Eigenstates 2.3.2

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eigenstates

~ classical states

~ definite states

— Me@2012-04-15 11:42:10 PM

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The concept of eigenstate is relative.

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First, you have to specify the eigenstate is of which physical observable.

A physical system can be at an eigenstate of one observable but at a superposition state of another observable.

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Second, you have to specify the state of that observable is eigen with respect to which observer.

— Me@2018-06-16 7:27 AM

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eigenstates

~ of which observable?

~ with respect to which observer?

— Me@2018-06-19 10:54:54 AM

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2018.06.19 Tuesday (c) All rights reserved by ACHK

大學經濟

這段改編自 2010 年 4 月 18 日的對話。

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我猜想,當一個人改變存在型態時,會立刻或者將會,知道很多生時不知道的東西。但是,那些新知識,未必包括你想知道的東西。

比喻說,由中學升到大學,你將會學到,很多超過中學程度的知識。但是,如果你中學時,沒有讀過經濟科的話,單單是「升大學」本身,並不會令你,立刻獲得經濟科的知識。

大學生「由零開始學經濟學」,都同樣要花時間;分別是,通常而言,比中學生「由零開始學經濟學」,速度會高一點。

(問:不一定呀。中學生比較年青,腦袋理應高速一點。)

無錯。

方便起見,暫時用同一個人來比較,例如你。

「中學的你」可以因為腦袋較年青,學習新事物比「大學的你」較快。「大學的你」可能因為知識和經驗較多,學習新事物比「中學的你」較快。

視乎情況,因人而異,沒有一定的答案。

但是,至少你會同意一點:

如果你中學時,沒有讀過經濟科,在大學時要「由零開始學經濟學」的話,你會立刻看大學程度的經濟書,而不是由中學教科書開始學。

— Me@2018-06-05 11:54:51 AM

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2018.06.07 Thursday (c) All rights reserved by ACHK

Plato

trowawee 4 months ago

I’m a little frustrated at the tossed-off reference to Plato and Aristotle at the beginning – “The good life may have sufficed for Plato and Aristotle, but it is no longer enough.” – because I feel like that ignores the fact that both Plato and Aristotle, along with a lot of philosophers, actually had a lot to say about physical fitness. Plato was a champion wrestler, and both he and Aristotle viewed physical education as a fundamental component to living the good life. Xenophon quotes Socrates saying this:

“For in everything that men do the body is useful; and in all uses of the body it is of great importance to be in as high a state of physical efficiency as possible. Why, even in the process of thinking, in which the use of the body seems to be reduced to a minimum, it is matter of common knowledge that grave mistakes may often be traced to bad health.”

The whole article feels a little too mired in presentism, and ignorant of the history of self-improvement ideas.

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coldtea 4 months ago

>Plato was a champion wrestler

And the name Plato is a nickname — meaning “the broad/wide one” given to him for his broad shoulders because of that training and physical appearance. Real name: Aristocles.

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kernelbandwidth 4 months ago

It’s funny to consider that one of the canonically great philosophers in history is known essentially by the equivalent of his WWE wrestling name. It’s like if in the future there were classes taught on the philosophical ideas of “The Rock”.

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coldtea 4 months ago

Some other amusing related stuff: so, Plato, was called for for the ancient greek word for broad/wide.

Modern [English] words that stem from the same root: plateau, platitude, plat, plate — via French and Latin (plattus) from Greek (platis “flat, wide, broad”).

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danohu 4 months ago

Well, the first Pope was literally called The Rock (Peter). Jesus appointed him by saying “you are The Rock, and I’ll build my church on this rock”.

Exactly what he meant has led to centuries of debate between protestants and catholics.

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acangiano 4 months ago

“No man has the right to be an amateur in the matter of physical training. It is a shame for a man to grow old without seeing the beauty and strength of which his body is capable.”

― Socrates

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— Improving Ourselves to Death

— Hacker News

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2018.06.01 Friday ACHK

Problem 14.4b1.4

Closed string degeneracies | A First Course in String Theory

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What is the meaning of “With a = 1, ..., 8 and \bar b = \bar 1, ..., \bar 8, …”?

— Me@2015.09.14 12:11 PM

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p.315 “Explicitly, the eight states | R_a \rangle, a = 1, 2, ..., 8, with an even number of creation operators are … ”

p.316 “The eight states |R_{\bar{a}} \rangle, \bar a = \bar 1, \bar 2, ..., \bar 8, with an odd number of creation operators are … ”

— Me@2018-05-24 11:41:34 AM

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2018.05.24 Thursday (c) All rights reserved by ACHK

Life, 3

生命 3

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We exist in time because time is change.

Growing is part of the definition of life. Growing is a kind of change.

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Also, without time/change, there would be no thinking and no thoughts.

— Me@2017-12-26 11:42 am

— Me@2018-05-23 10:05:03 PM

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time ~ change

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Time is logically necessary if change is necessary.

— Me@2018-02-04 09:07:48 PM

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2018.05.23 Wednesday (c) All rights reserved by ACHK

多項選擇題 6

Multiple Choices 6

這段改編自 2010 年 8 月 24 日的對話。

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有時,一題物理 MC(多項選擇題)會,同時有數學做法和物理做法。

那時,你就先用物理方法做一次,再用數學方法做多一次,以作驗算。

(問:哪有那麼多的時間?)

之前講過,那些做法,必須透過考試前,平日多加收集和練習而來;並不是在考試中途,才花額外時間發明。

— Me@2018-05-22 06:02:40 PM

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2018.05.22 Tuesday (c) All rights reserved by ACHK

Quantum Computing, 2

stcredzero 3 months ago

A note for the savvy: A quantum computer is not a magic bit-string that mysteriously flips to the correct answer. A n-qubit quantum computer is not like 2^n phantom computers running at the same time in some quantum superposition phantom-zone. That’s the popular misconception, but it’s effectively ignorant techno-woo.

Here’s what really happens. If you have a string of n-qubits, when you measure them, they might end up randomly in [one] of the 2^n possible configurations. However, if you apply some operations to your string of n-qubits using quantum gates, you can usefully bias their wave equations, such that the probabilities of certain configurations are much more likely to appear. (You can’t have too many of these operations, however, as that runs the risk of decoherence.) Hopefully, you can do this in such a way, that the biased configurations are the answer to a problem you want to solve.

So then, if you have a quantum computer in such a setup, you can run it a bunch of times, and if everything goes well after enough iterations, you will be able to notice a bias towards certain configurations of the string of bits. If you can do this often enough to get statistical significance, then you can be pretty confident you’ve found your answers.

— An Argument Against Quantum Computers

— Hacker News

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2018.05.17 Thursday ACHK

Van der Waals equation 1.2

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}

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These are the reasons that

V_{\text{ideal}} < V_{\text{measured}}

P_{\text{ideal}} > P_{\text{measured}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT

— Me@2018-05-16 07:12:51 PM

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… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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2018.05.16 Wednesday (c) All rights reserved by ACHK

Tree rings, 2

69b73-growth_rings

This file is licensed under the Creative Commons Attribution 2.0 Generic license. Author: Lawrence Murray from Perth, Australia

Time-traveling to the past is like “making an outside ring more inside”, which is logically impossible.

— Me@2011.09.18

8fd2c-pastpresentfuture

Me@2010

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2018.05.16 Wednesday (c) All rights reserved by ACHK

Richard Stallman

I’ve always lived cheaply. I live like a student, basically. And I like that, because it means that money is not telling me what to do. I can do what I think is important for me to do. It freed me to do what seemed worth doing. So make a real effort to avoid getting sucked into all the expensive lifestyle habits of typical Americans. Because if you do that, then people with the money will dictate what you do with your life. You won’t be able to do what’s really important to you.

— Richard Stallman

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There is nothing wrong to be a student-having a lot of new learnings and new young friends, as long as you can earn enough money.

— Me@2011.08.20

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2018.05.15 Tuesday (c) All rights reserved by ACHK

On Keeping Your Soul, 3

這段改編自 2010 年 4 月 18 日的對話。

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「學術研究的論文」和「網誌文章」的主要分別是,論文要花很多時間,才可寫成一篇,然後只有極少人去閱讀:

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時間 ~ 1 至 2 年

潛在讀者人數 ~ 10 至 100 人

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網誌文章則相反:

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時間 ~ 1 至 2 星期

潛在讀者人數 ~ > 100 人

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不過,曲高自然和寡:

如果一篇網誌文章的學術成份高的話,閱讀人數自然極少。

— Me@2018-05-15 05:00:47 PM

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後記:

英雄所見略同,物理學家 Lubos Motl,有近乎相同的見解。

— Me@2018-05-15 05:00:55 PM

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2018.05.15 Tuesday (c) All rights reserved by ACHK

Ask users

gallerdude 3 months ago

There’s a good moral here. Everytime they had a question, they asked their users. Users don’t lie.

atYevP 3 months ago

Yev from Backblaze here -> That’s right! For better or worse they usually tell it like they see it, and that helps us inform decisions!

— How Backblaze Got Started (2017)

— Hacker News

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2018.05.14 Monday ACHK

Problem 14.4b1.3

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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— This answer is my guess. —

Since for NS, the first 5 levels’ degeneracies are 8, 36, 128, 402, 1152, the degeneracies of (NS, NS) are 8^2, 36^2, 128^2, 402^2, 1152^2.

This is incorrect, for there are no (NS, NS) states. Instead, you should consider (NS+, NS+).

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Since for NS+, the first 5 levels’ degeneracies are 8, 128, 1152, 7680, 42112, the degeneracies of (NS+, NS+) are 8^2, 128^2, 1152^2, 7680^2, 42112^2.

p.317 Consider the relationship of the degeneracy of R+ and that of R-:

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How about the first 5 levels of R+?

The degeneracies are the same as those of R-.

p.317 Equation (14.54) “The appearance of an equal number of bosonic and fermionic states at every mass level is a signal of supersymmetry. This is, however, supersymmetry on the world-sheet.”

Equation (14.71):

f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...

p.321 “Indeed, the integer mass-squared levels in the NS generating function (14.67) have degeneracies that match those of (14.71) for the R- sector.”

— This answer is my guess. —

— Me@2018-05-14 02:51:55 PM

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2018.05.14 Monday (c) All rights reserved by ACHK

Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20
Ram Bharadwaj

— Physics Stackexchange

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Ideal gas law:

P_{\text{ideal}} V_{\text{ideal}} = nRT

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

P_{\text{measured}} = P_{\text{real}}

P_{\text{measured}} < P_{\text{ideal gas}}

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

V_{\text{measured}} > V_{\text{real}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute P_{\text{measured}} onto the LHS, since P_{\text{measured}} < P_{\text{ideal}}, the LHS will be smaller than the RHS:

P_{\text{measured}} V_{\text{ideal}} < nRT

So in order to maintain the equality, a correction term to the pressure must be added:

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute V_{\text{measured}} onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

P_{\text{ideal}} V_{\text{measured}} > nRT

So in order to maintain the equality, a correction term to the pressure must be subtracted:

P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT

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In other words,

V_{\text{measured}} > V_{\text{real}}

V_{\text{ideal}} = V_{\text{real}}

V_{\text{measured}} > V_{\text{ideal}}

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

P_{\text{measured}} = P_{\text{real}}

but

V_{\text{measured}} \ne V_{\text{real}}?

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How come

V_{\text{real}} = V_{\text{ideal}}

but

P_{\text{real}} \ne P_{\text{ideal}}?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” V_{\text{real}} has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

V_{\text{real}} \equiv V_{\text{measured}}

P_{\text{real}} \equiv P_{\text{measured}}

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Or even better, avoid the terms P_{\text{real}} and V_{\text{real}} altogether. Instead, just consider the relationship between (P_{\text{ideal}}, P_{\text{measured}}) and that between (V_{\text{ideal}}, V_{\text{measured}}).

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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2018.05.13 Sunday (c) All rights reserved by ACHK

時空兌換率

這段改編自 2015 年的對話。

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我的相對論教授說,所謂

E = m c^2

在某些意思之下,沒有那麼特別,因為,你可以把它看成,貨幣的兌換。

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E = c^2 m

能量 =(光速二次方)\times 質量

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1 \text{USD} \approx 8 \times 1 \text{HKD}

1 美元 \approx 8 \times 1 港元

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公式中的 c^2(光速平方),角色其實正正就是,能量 E 和質量 m 之間的「貨幣兌換率」。

(而光速 c,則是時間和空間的兌換率。)

— Me@2018-05-11 09:10:00 PM

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2018.05.11 Friday (c) All rights reserved by ACHK

Sleep apnea

erentz 3 months ago

Slightly tangential to the study here but I’ve been going down the sleep apnea rabbit hole in recent months and I strongly encourage folks to investigate their sleep. A lot of people have sleep apnea and don’t realize it. You don’t have to snore to have it. A surprising statistic I found was 20-30% of people with ADHD have sleep apnea. A lot of people may be treating symptoms of sleep apnea like ADHD and high blood pressure with medications while ignoring the root cause. My experience with this has been that doctors are surprisingly ignorant. They’ll happy prescribe you medications for anxiety, ADHD, blood pressure for years, and never think to ask you about your sleep. Do some of your own investigations or ask about it if you have any suspicions.

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copperx 3 months ago

I’m going to add myself as a data point. After suffering years of feeling sleepy regardless of how much I slept and frequently unmotivated and withdrawn, I was diagnosed with severe sleep apnea. A few years ago I was given Dexedrine, Ritalin, Adderall, and I even tried self-medicating with modafinil to ameliorate the symptoms of what doctors thought was ADHD with varying levels of success; but the drug-free treatment of sleep apnea with a BiPAP machine got rid of all of these issues; in addition, I feel about ten years younger.

— Sleep and Mortality: A Population-Based 22-Year Follow-Up Study (2007)

— Hacker News

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2018.05.09 Wednesday ACHK

Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

NS+ equations of (14.38):

\alpha'M^2=0, ~~N^\perp = \frac{1}{2}: ~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle,
\alpha'M^2=1, ~~N^\perp = \frac{3}{2}: ~~~~\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle,
\alpha'M^2=2, ~~N^\perp = \frac{5}{2}: ~~~~\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}
\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \} |NS \rangle \otimes |p^+, \vec p_T \rangle,

For N^\perp = \frac{5}{2}, the number of states is

8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2
+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}
= 1152

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Since \alpha' M^2 = N^\perp - \frac{1}{2}, when N^\perp = \frac{5}{2}, \alpha' M^2 = 2.

Equation (14.67):

f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

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

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p.321

If we take f_{NS} (x) in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

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

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

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\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8 is the boson contribution.

\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8 is the fermion contribution.

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Turning Equation (14.67) into

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

is equivalent to turning all \sqrt{x} into - \sqrt{x}:

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

Me@2015.08.29 12:49 PM: Somehow, \sqrt{x} represents “contribution from fermions”.

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Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all \sqrt{x} with y.

f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)

f_{NS} (x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

f_{NS+} (x)
= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)
= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...

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It is _not_ correct. Just consider it as \left(\sqrt{x} \to -\sqrt{x} \right) is not correct, since the beginning factor \frac{1}{\sqrt{x}} is not considered yet.

Instead, we should present in the following way:

f_{NS} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= \frac{1}{\sqrt{x}} g_{NS}(x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

g (\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

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g (\sqrt{x}) - g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...

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f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]
= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]
= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]
= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...

— Me@2018-05-08 08:50:32 PM

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2018.05.08 Tuesday (c) All rights reserved by ACHK