Self-information

Posted on December 31, 2015 by rpflee

The information entropy of a random event is the expected value of its self-information.

In information theory, self-information or surprisal is a measure of the information content [clarification needed] associated with an event in a probability space or with the value of a discrete random variable.

By definition, the amount of self-information contained in a probabilistic event depends only on the probability of that event: the smaller its probability, the larger the self-information associated with receiving the information that the event indeed occurred.

As a quick illustration, the information content associated with an outcome of 4 heads (or any specific outcome) in 4 consecutive tosses of a coin would be 4 bits (probability 1/16), and the information content associated with getting a result other than the one specified would be 0.09 bits (probability 15/16).

— Wikipedia on Self-information

2015.12.31 Thursday ACHK

Reality 4

Posted on December 30, 2015 by rpflee

“Real” has meanings other more than “lasting“.

For example, “pain is real” means “pain is objective“, instead of “pain is lasting“.

real

~ objective

lasting

~ independent of time (to a certain extent)

real

~ independent of most of the things

~ constant with respect to most of the things

— Me@2015-12-21 12:34 AM

注定外傳 2.2

Posted on December 29, 2015 by rpflee

Can it be Otherwise? 2.2

「

如果沒有明確指出，那個『必然』，是相對於哪個『觀測準確度』（觀察者解像度）而言的話，問一件事是不是『必然』，是沒有意思的，因為，無論那一件事，是在過去還是未來，往往既可以解釋成『必然』，又可以解釋為『非必然』。

」

對於未來之事，究竟注定與否，並不會指引到你，如何做決定。

例如，試想想，你下一次數學考試，成績是否注定，會怎樣影響你，現在的行動呢？

甲：如果並未注定，我就仍然有機會，透過努力來提升成績。那樣，我自然會選擇去溫習。如果已經注定，我溫不溫習，根本不會影響到成績。那樣，我自然會乾脆不溫習，節省時間。

乙：不可以是，注定你會溫習，從而成績大進嗎？

甲：都可以。但是，我不想溫習。

乙：那就即是話，你溫不溫習，是你的決定；跟成績是否注定，沒有關係。

「成績注定」和「主動溫習」，根本沒有矛盾。

如果你決定溫習，你可以說，那是因為你有自由，選擇溫習。亦可以說，那是因為命中注定，你會選擇溫習。

如果你決定不溫習，你可以說，那是因為成績如何，是命中注定的，溫習來也沒有影響。亦可以話，那是因為成績如何，不是必然的；即使我不溫習，也不代表成績一定差。

一方面，無論你的決定是哪一個，你總可以把，你決定的原因，講成「因為我覺得事情是注定的」；亦可以把，你決定的原因，說成「因為我覺得，我還有自由度，改進到事情的結果」，或者「因為我覺得，事情的結果，不是必然的」。

另一方面，如果從外評論你的決定，總可以把你說成有自由，亦可以把你說成沒有自由。

如果你覺得，一切皆為注定，我可以說，因為那是事實，所以你注定有這個想法；亦可以話，你有自由意志，去相信「一切皆為注定」。

如果你覺得，你有自由意志，我可以說，因為那是事實，所以你自然有這個想法；亦可以話，你的命中注定，會相信「我有自由意志」。

— Me@2015-12-29 03:12:39 PM

Ramond sector zero modes

Posted on December 12, 2015 by rpflee

Problem 14.3b4

A First Course in String Theory

What are $\xi_1, \xi_2, \xi_3, \xi_4$ in Equation (14.44)?

p.315 “Ramond fermions are more complicated than NS fermions because the eight fermionic zero mode $d_0^I$ must be treated with care. It turns out that these eight operators can be organized by simple linear combinations into four creation operators and four annihilation operators. Let us call the four creation operators …”

Since there are 8 possible transverse directions, there are 8 possible $d_0^I$ ‘s, where $I = 2,3, ..., 9$ .

What is the meaning of “… organized by simple linear combinations into four creation operators …”?

— Me@2015.11.01 03:53 AM

The $d_0^I$ operators are similar to but different from other $d_r^I$ operators.

$d_0^I$ ‘s and $d_r^I$ ‘s are similar in the sense that they all follow Equation (14.43):

$\{ d_m^I, d_n^J \} = \delta_{m+n, 0} \delta^{IJ}$

p.315 “Again, the negatively moded oscillators $d_{-1}^I, d_{-2}^I, d_{-3}^I, ...$ , are creation operators, while the positively moded ones $d_{1}^I, d_{2}^I, d_{3}^I, ...$ are annihilation operators.”

$d_0^I$ ‘s and $d_r^I$ ‘s are different in the sense that $d_0^I$ ‘s are neither creation nor annihilation operators.

(Based on the ideas from “Introduction to String Theory, A.N. Schellekens” and “A First Course in String Theory (Second Edition)” p.315:)

If we define $d_0 | 0 \rangle = 0$ ,

$\{ d_0^I, d_0^J \} | 0 \rangle$
$= \left( d_0^I d_0^J + d_0^I d_0^J \right) | 0 \rangle$
$= 0$

which does not match the requirement of

$\{ d_0^I, d_0^J \} = \delta^{IJ}$

So the definition $d_0 | 0 \rangle = 0$ does not work.

— Me@2015.11.12 11:30 AM

Instead, they are “organized by simple linear combinations into four creation operators”.

(Based on the idea from “Boundary Conformal Field Theory and the Worldsheet Approach to D-Branes, by Andreas Recknagel，Volker Schomerus” and “A First Course in String Theory (Second Edition)” p.315:)

Let

$c_0^i = d_0^{i+1}$
$e_i = \frac{1}{\sqrt{2}} \left( c_0^{2i} - i c_0^{2i - 1} \right)$
$e_i^\dagger = \frac{1}{\sqrt{2}} \left( c_0^{2i} + i c_0^{2i - 1} \right)$ .

Then

$\left\{ e_i, e_j^\dagger \right\}$
$= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$
$= \frac{1}{2} \delta^{ij} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2i} + i c_0^{2i - 1} \right) \right\}$

By p.315 Equation (14.43):

$\{ d_0^I, d_0^J \} = \delta^{IJ}$

In other words,

$\{ c_0^{I-1}, c_0^{J-1} \} = \delta^{I-1,J-1}$
$\{ c_0^{I}, c_0^{J} \} = \delta^{IJ}$

$\left\{ e_i, e_j^\dagger \right\}$
$= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$
$= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} - \left\{ i c_0^{2i - 1}, i c_0^{2i - 1} \right\} \right]$
$= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} + \left\{ c_0^{2i - 1}, c_0^{2i - 1} \right\} \right]$
$= \frac{1}{2} \delta^{ij} \left[1 + 1 \right]$
$= \delta^{ij}$