自由決定

Posted on May 30, 2015 by rpflee

這段改編自 2010 年 4 月 18 日的對話。

地球表面，視為「一塊土地」的話，就是正正「有限而無邊」的，因為它是一個球體的表面。

我懷疑，「自由」和「注定」，雖然表面上，像「有限」和「無邊」般，貌似有衝突，但是在它們的內心深處，其實互相兼容。

至於怎樣兼容法，我暫時不知道。

具體怎樣運作，並不是那麼容易想像得到。不過至少，「自由」和「注定」兼容，有不少的證據。

甚至，再有一個可能是，它們極端地兼容 — 其中一方直情建基於，另外一方。

前人提出過，如果沒有「注定」（物理等自然定律），人或其他生命體，就根本不可能有「自由意志」。

例如，你想拿起一隻茶杯。因為你有自由意志，所以可以由腦部下指令，訊號由神經線傳達到手部，拿起茶杯。手部正正是因為是「注定」的，即是受制於自然定律，才保證必會執行，腦部的指令。

試想想，如果手部未必根據自然定律來行事，它就不一定會執行，你心中的目標。那樣，你（腦部）反而就沒有自由意志，因為手部的動作根本是隨機的，不一定會把你（例如「拿起茶杯」）的意志，化成現實。

如果所有東西也是注定的，你就沒有自由。如果所有東西也是隨機（不注定）的，你也沒有自由。

— Me@2015.05.30

Euler problem 24

Posted on May 27, 2015 by rpflee

Haskell

------------------------------


deleteNth n xs | n>0 = take (n-1) xs ++ drop n xs
removeItem x xs = filter (/= x) xs
sub_lexico x [] = [[]]
sub_lexico x xs = map (x:) (lexico (removeItem x xs))
merge_sub_lexico [] = []
merge_sub_lexico (x:xs) = x ++ (merge_sub_lexico xs)
lexico [] = [[]]
lexico xs = (merge_sub_lexico (map (\x -> (sub_lexico x xs)) xs))
p24 = (lexico [0,1,2,3,4,5,6,7,8,9]) !! (1000000 - 1)

------------------------------

— Me@2015-05-27 12:09:33 PM

Quick Calculation 14.3

Posted on May 26, 2015 by rpflee

A First Course in String Theory

To test the $N^\perp$ includes the fermionic contribution to the number, show that the eigenvalue of $N^\perp$ on $b_{-r_1}^I b_{-r_2}^J | NS \rangle$ , with $r_1, r_2 > 0$ , is $r_1 + r_2$ .

~~~

$N^\perp b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \left[ \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^I b_r^I \right) \right] b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^{I'} \alpha_p^{I'} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} \right)$ $|n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} [ r_1 b_{-r_1}^{I'} b_{r_1}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I'} b_{r_2}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle]$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I} b_{r_2}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$+ r_1 b_{-r_1}^{J} b_{r_1}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + 0 + 0 + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

— Me@2015-05-25 12:52:23 AM

2015.05.26 Tuesday (c) All rights reserved by ACHK

Bell’s theorem, 6

Posted on May 23, 2015 by rpflee

E. T. Jaynes pointed out two hidden assumptions in Bell Inequality that could limit its generality. According to him:

1. Bell interpreted conditional probability P(X|Y) as a causal inference, i.e. Y exerted a causal inference on X in reality. However, P(X|Y) actually only means logical inference (deduction). Causes cannot travel faster than light or backward in time, but deduction can.

— Wikipedia on Bell’s theorem

However, it should be interpreted as causal inference because before measurement, there is no Y, such a classical state.

— Me@2012-04-07 2:35:55 PM

Action | Uncertainty, 2

Posted on May 18, 2015 by rpflee

Information is the decreaser of uncertainties.

Actions are great devices to decrease uncertainties, for they gather feedback.

— Me@2011.08.01

Speaking 6.2

Posted on May 18, 2015 by rpflee

這段改編自 2010 年 8 月 11 日的對話。

如果你現在有空泛題目的例子，你可以叫我示範一下，如何透過它們，去展示自己的腦袋實力。

（CYM：「冰、火、人」。）

（TK：應該是「火、水、人」。）

（CYM：是嗎？）

（TK：那是「個人短講」部分的一道題目。）

那明顯是剛才所講，天馬行空的題目。你自由發揮就可以。

亦即是話，你在準備時，想像到「火、水、人」之中，任何兩者之間，有什麼關係，就即管（只管）收集起來。然後，把那堆關係排列妥當，形成一個上文下理。

（記住，任何兩樣東西（事或物），無論如何，你總有方法講到它們有關係，因為，至起碼，大家同樣都是「東西」。）

例如，你可以這樣開始：

「

火，代表熱情。

水，代表冷漠。

人，就代表人生。

人生，就正正處於冷漠和熱情之間。

」

然後，你就加以舉例。只要你舉例舉得動聽就可以。

最後，講出結論：

…

— Me@2015.05.16

Quantum Gravity

Posted on May 6, 2015 by rpflee

We sometimes say that string theory is the only consistent theory of quantum gravity. It’s the only game in town. This is an observation mostly based on various types of circumstantial evidence. Whenever you try something that deviates from string/M-theory, you run into inconsistencies. Sometimes you don’t run into inconsistencies but something else happens. Many good ideas that were thought to be “competitors” to string theory were shown to be just aspects of some (usually special) solutions to string theory (noncommutative geometry, CFT, matrix models, and even the Hořava-Lifshitz class of theories have been found to be parts of string theory), and so on. And decades of attempts to find a truly inequivalent competing theory have utterly failed. That’s not a complete proof of their absence, either, but it is evidence that shouldn’t be completely ignored.

But that doesn’t mean that the statement that every consistent theory of quantum gravity has to be nothing else than another approach to string/M-theory is just an expression of vague feelings, a guesswork, or a partial wishful thinking. We don’t have the “most complete proof” of this assertion yet – this fact may be partly blamed on the absence of the completely universal, most rigorous definition of both “quantum gravity” and “string theory”. But there exist partial proofs and this paper is an example.

— Every theory of quantum gravity is a part of string theory: a partial proof

— A successful test in $AdS_3$

— Lubos Motl

2015.05.06 Wednesday ACHK

魔間傳奇 2.5.2

Posted on May 6, 2015 by rpflee

gnarbarian 146 days ago

Nobody gets what they deserve. You only get what you negotiate.

— Hacker News

2015.05.06 Wednesday ACHK

玄悟慧能 1.4

Posted on May 5, 2015 by rpflee

這段改編自 2010 年 4 月 18 日的對話。

剛才我講「人生注定」問題時，猜想答案是「兩者之間」，但卻仍未知道，是「兩者之間」的哪一處。

或者，現實是以某種特別形式，存在於「完全自由」和「完全注定」之間 ——　特別到根本不是「之間」。其中一個可能是，「自由」和「注定」是同一樣「東西」的兩個方面。

而另一個可能是，它們是相容重疊的。比喻說，如果我宣稱，有一塊有限大的田地，是沒有邊緣的，你會覺得沒有可能。「有限大」的土地，又怎可能「沒有邊緣」呢？

原來是有可能的。地球表面，視為「一塊土地」的話，就是正正「有限而無邊」的，因為它是一個球體的表面。

我懷疑，「自由」和「注定」，雖然表面上，像「有限」和「無邊」般，貌似有衝突，但是在它們的內心深處，其實互相兼容。

至於怎樣兼容法，我暫時不知道。

— Me@2015.05.03

Quick Calculation 14.2

Posted on May 3, 2015 by rpflee

A First Course in String Theory

$\sum_{s_1, s_2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

Prove that “only the part of the wavefunction $\psi$ that is antisymmetric under the simultaneous exchange $p_1 \leftrightarrow p_2$ and $s_1 \leftrightarrow s_2$ contributes…”

~~~

We divide the proof into two parts. In the first part, we prove that being antisymmetric under $p_1 \leftrightarrow p_2$ is a must. Then, in the second part, we use the same method to prove that being antisymmetric under $s_1 \leftrightarrow s_2$ is also a must.

First, we separate $\psi$ into two parts:

$\psi (p_1, s_1; p_2, s_2) = \psi_{sp} + \psi_{ap}$ ,

where $\psi_{sp}$ is symmetric under $p_1 \leftrightarrow p_2$ and $\psi_{ap}$ is anti-symmetric.

Wavefunctions $\psi_{sp}$ and $\psi_{ap}$ can be defined as

$\psi_{sp} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) + \psi (p_2, s_1; p_1, s_2) \right]$
$\psi_{ap} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) - \psi (p_2, s_1; p_1, s_2) \right]$

Now, we consider the contribution of the symmetric part, $\psi_{sp} (p_1, s_1; p_2, s_2)$ .

$\int d \vec p \psi_{sp} (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_2, s_1; p_1, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

By interchanging the dummy variables $p_1$ and $p_2$ in the second term, we have

$...$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger | \Omega \rangle$
$=\frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) \left[ f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger + f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger \right] | \Omega \rangle$
$= 0$

So we have proved that the part symmetric under $p_1 \leftrightarrow p_2$ does not contribute.

Using the same method, we can also prove that the part symmetric under $s_1 \leftrightarrow s_2$ also does not contribute.

Intellectual, 4

Posted on May 3, 2015 by rpflee

When you are a student, you are treated as an 100 percent intellectual; when you are an adult, you are treated as an 100 percent non-intellectual. That’s not correct.

— Me@2011.08.01

Speaking 6

Posted on May 2, 2015 by rpflee

這段改編自 2010 年 8 月 11 日的對話。

至於口試，很大程度上，是要觀察那個考生，有沒有頭腦。其實，即使不是考試，而是日常生活，你只要花大概五分鐘，跟一個人對話，就可以大概感受到，他的智力或學養，暫時達到哪個水平。

換句話說，只要你智力不太低，你的口試表現，自然不會太差，所以不用太擔心。

當然，你可能會說，高考中，「個人短講」的題目，有時十分天馬行空，不知如何是好。

但是，正正是因為這個問題，我才會事先強調，在考試時，考官的主要目標，並不是要透過你的說話，去學到些什麼；而是要透過你的說話，去評價你的智力學養，從而給予分數。所以，面對空泛題目時，不單不要太過害怕，而且更反而要，深明其好處。