# The Sixth Sense, 3

Mirror selves, 2 | Anatta 3.2 | 無我 3.2

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You cannot feel your own existence or non-existence. You can feel the existence or non-existence of (such as) your hair, your hands, etc.

But you cannot feel the existence or non-existence of _you_.

— Me@2018-03-17 5:12 PM

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Only OTHER people or beings can feel your existence or non-existence.

— Me@2018-04-30 11:29:08 AM

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# Story, 7

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story ~ a storage device for pieces of experience

— Me@2017-09-25 11:37:10 PM

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# 機遇再生論 1.10

「機遇再生論」有兩大（潛）假設：

1. 宇宙，有無限長的未來。

（這對應於撲克比喻中，「可以洗牌無限次」的假設。）

2. 宇宙中的粒子數目有限；而它們的組合及排列數目，都是有限的。

（這對應於撲克比喻中，「只有 52 隻牌」和「只有有限個排列」($52! \approx 8.07 \times 10^{67}$)的假設。）

「機遇再生論」的第二個假設，同第一個假設一樣，都是疑點重重。

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「狹義相對論」加「量子力學」，等於「量子場論」。如果「量子場論」是正確的，真空中不斷有粒子生滅。

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（這裡，「經典」的意思，並不是「歷史悠久」，而是「非量子」。「經典物理」即是「不是建基於量子力學架構的，物理定律」，例如牛頓力學。）

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— Me@2015.04.08

— Me@2017-12-09

— Me@2018-04-28

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# Importance, 3

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People never understand what you do for them …

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until you stop doing it!

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2018.04.26 Thursday ACHK

# Problem 14.4a4

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

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 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$ ${a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)^2$

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Can we create a formula for the number of states?

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$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2) + (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$
$= ...$
$= (D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$
$= 104976$
$= 324^2$
$= \left[ \frac{(D-2)(D-1)}{2} + (D-2) \right]^2$

The result is the same as the square of the coefficients of $x$ in Equation (14.63) on page 318.

 $\frac{1}{2} \alpha' M^2~|$ $N~|$ $~\bar N~$ $|~\text{Number of states}$ $-2~|$ $0~|$ $~0~$ $|~1$ $0~|$ $1~|$ $~1~$ $|~(D-2)^2$ $2~|$ $2~|$ $~2~$ $|~(D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$ $4~|$ $3~|$ $~3~$ $|~3200^2$ $8~|$ $4~|$ $~4~$ $|~25650^2$

— Me@2018-04-25 05:13:04 PM

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# Quantum decoherence 8

12. On the other hand, consistent histories are just a particular convenient framework to formulate physical questions in a certain way; the only completely invariant consequence of this formalism is the Copenhagen school’s postulate that physics can only calculate the probabilities, they follow the laws of quantum mechanics, and when decoherence is taken into account, to find both the quantum/classical boundary as well as the embedding of the classical limit within the full quantum theory, some questions about quantum systems follow the laws of classical probability theory (and may be legitimately asked) while others don’t (and can’t be asked)[.]

— Decoherence is a settled subject

— Lubos Motl

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2018.04.24 Tuesday ACHK

# Intuition

Unsourced variant:

The intellect has little to do on the road to discovery. There comes a leap in consciousness, call it intuition or what you will, and the solution comes to you and you do not know how or why. All great discoveries are made in this way.

The earliest published version of this variant appears to be The Human Side of Scientists by Ralph Edward Oesper (1975), p. 58, but no source is provided, and the similarity to the “Life Magazine” quote above suggests it’s likely a misquote.

In response to statement “You once told me that progress is made only by intuition, and not by the accumulation of knowledge.”

It’s not as simple as that. Knowledge is necessary, too. An intuitive child couldn’t accomplish anything without some knowledge.

There will come a point in everyone’s life, however, where only intuition can make the leap ahead, without ever knowing precisely how. One can never know why, but one must accept intuition as a fact.

— Albert Einstein

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2018.04.24 Tuesday ACHK

# Inception 16.4.2

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（CPK：試過臨知道答案前，就醒了。）

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— Me@2018-04-24 11:36:31 AM

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Some pundits have argued that the top was not in fact Cobb’s totem, rendering the discussion irrelevant. They say that the top was Mal’s totem; Cobb’s was his wedding ring, as he can be seen wearing it whenever he is in a dream and without it whenever he isn’t. As he hands his passport to the immigration officer, his hand is shown with no ring; thus he was conclusively in reality when seeing his children. Furthermore, the children were portrayed by different actors, indicating they had aged.

— Wikipedia on Inception

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# Problem 14.4a3

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

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But for non-massless states, this probably is not true anymore:

 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$ ${a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $(D-2)^2$

So the total number of states for $\frac{1}{2} \alpha' M^2 = 2$ ($N = \bar N = 2$) is

$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
$+ (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$

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Should $D$ be 10 or 26?

p.324 “Out of 26 left-moving bosonic coordinates of the bosonic factor only ten of them are matched by the right-moving bosonic coordinates of the superstring factor.”

$D$ should be 26 for bosonic strings. So the total number of states is

 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $2~|$ $~104976$

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What does the difference of this part and Section 14.6 come from?

This part is for bosonic closed string, while Section 14.6 is for bosonic open string. There is no $\bar N$ to consider in Section 14.6.

p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space if and only if it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$.

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Can we create a formula for the number of states?

— Me@2018-04-23 11:31:16 AM

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# Is logic empirical?

Each of the logics is analytic.

Which logic is the best for describing the world” is synthetic.

— Me@2012-04-14 11:32:36 AM

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# Almost nothing new

The most valuable insights are both general and surprising. F = ma for example. But general and surprising is a hard combination to achieve. That territory tends to be picked clean, precisely because those insights are so valuable.

Because these start out so general, you only need a small delta of novelty to produce a useful insight.

A small delta of novelty is all you’ll be able to get most of the time. Which means if you take this route your ideas will seem a lot like ones that already exist. Sometimes you’ll find you’ve merely rediscovered an idea that did already exist. But don’t be discouraged. Remember the huge multiplier that kicks in when you do manage to think of something even a little new.

It’s not true that there’s nothing new under the sun. There are some domains where there’s almost nothing new. But there’s a big difference between nothing and almost nothing, when it’s multiplied by the area under the sun.

— General and Surprising

— Paul Graham

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2018.04.22 Sunday ACHK

# 機遇再生論 1.9.2

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（問：那和宇宙壽命有無限，沒有直接關係。）

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$10^{10^{50}}$ 大概是，宇宙現時年齡的$10^{10^{50} - 10}$ 倍。

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$10^{36} \ll 10^{10^{50}}$

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（問：「宇宙」這個詞語的定義是「一切」。我們現時以為的「宇宙」，未必是真正的「宇宙」，因為，我們已知的「一切」，並非必定是，真正的「一切」。真正的「宇宙」，真正的「一切」，應連未知的部分，也包括在內。

「機遇再生論」有兩大（潛）假設：

1. 宇宙，有無限長的未來。

（這對應於撲克比喻中，「可以洗牌無限次」的假設。）

2. 宇宙中的粒子數目有限；而它們的組合及排列數目，都是有限的。

（這對應於撲克比喻中，「只有 52 隻牌」和「只有有限個排列」($52! \approx 8.07 \times 10^{67}$)的假設。）

「機遇再生論」的第二個假設，同第一個假設一樣，都是疑點重重。

— Me@2018-04-22 02:48:21 PM

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# Inspiring words

If you need inspiring words [for you to do something],

don’t do it.

— Elon Musk

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2018.04.20 Friday ACHK

# Quick Calculation 14.6b

A First Course in String Theory

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Construct explicitly all the states with $\alpha' M^2=2$ and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

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Let $N(n, k) = {n + k - 1 \choose k - 1}$, the number of ways to put n indistinguishable balls into k boxes.

p.318 “For open bosonic strings $\alpha' M^2 = N^\perp - 1$, …”

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When $\alpha' M^2 = 2, N^\perp = 3$, the cases are:

1. three $a_1^\dagger$‘s:

$N(3,24) = 2600$

2. one $a_2^\dagger$ and one $a_1^\dagger$:

$24 \times 24 = 576$

3. one $a_3^\dagger$:

24

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Total number of possible states for $N^\perp = 3$ is 3200.

— Me@2018-04-20 02:45:35 PM

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# Life as a recursion, 5

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recursion ~ act as if

self fulfilling ~ act as if

— Me@2011

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# Inception 16.4

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（CPK：在自己的夢裡面，是不是真的可以，想怎樣就怎樣？）

《潛行凶間》中的意念，你可以假想，有七成是真的。

（CPK：試過臨知道答案前，就醒了。）

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— Me@2018-04-18 02:48:05 PM

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# Sharp

It’s easy to look sharp when you haven’t done any work.

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2018.04.17 Tuesday ACHK

# Problem 14.4a2

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

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p.288 Equation (13.48):

$M^2 = \frac{2}{\alpha'} \left( N^\perp + \bar N^\perp - 2 \right)$

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p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space _if and only if_ it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$

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 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $-2~|$ $~1$ $0~|$ $~(D-2)^2$

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 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $2~|$ $...$

${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$

Assuming the signs of the wave functions do not matter:

 $|$ $|~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger}~|$ $\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$

p.291 How come the number of the components of the matrix represents the number of states?

p.292 For massless states, we have only one ${a_1^{I}}^\dagger$ and $\bar a_1^{J\dagger}$, where $a$ and $\bar a$ cannot interchange. So $I$ and $J$ also cannot interchange. In other words, in Equation (13.69), there is no double count. All the states are independent.

But for non-massless states, this probably is not true anymore:

 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$

— Me@2018-04-17 04:47:31 PM

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