# duplicate

If (member o l) finds o in the list l, it also returns the cdr of l beginning with o. This return value can be used, for example, to test for duplication. If o is duplicated in l, then it will also be found in the cdr of the list returned by member. This idiom is embodied in the next utility, duplicate:
 >(duplicate ’a ’(a b c a d)) (A D) 

(defun duplicate (obj lst &key (test #’eql))
(member obj (cdr (member obj lst :test test))
:test test))


— p.51

— On Lisp

— Paul Graham

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Exercise 4.4

Without using the existing function member, define duplicate as in
 >(duplicate ’a ’(a b c a d)) (A D)

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— This answer is my guess. —


(defun my-member (obj lst)
(cond ((not lst) NIL)
((eq obj (car lst)) lst)
(t (my-member obj (cdr lst)))))



— This answer is my guess. —

— Me@2019-01-21 06:34:46 AM

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# Problem 14.5d2

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

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d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$.

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$.

~~~

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— This answer is my guess. —

~~~

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The left R’+ sector:

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\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}}

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

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

$\displaystyle{N^\perp:}$

\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-1} x^{1} \right)^{32} \left( 1 + \lambda_{-2} x^{2} \right)^{32} ... \\ \end{aligned}}

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However, there are $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha\rangle_L}$ and $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha \rangle_R}$:

\displaystyle{\begin{aligned} (2^{15} + 2^{15}) \left[ \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 ... \left( 1 + \lambda_{-1} x^{1} \right)^{32} ... \right] \\ \end{aligned}}

\displaystyle{\begin{aligned} 2^{16} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r})^{32} \\ \end{aligned}}

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“Keep only states with $\displaystyle{(-1)^{F_L} = +1}$; this defines the left R’+ sector.”

\displaystyle{\begin{aligned} \frac{2^{16}}{2} \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned} &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ & \\ &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5}+39739981824 \, x^{6} + ... \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

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# Logical arrow of time, 6.4

The source of the macroscopic time asymmetry, aka the second law of thermodynamics, is the difference of prediction and retrodiction.

In a prediction, the deduction direction is the same as the physical/observer time direction.

In a retrodiction, the deduction direction is opposite to the physical/observer time direction.

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— guess —

If a retrodiction is done by a time-opposite observer, he will see the entropy increasing. For him, he is really doing a prediction.

— guess —

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— Me@2013-10-25 3:33 AM

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The existence of the so-called “the paradox of the arrow of time” is fundamentally due to the fact that some people insist that physics is about an observer-independent objective truth of reality.

However, it is not the case. Physics is not about “objective” reality.  Instead, physics is always about what an observer would observe.

— Lubos Motl

— paraphrased

— Me@2019-01-19 10:25:15 PM

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# Good intentions

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The road to hell is paved with good intentions.

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Hell is full of good meanings, but heaven is full of good works.

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2019.01.19 Saturday ACHK

# （反對）開夜車 2.1

Ken Chan 時光機 1.4.2.1

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（我暫時不記得，他以下說話的上半句是什麼。）

… 如何不是那樣，我就毋須於，放榜當天的晚上，在家裡哭。唉！還是不說了。

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— Me@2019-01-18 03:47:50 PM

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# Equality Predicates

6.3. Equality Predicates

Common Lisp provides a spectrum of predicates for testing for equality of two objects: eq (the most specific), eql, equal, and equalp (the most general).

eq and equal have the meanings traditional in Lisp.

eql was added because it is frequently needed, and equalp was added primarily in order to have a version of equal that would ignore type differences when comparing numbers and case differences when comparing characters.

If two objects satisfy any one of these equality predicates, then they also satisfy all those that are more general.

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[Function]
eq x y

(eq x y) is true if and only if x and y are the same identical object. (Implementationally, x and y are usually eq if and only if they address the same identical memory location.)

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The predicate eql is the same as eq, except that if the arguments are characters or numbers of the same type then their values are compared. Thus eql tells whether two objects are conceptually the same, whereas eq tells whether two objects are implementationally identical. It is for this reason that eql, not eq, is the default comparison predicate for the sequence functions defined in chapter 14.

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[Function]
eql x y

The eql predicate is true if its arguments are eq, or if they are numbers of the same type with the same value, or if they are character objects that represent the same character.

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[Function]
equal x y

The equal predicate is true if its arguments are structurally similar (isomorphic) objects. A rough rule of thumb is that two objects are equal if and only if their printed representations are the same.

Numbers and characters are compared as for eql. Symbols are compared as for eq. This method of comparing symbols can violate the rule of thumb for equal and printed representations, but only in the infrequently occurring case of two distinct symbols with the same print name.

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[Function]
equalp x y

Two objects are equalp if they are equal; if they are characters and satisfy char-equal, which ignores alphabetic case and certain other attributes of characters; if they are numbers and have the same numerical value, even if they are of different types; or if they have components that are all equalp.

— Common Lisp the Language, 2nd Edition

— Guy L. Steele Jr.

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Conrad’s Rule of Thumb for Comparing Stuff:

1. Use eq to compare symbols

2. Use equal for everything else

— Land of Lisp, p.63

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2019.01.16 Wednesday ACHK

# Problem 14.5d1.1.2

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

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d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$.

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$.

~~~

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— This answer is my guess. —

~~~

p.322

$\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}$

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}$

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The left NS’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

$\displaystyle{N^\perp:}$

\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-\frac{1}{2}} x^{\frac{1}{2}} \right)^{32} \left( 1 + \lambda_{-\frac{3}{2}} x^{\frac{3}{2}} \right)^{32} ... \\ \end{aligned}}

\displaystyle{\begin{aligned} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r-\frac{1}{2}})^{32} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned} &f_{L, NS'}(x) \\ &= a_{NS'} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}

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“The left NS’ sector is built with oscillators $\displaystyle{\bar \alpha_{-n}^I}$ and $\displaystyle{\lambda_{-r}^A}$ acting on the vacuum $\displaystyle{|NS' \rangle_L}$, declared to have $\displaystyle{(-1)^{F_L} = + 1}$:”

$\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}$

So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = +1}$.

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\displaystyle{ \begin{aligned} &f_{L, NS'}(x) \\ \end{aligned}}

$\displaystyle{ = \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

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Let

\displaystyle{ \begin{aligned} &g (\sqrt{x}) \\ &= \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ &= 1+32 \, \sqrt{x}+504 \, x+5248 \, x^{\frac{3}{2}}+40996 \, x^{2}+258624 \, x^{\frac{5}{2}}+1384320 \, x^{3}+6512384 \, x^{\frac{7}{2}} + ... \\ \end{aligned}}

Then

\displaystyle{ \begin{aligned} &g (-\sqrt{x}) \\ &= \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ &= 1 - 32 \sqrt{x} + 504 x - 5248 \, x^{\frac{3}{2}} + 40996 \, x^{2} - 258624 \, x^{\frac{5}{2}}+1384320 \, x^{3} - 6512384 x^{\frac{7}{2}} + ... \\ \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &f_{L, NS'+}(x) \\ &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\ &= \frac{1}{2x} \left[ g(\sqrt{x}) + g(-\sqrt{x}) \right] \\ &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\ \end{aligned}}

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The left R’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

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According to special relativity, in EPR, which of Alice and Bob collapses the wavefunction is not absolute. In other words, they do not have any causal relations.

— Me@2012-04-12 10:42:22 PM

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Publish! 11

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— Me@2011.07.03

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# PhD, 3.1

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— Me@2019-01-13 06:22:43 PM

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# Problem 14.5d1.2 | SageMath

The generating function is an infinite product:

\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}

To evaluate the infinite product, you can use SageMath with the following commands:

typeset_mode(True)

 (1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..oo)) a = (1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..200)) F = a.taylor(x,0,6) g = "+".join(map(latex, sorted([f for f in F.operands()], key=lambda exp:exp.degree(x)))) 

g

\displaystyle{ \begin{aligned} &f_{L, NS+}(x) \\ \end{aligned}}

$\displaystyle{ \approx \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

— Me@2019-01-11 11:52:33 AM

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# Problem 14.5d1

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

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d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$.

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$.

~~~

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— This answer is my guess. —

~~~

p.322

$\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}$

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}$

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The left NS’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

$\displaystyle{N^\perp:}$

\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-\frac{1}{2}} x^{-\frac{1}{2}} \right)^{32} \left( 1 + \lambda_{-\frac{3}{2}} x^{-\frac{3}{2}} \right)^{32} ... \\ \end{aligned}}

\displaystyle{\begin{aligned} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r-\frac{1}{2}})^{32} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}

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The left R’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-10 01:49:43 PM

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# Consistent histories, 6

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an observer ~ a consistent history

— Me@2019-01-05 04:02:43 PM

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# Ken Chan 時光機 2.2

— Me@2019-01-06 02:18:47 PM

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# 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.

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I followed the official instructions to build Clasp:

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

— Me@2019-01-04 10:11:43 PM

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# 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. —

~~~

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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# 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

# Amazing University

The meaning of life is “entering university”.

Turn your real life into an amazing university, for otherwise, your life has no meaning.

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[At this amazing university,]

Which year (of study) are you in?

— Me@2011.07.01

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# 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。

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（問：那樣，如果要「複雜地說」呢？）

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— Me@2019-01-01 11:20:57 PM

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