Quick Calculation 14.7

A First Course in String Theory

Count the number of graviton, Kalb-Ramond, and dilation states in ten dimensions. Add these numbers up and confirm you get 64.

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Equation (13.69):

$\sum_{I,J} \hat S_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

p.292 “… the states (13.69) represent one-particle graviton states … ”

Equation (13.64):

In the closed string state space, the general state of fixed momentum at the massless level is

$\sum_{I,J} R_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

Equation (13.68):

$R_{IJ} = \hat S_{IJ} + A_{IJ} + S' \delta_{IJ}$

$S' = \frac{S}{D-2}$

Equation (10.108):

For the symmetric traceless part, $\hat S_{IJ}$ (with the size $(D-2) \times (D-2)$), the number of independent components is

$n(D) = \frac{1}{2} (D - 2) (D - 1) - 1$

The $-1$ at the end reflects the fact that if the trace is zero, the last component of the trace is not independent anymore.

In ten dimensions ($D = 10$),

$n(D) = \frac{1}{2} (8) (9) -1 = 35$

p.292 For $A_{IJ}$ (a skew-symmetric matrix), the number of independent components is

$n(D) = \frac{1}{2} (D - 3) (D - 2) = 28$

p.293 “The oscillator part of (13.71) has no free indices ($I$ is summed over), so it represents one state. It corresponds to [an] one-particle state of a massless scalar field. This field is called the dilation.”

— Me@2015-06-29

Predestination

Predestination is a 2014 Australian science fiction film, written and directed by Michael and Peter Spierig.

— Wikipedia on Predestination (film)

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2015.06.25 Thursday ACHK

避免犯錯 2

2015.06.23 Tuesday ACHK

— Me@2015-06-21

Euler problem 27

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 primes = 2 : filter (null . tail . primeFactors) [3,5..] primeFactors n = factor n primes                 where                   factor n (p:ps)                       | p*p > n = [n]                       | n mod p == 0 = p : factor (n div p) (p:ps)                       | otherwise = factor n ps list_max (x:xs) = lmf x xs    where       lmf x []                    = x      lmf x xs | (x >= (head xs)) = lmf x (tail xs)               | otherwise        = lmf (head xs) (tail xs) elemInd y [] = -1elemInd y (x:xs) = ei 0 y (x:xs)                    where                       ei n y (x:xs)                           | (x == y)  = n                          | (null xs) = -1                          | otherwise = ei (n+1) y xs                                    b_list = takeWhile (<= 1000) primes isPrime x | x <= 1    = False          | otherwise = null $tail (primeFactors x) cPrimeLen [a, b] = [m, a, b] where m = length$ takeWhile (== True) $map (isPrime . (\n -> n^2 + a*n + b)) [0..] p27_list = [cPrimeLen [a, b] | a <- [-n..n], b <- b_list] where n = 1000 p27_n = map head p27_list p27 = p27_list !! (elemInd (list_max p27_n) p27_n) -- 71, a == -61, b == 971  ------------------------------ — Me@2015-06-19 10:01:22 PM 2015.06.20 Saturday (c) All rights reserved by ACHK Quick Calculation 14.4 A First Course in String Theory How many states are there at $N^\perp = \frac{3}{2}$? ~~~ This answer is my guess. There are $(D-2)$ transverse (i.e. non-lightcone) coordinates. When $D = 10$, the number of transverse coordinates is 8. So there are 8 states with $N^\perp = \frac{1}{2}$. For $N^\perp = \frac{3}{2}$, there are 3 indices $I$, $J$, and $K$. So there are totally 24 states. — Me@2015.05.26 03:56 PM The above answer is incorrect. Equation (14.38) $\left(\alpha'M^2 = 1,~~N^\perp = \frac{3}{2}\right)$: $\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$ There are $8 \times 8$ states for $\alpha_{-1}^I b_{-1/2}^J |NS \rangle \otimes |p^+, \vec p_T \rangle$ There are $8$ states for $b_{-3/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle$ There are $\frac{8 \times 7 \times 6}{3!}$ states for $b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K |NS \rangle \otimes |p^+, \vec p_T \rangle$. So the total number of states is 64 + 8 + 56 = 128. You can check this answer against Equation (14.67): $f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$ — Me@2015-08-21 08:15:50 AM 2015.06.13 Saturday (c) All rights reserved by ACHK EPR paradox, 5.2 With a spacelike separation, it makes no absolute sense to say which of Alice and Bob causes the wave function to collapse. — Me@2012.04.10 2015.06.13 Saturday (c) All rights reserved by ACHK 天空堤壩 Talent at its best and character at its worst . Lord Acton said we should judge talent at its best and character at its worst. For example, if you write one great book and ten bad ones, you still count as a great writer — or at least, a better writer than someone who wrote eleven that were merely good. Whereas if you’re a quiet, law-abiding citizen most of the time but occasionally cut someone up and bury them in your backyard, you’re a bad guy. — Paul Graham . 才能方面，如果一個人的最高點是可以接受的話，你就可以錄用他，即使他有其他任何才能缺點； 品德方面，如果一個人的最低點是不可接受的話，你就千萬不要錄用他，即使他有其他任何品德優點。 — Me@2010.03.06 . . . . . . . . . . . . . . . . . . 才能方面，一個太陽可以照亮整個天空； 品德方面，一條裂縫可以摧毀整個堤壩。 — Me@2010.03.06 . . . 2010.03.07 Sunday (c) All rights reserved by ACHK Speaking 6.3 這段改編自 2010 年 8 月 11 日的對話。 . 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。 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，令人覺得你是「有腦人士」就可以。 其實，即使你內心十分熱情，如果你外表太有熱情，反而會阻礙工作進展。例如，即使你十分重視考試成績，你在考試作答期間，也應保持冷靜。太過激動，反而會傷害成績。 所以有些情況下，你需要『內熱外冷』。 但是，亦有些情況下，你應該相反，要做到『內冷外熱』。例如，雖然你平時內心十分平靜，但是如果遇到一些情境，你發現如果不發脾氣，就不能立刻制止，某些人做壞事的話，你就應該發脾氣，即使你不喜歡，因為那是劇情的需要，兩害取其輕。 最後，講出結論： 所以，在人生遇到的眾多處境之中，有時應該『外熱內熱』；有時應該『外冷內熱』；有時應該『外熱內冷』；有時應該『外冷內冷』。 至於什麼時候需要怎麼樣，不能一概而論，要隨機應變。 你不能期望，可以在年輕時，就準確地掌握到，因為，那需要人生的磨練。 你發覺，我並沒有講什麼實質的東西。 （HYC：但那假扮到十分有文有路。） 那只是一些簡單的技巧。只要看多幾題例子，你都可以做到。 — Me@2015.06.10 . . 2015.06.10 Wednesday (c) All rights reserved by ACHK Euler problem 26 Haskell ——————————  rems n d xs | (r == 0) = xs ++ [0] | (elem r xs) = xs ++ [r] ++ [-1] | otherwise = (rems (10*r) d (xs ++ [r])) where r = n rem d remainders n d = rems n d [] recur_cyc (x:xs) | (elem 0 ys) = 0 | (not (elem (-1) ys)) = -2 | (not (elem x xs)) = recur_cyc xs | otherwise = (length ys) - 2 where ys = (x:xs) elemInd y [] = -1elemInd y (x:xs) = ei 0 y (x:xs) where ei n y (x:xs) | (x == y) = n | (null xs) = -1 | otherwise = ei (n+1) y xs list_max [x] = xlist_max (x:xs) = max x$ list_max xs recur_list = map (\x -> (recur_cyc (remainders 1 x))) [1..1000] p26 = 1 + (elemInd (list_max recur_list) recur_list) 

-- 983

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— Me@2015-06-05 03:25:44 PM

Quick Calculation 14.6

A First Course in String Theory

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Construct explicitly all the states with $\alpha' M^2 =2$ and count them, verifying that there are indeed a total of 3200 states.

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$\alpha' M^2 = 2~~~\to~~~N^\perp = 3$

The first possibility group includes the cases of one $a_3^\dagger$. Since there are 24 species, this group has 24 possibilities.

The second possibility group includes the cases of one $a_1^\dagger$ with one $a_2^\dagger$. (Their order is not important because interchanging the order would create the same state anyway.) Since there are 24 species for each of $a_1^\dagger$ and $a_2^\dagger$, this group has $24 \times 24$ possibilities.

The third possibility group includes the cases of three $a_1^\dagger$‘s. We need to further divide this group into 3 sub-groups.

For the first sub-group, three $a_1^\dagger$‘s are all different:

$a_{1A}^\dagger a_{1B}^\dagger a_{1C}^\dagger$

Since there are 24 species for each $a_1^\dagger$, there should be $24 \times 23 \times 22$ possibilities.

However, since their order is not important, there are repetitions among those possibilities. To eliminate repetitions, we divide the number with $3!$. So, this subgroup has

$\frac{24 \times 23 \times 22}{6}$

possibilities.

For the second sub-group, two of the three $a_1^\dagger$‘s are identical:

$a_{1A}^\dagger a_{1B}^\dagger a_{1B}^\dagger$

There should be $24 \times 24$ possibilities.Me@2018-04-20 12:18:38 PM

There should be $24 \times 23$ possibilities.

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For the third sub-group, all three are identical. In other words, there are 24 possibilities.

— Me@2015-06-04 10:19:30 PM

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Truth values in Quantum Mechanics

It’s pretty much the same mistake that all the “realists” are doing all the time. They are imagining that they can assign truth values to all questions that could be in principle answered by experiments. (In some way, they give answers “No” or “zero” to all the other questions that were actually not addressed by the measurement.) But quantum mechanics prohibits that. If one assigns classical truth values (or real values) to some operators, one can no longer assign truth values (or real values) to “complementary” (not mutually commuting) operators and questions they represent. Instead of the correct statement that “the value of $$N_a$$ isn’t determined if one measures $$L$$ instead”, they say that it is zero which is just wrong.

— No particle upon a quantum field means no information

— Lubos Motl

2015.06.03 Wednesday ACHK

注定外傳 1.1

（問：那如果有某一件將來的事，改變到的機會率極小，近乎是零，該件事算不算是注定的呢？

「過去」和「將來」的分別，並不是

「過去」和「將來」的真正的分別，應該是

「不完全注定」有兩個意思。

（ 問：但是，有沒有一個可能是，其實，將來所有事情的所有方面，都是注定的。所謂的「將來」，其實和「過去」一樣，都完全是固定的。

— Me@2015.05.26