# Light cone, 2

Light is the interface of space and time.

— Me@2012-04-11 9:49:53 PM

# Inception 4.2

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What’s the most resilient parasite? An Idea. A single idea from the human mind can build cities. An idea can transform the world and rewrite all the rules. Which is why I have to steal it.

— Inception (film)

— Me@2010.07.30

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2011.01.20 Thursday ACHK

# 注定外傳 1.10

Can it be Otherwise? 1.10

（問：如果只是「類似」，當然可以有不同結果。你應該直接問：

』）

（問：不是呀。在量子力學中，即使有兩組百分百一樣的物理系統，即使它們獲得完全相同的輸入，都可能有不同的輸出。）

「相同」的意思，並不是指「沒有可能找到任何分別」。

「相同」的意思是「分別小到不易察覺」。

」，

— Me@2015-10-29 10:12:16 PM

# Exercise 6a (Corrected version)

Show that f' * unit = unit * f' = bind f'

——————————

f :: a -> bf' :: a -> m aunit :: a -> m a

lift f = unit . ff' = lift f

The lift function in this tutorial is not the same as the liftM in Haskell. So you should use lift (but not liftM) with bind.

— Me@2015-10-13 11:59:53 AM

(f' * g') xs
= ((bind f') . (bind g')) xs

bind f' xs = concat (map f' xs)
unit x = [x]

bind unit xs = concat (map unit xs)= concat (map unit [x1, x2, ...])= concat [unit x1, unit x2, ...]= concat [[x1], [x2], ...]= [x1, x2, ...]= xs

(f' * unit) (x:xs)= bind f' (bind unit (x:xs))= bind f' (concat (map unit (x:xs)))= bind f' (concat (map unit [x1, x2, ...]))= bind f' (concat [[x1], [x2], ...])= bind f' [x1, x2, ...]= concat (map f' [x1, x2, ...])= concat [f' x1, f' x2, ...]= concat [(unit . f) x1, (unit . f) x2, ...]= concat [(unit (f x1)), (unit (f x2)), ...]= concat [[f x1], [f x2], ...]= [f x1, f x2, ...]

(unit * f') (x:xs)= ((bind unit) . (bind f')) (x:xs)= bind unit (bind f' (x:xs))= bind unit (concat (map f' (x:xs)))= bind unit (concat (map f' [x1, x2, ...]))= bind unit (concat [f' x1, f' x2, ...])= bind unit (concat [(unit . f)  x1, (unit . f) x2, ...])= bind unit (concat [(unit (f x1)), (unit (f x2)), ...])= bind unit (concat [[f x1], [f x2], ...])= bind unit [f x1, f x2, ...]= concat (map unit [f x1, f x2, ...])= concat [[f x1], [f x2], ...]= [f x1, f x2, ...]

— Me@2015-10-15 07:19:18 AM

If we use the identity bind unit xs = xs, the proof will be much shorter.

(f' * unit) (x:xs)= ((bind f') . (bind unit)) (x:xs)= bind f' (bind unit (x:xs))= bind f' (x:xs)

(unit * f') (x:xs)= ((bind unit) . (bind f')) (x:xs)= bind unit (bind f' (x:xs))= bind f' (x:xs)

— Me@2015-10-15 11:45:44 AM

# Problem 14.3b3

A First Course in String Theory

14.3 Massive level in the open superstring.

~~~

What is a zero mode?

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”

How come there are creation operators that do not contribute to the mass-squared of the states?

Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.

If an index can be zero, the corresponding term can be zero.

— Me@2015.09.26 08:44 PM

Consider the NS-sector:

Equation (14.37):

$M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)$

$N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I$

For the NS sector, the $r$ values in $b^I_{r}$ are half-integers, thus cannot be zero. So every creation operator $b_{-r}^I$ contributes to the mass-squared.

(p.312 “… the negatively moded coefficients $b_{-1/2}^I$, $b_{-3/2}^I$, $b_{-5/2}^I$, …, are creation operators, …”)

Consider the R-sector:

Equation (14.53):

$M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)$

For the R sector, the $n$ values in $d^I_n$ are integers, thus can be zero. So some of the creation operators $d^I_{-n}$ are zero modes.

p.315 “… the eight fermionic zero modes $d_0^I$…”

— Me@2015-10-11 11:01:44 AM

# 同情地理解

「同情地理解」的意思是，從對方的情境中，去理解對方的言論。

— Me@2015-10-07 08:41:51 AM

# 注定外傳 1.9

Can it be Otherwise? 1.9

（這裡的「東西」，是指宏觀的物理系統。至於兩粒微觀粒子，則有可能「全同」。但那是另一個話題，容後再談。）

（問：那如果連位置都相同呢？）

— Me@2015-10-07 02:52:21 PM

# Exercise 6a

Show that f * unit = unit * f

——————————

(f * g) (x, xs)= ((bind f) . (bind g)) (x, xs)

bind f x = concat (map f x)

(f * unit) (x:xs)= bind f (bind unit (x:xs))= bind f (concat (map unit (x:xs)))= bind f (concat (map unit [x1, x2, x3, ...]))= bind f (concat ([[x1], [x2], [x3], ...]))= bind f [x1, x2, x3, ...]= concat (map f [x1, x2, x3, ...])= concat [f x1, f x2, f x3, ...]= [f x1, f x2, f x3, ...]

(unit * f) (x:xs)= ((bind unit) . (bind f)) (x:xs)= bind unit (bind f (x:xs))= bind unit (concat (map f (x:xs)))= bind unit (concat (map f [x1, x2, ...]))= bind unit (concat [f x1, f x2, ...])= bind unit [f x1, f x2, ...]= concat (map unit [f x1, f x2, ...])= concat [[f x1], [f x2], ...]= [f x1, f x2, ...]

— Me@2015.07.20 09:00 PM

# Problem 14.3b2

A First Course in String Theory

14.3 Massive level in the open superstring

~~~

How come R sector has a factor 16 while NS sector has not?

Equation (14.66):

$f_{NS}(x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n - \frac{1}{2}}}{1-x^n} \right)^8$

Equation (14.68):

$f_R(x) = 16 \prod_{n=1}^\infty \left( \frac{1+x^n}{1-x^n} \right)^8$

p.319 “The overall multiplicative factor appears because each combination of oscillators gives rise to 16 states by acting on each of the available ground states.”

p.319 “We note that the R coefficients are actually double the corresponding NS coefficients. This is not a coincidence, as we will see in the following section.”

p.320 “We have seen that the Ramond sector has world-sheet supersymmetry: there are equal numbers of fermionic and bosonic states at each mass level.”

With the factor 16, how come the R coefficients are only double, but not 16 times as big as the corresponding NS coefficients?

It is caused by the difference of $x^n$ and $x^{n-\frac{1}{2}}$.

— Me@2015.10.06 08:23 AM

# Person of Interest

Person of Interest is an American science fiction crime drama television series created by Jonathan Nolan that premiered on September 22, 2011, on CBS. It is executive produced by Nolan, alongside J. J. Abrams, Bryan Burk, and Greg Plageman. It stars Jim Caviezel as John Reese, a former CIA agent who is presumed dead. He is approached by a mysterious billionaire named Harold Finch (Michael Emerson), who is trying to prevent violent crimes before they happen by using an advanced surveillance system dubbed “The Machine”, which turns out to have evolved into a sentient AI.

— Wikipedia on Person of Interest (TV series)

2015.10.05 Monday ACHK

# Inception 4.1

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What’s the most resilient parasite? A bacteria? A virus? An intestinal worm?

An idea.

Resilient, highly contagious. Once an idea’s taken hold in the brain it’s almost impossible to eradicate. A person can cover it up, ignore it – but it stays there.

Information, yes. But an idea? Fully formed, understood? That sticks… (taps forehead) In there, somewhere.

— Inception (film)

— Me@2011.01.18

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2011.01.18 Tuesday ACHK

# 注定外傳 1.8

Can it be Otherwise? 1.8

（問：那如果是數數目（使用整體）的情況呢？

（問：那為什麼不可以說「絕對相同」？）

「絕對」，應該用作「相對」的相反。而「近似」的相反，則應該用「確切」。

（這裡的「東西」，是指宏觀的物理系統。至於兩粒微觀粒子，則有可能「全同」。但那是另一個話題，不宜在這裡詳述。）

— Me@2015-10-04 07:32:32 AM

# Exercise 3.2

Show that lift f * lift g = lift (f.g)

——————————

The meaning of f' * g' should be (bind f') . (bind g') instead.

f' * g' = (bind f') . (bind g')lift f = unit . ff' = lift f

(lift f * lift g) (x, xs)= (bind (lift f)) . (bind (lift g)) (x, xs)= bind (lift f) (bind (lift g) (x, xs))= bind (lift f) (gx, xs++gs)  where    (gx, gs) = (lift g) x

 = bind (lift f) (gx, xs++gs)  where    (gx, gs) = (g x, "") = bind (lift f) (g x, xs) = (fx, xs++fs)  where    (fx, fs) = (lift f) gx  = (fx, xs++fs)  where    (fx, fs) = (f gx, "") = (fx, xs)  where    (fx, fs) = (f (g x), "") 

= (f (g x), xs)

bind f' (gx,gs) = (fx, gs++fs)                  where                    (fx,fs) = f' gx

 bind (lift (f.g)) (x, xs)= (hx, xs++hs)  where    (hx, hs) = (lift (f.g)) x = (hx, xs++hs)  where    (hx, hs) = ((f.g) x, "") 

= ((f (g x)), xs)

— Me@2015.07.19 11:04 PM

# Problem 14.3b1

A First Course in String Theory

14.3 Massive level in the open superstring.

~~~

What is the difference between “open bosonic string” and “open superstring”?

p.320 “With our conventions, these are world-sheet fermionic states that are now recognized to be states of spacetime fermions. The resulting, truncated sector is called the R- sector.”

NS(+): Equation (14.38) $\left(\alpha' M^2 = 1, N^\perp = \frac{3}{2}\right)$:

$\left\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b^I_{-1/2} b^J_{-1/2} b^K_{-1/2} \right\} | NS \rangle \otimes | p^+, \vec p_R \rangle$

There are 128 states (cf. Quick Calculation 14.4).

R: Equation (14.54) $\left(\alpha' M^2 = 1\right)$:

$\alpha^I_{-1} | R_a \rangle, d_{-1}^I | R_{\bar a} \rangle~||~\alpha^I_{-1} | R_{\bar a} \rangle, d_{-1}^I | R_{a} \rangle$

Only the left hand side represents the R- states.

Each of index $I$ and index $a$ has 8 possible values. So there are totally 64 states in the form of $\alpha^I_{-1} | R_a \rangle$.

Similarly, there are totally 64 states in the form of $d_{-1}^I | R_{\bar a} \rangle$.

Totally, for $\alpha' M^2 = 1$, there are 128 R- states.

— Me@2015-10-02 02:16:23 PM

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–– Me@2010.03.06

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— Me@2010.06.01

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