Fallacy of infinity thinking, prequel

Why should you NOT murder one innocent person in order to save millions of people?

(Note: This question is NOT the same as “Could we give up one innocent person’s life in order to save millions of people’s?”)

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Could we murder one innocent person in order to save millions of people?

— Me@2017-06-20 01:04:56 PM

— Me@2021-01-19 06:15:54 PM

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NO.

If we can murder one innocent person in order to save all other people, then no one is safe after all, because anyone could be THAT innocent person, being sacrificed at any time.

(In the situation that we cannot save all the people at the same time, which person or which group has higher or lower priorities depends on context. There is no universal answer.)

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Instead, if we protect each person’s life, then all the people’s lives are protected.

— Me@2021-01-19 06:01:24 PM

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If you start with protecting each one, then every one person and thus the whole society will be protected.

If you start with protecting the whole society at all costs, then no one will be safe, because anyone could be that cost; any innocent person could be sacrificed at any time.

— Me@2021-01-20 6:48 AM

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2021.01.20 Wednesday (c) All rights reserved by ACHK

鐵達尼極限 2.2

尋覓 1.5

這段改編自 2010 年 10 月 14 日的對話。

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但是,如果有一個人,心中有一個太陽,自己會發熱發亮的話,他就不怕人情冷暖。亦即是話,如果你一個人時,生活已經十分精采的話,你就不需要愛情。

留意,「不需要」不代表「不應有」。一個人不需要愛情,而仍然選擇愛情的話,可以是因為技術上的問題,例如,剛巧對象與他相愛,而他又想有自己的兒女。

Enjoy everything, need nothing, …

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… especially for human relationships.

– Conversations with God

然後,可以再追問,為何有些人會,想有自己的兒女?

在這之前,你反而應該先追問,「愛情的感覺」從何而來?

我現在是講普通人,而不是講外星人。

大部分情況下,愛情的感覺,其實是大自然對人的戲弄。這就是傳說中的「愛情陷阱」。

某些原因,大自然想人類(或其他生物)繁殖。

(那某些原因,其實就是「基因」。「基因」只顧自己生命的延續,企圖不斷複製自己,即使犧牲「基因載體」的利益、幸福,甚至生命,也在所不惜。「基因載體」者,人或其他生物也。)

所以,大自然令到,年輕男子和年輕女子,有愛情的感覺,從而互相吸引。不幸的是,那愛情感覺,大自然不會提供一生,只會提供暫時。你今天的分手,就是其中一個例子,大自然中止愛情感覺。

更常的例子是,在你有了子女後,大自然覺得,你已經失去了利用價値。所以,祂不再需要,提供「愛情感覺」給你。你和當時另一半,再不會有愛情的感覺,暫時。

— Me@2021-01-03 04:18:55 PM

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2021.01.19 Tuesday (c) All rights reserved by ACHK

Problem 2.5b1

A First Course in String Theory

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2.5 Constructing simple orbifolds

~~~

The wikipedia page “Fundamental polygon”, specifically the subsection entitled “group generators”, has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are identified to the same point. In the picture you provided and which can be seen on that page, one opposite pair of vertices of the square is identified to one point on the sphere, the other opposite pair of vertices is identified to a different point on the sphere.

There is still a way to derive a presentation for the fundamental group from a fundamental polygon, but it is not the way described on the wikipedia page. In the sphere example of your question, you have to ignore one of the two letters \displaystyle{A}, \displaystyle{B}, keeping only the other letter. For example, ignoring \displaystyle{A} and keeping \displaystyle{B}, you get a presentation \displaystyle{ \langle B \mid B B^{-1} = 1 \rangle }, which is a presentation of the trivial group. The way you tell which to ignore and which to keep is by taking the quotient of the boundary of the polygon which is a graph with vertices and edges, choosing a maximal tree in that graph, ignoring all edge labels in the maximal tree, and keeping all edge labels not in the maximal tree.

On that wikipedia page, the Klein bottle and the torus examples are correct and you do not have to ignore any edge labels: all vertices are identified to a single point and the maximal tree is just a point. The sphere and the projective plane examples are incorrect: the four vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label. The example of a hexagon fundamental domain for the torus is also incorrect: the six vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label.

edited Jul 23 ’14 at 17:17

answered Jul 23 ’14 at 17:11

Lee Mosher

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yes, i thought that the fundamental polygon is this quotient space. – user159356 Jul 23 ’14 at 17:28

That’s backward: in your example, the sphere is the quotient space of the fundamental polygon, not the other way around. – Lee Mosher Jul 23 ’14 at 17:30

— Mathematics Stack Exchange

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2021.01.18 Monday ACHK

Summing over histories, 2

The square root of the probability, 5

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If there is more than one way to achieve the present state, present == sum over all possible pasts, with weightings.

— Me@2011.06.26

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This is false for a physical state. This is only true for wave functions, which are NOT probabilities.

Wave functions are used for calculating probabilities; but they are not themselves probabilities.

Wave functions are quantum states, but not physical states.

Wave functions are logical and mathematical, but not physical.

A physical state is something observable, something can be measured, at least in principle.

A physical state is something that exists in spacetime, a wave function is not.

— Me@2021-01-16 06:12:08 PM

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2021.01.17 Sunday (c) All rights reserved by ACHK

Zooming out, 2

In this morning’s dream, I could hold my lucid dream without feeling a headache. I felt comfortable as awake.

I was in a city and listening to an Inception-style music.

I tried to fold the city as in Inception so that I could have walked on a vertical land.

However, it did not work. The city got folded together, not as an L shape, [?]but as an L with a horizontal line head shape.[?]

Then I got myself waking up by flying-falling downwards.

— Me@2011.08.03

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2021.01.16 Saturday (c) All rights reserved by ACHK

機遇創生論, 2.1

這段改編自 2010 年 4 月 18 日的對話。

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不行。雖然尚算準確,但是不夠精采。

還有,「人生攻略理論」令人聯想到很多東西,而大部分也不是,我們那個合體大理論的內容。

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「緣份攻略」都不行,因為感覺有點怪。

(安:那就不如叫做「緣份理論」。)

「理論」很空泛。不應把「理論」,視為名字的一部分。

(安:不如叫做「超級種子理論」,或者「廣義種子理論」?)

種子論最終定版?

種子論傳 …

種子論 X 傳 …

字母 X 可代表,前中後左中右上下東南西北,例如,種子論前傳、種子論左傳 等等。原本的種子論,就為之正傳,種子正傳。即是好像 Microsoft DirectX 可以代表 Direct3D、DirectDraw、DirectMusic、DirectPlay、DirectSound 等等。.

但是「種子 X 傳」相當「拗口」,讀法不暢順。

(安:無錯,那相當「拗口」。)

大種子論?

(安:我剛才都是想到「大種子論」。)

但是,大的是理論,而不是種子。

種子大論?

(安:但「種子大論」比較奇怪。)

都是「大種子論」比較好一點。還有,遲一些有更大的理論時,可以叫做,巨大種子論、超巨大種子論、特超巨大種子論 等等。

— Me@2021-01-13 04:52:02 PM

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2021.01.15 Friday (c) All rights reserved by ACHK

Ex 1.14 Lagrange equations for L’, 2

Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for \displaystyle{L'} are satisfied if the Lagrange equations for \displaystyle{L} are satisfied.

~~~

\displaystyle{x = F(t, x')}

\displaystyle{x' = G(t, x)}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\   &= \frac{\partial}{\partial v} L(t, x, v) \\   &= \frac{\partial}{\partial v} L'(t, x', v') \\   &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\   &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\  \end{aligned}}

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\displaystyle{v' = \partial_0 G(t, x) + \partial_1 G(t,x) v}

\displaystyle{ \begin{aligned}   \frac{\partial v'}{\partial v} &= \partial_1 G(t,x) = \frac{\partial x'}{\partial x}  \\  \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\   &= \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x}  \\  \end{aligned}}

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The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

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\displaystyle{   \begin{aligned}   &\partial_1 L \circ \Gamma[q] \\  &= \partial_1 L' \circ \Gamma[q'] \\    &= \frac{\partial}{\partial x} L' (t, x', v') \\    &= \frac{\partial L'}{\partial x'}  \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \\    \end{aligned}}

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\displaystyle{ \begin{aligned}   D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\   D \left(  \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'}  \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\   D \left(  \frac{\partial L'}{\partial v'} \right) \frac{\partial x'}{\partial x}   + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right)   - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\   \end{aligned}}

\displaystyle{ \begin{aligned}   \left[ D \left(  \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x}  + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right)   - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\   \end{aligned}}

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\displaystyle{ \begin{aligned}   &D \left( \frac{\partial x'}{\partial x} \right) \\   &= \frac{d}{dt} \left( \frac{\partial x'}{\partial x} \right) \\   &=     \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial t}{\partial t}  + \frac{\partial}{\partial x'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial x'}{\partial t}  + \frac{\partial}{\partial v'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial v'}{\partial t} \\   &=     \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right)   + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial x'} \right) \frac{\partial x'}{\partial t}  + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial v'} \right) \frac{\partial v'}{\partial t} \\   &=     \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right)   + \frac{\partial}{\partial x} \left( 1 \right) \frac{\partial x'}{\partial t}  + \frac{\partial}{\partial x} \left( 0 \right) \frac{\partial v'}{\partial t} \\   &=     \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial t} \right)   + 0  + 0 \\   &=     \frac{\partial v'}{\partial x} \\   \end{aligned}}

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\displaystyle{ \begin{aligned}   \left[ D \left(  \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x}   + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x}     - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\   \left[ D \left(  \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} &= 0 \\   D \left(  \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} &= 0 \\   \end{aligned}}

— Me@2021-01-06 08:10:46 PM

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2021.01.09 Saturday (c) All rights reserved by ACHK

The square root of the probability, 4.3

Eigenstates 3.4.3

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The indistinguishability of cases is where the quantum probability comes from.

— Me@2020-12-25 06:21:48 PM

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In the double slit experiment, there are 4 cases:

1. only the left slit is open

2. only the right slit is open

3. both slits are open and a measuring device is installed somewhere in the experiment setup so that we can know which slit each photon passes through

4. both slits are open but no measuring device is installed; so for each photon, we have no which-way information

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For simplicity, we rephrase the case-3 and case-4:

1. only the left slit is open

2. only the right slit is open

3. both slits are open, with which-way information

4. both slits are open, without which-way information

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Case-3 can be regarded as a classical overlapping of case-1 and case-2, because if you check the result of case-3, you will find that it is just an overlapping of result case-1 and result case-2.

However, case-4 cannot be regarded as a classical overlapping of case-1 and case-2. Instead, case-4 is a quantum superposition. A quantum superposition canNOT be regarded as a classical overlapping of possibilities/probabilities/worlds/universes.

Experimentally, no classical overlapping can explain the interference pattern, especially the destruction interference part. An addition of two non-zero probability values can never result in a zero.

Logically, case-4 is a quantum superposition of go-left and go-right. Case-4 is neither AND nor OR of the case-1 and case-2.

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We can discuss AND or OR only when there are really 2 distinguishable cases. Since there are not any kinds of measuring devices (for getting which-way information) installed anywhere in the case-4, go-left and go-right are actually indistinguishable cases. In other words, by defining case-4 as a no-measuring-device case, we have indirectly defined that go-left and go-right are actually indistinguishable cases, even in principle.

Note that saying “they are actually indistinguishable cases, even in principle” is equivalent to saying that “they are logically indistinguishable cases” or “they are logically the same case“. So discussing whether a photon has gone left or gone right is meaningless.

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If 2 cases are actually indistinguishable even in principle, then in a sense, there is actually only 1 case, the case of “both slits are open but without measuring device installed anywhere” (case-4). Mathematically, this case is expressed as the quantum superposition of go-left and go-right.

Since it is only 1 case, it is meaningless to discuss AND or OR. It is neither “go-left AND go-right” nor “go-left OR go-right“, because the phrases “go-left” and “go-right” are themselves meaningless in this case.

— Me@2020-12-19 10:38 AM

— Me@2020-12-26 11:02 AM

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It is a quantum superposition of go-left and go-right.

Quantum superposition is NOT an overlapping of worlds.

Quantum superposition is neither AND nor OR.

— Me@2020-12-26 09:07:22 AM

When the final states are distinguishable you add probabilities:

\displaystyle{P_{dis} = P_1 + P_2 = \psi_1^*\psi_1 + \psi_2^*\psi_2}

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When the final state are indistinguishable,[^2] you add amplitudes:

\displaystyle{\Psi_{1,2} = \psi_1 + \psi_2}

and

\displaystyle{P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2}

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[^2]: This is not precise, the states need to be “coherent”, but you don’t want to hear about that today.

edited Mar 21 ’13 at 17:04
answered Mar 21 ’13 at 16:58

dmckee

— Physics Stack Exchange

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\displaystyle{ P_{ind} = P_1 + P_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2 }

\displaystyle{ P_{\text{indistinguishable}} = P_{\text{distinguishable}} + \text{interference terms} }

— Me@2020-12-26 09:07:46 AM

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interference terms ~ indistinguishability effect

— Me@2020-12-26 01:22:36 PM

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2021.01.05 Tuesday (c) All rights reserved by ACHK

鐵達尼極限 2

尋覓 1.4

這段改編自 2010 年 10 月 14 日的對話。

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對於沒有自己人生目標的人來說,

愛情有如患精神病,過不了「鐵達尼極限」(三日)。

— 黃子華

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然後,又再深一層。

如果你要愛情事業成功,首先要令到自己,不需要愛情。

比喻說,如果你吃雪糕的原因是,你上了癮的話,吃雪糕就只能令你避免,暫時的不安,但不可以令你得到,真正的快樂。相反,如果你根本沒需要吃雪糕,而選擇吃雪糕,純粹是出於自由意志的話,你就可以真正享受得到,吃雪糕的樂趣。

那如何訓練到,自己不需要愛情呢?

剛才所講,

各人的心靈腦海之中,其實都有超過一個自己。只不過,眾多自我被困於,同一個頭顱之中。

如果你自己的眾多自我,也相處融洽時,你就不太需要外在愛情。相反,如果他們水火不容時,你就極有需要有愛情,用來逃避自己心中,那些眾多不友善的自我。

自己都不愛 怎麼相愛 怎麼可給愛人好處 – 林夕

人為什麼要有同伴?

比喻說,人自己一個時,太過孤獨,太過冰冷,所以,期望透過有同伴,去感受溫暖。

根據哲學家叔本華所講,人有如「寒冬裡的刺蝟」:一方面,牠們要走近對方,互相取暖;另一方面,走得太近,又會刺傷對方。

但是,如果有一個人,心中有一個太陽,自己會發熱發亮的話,他就不怕人情冷暖。亦即是話,如果你一個人時,生活已經十分精采的話,你就不需要愛情。

留意,「不需要」不代表「不應有」。一個人不需要愛情,而仍然選擇愛情的話,可以是因為技術上的問題,例如,剛巧有對象與他相愛,而他又想有自己的兒女。

Enjoy everything, need nothing, …

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… especially for human relationships.

– Conversations with God

— Me@2021-01-03 04:18:55 PM

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2021.01.04 Monday (c) All rights reserved by ACHK

Problem 2.5a

A First Course in String Theory

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2.5 Constructing simple orbifolds

(a) Consider a circle \displaystyle{S^1}, presented as the real line with the identification \displaystyle{x \sim x + 2}. Choose \displaystyle{-1 < x \le 1} as the fundamental domain. The circle is the space \displaystyle{-1 < x \le 1} with points \displaystyle{x = \pm 1} identified. The orbifold \displaystyle{S^1/\mathbb{Z}_2} is defined by imposing the (so-called) \displaystyle{\mathbb{Z}_2} identification \displaystyle{x \sim -x}. Describe the action of this identification on the circle. Show that there are two points on the circle that are left fixed by the \displaystyle{\mathbb{Z}_2} action. Find a fundamental domain for the two identifications. Describe the orbifold \displaystyle{S^1/\mathbb{Z}_2} in simple terms.

~~~

Put point \displaystyle{x=0} and point \displaystyle{x=1} on the positions that they can form a horizontal diameter.

Then the action is a reflection of the lower semi-circle through the horizontal diameter to the upper semi-circle.

Point \displaystyle{x=0} and point \displaystyle{x=1} are the two fixed points.

A possible fundamental domain is \displaystyle{0 \le x \le 1}.

If a variable point \displaystyle{x} moves from 0 to 1 and then keeps going, that point will actually go back and forth between 0 and 1.

— Me@2020-12-31 04:43:07 PM

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2021.01.01 Friday (c) All rights reserved by ACHK

The square root of the probability, 4.2

Eigenstates 3.4.2

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The difference between quantum and classical is due to the indistinguishability of cases.

— Me@2020-12-26 01:25:03 PM

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Statistical effects of indistinguishability

The indistinguishability of particles has a profound effect on their statistical properties.

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The differences between the statistical behavior of fermions, bosons, and distinguishable particles can be illustrated using a system of two particles. The particles are designated A and B. Each particle can exist in two possible states, labelled \displaystyle{ |0 \rangle } and \displaystyle{|1\rangle}, which have the same energy.

The composite system can evolve in time, interacting with a noisy environment. Because the \displaystyle{|0\rangle} and \displaystyle{|1\rangle} states are energetically equivalent, neither state is favored, so this process has the effect of randomizing the states. (This is discussed in the article on quantum entanglement.) After some time, the composite system will have an equal probability of occupying each of the states available to it. The particle states are then measured.

If A and B are distinguishable particles, then the composite system has four distinct states: \displaystyle{|0\rangle |0\rangle}, \displaystyle{|1\rangle |1\rangle} , \displaystyle{ |0\rangle |1\rangle}, and \displaystyle{|1\rangle |0\rangle }. The probability of obtaining two particles in the \displaystyle{|0\rangle} state is 0.25; the probability of obtaining two particles in the \displaystyle{|1\rangle} state is 0.25; and the probability of obtaining one particle in the \displaystyle{|0\rangle} state and the other in the \displaystyle{|1\rangle} state is 0.5.

If A and B are identical bosons, then the composite system has only three distinct states: \displaystyle{|0\rangle |0\rangle}, \displaystyle{ |1\rangle |1\rangle }, and \displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle +|1\rangle |0\rangle)}. When the experiment is performed, the probability of obtaining two particles in the \displaystyle{|0\rangle} is now 0.33; the probability of obtaining two particles in the \displaystyle{|1\rangle} state is 0.33; and the probability of obtaining one particle in the \displaystyle{|0\rangle} state and the other in the \displaystyle{|1\rangle} state is 0.33. Note that the probability of finding particles in the same state is relatively larger than in the distinguishable case. This demonstrates the tendency of bosons to “clump.”

If A and B are identical fermions, there is only one state available to the composite system: the totally antisymmetric state \displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle -|1\rangle |0\rangle)}. When the experiment is performed, one particle is always in the \displaystyle{|0\rangle} state and the other is in the \displaystyle{|1\rangle} state.

The results are summarized in Table 1:

— Wikipedia on Identical particles

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2020.12.30 Wednesday (c) All rights reserved by ACHK

Cheer up?

What you encourage people, don’t say “don’t worry, be happy” or “cheer up”.

They are nonsense, because they are not actionable.

What I really need to know is HOW to be happy.

What I really need to know is what ACTION to take to solve my problem.

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When you see a doctor, what if, without giving any other advice, he just tells you to “cheer up”?

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在開解朋友時,千萬不要講「加油」和「睇開啲啦」等廢話。

那只會令當事人,更加不開心。

— Me@2011.06.26

— Me@2020-12-30

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2020.12.30 Wednesday (c) All rights reserved by ACHK

機遇創生論 1.7.2

因果律 2.2

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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

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。

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「自由意志」問題方面,如果要討論的話,要先釐清「人有沒有自由意志」的意思,因為,它有超過一個常用的詮釋:

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1. 思:

人可不可以控制到,自己的「思想意志」?

如果可以的話,

2. 因:

人(的自由思想,)可不可以控制到,自己身體的行動?

如果可以的話,

3. 果:

人(的自由行動,)可不可以控制到,自己人生(或者世界歷史)的發展大方向?

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詳細一點的版本是:

1. 思:

意志有沒有自由?

究竟人的思想是有自由?還是其實,人的思想受制於教育等外在因素,所以都是不由自主的呢?

那就有如機械人的思想般,其實只是根據程式碼來運行。

2. 因:

即使假設意志有自由,人的身體是物體,它的運行仍然,受著物理力學定律所主宰,理應沒有自由。

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在「Laplace 因果律」的觀點下,兩者其實又沒有分別,因為「意志」都是,受物理定律的支配。「思想」是腦部的狀態,而腦部的生物化學作用,最底層也其實是,由物理定律控制。

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如果「Laplace 因果律」正確的話,人或者任何其他東西,也沒有自由。

至於「Laplace 因果律」正確與否,和人有自由與否,在我們以前的討論中,已得出了結論:

「自由論」和「決定論」,其實沒有實質上的分別。

所以,我把這個理論,稱為「自由決定論」。

如果對「因果律」和「自由意志」話題有興趣,想知道這結論的理據的話,請在本網誌搜尋「因果律」。

— Me@2020-12-29 05:46:05 PM

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2020.12.29 Tuesday (c) All rights reserved by ACHK

1994

This was my Art result. I had a great art teacher Mr Lo in that year. He taught us a lot of design concepts.

— Me@2020-12-29 10:52:04 AM

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2020.12.29 Tuesday (c) All rights reserved by ACHK

Ex 1.14 Lagrange equations for L’

Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for \displaystyle{L'} are satisfied if the Lagrange equations for \displaystyle{L} are satisfied.

~~~

Equation (1.69):

\displaystyle{C \circ \Gamma[q'] = \Gamma[q]}

Equation (1.70):

\displaystyle{L' = L \circ C}

Equation (1.71):

\displaystyle{L' \circ \Gamma[q'] = L \circ C \circ \Gamma[q'] = L \circ \Gamma[q]}

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned}   &\partial_2 L \circ \Gamma[q] \\  &= \frac{\partial}{\partial v} L(t, x, v) \\   &= \frac{\partial}{\partial v} L'(t, x', v') \\   &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\   \end{aligned}}

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Since it is just a coordinate transformation \displaystyle{x = F(t, x')}, \displaystyle{x} has no explicitly dependent on \displaystyle{v'}. Similarly, if we consider the coordinate transformation \displaystyle{x' = G(t, x)}, \displaystyle{x'} has no explicitly dependent on \displaystyle{v}. So

\displaystyle{ \begin{aligned}   \frac{\partial x'}{\partial v} &= 0 \\   \end{aligned}}

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\displaystyle{ \begin{aligned}   &\partial_2 L \circ \Gamma[q] \\  &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\   \end{aligned}}

— Me@2020-12-28 04:03:24 PM

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2020.12.28 Monday (c) All rights reserved by ACHK

The square root of the probability, 4

Eigenstates 3.4

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quantum ~ classical with the indistinguishability of cases

— Me@2020-12-23 06:19:00 PM

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In statistical mechanics, a semi-classical derivation of the entropy that does not take into account the indistinguishability of particles, yields an expression for the entropy which is not extensive (is not proportional to the amount of substance in question). This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs who proposed this thought experiment in 1874‒1875. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the “mixing paradox”. If one takes the perspective that the definition of entropy must be changed so as to ignore particle permutation, the paradox is averted.

— Wikipedia on Gibbs paradox

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2020.12.27 Sunday (c) All rights reserved by ACHK