Ex 2.1-1 Particle in a Box

Remove["Global`*"]

hbar := \[HBar]

H[V_] @ psi_  := -hbar^2/(2m) D[psi,{x,2}] + V psi

SchD[V_] @  psi_ := H[V] @ psi - En psi

SchD[V[x]] @ psi[x] == 0 // TeXForm

\displaystyle{-\text{En} \psi (x)-\frac{\hbar ^2 \psi ''(x)}{2 m}+\psi (x) V(x)=0}

phi[n_, x_] := Cn Sin[n Pi x/L]

SchD[0] @ phi[n,x] == 0 // ExpandAll

SchD[0] @ phi[n,x] / phi[n,x] == 0 // Simplify // TeXForm

\displaystyle{2 \text{En} L=\frac{\pi ^2 n^2 \hbar ^2}{L m}}

Starting with a linear superposition A E^(Ikx) + B E^(-Ikx) of independent plane waves, where A and B are constants, verify the box eigen-functions and eigen-energies given above.

Thus, show that this superposition is a solution of the Schrodinger equation and, by invoking the boundary condition, that k -> n Pi/L.

— Quantum Methods with Mathematica

phi[k_, x_] := A E^(I k x) + B E^(- I k x)

eq1 := SchD[0] @ phi[k,x]/phi[k,x] == 0 // Simplify

Ek[k_] := En /. Solve[eq1, En] [[1]]

Ek[k] // TeXForm

\displaystyle{\frac{k^2 \hbar ^2}{2 m}}

.

The boundary conditions are \displaystyle{\phi_{x=0} = \phi_{x=L} = 0}.

\displaystyle{  \begin{aligned}  A+B&=0 \\   A e^{i k L}+B e^{-i k L}&=0 \\  \end{aligned}}

So

\displaystyle{  \begin{aligned}  e^{i k L} - e^{-i k L}&=0 \\  \end{aligned}}

Solve[E^(I k L) - E^(-I k L) == 0, k] // Simplify

\displaystyle{  \left\{k\to -\frac{2 \pi  c_1}{L}\text{ if }c_1\in \mathbb{Z}, ~~~k\to -\frac{2 \pi  c_1+\pi }{L}\text{ if }c_1\in \mathbb{Z} \right\}}

\displaystyle{  =\left\{k\to \frac{n \pi }{L}\text{ if }n \in \mathbb{Z}\right\}}

.

phi[n_, x_] := Cn Sin[n Pi x/L]

norm[n_] = Cn /.
    Solve[
        Integrate[ phi[n,x]^2, {x,0,L}] == 1 /.
            Sin[m_Integer n Pi] -> 0, 
        Cn
    ] [[1]] // TeXForm

\displaystyle{  -\frac{\sqrt{2}}{\sqrt{L}}  }

It’s interesting to note in passing that the 1D box eigenfunctions are also classically the eigenfunctions of a taut string. However, whereas the quantum mechanical energies scale as n^2, the classical eigenfrequencies of the string’s normal modes are linear in n. This is a consequence of the classical wave equation being second order in time in contrast to the quantum wave equation being first order.

— Quantum Methods with Mathematica

— Me@2022-12-16 10:23:45 AM

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.

2022.12.17 Saturday (c) All rights reserved by ACHK

Euler problem 11.5

A project created by using Lazarus on one platform can be compiled on any other one which Free Pascal compiler supports. For desktop applications a single source can target macOS, Linux, and Windows, with little or no modification. An example is the Lazarus IDE itself, created from a single code base and available on all major platforms including the Raspberry Pi.

— Wikipedia on Lazarus (software)

.

.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

program e11;          

uses
   SysUtils, math;

type
   T2d_int_array = array of array of integer;

const
   filename = 't.txt';

var
   current_dir: string;
   ptext: TextFile;
   
   line_string: string;
   s: string = '';
   lines: array of string;
   array_string_2d: array of array of string;
   array_int_2d: T2d_int_array;
   
   isbegin: boolean = true;
   i: integer;
   j: integer;
   lenx: integer = 0;
   leny: integer = 0;
   c: integer = 0;


function max4_iter(a2d: T2d_int_array;
                  i, j: integer;
                   acc: longint): integer;
var
   dia_l: longint = 0;
   dia_r: longint = 0;
   hori: longint = 0;
   vert: longint = 0;
   k: integer = 0;
   accm: longint = 0;
   x: integer = 0;
   y: integer = 0;

begin
   if i <= (lenx - 4) then begin
      hori := 1;
      for k := 0 to 3 do begin
         hori := hori*a2d[i+k, j];
      end;
   end;

   if j <= (leny - 4) then begin
      vert := 1;
      for k := 0 to 3 do begin
         vert := vert*a2d[i, j+k];
      end;
   end;

   if (i <= (lenx - 4)) and (j <= (leny - 4)) then
   begin
      dia_r := 1;
      for k := 0 to 3 do begin
         dia_r := dia_r*a2d[i+k, j+k];
      end;
   end;

   if (i >= 3) and (j <= (leny - 4)) then begin
      dia_l := 1;
      for k := 0 to 3 do begin
         dia_l := dia_l*a2d[i-k, j+k];
      end;
   end;

   if (i = (lenx - 1)) and (j = (leny - 1)) then
   begin
      max4_iter := acc;
   end else begin
      accm := max(acc,
                  max(hori,
                      max(vert,
                          max(dia_l, dia_r))));
      
      if i = lenx - 1 then begin
         x := 0;
         y := j + 1;
      end else begin
         x := i + 1;
         y := j;
      end;
      max4_iter := max4_iter(a2d, x, y, accm);
   end;
end;

begin

   current_dir := GetCurrentDir;

   writeln(filename, ':');
   writeln('==================================');

   if not FileExists(filename) then begin
      writeln(filename, ' does not exist.');
      exit;
   end;

   assign(ptext, filename);
   reset(ptext);

   setLength(lines, 0);
   s := '';
   isbegin := true;
   while not eof(ptext) do begin
      readln(ptext, line_string);
      if isbegin then begin
         s := s + line_string;
         isbegin := false;
      end else begin
         s := s + LineEnding + line_string;
      end;
   end;
   close(ptext);

   lines := s.split(LineEnding);
   lenx := length(lines);
   setLength(array_string_2d, lenx);

   for i := 0 to (lenx - 1) do begin
      array_string_2d[i] := lines[i].split(' ');
   end;
   leny := length(array_string_2d[0]);

   setLength(array_int_2d, lenx);
   for i := 0 to (lenx - 1) do begin
      setLength(array_int_2d[i], leny);
   end;

   for i := 0 to (lenx - 1) do begin
      for j := 0 to (leny - 1) do begin
         val(array_string_2d[i][j],
             array_int_2d[i][j],
             c);
      end;
   end;

   writeln(max4_iter(array_int_2d, 0, 0, 0));
   writeln('==================================');
end.

— Me@2022-12-14 03:56:23 PM

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2022.12.14 Wednesday (c) All rights reserved by ACHK

Light-like

Spacetime interval, 3 | How far away is tomorrow?, 2

.

Any two events in a spacelike region are not causally related.

Any two events in a timelike region can be causally related.

How about two events on the boundary (light-like)?

— Me@2016-06-30 06:38:06 PM

.

Their spacetime distance is zero, meaning that the communication signals/particles between the two events would have to travel at the speed of light.

In other words, the two events can only be marginally causally related.

— Me@2022-12-14 10:55:20 AM

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2022.12.14 Wednesday (c) All rights reserved by ACHK

Advanced course == course in advance

philip1209 on June 30, 2018

One of my favorite classes in college was a physics course taught in this way:

1. Before class, we were assigned to read a chapter from the textbook, understand the material, and complete two or three homework problems from the material we had just self-learned.

2. After submitting homework, lectures focused on discussing the concepts more in-depth. Everybody already had a baseline knowledge, so the professor would highlight the important takeaways, applications, live demonstrations of concepts, etc. I found these lectures engaging because I had already learned the material – and the lectures focused on mastering it.

3. Sometimes there would be follow-up homework problems focusing on advanced applications or derivations. These advanced problems were closest to exam questions.

Some takeaways for me:

  • If we didn’t have homework to do before class, I doubt I would have consistently learned the material before lecture.

  • Lectures taught us more than the “what” – it taught us the “why” and how these concepts related to other areas.

  • Lectures focused on answering questions, exploring curiosities (like “what if” questions), demonstrations/experiments, and mastery. The professor added value beyond the written material!

I hope this helps OP because it sounds like they have material prepared beforehand, which means that the lectures could go beyond the material.

— Hacker News

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2022.12.13 Tuesday ACHK

人生 Presentation, 2.1

這段改編自 2010 年 4 月 24 日的對話。

.

(安:雖然我不是從事教學工作,但是,在公司解釋東西給上司時,其實很多時也需要使用到,所謂的「教學技巧」。

你有沒有一些推介呢?)

假設每一節課,都是一小時以內。

每一節課應該,只有一個重點。在該課中,你要用不同的句子、字眼、例子,來重複釐清和闡述,那個重點。講課時,一切的細節,都要圍繞著,那個重點來運行。

.

(安:但是,除了這一點外,你應該還有其他技巧,因為,我覺得你很多時也會,在適當時候,講適當的東西。)

.

「在適當時候,講適當的東西」這個講法好得意,因為,你可以反問:

有沒有可能「在適當時候,講不適當的東西」,或者「在不適當時候,講適當的東西」呢?

.

的確還有其他技巧。

第一,要有能力說完一句完整句子。

很好像不是什麼「技巧」。但其實,大部分人,包括我,也沒有這個能力。大部分情況下,都靠斷句傾談。平時大家也不發覺,而一路以為正在說完整句子。其原因是,每人在聽自己或他人說話時,腦部也會自動,修補對方句子的不完整處。

上次我講:

我在之前的工作,訓練了表達技巧。我在中學教書時,需要每天講學三個多小時。講了一年半載後,說話才通順,可以完成完整句子而不「跳線」。

之後,我說話時就好像,口部自己有思想,內容自己有舖排,而毋須經過大腦的思考。

要鍊成這個「說話內容思路大致清晰」,唯有靠大規模的練習。

— Me@2022-12-12 02:40:07 PM

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2022.12.13 Tuesday (c) All rights reserved by ACHK

Direct from Dell

Euler problem 9.2

.

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

g p =
  [ [a, b, c]
    | m <- [2 .. limit],
      n <- [1 .. (m - 1)],
      let a = m ^ 2 - n ^ 2,
      let b = 2 * m * n,
      let c = m ^ 2 + n ^ 2,
      a + b + c == p
  ]
  where
    limit = floor . sqrt . fromIntegral $ p

— based on Haskell official

.

Euclid’s formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of integers m and n with m > n > 0. The formula states that the integers

\displaystyle{ a=m^{2}-n^{2},\ \,b=2mn,\ \,c=m^{2}+n^{2}}

form a Pythagorean triple. The triple generated by Euclid’s formula is primitive if and only if m and n are coprime and one of them is even. When both m and n are odd, then a, b, and c will be even, and the triple will not be primitive; however, dividing a, b, and c by 2 will yield a primitive triple when m and n are coprime.

Every primitive triple arises (after the exchange of a and b, if a is even) from a unique pair of coprime numbers m, n, one of which is even.

— Wikipedia on Pythagorean triple

— Me@2022-12-10 09:57:27 PM

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2022.12.11 Sunday (c) All rights reserved by ACHK

Ex 2.1 Curves

Functional Differential Geometry

.

a. The rectangular coordinate equation for the Lemniscate of Bernoulli is

\displaystyle{(x^2 + y^2)^2 = 2 a^2 (x^2 - y^2)}.

Find the expression in polar coordinates.

b. Describe a helix space curve in both rectangular and cylindrical coordinates.

~~~

(define-coordinates (up x y) R2-rect)

(define-coordinates (up r theta) R2-polar)

;

(define R2-rect-chi (chart R2-rect))

; R2-rect-chi
;     generates the rectangle coordinates of a point.

(define R2-rect-chi-inverse (point R2-rect))

; R2-rect-chi-inverse
;     gets the abstract representation.

(x (R2-rect-chi-inverse (up 'x0 'y0)))

; Function x
;     gets the x coordinate
;          of an (abstract-represented) point.

;

(define R2-polar-chi (chart R2-polar))

(define R2-polar-chi-inverse (point R2-polar))

(x (R2-polar-chi-inverse (up 'r0 'theta0)))

(r (R2-polar-chi-inverse (up 'r0 'theta0)))

(r (R2-rect-chi-inverse (up 'x0 'y0)))

(theta (R2-rect-chi-inverse (up 'x0 'y0)))

;

(define h (+ (* x (square r)) (cube y)))

(define R2-rect-point
  (R2-rect-chi-inverse (up 'x_0 'y_0)))

(show-expression
 (h R2-rect-point))

(show-expression
 (h (R2-polar-chi-inverse (up 'r_0 'theta_0))))

(show-expression
 ((- r (* 2 'a (+ 1 (cos theta))))
  ((point R2-rect) (up 'x 'y))))

; Ex 2.1 a

(show-expression
 ((- (square (+ (square x) (square y)))
     (* 2 (square 'a) (- (square x) (square y))))
  ((point R2-rect) (up 'x 'y))))

(show-expression
 ((- (square (+ (square x) (square y)))
     (* 2 (square 'a) (- (square x) (square y))))
  ((point R2-polar) (up 'r 'theta))))

; Ex 2.1 b

(define-coordinates (up r theta z) R3-cyl)

(define-coordinates (up x y z) R3-rect)

(show-expression
 ((- (up r z) (up 'R (* 'a theta)))
  ((point R3-cyl) (up 'r 'theta 'z))))

(show-expression
 ((- (up r z) (up 'R (* 'a theta)))
  ((point R3-rect) (up 'x 'y 'z))))

— Me@2022-12-10 10:29:59 AM

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2022.12.10 Saturday (c) All rights reserved by ACHK

Posted in FDG

Why does the universe exist? 7.0

ParEdit, 2

.

What is there before the beginning of the universe?

What does the universe change from?

— Me@2016-07-20 06:16:34 AM

.

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

(autoload 'enable-paredit-mode
  "paredit" "Turn on pseudo-structural editing." t)

;;;;;;;;;;;

(defun add-hooks (hooks fn)
  (mapc (lambda (hook)
      (add-hook hook fn))
    hooks))

(add-hooks (list
        'emacs-lisp-mode-hook
        'eval-expression-minibuffer-setup-hook
        'ielm-mode-hook
        'lisp-mode-hook
        'lisp-interaction-mode-hook
        'scheme-mode-hook)

       #'enable-paredit-mode)

;;;;;;;;;;;

(add-hook 'slime-repl-mode-hook
      (lambda () (paredit-mode +1)))

(defun override-slime-repl-bindings-with-paredit ()
   (define-key slime-repl-mode-map
     (read-kbd-macro paredit-backward-delete-key) nil))

(add-hook 'slime-repl-mode-hook
      'override-slime-repl-bindings-with-paredit)

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

— Me@2022-12-09 01:24:55 PM

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.

2022.12.09 Friday (c) All rights reserved by ACHK

Computing Note, 3


Summer 2002

1. Perhaps the Encyclopedia Britannica 
                    Research Assistant 
                        is written by[in] Java.

2. Work Hard Play Hard 
           Live your life with[in] all [its] fullness.

3. H(Learning): HyperLearning 真切學習:
[]
Be a professional professional: 比專業還專業

.

To edit Lazarus code in Emacs:

1. Open Emacs’ initialization file, whose location should be

~/.emacs

2. Add the follwing code:

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

(defmacro compile-with (command-string)

  `(progn

    (save-buffer)

    (unless visual-line-mode
      (visual-line-mode 1))

    (universal-argument)

    (compile ,command-string)))

(defmacro defcomp (fn command-string kbd-string)

  `(progn

     (defun ,fn ()

       (interactive)

       (compile-with ,command-string))

     (global-set-key (kbd ,kbd-string) ',fn)))

;;;;;;;;;;;;;;;;;;;;;

(defcomp g-la-com

    (concat
     "lazbuild "
     (replace-regexp-in-string
      "pas" "lpi" buffer-file-name))

    "C-x C-p")

;;;;;;;;;;;;;;;;;;;;;

(defcomp g-la-run

    (replace-regexp-in-string
     ".pas" "" buffer-file-name)

    "C-x C-[")

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

3. Close Emacs.

4. Use the Lazarus IDE to create a console project named p11.

5. Close Lazarus.

6. Use Emacs to open the file p11.pas.

— Me@2022-12-07 08:35:48 PM

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.

2022.12.08 Thursday (c) All rights reserved by ACHK

Pier, 2.2

Euler problem 9.1

.

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

(defun e9c ()
  (loop :for a :from 1 :to 1000 :do
    (loop :for b :from 1 :to a :do
      (let ((c (- 1000 (+ a b))))
        (if (= (+ (* a a) (* b b)) (* c c))
            (return-from e9c
                (list a b c (* a b c))))))))

— colorized by palette fm

— Me@2022-12-05 05:59:49 PM

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2022.12.05 Monday (c) All rights reserved by ACHK

Fortran Package Manager

Haskell mode, 3

.

The goal of this blog post is to set up an integrated programming environment for Fortran.

.

1. Read and follow the exact steps of my post titled “Haskell mode“.

2. Read and follow the exact steps of my post titled “Haskell mode, 2“.

3. Install the package manager Anaconda.

4. Use Anaconda to install the Fortran Package Manager (fpm), by following the fpm installation guide.

4.1. Add the additional channel mentioned in the fpm installation guide.

4.2. Install the fpm itself:

conda create -n fpm_env fpm

conda activate fpm_env

5. Install the Fortran language server:

conda install fortls

6. Install the Emacs plugin:

sudo apt-get install elpa-pyvenv

7. Open Emacs’ initialization file, whose location should be

~/.emacs

8. Add the following code to the file.


;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

(defun gfortran-run ()

  (interactive)

  (save-buffer)

  (unless visual-line-mode
    (visual-line-mode 1))

  (universal-argument)

  (compile "fpm run"))

(global-set-key (kbd "C-x C-r") 'gfortran-run)

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

(setenv "WORKON_HOME" "~/anaconda3/envs")

(pyvenv-mode 1)

(pyvenv-workon "fpm_env")

;;;;;;;;;;;;;;;;;;;;;;;;;;;

(add-hook 'f90-mode-hook #'lsp)

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

— Me@2022-12-01 09:18:59 AM

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2022.12.04 Sunday (c) All rights reserved by ACHK

Why does the universe exist? 7.3

Euler problem 10.4

.

program main

  implicit none

  integer :: p_max, p_count, i, j
  integer(kind = 16) :: p_sum
  logical, ALLOCATABLE, DIMENSION(:) :: is_prime

  p_max = 2000000

  ALLOCATE(is_prime(p_max))

  is_prime = .true.

  is_prime(1) = .false.

  do i = 2, ceiling(sqrt(real(p_max)))
     if (is_prime(i)) then
        do j = i*i, p_max, i
           is_prime(j) = .false.
        end do
     end if
  end do

  p_count = 0
  p_sum = 0
  do i = 1, p_max
     if (is_prime(i)) then
        p_count = p_count + 1
        p_sum = p_sum + i
     end if
  end do

  print *, "p_count == ", p_count
  print *, "p_sum == ", p_sum

  DEALLOCATE(is_prime)

end program main

For a universe part, which is partial in space or in time, you can ask for its cause.

But for the universe as a whole, you cannot.

If the big bang is the first cause, you cannot ask for the cause of its existence.

Asking for the cause of the existence of the universe is the same as asking for the cause of the first cause.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-27 09:09:53 PM

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.

2022.12.04 Sunday (c) All rights reserved by ACHK

Hacking Reality

Good marriage is better than no marriage.

No marriage is better than bad marriage.

— John T Reed

.

Your wife should be the exception of the (brutal) reality of this world.

If someone is just another reality, better not to marry her.

— Me@2022-11-10 01:21:15 AM

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.

2022.12.03 Saturday (c) All rights reserved by ACHK

大恩養仇人

小恩養貴人

這段改編自 2021 年 12 月 15 日的對話。

.

no good deed goes unpunished

Beneficial actions often go unappreciated or are met with outright hostility.

If they are appreciated, they often lead to additional requests.

— Wiktionary

.

記住,起點是,你有責任,去解決你自己的問題,但是,你並沒有責任,去解決其他人的問題。有時,間中幫助人,並不是因為責任,而是基於愛心或情義。而「有時候,在你沒有責任幫時,仍然選擇幫」,正正是人性光輝之處。

精確一點用字的話:必須做的工作,為之「責任」;可做可不做的,為之「愛心」或「情義」。

.

既然可幫可不幫,那樣,什麼時候應該幫呢?

大原則是:

1. 那是你自己的能力範圍以內。

2. 當事人有明確要求,或者情境預設,例如,當事人已昏迷,需立刻送院。

一個頭腦清醒的大人,如果沒有明確要求,你就沒有權力去幫助。越權就是無禮。

發乎情 止乎禮

禮者,人與人間之距離也。

.

那樣,在「有必要、應該幫」時,又應該幫到哪個程度呢?

「幫助別人」就有如「責備別人」,雖然有時必須,但是越小越好,越少越好。

「幫助別人」就有如「服用藥物」,雖然有時必須,但是越小越好,越少越好。

.

助人最小化,效果最大化。

— Me@2022-12-02 04:34:36 PM

.

.

2022.12.02 Friday (c) All rights reserved by ACHK

ParEdit

.

(autoload 'enable-paredit-mode
  "paredit" "Turn on pseudo-structural editing." t)

(add-hook 'emacs-lisp-mode-hook
      #'enable-paredit-mode)

(add-hook 'eval-expression-minibuffer-setup-hook   
      #'enable-paredit-mode)

(add-hook 'ielm-mode-hook
      #'enable-paredit-mode)

(add-hook 'lisp-mode-hook
      #'enable-paredit-mode)

(add-hook 'lisp-interaction-mode-hook
      #'enable-paredit-mode)

(add-hook 'scheme-mode-hook
      #'enable-paredit-mode)


(add-hook 'slime-repl-mode-hook
      (lambda () (paredit-mode +1)))

(defun override-slime-repl-bindings-with-paredit ()
   (define-key slime-repl-mode-map
     (read-kbd-macro paredit-backward-delete-key) nil))

(add-hook 'slime-repl-mode-hook
      'override-slime-repl-bindings-with-paredit)

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— Me@2022-11-29 10:03:49 PM

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2022.11.29 Tuesday (c) All rights reserved by ACHK

Ex 1.2-1 Stationary States

Quantum Methods with Mathematica

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Assume a wavefunction of the form psi[x, t] == f[t] psi[x] and perform a separation of variables on the wave equation.

Show that f[t] = E^(-I w t) where h w is the separation constant. Try the built-in function DSolve.

Equate h w to the Energy by evaluating the [expected] value of hamiltonian[V] in the state psi[x, t].

~~~

Remove["Global`*"]


hbar := \[HBar]

H[V_] @ psi_  := -hbar^2/(2m) D[psi,{x,2}] + V psi



psi[x_,t_] := f[t] psi[x]

I hbar D [psi[x,t],t] == H[V] @ psi[x, t]

I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]

\displaystyle{i \hbar  \psi (x) f'(t)=V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}

\displaystyle{\frac{i \hbar  f'(t)}{f(t)}=\frac{V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}{f(t) \psi (x)}}

E1 := I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]

Simplify[E1]

\displaystyle{\frac{1}{2} \hbar  \left(\frac{\hbar  \psi ''(x)}{m \psi (x)}+\frac{2 i f'(t)}{f(t)}\right)=V}

E2 := - 1/2 hbar hbar (D[D[psi[x],x],x]/(m psi[x])) == hbar omega

DSolve[E2, psi[x], x]


E3 := 1/2 hbar 2 i D[f[t],t] / f[t] == hbar omega

DSolve[E3, f[t], t]

\displaystyle{\left\{\left\{\psi (x)\to c_1 \cos \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)+c_2 \sin \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)\right\}\right\}}

\displaystyle{\left\{\left\{f(t)\to c_1 e^{\frac{\omega  t}{i}}\right\}\right\}}


k

psi[x_] := c E^(I k x)

psi[x]

f[t_] := E^(-I omega t)

f[t]

psi[x_,t_] := f[t] psi[x]

psi[x,t]

\displaystyle{  \left\{k,c e^{i k x},e^{-i \omega  t},c e^{i k x-i \omega  t}\right\}  }

E4 := Conjugate[psi[x,t]] H[0] @ psi[x,t]

E4

E5 := Simplify[E4]

E5

k := Sqrt[2 m omega / hbar]

Refine[E5, {Element[{c, omega, m, t, hbar, k, x}, Reals]}]

\displaystyle{  \frac{c k^2 \hbar ^2 c^* \exp \left(-i \left(-(\omega  t-k x)^*-k x+\omega  t\right)\right)}{2 m}  }

\displaystyle{  = c^2 \omega  \hbar  }

E6 := Conjugate[psi[x,t]] psi[x,t]

Simplify[E6]

\displaystyle{  c c^* \exp \left(-i \left(-(\omega  t-k x)^*-k x+\omega  t\right)\right)  }

\displaystyle{  = c^2  }

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\displaystyle{\begin{aligned}            \langle E \rangle     &= \frac{\int_{-\infty}^{\infty} \psi^* H_{V=0} \psi dx}{\int_{-\infty}^{\infty} \psi^* \psi dx} \\ \\     &= \frac{c^2 \omega  \hbar \int_{-\infty}^{\infty} dx}{c^2 \int_{-\infty}^{\infty} dx} \\ \\    &= \omega  \hbar \\    \end{aligned}}

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— Me@2022-11-26 07:17:29 PM

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2022.11.28 Monday (c) All rights reserved by ACHK

Why does the universe exist? 7.2

“There is nothing in that region of space”

and

“there is nothing outside the universe”

have different meanings.

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there is nothing except quantum fluctuations in that region of space

= the best detector detects nothing but quantum fluctuations

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there is nothing outside the universe

= whatever detected, label the whole collection as “the universe”

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“There is nothing outside the universe” does not (!!!) mean that “we go outside the universe to keep searching, but find nothing”.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-27 09:09:53 PM

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2022.11.28 Monday (c) All rights reserved by ACHK