— Me@2015.05.30

# Euler problem 24

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 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

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

# Bell’s theorem, 6

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

Information is the decreaser of uncertainties.

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

— Me@2011.08.01

# Speaking 6.2

.

（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。）

（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」。）

— Me@2015.05.16

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# Quantum Gravity

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

gnarbarian 146 days ago

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

— Hacker News

2015.05.06 Wednesday ACHK

— Me@2015.05.03

# Quick Calculation 14.2

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.

— Me@2015-05-02 01:10:27 PM

# Intellectual, 4

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

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— Me@2015.05.02

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