Inception 4.2

潛行凶間 4.2 | 如何拯救眾生 5.2 | 無限複製 5.2

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What’s the most resilient parasite? An Idea. A single idea from the human mind can build cities. An idea can transform the world and rewrite all the rules. Which is why I have to steal it.

— Inception (film)

— Me@2010.07.30

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2011.01.20 Thursday ACHK

注定外傳 1.10

Can it be Otherwise? 1.10

我們先回顧一下,今天的討論。首先,我們提到:

當一個人問一件事是不是注定時,意思往往是問:

下次如果遇到類似的情境,可不可以有不同的結果?

(問:如果只是「類似」,當然可以有不同結果。你應該直接問:

下次如果遇到相同的情境,可不可以有不同的結果?

』)

我不能話你這個講法錯。但是,如果你真是這樣問,我大概只可以答「不可以」,因為,如果真的是「百份百相同」的情境,又怎可能有不同的結果呢?

(問:不是呀。在量子力學中,即使有兩組百分百一樣的物理系統,即使它們獲得完全相同的輸入,都可能有不同的輸出。)

你大概正確。但是,你要留意,量子力學中的『百分百一樣』物理系統,未必是你心目中的『百分百一樣』。

然後,我用了四個要點,解釋了為什麼,量子力學中的「百分百一樣」物理系統,未必是你心目中的「百分百一樣」。

綜合以上解釋,你會知道,兩個物件,或者兩個物理系統的「相同」,不會是絕對的;而是相對於某個準確度,或者相對個別性質而言。

「相同」的意思,並不是指「沒有可能找到任何分別」。

「相同」的意思是「分別小到不易察覺」。

而「類似」,則是指「分別不大」。

在這個背景之下,在討論「注定問題」時,

下次如果遇到類似的情境,可不可以有不同的結果?

下次如果遇到相同的情境,可不可以有不同的結果?

」,

其實意思一樣。

當你問前者時,我可以追問:「情境有多類似?類似到什麼程度?」

當你問後者時,我亦可以反問:「那個『相同』,是相對於哪個『觀測準確度』而言?」

既然在這個上文下理中,意思一樣,方便起見,我把這兩個講法,統一為後者:

下次如果遇到相同的情境,可不可以有不同的結果?

只要答到這個問題,你就會知道,某一件事件是否「注定」,或者「必然」。

但是,這個問題的答案,取決於「相同」的意思;而兩個情境「是否相同」,又取決於「相對於哪個『觀測準確度』而言」。

換句話說,某一件事件是否「必然」,不會是絕對的;而是相對於某個「觀測準確度」而言。

— Me@2015-10-29 10:12:16 PM

2015.10.29 Thursday (c) All rights reserved by ACHK

Exercise 6a (Corrected version)

You Could Have Invented Monads! (And Maybe You Already Have.)

Show that f' * unit = unit * f' = bind f'

——————————

f :: a -> b
f' :: a -> m a
unit :: a -> m a

lift f = unit . f
f' = lift f

The lift function in this tutorial is not the same as the liftM in Haskell. So you should use lift (but not liftM) with bind.

— Me@2015-10-13 11:59:53 AM

(f' * g') xs
= ((bind f') . (bind g')) xs

bind f' xs = concat (map f' xs)
unit x = [x]

bind unit xs
= concat (map unit xs)
= concat (map unit [x1, x2, ...])
= concat [unit x1, unit x2, ...]
= concat [[x1], [x2], ...]
= [x1, x2, ...]
= xs

(f' * unit) (x:xs)
= bind f' (bind unit (x:xs))
= bind f' (concat (map unit (x:xs)))
= bind f' (concat (map unit [x1, x2, ...]))
= bind f' (concat [[x1], [x2], ...])
= bind f' [x1, x2, ...]
= concat (map f' [x1, x2, ...])
= concat [f' x1, f' x2, ...]
= concat [(unit . f) x1, (unit . f) x2, ...]
= concat [(unit (f x1)), (unit (f x2)), ...]
= concat [[f x1], [f x2], ...]
= [f x1, f x2, ...]

(unit * f') (x:xs)
= ((bind unit) . (bind f')) (x:xs)
= bind unit (bind f' (x:xs))
= bind unit (concat (map f' (x:xs)))
= bind unit (concat (map f' [x1, x2, ...]))
= bind unit (concat [f' x1, f' x2, ...])
= bind unit (concat [(unit . f)  x1, (unit . f) x2, ...])
= bind unit (concat [(unit (f x1)), (unit (f x2)), ...])
= bind unit (concat [[f x1], [f x2], ...])
= bind unit [f x1, f x2, ...]
= concat (map unit [f x1, f x2, ...])
= concat [[f x1], [f x2], ...]
= [f x1, f x2, ...]

— Me@2015-10-15 07:19:18 AM

If we use the identity bind unit xs = xs, the proof will be much shorter.

(f' * unit) (x:xs)
= ((bind f') . (bind unit)) (x:xs)
= bind f' (bind unit (x:xs))
= bind f' (x:xs)

(unit * f') (x:xs)
= ((bind unit) . (bind f')) (x:xs)
= bind unit (bind f' (x:xs))
= bind f' (x:xs)

— Me@2015-10-15 11:45:44 AM

2015.10.15 Thursday (c) All rights reserved by ACHK

Problem 14.3b3

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.
 
~~~
 
What is a zero mode?
 
p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”
 
 
How come there are creation operators that do not contribute to the mass-squared of the states?
 
Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.
 
If an index can be zero, the corresponding term can be zero.
 
— Me@2015.09.26 08:44 PM
 
 
Consider the NS-sector:
 
Equation (14.37):
 
M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)
 
N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I
 
For the NS sector, the r values in b^I_{r} are half-integers, thus cannot be zero. So every creation operator b_{-r}^I contributes to the mass-squared.
 
(p.312 “… the negatively moded coefficients b_{-1/2}^I, b_{-3/2}^I, b_{-5/2}^I, …, are creation operators, …”)
 
 
Consider the R-sector:
 
Equation (14.53):

M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)
 
For the R sector, the n values in d^I_n are integers, thus can be zero. So some of the creation operators d^I_{-n} are zero modes.
 
p.315 “… the eight fermionic zero modes d_0^I…”
 
— Me@2015-10-11 11:01:44 AM
 
 
 
2015.10.11 Sunday (c) All rights reserved by ACHK

同情地理解

「同情地理解」的意思是,從對方的情境中,去理解對方的言論。

當別人的言論錯漏百出時,未必需要立刻否定;你可以易地而處,嘗試估計對方為何要那樣說,企圖要表達的真正意思是什麼。

但是,自己發表言論時,則不應假設,別人會同情地理解;因為,如果自己的言論,時常要別人同情地理解,才能有機會理解到的話,那會反映你的語言邏輯能力,十分有問題。

— Me@2015-10-07 08:41:51 AM

2015.10.08 Thursday (c) All rights reserved by ACHK

注定外傳 1.9

Can it be Otherwise? 1.9

換而言之,兩樣東西,不會「絕對相同」。

(這裡的「東西」,是指宏觀的物理系統。至於兩粒微觀粒子,則有可能「全同」。但那是另一個話題,容後再談。)

第四,即使假設了,有一個情況是,你有一件超特級的量度儀器,可以準確到小數後無限個位;而用它來量度兩樣東西(甲和乙)時,發現它們竟然,真的完全百分百,在任何方面也「絕對相同」,邏輯問題仍然存在,因為,它們至少,會佔有不同的空間,即是處於不同的位置。

(問:那如果連位置都相同呢?)

那甲和乙就再不是「兩」件東西,而根本是同一件東西。只不過,那一件東西,有兩個名字或者身份而已。

叫得做「兩」個物理系統,就總有不同之地方。

— Me@2015-10-07 02:52:21 PM

2015.10.07 Wednesday (c) All rights reserved by ACHK

Exercise 6a

You Could Have Invented Monads! (And Maybe You Already Have.)

Show that f * unit = unit * f

——————————

(f * g) (x, xs)
= ((bind f) . (bind g)) (x, xs)

bind f x = concat (map f x)

(f * unit) (x:xs)
= bind f (bind unit (x:xs))
= bind f (concat (map unit (x:xs)))
= bind f (concat (map unit [x1, x2, x3, ...]))
= bind f (concat ([[x1], [x2], [x3], ...]))
= bind f [x1, x2, x3, ...]
= concat (map f [x1, x2, x3, ...])
= concat [f x1, f x2, f x3, ...]
= [f x1, f x2, f x3, ...]

(unit * f) (x:xs)
= ((bind unit) . (bind f)) (x:xs)
= bind unit (bind f (x:xs))
= bind unit (concat (map f (x:xs)))
= bind unit (concat (map f [x1, x2, ...]))
= bind unit (concat [f x1, f x2, ...])
= bind unit [f x1, f x2, ...]
= concat (map unit [f x1, f x2, ...])
= concat [[f x1], [f x2], ...]
= [f x1, f x2, ...]

— Me@2015.07.20 09:00 PM

2015.10.06 Tuesday (c) All rights reserved by ACHK

Problem 14.3b2

A First Course in String Theory
 
 
14.3 Massive level in the open superstring

~~~

How come R sector has a factor 16 while NS sector has not?

Equation (14.66):

f_{NS}(x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n - \frac{1}{2}}}{1-x^n} \right)^8

Equation (14.68):

f_R(x) = 16 \prod_{n=1}^\infty \left( \frac{1+x^n}{1-x^n} \right)^8

p.319 “The overall multiplicative factor appears because each combination of oscillators gives rise to 16 states by acting on each of the available ground states.”

p.319 “We note that the R coefficients are actually double the corresponding NS coefficients. This is not a coincidence, as we will see in the following section.”

p.320 “We have seen that the Ramond sector has world-sheet supersymmetry: there are equal numbers of fermionic and bosonic states at each mass level.”
 
 
With the factor 16, how come the R coefficients are only double, but not 16 times as big as the corresponding NS coefficients?
 
It is caused by the difference of x^n and x^{n-\frac{1}{2}}.

— Me@2015.10.06 08:23 AM
 
 
 
2015.10.06 Tuesday (c) All rights reserved by ACHK

Person of Interest

Person of Interest is an American science fiction crime drama television series created by Jonathan Nolan that premiered on September 22, 2011, on CBS. It is executive produced by Nolan, alongside J. J. Abrams, Bryan Burk, and Greg Plageman. It stars Jim Caviezel as John Reese, a former CIA agent who is presumed dead. He is approached by a mysterious billionaire named Harold Finch (Michael Emerson), who is trying to prevent violent crimes before they happen by using an advanced surveillance system dubbed “The Machine”, which turns out to have evolved into a sentient AI.

— Wikipedia on Person of Interest (TV series)

2015.10.05 Monday ACHK

Inception 4.1

潛行凶間 4.1 | 如何拯救眾生 5.1 | 無限複製 5.1

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What’s the most resilient parasite? A bacteria? A virus? An intestinal worm?

An idea.

Resilient, highly contagious. Once an idea’s taken hold in the brain it’s almost impossible to eradicate. A person can cover it up, ignore it – but it stays there.

Information, yes. But an idea? Fully formed, understood? That sticks… (taps forehead) In there, somewhere.

— Inception (film)

— Me@2011.01.18

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2011.01.18 Tuesday ACHK

注定外傳 1.8

Can it be Otherwise? 1.8

凡是量度,都只會得到近似值,所以,只能討論「近似同一性」(量度準確度)。亦即是話,當我們說,那兩支筆的長度「相同」時,是指它們的長度,相近到以當時的肉眼和儀器,暫時感受不到分別而已。

(問:那如果是數數目(使用整體)的情況呢?

例如:你有 10 隻手指,我又有 10 隻手指。那樣,這兩個 10,不就是「絕對相同」嗎?)

應該是「確切相同」,而不是「絕對相同」。在這裡,「確切」的意思是,不再只是近似。

那兩個數字,不再只是「相似」(近似相同),而根本是同一個數字。

凡是量度,都只會得到近似值,所以,只能討論「近似同一性」(量度準確度)。凡是數數,則有可能得到確切值,所以,可以討論「確切同一性」(數數準確度)。

(問:那為什麼不可以說「絕對相同」?)

幾乎可以,但未臻完善。

如果只是討論那兩個整數,兩者的確是「絕對相同」。

但是,根據現在的上文下理,我們要考慮的,不只那兩個整數。我們還要考慮的是,「兩個物件,或者兩個物理系統,有沒有可能完全相同?」

在量子力學中,即使有兩組百分百一樣的物理系統,即使它們獲得完全相同的輸入,都可能有不同的輸出。

下次如果遇到相同的情境,可不可以有不同的結果?

換句話說,正式要比較的,不只是(例如)兩個人的手指數目,而是那兩個人。

在這個情況下,相對於手指數目而言,他們就絕對相同。但是,相對於整體而言,他們就沒有可能,在所有方面,都百分百相同。例如,他們的左手食指長度,只可能近似相同,不可能確切相同。

由於二人只可能「相對於某些方面」而言,絕對相同。這個「絕對」,並不是真的那麼「絕對」。那樣,用字就應嚴格一點。

「絕對」,應該用作「相對」的相反。而「近似」的相反,則應該用「確切」。

兩件物件,或者兩個物理系統,不可能在所有方面,都確切相同、完全一樣,因為,比較兩者時,總會有些量度(例如左手食指長度)的成份。

凡是量度,都只會得到近似值,所以,只能討論『近似同一性』(量度準確度)。

換而言之,兩樣東西,不會「絕對相同」。

(這裡的「東西」,是指宏觀的物理系統。至於兩粒微觀粒子,則有可能「全同」。但那是另一個話題,不宜在這裡詳述。)

— Me@2015-10-04 07:32:32 AM

2015.10.04 Sunday (c) All rights reserved by ACHK

Exercise 3.2

You Could Have Invented Monads! (And Maybe You Already Have.)

Show that lift f * lift g = lift (f.g)

——————————

The meaning of f' * g' should be (bind f') . (bind g') instead.

f' * g' = (bind f') . (bind g')
lift f = unit . f
f' = lift f

(lift f * lift g) (x, xs)
= (bind (lift f)) . (bind (lift g)) (x, xs)
= bind (lift f) (bind (lift g) (x, xs))
= bind (lift f) (gx, xs++gs)
  where
    (gx, gs) = (lift g) x

= bind (lift f) (gx, xs++gs)
  where
    (gx, gs) = (g x, "")

= bind (lift f) (g x, xs)

= (fx, xs++fs)
  where
    (fx, fs) = (lift f) gx
 
= (fx, xs++fs)
  where
    (fx, fs) = (f gx, "")

= (fx, xs)
  where
    (fx, fs) = (f (g x), "")

= (f (g x), xs)

bind f' (gx,gs) = (fx, gs++fs)
                  where
                    (fx,fs) = f' gx

bind (lift (f.g)) (x, xs)
= (hx, xs++hs)
  where
    (hx, hs) = (lift (f.g)) x

= (hx, xs++hs)
  where
    (hx, hs) = ((f.g) x, "")

= ((f (g x)), xs)

— Me@2015.07.19 11:04 PM

2015.10.02 Friday (c) All rights reserved by ACHK

Problem 14.3b1

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.

~~~

What is the difference between “open bosonic string” and “open superstring”?

p.320 “With our conventions, these are world-sheet fermionic states that are now recognized to be states of spacetime fermions. The resulting, truncated sector is called the R- sector.”

NS(+): Equation (14.38) \left(\alpha' M^2 = 1, N^\perp = \frac{3}{2}\right):

\left\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b^I_{-1/2} b^J_{-1/2} b^K_{-1/2} \right\} | NS \rangle \otimes | p^+, \vec p_R \rangle

There are 128 states (cf. Quick Calculation 14.4).

R: Equation (14.54) \left(\alpha' M^2 = 1\right):

\alpha^I_{-1}  | R_a \rangle, d_{-1}^I | R_{\bar a} \rangle~||~\alpha^I_{-1}  | R_{\bar a} \rangle, d_{-1}^I | R_{a} \rangle

Only the left hand side represents the R- states.

Each of index I and index a has 8 possible values. So there are totally 64 states in the form of \alpha^I_{-1} | R_a \rangle.

Similarly, there are totally 64 states in the form of d_{-1}^I | R_{\bar a} \rangle.

Totally, for \alpha' M^2 = 1, there are 128 R- states.

— Me@2015-10-02 02:16:23 PM
 
 
 
2015.10.02 Friday (c) All rights reserved by ACHK