# Bus Stop 5

I am the first in the queue because I have missed the last bus.

— Me@2009.09.13

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If you are the smartest person in the room, you are in the wrong room,

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for you should have gone to a better room.

— Me@2019-12-21 07:11:21 PM

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for you deserve a better room.

— Me@2020-01-25 04:47:13 PM

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# Ken Chan 時光機 3.3

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（問：你研究院碩士畢業時，才二十四歲。為什麼要在二十九歲時，才有所領悟？）

「二十九歲時才有所領悟」的意思是，我在那時才有一個，（我覺得）完整的讀書策略體系，適合大部分人使用。

（問：你又怎樣知道，你的讀書策略體系完整，適合大部分人使用？）

What is most personal is most general.

— Carl Rogers

（問：那些是什麼問題？）

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— Me@2020-01-20 10:26:20 PM

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# 1990

First photo

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— Me@2020-01-08 12:11:16 AM

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# Quick Calculation 13.2

A First Course in String Theory

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Verify that

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I + {X^I}' \right)}$,

$\displaystyle{\left[ L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I - {X^I}' \right)}$,

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Equation (13.24):

$\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

Equation (13.39):

$\displaystyle{{\dot{X}}^- + {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \bar \alpha_n^- e^{-in (\tau + \sigma)}}$

$\displaystyle{{\dot{X}}^- - {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^- e^{-in (\tau - \sigma)}}$

Equation (13.51):

$\displaystyle{\left[\bar L_m^\perp, \bar \alpha_n^J \right] = - n \bar{\alpha}_{m+n}^J}$

$\displaystyle{\left[L_m^\perp, \alpha_n^J \right] = - n \alpha_{m+n}^J}$

Equation (13.52):

$\displaystyle{\left[L_m^\perp, \bar \alpha_n^J \right] = 0}$

$\displaystyle{\left[\bar L_m^\perp, \alpha_n^J \right] = 0}$

Equation (13.53):

$\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m}$

$\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}$

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$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right]}$,

$\displaystyle{= \left[ \bar L_0^\perp, x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^I e^{i n \sigma} + \bar \alpha_n^I e^{-in \sigma}) \right]}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} e^{-in \sigma} (-n \bar \alpha_n^I)}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 - i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} e^{-in(\tau + \sigma)} ( \bar \alpha_n^I)}$

$\displaystyle{= - \frac{i}{2} \sqrt{2\alpha'} \sum_{n \in \mathbb Z} \bar \alpha_n^I e^{-in(\tau + \sigma)}}$

— Me@2020-01-06 11:30:38 PM

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# Compare results

Within a universe, any two observers can, at least in principle, compare results.

When they compare, they will have consistent results for any observables/measurables.

— Me@2018-02-06 08:48:24 PM

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being entangled ~ being consistent, with respect to any two observers, when they compare the results

— Me@2018-02-05 10:11:45 PM

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# 時間止血 1.6.2

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— Me@2013.06.25
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— Me@2019-12-24 11:15:13 AM

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