Ken Chan 時光機 1.2

多項選擇題 7.2 | Multiple Choices 7.2

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重點是,無論「勤力」還是「懶惰」,也只是手段,不是目的本身。

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二來,即使有時,「勤力」是正確的;那又應該如何「勤力」呢?

我的老師中,大概只有兩位,曾經講過一點讀書方法。其中一位,就正正是 Ken Chan。(另一位則是,我日校的「純數學」的老師(程兆海)。)

我記憶所及,Ken Chan 教過的溫習技巧,又不是真的很多,因為他並沒有,(例如)花一個課堂的時間去講。但是,就憑他有講的一點點,就已經令我,受用無窮。

Ken Chan 所提及,其中一個技巧是,在物理科,如果新學一個課題,就應該先做,大量該個課題的 MC(多項選擇題)。那樣,你就可以極速釐清,該個課題中的新概念。

註:必須為考試局所出的,以往公開試題目,而不是坊間出版社的練習。

— Me@2018-10-20 11:46:04 AM

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2018.10.20 Saturday (c) All rights reserved by ACHK

如果需要聯絡本人,可用右邊的電郵地址。

SLIME

SLIME, the Superior Lisp Interaction Mode for Emacs, is an Emacs mode for developing Common Lisp applications. SLIME originates in an Emacs mode called SLIM written by Eric Marsden. It is developed as an open-source public domain software project by Luke Gorrie and Helmut Eller. Over 100 Lisp developers have contributed code to SLIME since the project was started in 2003. SLIME uses a backend called Swank that is loaded into Common Lisp.

— Wikipedia on SLIME

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C-x o

Window-Move to other

C-x C-e

Evaluate last expression

C-c C-r

Evaluate region

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2018.10.19 Friday (c) ACHK

Problem 14.5a4

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

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

The GSO projection here keeps the states with \displaystyle{(-1)^{F_L} = + 1}; this defines the left NS’+ sector.

Write explicitly and count the states we keep for the three lowest mass levels, indicating the corresponding values of \displaystyle{\alpha' M_L^2}. [This is a long list.]

~~~

p.314 “Let us declare that number to be minus one, thus making the ground states fermionic:”

Equation (14.39):

\displaystyle{(-1)^F |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle = - |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle}

Equation (14.40):

\displaystyle{(-1)^F |\lambda \rangle = -(-1)^{\sum_{r,J} \rho_{r,J}} |\lambda \rangle}

p.315 “So all the states with integer \displaystyle{N^{\perp}} have \displaystyle{(-1)^F = -1}; they are fermionic states.”

However, in this problem:

“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}.

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

If we define N^\perp in the way similar to equation (14.37), we have

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

\displaystyle{\begin{aligned}  \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\  \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \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 \\  \end{aligned}}

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}  \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&1 \\  \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&8 + \frac{32 \times 31}{2} = 504 \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\left( \frac{8 \times 8}{2} + \frac{8}{2} \right) = 36, 8 \times \left( \frac{32 \times 31}{2} \right) = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\  \end{aligned}}

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&N(2,8) = 36, 8 \times {32 \choose 2} = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~& 36 + 3968 + 1024 + 35960 = 40988 \\  \end{aligned}}

— This answer is my guess. —

— Me@2018-10-14 03:25:08 PM

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2018.10.16 Tuesday (c) All rights reserved by ACHK

How far away is tomorrow?

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The cumulative part of spacetime is time.

It is the cumulative nature of time [for an macroscopic scale] that makes the time a minus in the spacetime interval formula?

\displaystyle{\Delta s^{2} = - (c \Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}}

— Me@2011.09.21

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Space cannot be cumulative, for two things at two different places at the same time cannot be labelled as “the same thing”.

— Me@2013-06-12 11:41 am

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There is probably no directly relationship between the minus sign and the cumulative nature of time.

Instead, the minus sign is related to fact that the larger the time distance between two events, the causally-closer they are.

— Me@2018-10-13 12:46 am

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Recommended reading:

d_2018_10_13__20_54_50_PM_

— Distance and Special Relativity: How far away is tomorrow?

— minutephysics

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2018.10.13 Saturday (c) All rights reserved by ACHK

凌晨舊戲

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。

我當年閱讀,有關「瀕死經驗」的文章,是因為有極大的興趣。

而我當年閱讀,很多有關「瀕死經驗」的文章,則是因為那時是,我的研究生時代;在工作上,有著極大的不幸。閱讀「瀕死經驗」的文章,可以暫時抽離當時的生活,以作減壓。

那段時期,我大部晚上,也不太願睡覺;往往拖到凌晨三點鐘才睡。我十分不想,明日的來臨。我不願面對,明日的事務。

當然,那十分不健康,不宜鼓勵。

— Me@2018-10-12 05:48:23 AM

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2018.10.12 Friday (c) All rights reserved by ACHK

Insomnia

如夢初醒 4.2 | 讀書與睡眠 4.2

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DrPhish 5 months ago [-]

This is exactly the Zen moment that fixed my lifelong insomnia: lying in bed relaxed with my eyes closed, even if I’m not asleep, leaves me feeling worlds better in the morning than fretting about not being able to sleep. It’s something that happened when I forgot about it completely. Once I realized I could literally just lie in bed for 8 hours and be OK, I was able to let go.

It’s like a trap that closes tighter about you the more you struggle, and releases if you allow yourself to relax[.]

Once I really got that, sleep became regular and uninterrupted[.]

laumars 5 months ago [-]

I discovered this recently too. Even in the instances when I still don’t get to sleep (eg my 1 year old coming into bed and wriggling in her sleep waking me) I’ve still felt more refreshed when I’ve relaxed rather than laid in bed wound up. I’m not into new age / whatever stuff but I like to think of it as night time meditation as while it’s still not as good as sleep it’s still a great deal better than full blown insomnia! And best of all, sometimes it clears your mind enough to actually sleep.

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— Thinking of yourself as an insomniac may be a part of the problem

— Hacker News

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2018.10.11 Thursday ACHK

Problem 14.5a4

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

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

~~~

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\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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— This answer is my guess. —

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\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{24} \left[ (D - 2) \right]_{I= 2, 3, ..., 9} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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\displaystyle{\alpha' M_L^2 = \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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Equation (14.34):

\displaystyle{\frac{1}{2} \sum_{r=- \frac{1}{2}, -\frac{3}{2}} r b_{-r}^I b_{r}^I = \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r b_{-r}^I b_{r}^I - \frac{1}{48} (D - 2)}

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\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \frac{1}{2} \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \left[ \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r \lambda_{-r}^A \lambda_{r}^A - \frac{1}{48} \left[D - 2\right]_A \right]}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A - \frac{32}{48}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-10 05:38:08 PM

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2018.10.10 Wednesday (c) All rights reserved by ACHK

Existence and Description

Bertrand Russell, “Existence and Description”

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§1 General Propositions and Existence

“Now when you come to ask what really is asserted in a general proposition, such as ‘All Greeks are men’ for instance, you find that what is asserted is the truth of all values of what I call a propositional function. A propositional function is simply any expression containing an undetermined constituent, or several undetermined constituents, and becoming a proposition as soon as the undetermined constituents are determined.” (24a)

“Much false philosophy has arisen out of confusing propositional functions and propositions.” (24b)

A propositional function can be necessary (when it is always true), possible (when it is sometimes true), and impossible (when it is never true).

“Propositions can only be true or false, but propositional functions have these three possibilities.” (24b)

“When you take any propositional function and assert of it that it is possible, that it is sometimes true, that gives you the fundamental meaning ‘existence’…. Existence is essentially a property of a propositional function. It means that the propositional function is true in at least one instance.” (25a)

— Brandon C. Look

— University Research Professor and Chair

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2018.10.07 Sunday ACHK

Simpson’s paradox

d_2018_10_06__21_40_12_PM_

Simpson’s paradox, or the Yule–Simpson effect, is a phenomenon in probability and statistics, in which a trend appears in several different groups of data but disappears or reverses when these groups are combined.

— Wikipedia on Simpson’s paradox

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d_2018_10_05__19_25_22_PM_

— MinutePhysics

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2018.10.06 Saturday ACHK

Ken Chan 時光機 1.1

多項選擇題 7 | Multiple Choices 7

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我在中學時代,很想知道讀書方法。但來,近乎從來沒有老師,教授求學攻略;彷彿那完全不重要。

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大部分老師,大概只會說:「勤力一點。」

那是答非所問。

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一來,勤力一定有用嗎?

如要只要勤力,就解決到問題的話,世界會簡單很多。實情是,那不會。

試想想,人的科技發展幅度,遠大於其他動物,並不是因為人勤力過其他動物。有時候,剛剛相反。正正是因為人想「偷懶」,才會發展各式各樣的科技,以逸代勞。

重點是,無論「勤力」還是「懶惰」,也只是手段,不是目的本身。

— Me@2018-10-03 03:59:04 PM

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2018.10.05 Friday (c) All rights reserved by ACHK

如果需要聯絡本人,可用右邊的電郵地址。

Length Contraction and Time Dilation

d_2018_09_26__20_58_04_PM_

Length Contraction and Time Dilation | Special Relativity Ch. 5

minutephysics

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I recommend this video.

Without it, I would have never realized that besides length contraction and time dilation, there are also distance dilation and time contraction.

— Me@2018-09-26 10:12:19 PM

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2018.09.26 Wednesday (c) All rights reserved by ACHK

Problem 14.5a3

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

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Paraphrasing the description of heterotic (closed) string theory:

  • right-moving part \sim an open superstring theory
    • NS sector: \alpha_{-r}^I, b_{-r}^I,~~~I = 2,3,...,9
    • R sector: \alpha_{-n}^I, d_{-n}^I,~~~I=2,3,...,9
    • “The standard GSO projection down to NS+ and R- applies.”

    .

  • left-moving part \sim a peculiar bosonic openstring theory
    • I = 2, 3, ..., 23: There are totally 24 transverse coordinates
      • 8 bosonic coordinates X^I with oscillators \bar \alpha_{-n}^I
      • 16 peculiar bosonic coordinates

        • can be replaced by 32 two-dimensional left-moving fermion fields, \lambda^A
        • \lambda^A (anti-commuting) fermion fields \to has NS' and R' sectors
    • .

    • NS': oscillators \bar \alpha_{-n}^I, \lambda_{-r}^A act on the vacuum |NS' \rangle

      given (-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L \to defines the left NS'+ sector

      \displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbb{Z} + \frac{1}{2}} r \lambda_{-r}^A \lambda_r^A}

    • R': oscillators \bar \alpha_{-n}^I, \lambda_{-n}^A act on R' ground states

      \displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \left( \bar \alpha_{-n}^I \bar \alpha_n^I + n \lambda_{-n}^A \lambda_n^A \right)}

    • .

— Me@2018-09-20 09:51:17 PM

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2018.09.21 Friday (c) All rights reserved by ACHK

神的旨意 2.4

魔:為什麼「全能」者不可「全惡」?

甲:你如果是「全能」,就毋須問我這個問題。

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如果你是「全惡」,你的構成部分,就不能相處。而「你」,作為一個整體,並不會存在;必須散落成一大堆,獨立的部分而存在。而各個分部,各自內裡必有善部,才可能凝聚,不再分裂。

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魔:即使我不是「全惡」;即使我有所謂「善部」,你難保我「惡部」的法力,大於「善部」?

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甲:

第一,即使假設是那樣,那也沒有大意義,因為,你總不能完全忽略,你善部的旨意。

如果你的惡部,完全不理善部地作惡,那就即是,你的善部名存實亡。沒有善部,「你」必會分裂。

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第二,除了你存在以外,我也存在;其他生命體也存在。

善的會合作,那是定義。善的會合作,去抵擋你的惡。

雖然,惡部有「破壞容易過建設」的優勢,但是,善部也可以應用「破壞容易過建設」,去破壞惡部的破壞計劃 。

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第三,惡人自有惡人磨:

相似的人,因為各種原因,傾向身處相近的地方,簡稱「物以類聚」。

壞人的身邊,通常是其他壞人。壞人最怕的,往往是其他壞人。

最終對付你的人,是你自己。最終對付你惡部的人,是你自己的惡部。

— Me@2018-09-02 03:05:45 PM

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2018.09.02 Sunday (c) All rights reserved by ACHK

Illusions destroyed, 2

Ask HN: Why is nearing completion so demotivating?

534 points by danschumann 3 months ago

So I’ve been working on animation software for over two years. Part of me is very excited for launch so I can have money again (I’ve been freelancing a minimum amount these last two years, and went car-less, moved, cut lifestyle into a third). I should be wholeheartedly excited, but I’m feeling tired and generally sluggish regarding the project. I still make consistent progress, but it takes a lot of will power.

Part of me thinks it might be an aversion to sales. Part of me thinks this could have been built up so much in my head that anything short of overnight millions would be a disappointment (though I would be happy with 1500 bucks a month), part of me thinks I might be scared of success (or scared of surpassing my parents) (media attention), part of me fears the attacks that might come with success (having something to lose), part of it is the un-fun-ness of mature projects where the focus is on polish and bugs rather than broad new features, and part of me is scared of commitment: if I succeed I have to stick with this (freedom value), part of me wonders what will happen when more people become involved, if I will be able to maintain my creative direction, since I’m scratching my own itch. Part of me wonders if diet and exercise isn’t a factor.

A combination, likely…

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mikekchar 3 months ago [-]

When your project is finished, the dream is dead and the reality is born. The death of a dream is like the death of a friend. It’s probably been with you for a long time — longer even than the length of the project. A dream is the manifestation of what’s possible. When it is over, the possible diminishes very quickly and you are left with what actually is. Will people respond well to your project — in the dream stage it is possible; everything is possible. In the reality stage, it will only be what it is.

So while it’s common to think of a release as a birth of something new, realise that you also have a significant loss. You will mourn that loss. Give yourself some emotional space to deal with the mourning.

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riantogo 3 months ago [-]

This is exactly it. Tens of my personal projects have died in this stage. It was always much easier to move on to the next dream. There is always the next big problem that could use a solution. Why not build when it is what we do best? Rinse, repeat.

I took a break from side projects for several years but recently got back to it and couple weeks back finished building. It is the same story all over again. Same feeling. I’m dreading what comes next.

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— Why is nearing completion so demotivating?

— Hacker News

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2018.09.01 Saturday ACHK

Problem 14.5a2

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

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

~~~

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\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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— This answer is my guess. —

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Equation at Problem 14.5:

\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}
\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}

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Equation (13.116):

\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}

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\displaystyle{\begin{aligned} &\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A \\  &= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right] \\ &= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\  &= - \frac{1}{2} \sum_{r = 1, 3, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{24} + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= \frac{-7}{48} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\  \end{aligned}}

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If we define N^\perp in the way similar to equation (14.37), we have

\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= \frac{-7}{48} + N^\perp \\  \end{aligned}}

— This answer is my guess. —

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— Me@2018-09-01 06:05:29 AM

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2018.09.01 Saturday (c) All rights reserved by ACHK

The square root of the probability

Probability amplitude in Layman’s Terms

What I understood is that probability amplitude is the square root of the probability … but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

edited Mar 1 at 16:31
nbro

asked Mar 21 ’13 at 15:36
Deepu

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Part of you problem is

“Probability amplitude is the square root of the probability […]”

The amplitude is a complex number whose amplitude is the probability. That is \psi^* \psi = P where the asterisk superscript means the complex conjugate.{}^{[1]} It may seem a little pedantic to make this distinction because so far the “complex phase” of the amplitudes has no effect on the observables at all: we could always rotate any given amplitude onto the positive real line and then “the square root” would be fine.

But we can’t guarantee to be able to rotate more than one amplitude that way at the same time.

More over, there are two ways to combine amplitudes to find probabilities for observation of combined events.

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When the final states are distinguishable you add probabilities:

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:

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

and

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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The terms that mix the amplitudes labeled 1 and 2 are the “interference terms”. The interference terms are why we can’t ignore the complex nature of the amplitudes and they cause many kinds of quantum weirdness.

{}^1 Here I’m using a notation reminiscent of a Schrödinger-like formulation, but that interpretation is not required. Just accept \psi as a complex number representing the amplitude for some observation.

{}^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

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