Trajectory

It is not possible to derive Schrödinger’s equation from “anything we know”.

— R. P. Feynman

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The most confusing part in the Quantum Mechanics is [the concept of] Trajectory.

There exist[s] no fixed path for a particle to go from Point A to Point B. This is clearly visible from [the] Interference Experiment.

So, the approach here is to work with deductive reasoning. We eliminate the possible region/paths which [are] impossible to be followed.

To do this we assume that Energy Conservation Relation is valid for Quantum Mechanics too. So, those regions where particle[s] [violate] this law automatically [get] eliminated.

Then, we guess [the] State Function[s] for certain conditions i.e. how it should be in certain cases, then build an energy conservation equation with that. We will shortly demonstrate how Schrodinger itself reached the conclusion.

— Why can’t the Schrödinger equation be derived?

— Abhas Kumar Sinha

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2021.02.28 Sunday ACHK

Deus ex machina 4

This [Hegel’s philosophy] illustrates an important truth, namely, that the worse your logic, the more interesting the consequences to which it gives rise.

― A History of Western Philosophy

― Bertrand Russell

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A good way to have an intellectual stimulation is to let go of a part of logic and see how the story evolves.

For example, when you let go of the time logic, you will get a time travel story.

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let go of a part of logic

~ perturb the reality

~ be thought provocative

~ follow a new set of rules/physical laws

~ 迷魂

— Me@2016-01-29 8:47 AM

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We need to distinguish between real philosophical insights and bewitchments.

— Me@2016-02-02 04:29:34 PM

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… because both can give you the same feeling of thought-provocative-ness.

— Me@2021-02-27 08:55:27 PM

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2021.02.27 Saturday (c) All rights reserved by ACHK

如此仙子 1.3

鐵達尼極限 2.5 | 尋覓 1.8 | 已婚單身 1.4 | Singles 1.4

這段改編自 2010 年 10 月 14 日的對話。

.

我這樣說,是不是很灰呢?(是不是十分令人灰心呢?)

你覺得很灰的話,是因為你還沒有我提供,最先進的心靈科技。放心,那正正是我一路會講落去的東西。

其實,喺我來講,以上的擔憂,都不緊要,都不是問題,因為,「愛情」是在生時候的事情,但我思考的,是超過在生時候的東西。

試想想,為什麼要有愛情呢?

.

學生時代幾乎每天也會在學校,見到一大班同學,當中大概有一小班,是你的朋友,所以不太孤獨。但是,工作時代時,每天也只會見到幾位同事,而又因為工作場合有利害關係,大家也必須有機心,所以不易識到真心朋友。

然後,無論是在學生時代,或工作時代,相識的朋友,通常,在他們在單身時,就會有時間與你相約聚會。只要他們有另一半,甚至再有子女時,他們可以見你的時間,就少了很多。即使他們有動機、有時間,都未必有心神去見你。平時工作和家事太忙,有時間的話,正常人也會、也應該走去睡覺,安全第一,避免過勞猝死。

所以,自己如果沒有另一半的話,基本上幾乎沒有可能,有人陪伴你左右。

可能你會話,朋友當中,總有些和自己一樣是單身。那樣,只要和他或他們,相約時常在一起,甚至一起住,分擔租金。那不就解決了問題嗎?

我會反問,那和尋找另一半,有可分別?

或者說,那會容易過尋找老公/老婆嗎?

所以,「愛情」這個概念,在某些意思之下,其實是搵笨的。搵笨者,騙人也。

愛情可能不需要愛情。 

配偶,可能只是一個有責任要陪你的朋友。

— Me@2010.01.28

— Me@2021-02-27 08:40:29 AM

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2021.02.27 Saturday (c) All rights reserved by ACHK

Spinister

~ 1988 – 1990

The Transformers of this size were sold 37 HKD each at the Watsons near my home. That was much more expensive than an ordinary size one, which costed only around 25 HKD.

One time, mom brought me to ToyRUs. I found that the Transformers there were much cheaper. This one costed only about 29 HKD.

I did not mind buying a Decepticon.

A much more beautiful one was released last year.

— Me@2021-02-25 12:38:38 PM

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2021.02.25 Thursday (c) All rights reserved by ACHK

Problem 2.6c

A First Course in String Theory

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2.5 Constructing \displaystyle{T^2/\mathbb{Z}_3} orbifold

(c) Determine the three fixed points of the \displaystyle{\mathbb{Z}_3} action on the torus. Show that the orbifold \displaystyle{T^2/\mathbb{Z}_3} is topologically a two-dimensional sphere, naturally presented as a triangular pillowcase with seamed edges and corners at the fixed points.

~~~

[guess]

To find the fixed points, we consider the cases when

\displaystyle{z + m + n e^{i \pi/3} = e^{2 \pi i/3} z},

where \displaystyle{m,n \in \mathbb{Z}}.

\displaystyle{(e^{2 \pi i/3} - 1) z = m + n e^{i \pi/3}}

\displaystyle{z = \frac{m + n e^{i \pi/3}}{e^{2 \pi i/3} - 1}}

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When \displaystyle{m, n = 0},

\displaystyle{z = 0}

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When \displaystyle{m = 0; n = 1},

\displaystyle{z = \frac{e^{i \pi/3}}{e^{2 \pi i/3} - 1} = \frac{-i}{\sqrt{3}}}

When \displaystyle{m = 1; n = 0},

\displaystyle{z = \frac{1}{e^{2 \pi i/3} - 1} = \frac{1}{\sqrt{3}} e^{- 5 i \pi/6}}

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When \displaystyle{m = -2; n = 1},

\displaystyle{z = \frac{-2 + 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = 1}

When \displaystyle{m = -1; n = -1},

\displaystyle{z = \frac{-1 - 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = e^{i \pi/3}}

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In the fundamental domain, the 3 fixed points are:

\displaystyle{z = 0}

when \displaystyle{z = R(z)};

\displaystyle{z = 1}

when \displaystyle{T_2 \circ T_1^{-1} \circ T_1^{-1} (z) = R(z)};

\displaystyle{z = e^{i \pi/3}}

when \displaystyle{T_2^{-1} \circ T_1^{-1} (z) = R(z)}.

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Duplicate the fundamental triangle to create a fundamental parallelogram.

If we label the some edges as B instead of A, the fundamental parallelogram will have a sphere topology \displaystyle{ ABB^{-1}A^{-1} }.

However, it is not exactly the same as a sphere topology, because a sphere topology would not have the A=B identification.

[guess]

— Me@2021-02-23 03:44:57 PM

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2021.02.24 Wednesday (c) All rights reserved by ACHK

Particle indistinguishability is the major source of quantum effects, 1.2

However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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This question is exactly what the Bell tests designed for.

No. It is not correct. A Bell test can check whether the trajectory concept is well-defined, but not whether “the trajectory concept is broken” is the major source of quantum randomness.

However, it is the undefinable trajectory concept that makes the superposition, which is a unique and major feature of quantum mechanics.

— Me@2021-02-07 06:03:53 PM

— Me@2021-02-15 10:24:17 PM

— Me@2021-02-21 05:14:55 PM

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To date, all Bell tests have found that the hypothesis of local hidden variables is inconsistent with the way that physical systems behave.

— Wikipedia on Bell test

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Source of quantumness

~ the indistinguishability of cases

~ the individual trajectory of individual particles cannot be well-defined

~ the indistinguishability of particles

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~ “individual” particle has no individuality

~ “individual” particle has no individual identity

— Me@2021-02-06 4:03 PM

— Me@2021-02-15 9:14 PM

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2021.02.21 Sunday (c) All rights reserved by ACHK

Logan

Logan is a 2017 American superhero film starring Hugh Jackman as the titular character.

— Wikipedia on Logan (film)

— Me@2017-06-12 09:24:16 PM

— Me@2021-02-21 12:43:45 PM

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The movies X-Men (2000) and X2 (2003) are prerequisites.

— Me@2021-02-21 12:43:45 PM

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2021.02.21 Sunday ACHK

大種子論, 2.4

機遇創生論, 2.4

這段改編自 2010 年 4 月 18 日的對話。

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這個大統一理論的成員,包括(但不止於):

精簡圖:

種子論
反白論
間書原理
完備知識論

 心靈作業系統

 多重自我
 虛擬機器

自由決定論

它們可以大統一的成因,在於它們除了各個自成一國外,還可以合體理解和應用。

.

一部電腦,安裝兩個作業系統。需要用系統 A 時,就載入系統 A;需要用系統 B 時,就重新開機成系統 B。

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這就兼容於「多重自我」理論,因為在我們以前的討論中,其實假設了,每個人的腦,在很少情況下,需要同時啓用兩個自我。

如果要兩者也常用的話,你電腦就要時常切換系統,十分費時,因為,每次也要重新開機。不要那樣麻煩的話,你可以考慮使用「虛擬機器」程式。

例如,在你的蘋果作業系統上,先安裝一個「虛擬機器」程式。然後,在那一個「虛擬機器」程式之中,安裝一個(例如)視窗系統。那樣,你的一部電腦,就可以同時運行,蘋果系統和視窗系統。

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「心靈作業系統」和「多重自我」的關係是:一個人腦對應於一部電腦,而「多重自我」中的每個「人格」,就是一個「心靈作業系統」,對應於該電腦的其中一個作業系統。

正如一部電腦可以安裝,超過一個「作業系統」,一個人腦可以安裝超過一個「心靈」。「心靈」者,「自我人格」也。

— Me@2021-02-19 07:25:39 PM

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2021.02.20 Saturday (c) All rights reserved by ACHK

Ex 1.18 Bead on a triaxial surface

Structure and Interpretation of Classical Mechanics

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A bead of mass m moves without friction on a triaxial ellipsoidal surface. In rectangular coordinates the surface satisfies

\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}

for some constants a, b, and c. Identify suitable generalized coordinates, formulate a Lagrangian, and find Lagrange’s equations.

~~~

[guess]

The generalized coordinates:

{\displaystyle {\begin{aligned}x&=a\sin(\theta )\cos(\varphi),\\y&=b\sin(\theta )\sin(\varphi),\\z&=c\cos(\theta),\end{aligned}}\,\!}

where

{\displaystyle 0\leq \theta \leq \pi ,\qquad 0\leq \varphi <2\pi .}

.

{\displaystyle {\begin{aligned}  \dot x &= a \dot \theta \cos(\theta) \cos(\varphi) - a \dot \varphi\sin(\theta ) \sin(\varphi) \\  \dot y &= b \dot \theta \cos(\theta) \sin(\varphi) + b \dot \varphi \sin(\theta) \cos(\varphi) \\  \dot z &= - c \dot \theta \sin(\theta)   \end{aligned}}}


(define theta (literal-function 'theta))

(define phi (literal-function 'phi))

; 

(define (x t) (* 'a (sin (theta t)) (cos (phi t))))

((D x) 't)

(show-expression ((D x) 't))

;

(define (y t) (* 'b (sin (theta t)) (sin (phi t))))

(show-expression (y 't))

((D y) 't)

(show-expression ((D y) 't))

[guess]

— Me@2021-02-16 07:20:25 AM

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2021.02.18 Thursday (c) All rights reserved by ACHK

Particle indistinguishability is the major source of quantum effects, 1.1

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If particles are distinguishable, there is no quantum-ness.

Why?

— Me@2021-02-06 4:00 PM

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If there is no particle indistinguishability, all trajectories are distinguishable, then there is no case indistinguishability.

In other words, if every trajectory is well-defined, there is no indistinguishability of cases, even when no detector is installed.

Why?

— Me@2021-02-06 4:01 PM

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In other words, how to define “every trajectory is well-defined” when no detector is installed?

— Me@2021-02-15 5:00 PM

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Thought experiment:

In the double-slit experiment, turn on the detector. Then observe the pattern on the final screen.

Next, tune down the detector’s accuracy/resolution a little bit. Repeat the experiment. Observing the pattern again.

Keep repeating the experiment with a little bit lower detector accuracy/resolution at each iteration.

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If a non-classical pattern never appears on the final screen, we can say that each trajectory is well-defined.

In other words, if the trajectory concept can predict correct experiment results, we say that the trajectory concept is well-defined.

— Me@2021-02-06 4:02 PM

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It is because the final screen itself is a kind of detector, although not a position detector.

— Me@2021-02-06 05:07:21 PM

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So it is a kind of Bell-type experiment.

— Me@2021-02-07 06:03:53 PM

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However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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2021.02.15 Monday (c) All rights reserved by ACHK

Publish! 11

Why do we like to share insights?

So that those insights can have a future.

— Me@2011.08.08

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So that those ideas can continue to exist.

So that those ideas can be upgradable.

— Me@2021-02-15 09:00:18 AM

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2021.02.15 Monday (c) All rights reserved by ACHK

如此仙子 1.2

鐵達尼極限 2.4 | 尋覓 1.7 | 已婚單身 1.3 | Singles 1.3

這段改編自 2010 年 10 月 14 日的對話。

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人,其實就是處於,動物和靈性之間的物體。身體是動物,心靈如神明。

所以,最理想的愛情是,相遇如動物,相處如精靈。

雖然相互吸引,是因為大自然基因驅動,但是相遇以後,開始由動物傾向,慢慢轉化為,神明傾向。那就變成為精神上的愛情,向不受大自然的設局所規限。例如,雖然年青時相遇是,因為對方美貌吸引,但愛情於年紀大時,亦不會隨外表的衰落而減退。

如果知道這些知識,你人生可以少了很多,標準指定災難。例如,一個典型劇情是,假設你今天沒有事,沒有分手,但到了你們 22 歲時,你倆其中一方,要去外國升學。

那樣,你倆需要分手嗎?

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那樣,指定的劇情,就是會分手。

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不分手不可以嗎?

不分手又有什麼後果呢?

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後果就是,感情很難維持。

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上次我講過,人生是一個解決問題的過程;感情是一個共同解決問題的經歷。

換句話說,例如,他會過外國四年,而你就留在香港四年的話,你倆再沒有共同解決問題(或者解決共同問題)的機會,而這個狀態會維持四年。那樣,在那四年間,感情會逐漸銷聲匿跡,通常。這是第一點。

第二點,剛才講過:「

除了相遇那年要夾(融洽)外,還要在之後的每一年,也是那樣夾。換句話說,即是兩人的變法要配合。所以,難度其實深了一層。

如果一個在香港,一個在(例如)英國四年的話,很難會雙方都變到,二人四年後仍然會夾。換句話說,二人也各自變成了不同的人了。你,不再是四年前的你;而他,亦不是四年前的他了。

更加重要的是,第三點,如果為了對方而放棄理想,不去升學的話,十年後會埋怨對方。剛才所講,地球人的愛情感覺,是暫時的。而理想人生目標,卻是永久的。十年後當再沒有熱戀的感覺時,每當你見到對方,都可能會回想起,當年可能的未來。

If you begin by sacrificing yourself to those you love, you will end by hating those to whom you have sacrificed yourself.

— George Bernard Shaw

我這樣說,是不是很灰呢?(是不是十分令人灰心呢?)

你覺得很灰的話,是因為你還沒有我提供,最先進的心靈科技。放心,那正正是我一路會講落去的東西。

其實,喺我來講,以上的擔憂,都不緊要,都不是問題,因為,「愛情」是在生時候的事情,但我思考的,是超過在生時候的東西。

試想想,為什麼要有愛情呢?

— Me@2021-01-31 05:29:02 PM

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2021.02.14 Sunday (c) All rights reserved by ACHK

Problem 2.6

A First Course in String Theory

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2.5 Constructing \displaystyle{T^2/\mathbb{Z}_3} orbifold

(a) A fundamental domain, with its boundary, is the parallelogram with corners at \displaystyle{z = 0, 1} and \displaystyle{e^{i \pi/3}}. Where is the fourth corner? Make a sketch and indicate the identifications on the boundary. The resulting space is an oblique torus.

(b) Consider now an additional \displaystyle{\mathbb{Z}_3} identification

\displaystyle{z \sim R(z) = e^{2 \pi i/3} z}

To understand how this identification acts on the oblique torus, draw the short diagonal that divides the torus into two equilateral triangles. Describe carefully the \displaystyle{{Z}_3} action on each of the two triangles (recall that the action of \displaystyle{R} can be followed by arbitrary action with \displaystyle{T_1}, \displaystyle{T_2}, and their inverses).


[guess]

(a)

\displaystyle{z = 1 + e^{\frac{i \pi}{3}}}

(b)

[guess]

— Me@2021-02-11 06:03:36 PM

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2021.02.12 Friday (c) All rights reserved by ACHK

Quantum information makes classical information consistent, 1.2

Consistent histories, 10.2 | Cosmic computer, 2.2

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Quantum mechanics is a theory of classical information.

— Me@2021-02-03 07:48:01 AM

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Quantum mechanics is a theory of measurement results.

Quantum mechanics explains why measurement results are always consistent with each other.

— Me@2021-02-11 11:10:17 AM

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2021.02.11 Thursday (c) All rights reserved by ACHK

大種子論

機遇創生論, 2.3

這段改編自 2010 年 4 月 18 日的對話。

.

這個大統一理論的成員,包括(但不止於):

精簡圖:

種子論
反白論
間書原理
完備知識論

 心靈作業系統

多重自我
虛擬機器

自由決定論

它們可以大統一的成因,在於它們除了各個自成一國外,還可以合體理解和應用。

.

(安:我們原本的「心靈作業系統」,似乎不是講這些東西。

我們以前討論的是,人既有的心理結構,而不是研究「如何編寫,一個良好的『心靈作業系統』」。如果根據前者,即是「原著」的話,「大種子論」並不是,「心靈作業系統」的一部分。

反而,「心靈作業系統」應該是「大種子論」一部分。了解普遍人性,才有機會駕馭人生。)

.

(安:你那個「多重自我」理論,和「心靈作業系統」理論,又可不可以合到體呢?)

合到體,因為那就即是「多重開機」。

例如,你電腦雖然通常用視窗系統(簡稱系統 A),但是,有時需要使用,另一套作業系統(簡稱系統 B)。 最簡單的方法是,你買兩部電腦,一部裝 A,另一部裝 B。

雖然電腦不貴,但是,一般人的家裡,不會有那麼多地方,負擔得起擺放兩部電腦。所以,通常會用同一部電腦,安裝兩個作業系統。需要用系統 A 時,就載入系統 A;需要用系統 B 時,就重新開機成系統 B。

.

(安:但那不能同時運行兩個系統。)

無錯。

但是,兩個系統需要同時運行的機會不大。通常是你電腦有一個主要用的系統,而另一套系統則,只需要間中使用。例如,工作時系統 A,遊戲時系統 B。

曾經有一段時期的蘋果手提電腦,可以同時安裝視窗系統。所以,有些人會日間上班時,把蘋果手提電腦,開機成運行視窗系統;而晚間回到家,則會把其開機成,運行蘋果作業系統。

這就兼容於「多重自我」理論,因為在我們以前的討論中,其實假設了,每個人的腦,在很少情況下,需要同時啓用兩個自我。

如果要兩者也常用的話,你電腦就要時常切換系統,十分費時,因為,每次也要重新開機。不要那樣麻煩的話,你可以考慮使用「虛擬機器」程式。

例如,在你的蘋果作業系統上,先安裝一個「虛擬機器」程式。然後,在那一個「虛擬機器」程式之中,安裝一個(例如)視窗系統。那樣,你的一部電腦,就可以同時運行,蘋果系統和視窗系統。

— Me@2021-02-08 12:34:31 AM

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我估計那齣戲所講的是「多重人格障礙」(multiple personality disorder)。那是一種病患:同一個身體的多重人格之間,不能互相溝通。

而我所講的,是「多重人格」。那不是一種病患,那是正常的。每一個人也會有多重人格,而不同人格之間可以互相溝通。互相溝通的速度高到眾多人格,長年累月以來,一直以為他們是同一個「人」。

— Me@2011.02.24

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… 後來改稱「解離性身分疾患」(dissociative identity disorder,DID) …

— 維基百科

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2021.02.10 Wednesday (c) All rights reserved by ACHK

Ex 1.17 Bead on a helical wire

Structure and Interpretation of Classical Mechanics

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A bead of mass m is constrained to move on a frictionless helical wire. The helix is oriented so that its axis is horizontal. The diameter of the helix is d and its pitch (turns per unit length) is h. The system is in a uniform gravitational field with vertical acceleration g. Formulate a Lagrangian that describes the system and find the Lagrange equations of motion.

~~~

[guess]

The coordinates of the bead is

\displaystyle{\left( \frac{d}{2} \cos \theta, \frac{d}{2} \sin \theta, \frac{\theta}{2 \pi} h \right)},

where the \displaystyle{x}-direction is horizontal, the \displaystyle{y}-direction points upwards, and the \displaystyle{z}-direction is along the axis of the helix.

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Lagrangian \displaystyle{L = T - V}, where the kinetic energy, \displaystyle{T = \frac{1}{2} m \left( \dot x^2 + \dot y^2 + \dot z^2 \right)} and the potential energy, \displaystyle{V = m g y}.

\displaystyle{ (\dot x, \dot y, \dot z) = \left( \frac{-d}{2} (\sin \theta) \dot \theta, \frac{d}{2} (\cos \theta) \dot \theta, \frac{h}{2 \pi} \dot \theta \right)}

\displaystyle{L = \frac{1}{2} m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \dot \theta^2 - \frac{d}{2} m g \sin \theta}

\displaystyle{\frac{\partial L}{\partial \dot \theta} = m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \dot \theta}
\displaystyle{\frac{\partial L}{\partial \theta} = - \frac{d}{2} m g \cos \theta}

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The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\   \frac{d}{dt} \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} &= 0 \\    m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \ddot{\theta} + \frac{d}{2} m g \cos \theta &= 0 \\   \end{aligned}}

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(define ((T-hl d h m g) local)
  (let ((t (time local))
        (thetadot (velocity local)))    
    (* 1/8 m (square thetadot) 'H)))

;; (+ (square d) 
;; (/ (square h) (square 'pi))))))

(show-expression
  ((T-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(define ((V-hl d h m g) local)
  (let ((t (time local))
        (theta (coordinate local)))
    (let ((y (* 1/2 d (sin theta))))
      (* m g y))))


(show-expression
  ((V-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(define L-hl (- T-hl V-hl))

(show-expression
  ((L-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(show-expression
 (((Lagrange-equations
    (L-hl 'd 'h 'm 'g))
   (literal-function 'theta))
  't))

[guess]

— Me@2021-02-05 04:23:02 PM

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2021.02.06 Saturday (c) All rights reserved by ACHK