Double slit experiment, 8

.

Although the screen itself is a photon position detector, it gets no which-way information. So it can get an interference pattern.

— Me@2012-04-09 7:24:23 PM

.

.

2018.11.27 Tuesday (c) All rights reserved by ACHK

PhD, 2.1

故事連線 1.1.3 | 碩士 3.1

這段改編自 2010 年 4 月 18 日的對話。

.

(安:你的意思時,不鼓勵年青人,攻讀研究院?)

不是。

研究工作,有其獨特的至尊好處。我反而覺得,每個人都應花兩年時間,經歷一次。換句話說,你可以考慮,先讀一個研究式的碩士,然後才盤算,自己適不適合再攻讀博士。

不過,你要留意,無論是攻讀碩士還是博士,你在之前選擇指導教授時,都要格外小心。你要保證,你的指導教授,有品德和有才能,幫你把與論文課題沒有直接關係的工作,全部推開。當然,那樣可靠的教授,萬中無一。

— Me@2012.08.20

.

在從來未與一位教授共事過的情況下,就選他為自己的碩士或博士論文導師,是十分危險的事。

那足以令你,自此不能再於學術界發展,抱憾終生。

(問:在那位教授成為你的論文導師前,又怎會共事過呢?)

可以在之前,有過其他形式的工作關係,直接或間接地,知道他的才德如何。

(問:你意思時,在本科時,就選修他的課?)

可以那樣說。聽他的講課和做他所予之功課,就可以知道他,工作態度認不認真、思考清不清晰 和 顧不顧及他人感受。但是,那也只是第一重的「測試」,先決而未充分;因為,那只是他的「公眾形象」而已。當他作為你的論文導師,你作為他的研究生時,你再也不是,和他的「公眾形象」相處了。

有很多人的「個人形象」和「公眾形象」,都是差天共地的。

(問:即是表裡不一?)

有時,甚至是「表裡相反」。

(問:那應該怎麼辦?)

— Me@2018-11-24 08:17:17 PM

.

.

2018.11.25 Sunday (c) All rights reserved by ACHK

Common Lisp for Android App Development 2018

An REPL called “CL REPL” is available in the Google Play Store. But itself is not for developing standalone Android apps, unless those apps are Common Lisp source code files only.

However, “CL REPL” itself is an open source GUI app using Common Lisp and Qt. So by learning and using its source, in principle, we can create other Android apps using Common Lisp with Qt.

The library that “CL REPL” uses is EQL5-Android.

— Me@2018-11-23 04:07:54 PM

.

.

2018.11.23 Friday (c) All rights reserved by ACHK

Problem 14.5c2

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

.

At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = 4k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

~~~

.

— This answer is my guess. —

.

The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \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 \\ \end{aligned}}

.

The left R’+ sector:

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}

.

The right-moving NS+ states:

NS+ equations of (14.38):

\displaystyle{\begin{aligned}  \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 \end{aligned}}

.

The R- states (that used as right-moving states):

Mass levels of R- and R+ (Equations 14.54):

\displaystyle{\begin{aligned}  \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\  \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\  \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\  &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}

.

There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with \displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}.

.

— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

.

.

2018.11.22 Thursday (c) All rights reserved by ACHK