Clasp

Overview

Clasp is a new Common Lisp implementation that seamlessly interoperates with C++ libraries and programs using LLVM for compilation to native code. This allows Clasp to take advantage of a vast array of preexisting libraries and programs, such as out of the scientific computing ecosystem. Embedding them in a Common Lisp environment allows you to make use of rapid prototyping, incremental development, and other capabilities that make it a powerful language.

— Clasp README.md

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I followed the official instructions to build Clasp:

d_2019_01_04__12_06_48_pm_

The building process had been going on for about an hour; and then I got this error:

d_2019_01_04__17_53_17_pm_

d_2019_01_04__23_07_39_pm_

— Me@2019-01-04 10:11:43 PM

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2019.01.04 Friday (c) All rights reserved by ACHK

Problem 14.5c9

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) Calculate the total number of states in heterotic string theory (bosons plus fermions) at \displaystyle{\alpha' M^2 =4}.

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— This answer is my guess. —

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spacetime bosons:

\displaystyle{NS'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-2}^I, \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

\displaystyle{\begin{aligned}  I, J, I' &= 2, 3, ..., 9 \\  A, B, C, D &= 1, 2, ..., 32 \\  \end{aligned}}

Number of states:

Let \displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}, the number of ways to put n indistinguishable balls into k boxes.

\displaystyle{\begin{aligned}  &\left( 8+ N(2,8) +8 \times {32 \choose 2} + 32^2 + {32 \choose 4} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 40996 \times 128  \end{aligned}}

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\displaystyle{R'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

Following the same logic:

Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{2^{16}} degenerate Ramond ground states.

Therefore, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_L}.

— Me@2018-10-29 03:11:07 PM

\displaystyle{\begin{aligned}  I, J, K &= 2, 3, ..., 9 \\  \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 32768 \times 128  \end{aligned}}

~~~

spacetime fermions:

\displaystyle{NS'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 40996 \right) \times \left( 8^2 + 8^2 \right) \\  &= 40996 \times 128  \end{aligned}}

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\displaystyle{R'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15}\right) \times \left( 8^2 + 8^2 \right) \\  &= 32768 \times 128  \end{aligned}}

~~~

— This answer is my guess. —

— Me@2019-01-03 05:26:59 PM

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2019.01.03 Thursday (c) All rights reserved by ACHK

Photon dynamics in the double-slit experiment, 5

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What is the relationship between a Maxwell photon and a quantum photon?

— Me@2012-04-09 7:38:06 PM

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The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence

Maxwell’s theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

That caught me by surprise, because I always thought, Maxwell’s equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967):

The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

Isn’t the view of Sakurai in contradiction to Gloge? Do Maxwell’s equation describe a single photon or an infinite number of photons? Or do Maxwell’s equations describe a single photon and also an infinite number of photons at the same time? But why do we need QED then at all?

— edited Nov 28 ’16 at 6:35
— tparker

— asked Nov 20 ’16 at 22:33
— asmaier

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Because photons do not interact, to very good approximation for frequencies lower than \displaystyle{m_e c^2 / h} (\displaystyle{m_e} = electron mass), the theory for one photon corresponds pretty well to the theory for an infinite number of them, modulo Bose-Einstein symmetry concerns. This is similar to most of the statistical theory of ideal gases being derivable from looking at the behavior of a single gas particle in kinetic theory.

Put another way, the single photon behavior \displaystyle{\leftrightarrow} Maxwell’s equations correspondence only holds if you look at the Fourier transform version of Maxwell’s equations. The real space-time version of Maxwell’s equations would require looking at a superposition of an infinite number of photons — one way to describe the taking [of] an inverse Fourier transform.

If you want to think of it in terms of Feynman diagrams, classical electromagnetism is described by a subset of the tree-level diagrams, while quantum field theory requires both tree level and diagrams that have closed loops in them. It is the fact that the lowest mass particle photons can produce a closed loop by interacting with, the electron, that keeps photons from scattering off of each other.

In sum: they’re both incorrect for not including frequency cutoff concerns (pair production), and they’re both right if you take the high frequency cutoff as a given, depending on how you look at things.

— edited Dec 3 ’16 at 6:28

— answered Nov 27 ’16 at 23:08

— Sean E. Lake

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Maxwells equations, which describe the wavefunction of a single noninteracting photon, don’t need Planck’s constant. I find that remarkable. – asmaier Dec 2 ’16 at 14:16

@asmaier : Maxwell’s equations predate the quantum nature of light, they weren’t enough to avoid the ultraviolet catastrophe. Note too that what people think of as Maxwell’s equations are in fact Heaviside’s equations, and IMHO some meaning has been lost. – John Duffield Dec 3 ’16 at 17:45

— Do Maxwell’s equations describe a single photon or an infinite number of photons?

— Physics StackExchange

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2019.01.03 Thursday ACHK

宇宙大戰 1.2

PhD, 2.4 | 故事連線 1.1.6 | 碩士 3.4

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(問:我也遇過類似的情境。

我和一位好朋友合作做小組習作時,雖然未至於反目,但總會有很多爭拗。和他合作前,明明和他感情要好。各自有什麼困難時,對方總會杖義相助。

為什麼人類會,那麼奇怪呢?)

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簡單地說,即使是同一個人,其實也有不同方面,各樣性格。

做朋友時,你只需要接受小部分—你可以選擇,只接受他,最好的優點。但是,做工作伙伴時,你卻要接收大部分—你未必可以選擇,不接受他,最壞的缺點。

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(問:那樣,如果要「複雜地說」呢?)

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複雜地說,每個個體也透過自己,在這宇宙間的經歷,形成一個「主觀宇宙」,簡稱「世界觀」。

大部分人,也不自覺地,以為他的主觀宇宙,就是客觀宇宙的全部。這個不幸,源於每個人的主觀宇宙,是他唯一能夠觀察到的「客觀宇宙部分」;每個人當時的主觀宇宙,是他當時唯一能夠,觀察到的「客觀宇宙部分」。

只有一些「被選擇的心靈」,簡稱「半神人」,才會想像到,他的主觀世界,只是客觀世界的極小部分。所以,如果兩個人也不是「半神人」,而又要在工作上合作的話,其實就相當於,把兩個(主觀)宇宙的大部分,重疊在一起。

每個宇宙原本,都有各自的運行法則;貿然要求兩個宇宙,互相干涉對方內政,自然會十分危險。

六千五百萬年前,單單是一個小行星與地球相撞,就足以令大部分恐龍滅絕。試想想,兩個宇宙相撞,殺傷力會大多少倍。

— Me@2019-01-01 11:20:57 PM

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2019.01.01 Tuesday (c) All rights reserved by ACHK