a def of a var to have a fn as its value, and then a call to .setMacro on that var, just like core.clj is doing on defmacro itself, manually, since it doesn’t have defmacro to use to define defmacro yet.

– dreish Aug 22 ’10 at 1:40

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2018.11.17 Saturday (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

Detecting a photon

In the double-slit experiments, how to detect a photon without destroying it?

— Me@2018-11-10 08:07:29 PM

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Artlav: I’ve been thinking about the double slit experiment – the one with single photons going thru two slits forming an interference pattern never the less. Now, one thing i was unable to find clarification for is the claim that placing a detector even in just one of the slits to find out thru which slit a photon passed will result in the disappearance of the interference pattern. The question is – how does such detector work? How can one detect a photon without destroying it?

Cthugha (Science Advisor): Well, in the kind of experiment you describe, the photon will usually be destroyed by detecting it. However, in some cases it is possible to detect photons without destroying them. Usually one uses some resonator, for example some cavity, in which photons go back and forth and prepare some atom in a very well defined spin state. Now the atom falls through the cavity perpendicular to the photons moving back and forth and the spin state of the atom after leaving the cavity will depend on the number of photons because the spin precession will be a bit faster in presence of photons. If you do this several times, you will get a nondestructive photon number measurement. However, these are so called weak measurements, so this means you do not change the photon states if you are in a photon number eigenstate already. The first measurement however might change the photon state from some undefined state to a photon number eigenstate.

Reference: physicsforums double-slit-experiment-counter.274914

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2018.11.10 Saturday ACHK

凌晨舊戲 2.2

這段改編自 2010 年 4 月 18 日的對話。

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繼而,甲版本的瀕死經驗,可信度亦會高一些;因為,如果甲所講述的「境界極高道理」,其實是他自己的創作的話,一般而言,他並沒有動機,去把功勞賦予一個,虛構的瀕死經驗。

(問:但是,道理又怎樣為之「境界極高」呢?)

這個問題,跟前一個問題性質一樣,所以,答案跟前一個答案相若,都是用「好樹結好果」這個大原則。

能結出好果的,就為之「好樹」。能治療(或舒緩)病患的,就為之「好醫生」。

同理,如果一個建議,或者個觀點,能大大改善,讀者的心靈世界,甚至實際生活的話,那就為之「境界極高道理」。

(以下的「瀕死經驗者」,是「自稱經歷瀕死經驗人仕」的簡稱。同理,「瀕死經驗後」,是「所宣稱的瀕死經驗後」的縮寫。)

有部分瀕死經驗者,在瀕死經驗後,性格太幅度地改善,道德超常地提升。那樣,他口中的道理,可信度自然甚高。

(問:那樣,你又如何呢?

雖然,你未曾有過瀕死經歷,但是,你有閱讀過,所謂由瀕死經歷中,領悟到的道理。你有沒有嘗試,應用過它們?結果是「好果」?還是「壞果」?)

有部分是好果,受用無窮。但是,不幸地,亦有部分是壞果,受害無窮。那就是為什麼,我剛才再三強調,

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇。

(問:那樣,你又可否舉例,哪些是帶來「好果」的真道理,而哪些卻是帶在「壞果」的假道理?)

不太可以,因為,沒有足夠的上文下理,個別道理說出來,很容易造成誤解。但是,要有足夠的上文下理的話,除了此刻需要,長篇大論外,聽者的人生閱歷,不能太少。

(問:即是話,要足夠老?)

可以這樣說。但是,我可以給你一個大方向。

年輕人,很容易以為,動聽的(所謂)道理,就是真確的道理。

只要你能夠提防這種錯覺,你就已經可以,避免大量的錯誤。

— Me@2018-11-08 11:24:23 AM

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2018.11.08 Thursday (c) All rights reserved by ACHK

defmacro

SLIME, 2

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Alt + Up/Down

Switch between the editor and the REPL

— Me@2018-11-07 05:57:54 AM

~~~

defmacro

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(defmacro our-expander (name) `(get ,name 'expander))

(defmacro our-defmacro (name parms &body body)
  (let ((g (gensym)))
    `(progn
       (setf (our-expander ',name)
	     #'(lambda (,g)
		 (block ,name
		   (destructuring-bind ,parms (cdr ,g)
		     ,@body))))
       ',name)))

(defun our-macroexpand-1 (expr)
  (if (and (consp expr) (our-expander (car expr)))
      (funcall (our-expander (car expr)) expr)
      expr))

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A formal description of what macros do would be long and confusing. Experienced programmers do not carry such a description in their heads anyway. It’s more convenient to remember what defmacro does by imagining how it would be defined.

The definition in Figure 7.6 gives a fairly accurate impression of what macros do, but like any sketch it is incomplete. It wouldn’t handle the &whole keyword properly. And what defmacro really stores as the macro-function of its first argument is a function of two arguments: the macro call, and the lexical environment in which it occurs.

— p.95

— A MODEL OF MACROS

— On Lisp

— Paul Graham

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(our-defmacro sq (x)
  `(* ,x ,x))

After using our-defmacro to define the macro sq, if we use it directly,


(sq 2)

we will get an error.

The function COMMON-LISP-USER::SQ is undefined.
[Condition of type UNDEFINED-FUNCTION]

Instead, we should use (eval (our-macroexpand-1 ':


(eval (our-macroexpand-1 '(sq 2)))

— Me@2018-11-07 02:12:47 PM

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2018.11.07 Wednesday (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}}

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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:~~~~~&\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

Monty Hall problem 1.6

Sasha Volokh (2015) wrote that “any explanation that says something like ‘the probability of door 1 was 1/3, and nothing can change that…’ is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance.”

— Wikipedia on Monty Hall problem

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2018.11.02 Friday ACHK

講道理 

情緒病 2 | 不思考之道 6

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年紀漸長,我越來越覺得「(太)喜歡講道理」,或者「(太)喜歡聽道理」,是一種情緒病。

— Me@2012.02.14

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2018.11.01 Thursday (c) All rights reserved by ACHK

Posted in OCD

Ken Chan 時光機 1.3

唔識就飛 8.4.2

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另一個技巧是,考試時「唔識就飛」:想不通的話,就立刻先做下一題,或者同一題的下一部分。

假設我現時記憶正確,Ken Chan 是第一個教我這個技巧的老師;我日校的純數學科老師,則是第二個(亦是最後一個)。

這個技巧,對我來說,實在太重要,因為,在兩年多後的純數學科高考中,如果我不是確切執行這個政策的話,我有極大機會會失手,繼而在當年,升不到大學。

那個故事,現不再詳述,因為已於另一篇文章「唔識就飛 8.4」中講過。請參考該文章。

— Me@2018-10-31 09:39:05 AM

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2018.10.31 Wednesday (c) All rights reserved by ACHK

如有 HKDSE physics 香港中學文憑試之物理問題,歡迎以右邊之電郵地址,聯絡本人。

funcall

In Common Lisp, apply can take any number of arguments, and the function given first will be applied to the list made by consing the rest of the arguments onto the list given last. So the expression

(apply #’+ 1 ’(2))

is equivalent to the preceding four. If it is inconvenient to give the arguments as
a list, we can use funcall, which differs from apply only in this respect. This expression

(funcall #’+ 1 2)

has the same effect as those above.

— p.13

— On Lisp

— Paul Graham

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Exercise 7.1

Define funcall.

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(defmacro our-funcall (f &rest p)
  `(apply ,f (list ,@p)))

— Me@2018-10-30 03:24:05 PM

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2018.10.30 Tuesday (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) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

We have 32 zero modes \displaystyle{\lambda_0^A} and 16 linear combinations behave as creation operators.

As usual half of the ground states have \displaystyle{(-1)^{F_L} = +1} and the other half have \displaystyle{(-1)^{F_L} = -1}.

Let \displaystyle{|R_\alpha \rangle_L} denote ground states with \displaystyle{(-1)^{F_L} = +1}.

How many ground states \displaystyle{|R_\alpha \rangle_L} are there?

~~~

— 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}}

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{16 = 2^4} degenerate Ramond ground states.”

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}.

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

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2018.10.29 Monday (c) All rights reserved by ACHK

Visualizing higher dimensions

The trick of visualizing higher dimension is: not to visualize it.

— Wikipedia

— Me@2011.08.19

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Besides trying to visualize, there are other methods to understand higher dimensions.

— Me@2018-10-28 04:28:01 PM

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What is the meaning of visualization?

— Me@2018-09-02 4:35 pm

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feel ~ receive all the data at once

(This definition is not totally correct, but is useful in the meantime.)

visual ~ feel at once through eyes

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you can visualize a 3D object ~ you can see all of a 3D object at once

you cannot visualize a 4D object ~ you cannot see all of a 4D object at once

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Actually, you can only visualize a 2D object, such as a square.

You cannot visualize a 3D object, such as a cube.

That’s why the screen of any computer monitor is 2 dimensional, not 3.

— Me@2018-10-28 04:32:41 PM

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2018.10.28 Sunday (c) All rights reserved by ACHK

A pretty girl, 2

Women are directly fitted for acting as the nurses and teachers of our early childhood by the fact that they are themselves childish, frivolous and short-sighted; in a word, they are big children all their life long–a kind of intermediate stage between the child and the full-grown man, who is man in the strict sense of the word. See how a girl will fondle a child for days together, dance with it and sing to it; and then think what a man, with the best will in the world, could do if he were put in her place.

— Of Women

— Arthur Schopenhauer

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When the elderly Schopenhauer sat for a sculpture portrait by the Prussian sculptor Elisabet Ney in 1859, he was much impressed by the young woman’s wit and independence, as well as by her skill as a visual artist. After his time with Ney, he told Richard Wagner’s friend Malwida von Meysenbug, “I have not yet spoken my last word about women. I believe that if a woman succeeds in withdrawing from the mass, or rather raising herself above the mass, she grows ceaselessly and more than a man.

— Wikipedia on Arthur Schopenhauer

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Anybody can look at a pretty girl and see a pretty girl. An artist can look at a pretty girl and see the old woman she will become. A better artist can look at an old woman and see the pretty girl that she used to be. But a great artist — a master — and that is what Auguste Rodin was — can look at an old woman, portray her exactly as she is… and force the viewer to see the pretty girl she used to be…. and more than that, he can make anyone with the sensitivity of an armadillo, or even you, see that this lovely young girl is still alive, not old and ugly at all, but simply prisoned inside her ruined body. He can make you feel the quiet, endless tragedy that there was never a girl born who ever grew older than eighteen in her heart…. no matter what the merciless hours have done to her. Look at her, Ben. Growing old doesn’t matter to you and me; we were never meant to be admired — but it does to them. Look at her! (UC)

— Robert A. Heinlein

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2018.10.27 Saturday ACHK

凌晨舊戲 2

這段改編自 2010 年 4 月 18 日的對話。

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(問:你閱讀過很多有關「瀕死經驗」的文章?)

可以這樣說。

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇

(問:那你怎樣分辨,「瀕死經驗」的文章之中,哪些是真,哪些為假?)

看看文中所說的,合不合理。

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例如,你怎樣知道,我說的話,是真還是假?

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合情合理,自圓其說的,就有機會真;無情無理,自相矛盾的,則必定為假。

——

如果,合情合理之餘,我感覺到它說的道理,遠超過我當時境界的話,那道理的可信程度,就再高一點。

(問:那亦可能只是你的誤會。按常理,程度較低的人不會知道,所謂「程度高的人或說話」的真假。

例如,如果你沒有學過物理,當一個物理學家,介紹核電廠如何運作時,其實,你並不會百分百肯定,他講的東西,是真還是假;除非,他的言論,明顯自相矛盾;那樣,你就可以立刻肯定,那是假的。)

所以,我只提及「可信度」,而沒有一口斷定「真假」。

其實,有更日常生活的例子:

我因為不是醫生,醫學知識遠低於醫生,有病時才會考慮去醫生。而正正是因為,我的醫學知識遠低於醫生,我並沒有能力,可以肯定一個醫生的建議是正確的。但是,我總不有能因為那樣,而任何情況下,都不去看醫生。

判斷一位醫生可不可信,唯有靠(他人或自己)的實證。

合理的做法是,如果一個醫生的建議,通常令你的病情舒緩的話,他的說話,就為之「可信」。相反,如果一個醫生的治療,只是「間中」令你病情舒緩,而「間中」到一個程度,令你覺得那跟隨機沒有分別的話,他的說話,就為之「不可信」。

——

當然,「瀕死經驗」並沒有所謂「實證」。你總不能叫我,刻意製造「瀕死經驗」,冒生命危險,去檢驗其真偽。

但是,留意,即使是剛才「判別醫生說話真偽」的例子,所講的「實證」,其實也只是,間接的實證。如果你要直接的實證,你就真的需要,先入醫學院,受訓成為醫生,才接近百分百地肯定,其他醫生的建議,是真還是假。

但是,那成本太高。在一般情況下,常人也不應那麼誇張。利用一個醫生的治療準繩度,來釐定其可信度,才是合理可行的方法。

——

同理,如果某人(甲)宣稱,經歷過「瀕死經驗」,而你又想「判別其真假」的話,你可以用以下的幾個方法。

(留意,這裡的「判別其真假」只是「評價其可信度」的簡寫。剛才你也提到,要百分百地證明,某些言論是真的,是沒有可能的。)

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第一,甲所講的瀕死經驗,和其他曾經瀕死人士的講法,有何異同。

第二,會不會甲其實根本未曾有過瀕死經驗,只是道聽途說,人云亦云而已。

這一點非常難考證。以下的則比較容易觀察。

第三,瀕死經驗中,他所領悟到的道理,合情合理之餘,我感覺到它說的道理,遠超過一般道理的境界的話,那道理的可信程度,就再高一點。

繼而,甲版本的瀕死經驗,可信度亦會高一些;因為,如果甲所講述的「境界極高道理」,其實是他自己的創作的話,一般而言,他並沒有動機,去把功勞賦予一個,虛構的瀕死經驗。

(問:但是,道理又怎樣為之「境界極高」呢?)

這個問題,跟前一個問題性質一樣,所以,答案跟前一個答案相若,都是用「好樹結好果」這個大原則。

— Me@2018-10-26 08:48:21 PM

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2018.10.26 Friday (c) All rights reserved by ACHK

A Road to Common Lisp

tumba 57 days ago [-]

My advice is this: as you learn Common Lisp and look for libraries, try to suppress the voice in the back of your head that says “This project was last updated six years ago? That’s probably abandoned and broken.” The stability of Common Lisp means that sometimes libraries can just be done, not abandoned, so don’t dismiss them out of hand.

I have found this to be true in my own experience. The perception of stagnation is, however, a common initial objection to folks working in CL for the first time.

mike_ivanov 57 days ago [-]

My personal problem with CL libraries is not that, but rather the lack of documentation. More often than not, there is no documentation at all, not even a readme file.. It feels like some library authors simply don’t care. I’d say this attitude has a negative impact on how people perceive viability of those libraries — and by extension, of the language.

armitron 56 days ago [-]

A lot of libraries that don’t have separate documentation in the form of HTML/PDF/README.. are actually well-documented at source level in the form of docstrings.
Since Common Lisp is an interactive programming language and is meant to be used interactively (think Smalltalk, not Python) it is common practice to (interactively) load a library in a Common Lisp image and explore it (interactively). One can see what symbols are exported from packages, what these symbols are used for and (interactively) retrieve their documentation. All of this takes place within the editing environment (ideally Emacs) in a rapid feedback loop (again think Smalltalk, not Python) that feels seamless and tremendously empowering.

stevelosh 56 days ago [-]

I agree with this — it’s a real problem. My only solution has been to try to be the change I want to see in the world and document all of my own libraries before I officially release them.

— A Road to Common Lisp

— Hacker News

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2018.10.24 Wednesday ACHK

Problem 14.5b

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

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

— This answer is my guess. —

The naive mass formula in the left R’ sector:

\displaystyle{ \begin{aligned}  \alpha' M_L^2 &= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \frac{1}{2} \sum_{n = -1, -2, ...} n \lambda_{-n}^A \lambda_n^A + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A  \\  \end{aligned}}

p.316 Equation (14.51):

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n = -1. -2, ...} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} (D - 2) \\  \end{aligned}}

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \left[ \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} \left[(D - 2)\right]_A \right] + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{32}{24} \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \\ \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \left[ \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \right] \\  &= 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}}

.

\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}}

— This answer is my guess. —

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2018.10.24 Wednesday (c) All rights reserved by ACHK

Relational quantum mechanics

EPR paradox, 10

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Relational quantum mechanics (RQM) is an interpretation of quantum mechanics which treats the state of a quantum system as being observer-dependent, that is, the state is the relation between the observer and the system. This interpretation was first delineated by Carlo Rovelli in a 1994 preprint, and has since been expanded upon by a number of theorists. It is inspired by the key idea behind special relativity, that the details of an observation depend on the reference frame of the observer, and uses some ideas from Wheeler on quantum information.

,,,

Relational solution

In RQM, an interaction between a system and an observer is necessary for the system to have clearly defined properties relative to that observer. Since the two measurement events take place at spacelike separation, they do not lie in the intersection of Alice’s and Bob’s light cones. Indeed, there is no observer who can instantaneously measure both electrons’ spin.

The key to the RQM analysis is to remember that the results obtained on each “wing” of the experiment only become determinate for a given observer once that observer has interacted with the other observer involved. As far as Alice is concerned, the specific results obtained on Bob’s wing of the experiment are indeterminate for her, although she will know that Bob has a definite result. In order to find out what result Bob has, she has to interact with him at some time {\displaystyle t_{3}} in their future light cones, through ordinary classical information channels.

The question then becomes one of whether the expected correlations in results will appear: will the two particles behave in accordance with the laws of quantum mechanics? Let us denote by {\displaystyle M_{A}(\alpha )} the idea that the observer {\displaystyle A} (Alice) measures the state of the system {\displaystyle \alpha} (Alice’s particle).

So, at time {\displaystyle t_{2}}, Alice knows the value of {\displaystyle M_{A}(\alpha )}: the spin of her particle, relative to herself. But, since the particles are in a singlet state, she knows that

{\displaystyle M_{A}(\alpha )+M_{A}(\beta )=0,}

and so if she measures her particle’s spin to be {\displaystyle \sigma }, she can predict that Bob’s particle ( {\displaystyle \beta } ) will have spin {\displaystyle -\sigma }. All this follows from standard quantum mechanics, and there is no “spooky action at a distance” yet. From the “coherence-operator” discussed above, Alice also knows that if at {\displaystyle t_{3}} she measures Bob’s particle and then measures Bob (that is asks him what result he got) — or vice versa — the results will be consistent:

{\displaystyle M_{A}(B)=M_{A}(\beta )}

Finally, if a third observer (Charles, say) comes along and measures Alice, Bob, and their respective particles, he will find that everyone still agrees, because his own “coherence-operator” demands that

{\displaystyle M_{C}(A)=M_{C}(\alpha )} and {\displaystyle M_{C}(B)=M_{C}(\beta )}

while knowledge that the particles were in a singlet state tells him that

{\displaystyle M_{C}(\alpha )+M_{C}(\beta )=0.}

Thus the relational interpretation, by shedding the notion of an “absolute state” of the system, allows for an analysis of the EPR paradox which neither violates traditional locality constraints, nor implies superluminal information transfer, since we can assume that all observers are moving at comfortable sub-light velocities. And, most importantly, the results of every observer are in full accordance with those expected by conventional quantum mechanics.

— Wikipedia on Relational quantum mechanics

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2018.10.22 Monday ACHK