The first order perturbation calculation for the helium atom ground state is treated in detail in the textbook “Quantum Mechanics Third Edition” by Leonard I. Schiff pages 257 to 259. I offer a numerical algorithm for computing the electron-electron repulsion interaction which is analytically determined by Schiff and other scientists. Next is the graphical user interface for the application and its output.
Application Graphical User Interface
The Ep text box is the ground state energy as found by a first order perturbation computation. The Ee text box is the experimental ground state energy. The IA text box is the analytic electron-electron repulsion interaction determined by Schiff and other quantum mechanics researchers. The IC text box is my numerical contribution. All the energies are in electron volts.
The application source code are the next items in this blog.
In my current return to my youthful dual interests in quantum chemistry and quantum mechanics that occupied much of my time in the 1960s, 1970s and 1980s, I am now using my knowledge of experimental numerical analysis. My interest in computer science and numerical analysis began in the summer of 1976 while I was a chemistry student at my local college namely LaGrange College in LaGrange, Georgia. As a child and teenager I was very interested in several disciplines of physics: classical mechanics, quantum mechanics, and the theories of special and general relativity. Later I added to my knowledge toolkit some tidbits of statistical mechanics and statistical thermodynamics.
This blog entry will explore the wonderful world of the hydrogenic atom which used to known by the moniker, hydrogen-like atom. The most well known isotope of hydrogen has one electron and one proton and its atomic number is 1 and it is sometimes denoted by the letter and numeral Z = 1. Of course, there are multiple other isotopes of hydrogen including deuterium (one proton and one neutron) and tritium (one proton and two neutrons). Hydrogen is the only atom whose wave functions both non-relativistic (see Erwin Schrödinger) and relativistic (view Paul Adrian Maurice Dirac) have analytic close formed solutions. Hydrogen is the most abundant chemical element on Earth and in the universe. The stars initially use a hydrogen plasma as a nuclear fuel to create more massive atomic ions and release massive amounts of nuclear fusion energy.
Way back in the 1920s Erwin Schrödinger decided to apply his work in wave hydrology to the newly found branch of physics known as quantum theory and quantum mechanics. From his work the branch of quantum mechanics known as wave quantum mechanics evolved. This branch was as important as another competing theory of quantum mechanics known as matrix quantum mechanics that was being concurrently developed by Werner Heisenberg. The key process in the derivation of a Schrödinger equation for any time independent scenario is to apply the first quantization rules to a valid classical Hamiltonian. The classical Hamiltonian is the total energy of a system and is the sum of the kinetic energy and the potential energy. The classical Hamiltonian for the hydrogen-like atom is shown in equation (1).
Equation (1), Many Sources
The first quantization rule is to apply the conversion from a classical momentum vector to a momentum quantum mechanical operator using the equation (2).
Equation (2), Several Sources
The lower case m is the mass of the electron and the upper case M is the mass of the atomic nucleus which is the Z times the proton mass plus the number of neutrons times the neutron mass. The Greek letter mu is the reduced mass of the hydrogen-like system. The italic i is the imaginary unit that is the square root of the number -1. The transcendental number pi is represented by the Greek letter pi and has the truncated real number value of 3.1415926535897932384626433832795. Schrödinger plugged Equation (2) into Equation (1) and found a three-dimensional Cartesian coordinate second order partial differential equation (3) that used the operator discovered by the mathematician Laplace.
Equation (3), Merzbacher, Messiah, Schiff Et. Al.Equation (4), Several Sources
In equation (4) the first partial differential operator is the Laplace operator which is the vector inner product of the three-dimensional Cartesian gradient operator from vector analysis. The scalar r in equation (4) is the Euclidean distance from the electron to the nucleus. The Greek letter psi (“pitchfork”) in equation (3) is the illustrious and elusive wave function.
The first thing that struck Schrödinger was that the equation (3) that he derived by much thought was unfortunately not a separable partial differential equation in three-dimensional Cartesian coordinates, however, he next applied a coordinate coordinate transformation from three-dimensional Cartesian coordinates to three dimensional spherical polar equations specified by the equations in the following PDF with some derivations.
The wave function for the hydrogen-like atom is dependent on the associated Laguerre polynomials and the spherical harmonics that dependent upon the associated Legendre functions.
using System;
namespace SteadyStateTempCylinder
{
public class PotPoint : IComparable
{
private double x, y, u;
public double X
{
get
{
return x;
}
set
{
x = value;
}
}
public double Y
{
get
{
return y;
}
set
{
y = value;
}
}
public double U
{
get
{
return u;
}
set
{
u = value;
}
}
public PotPoint(double x, double y, double u)
{
this.x = x;
this.y = y;
this.u = u;
}
public int CompareTo(object obj)
{
if (obj == null)
return 1;
PotPoint pp = (PotPoint)obj;
if (u > pp.u)
return 1;
else if (u == pp.u)
return 0;
else
return -1;
}
}
}
Sometimes in my group therapy, we play a game of taking an anagram and unscrambling the puzzle and determining all the words that can be created from the unscrambled anagram letters. Suppose we have the scrambled word “cimdteos“ then the following list is created using my new application.
1 demotics 2 domestic 3 ed 4 em 5 me 6 mo 7 om 8 to 9 ti 10 it 11 cs 12 med 13 mot 14 tom 15 tic 16 cit 17 sic 18 sci 19 demo 20 dome 21 mode 22 mote 23 tome 24 omit 25 tics 26 cits 27 stoic 28 sitcom 29 demotic 30 do 31 es 32 st 33 ts 34 mod 35 mes 36 ems 37 est 38 set 39 sit 40 tis 41 its 42 some 43 mets 44 stem 45 ties 46 site 47 domes 48 demos 49 modes 50 motes 51 tomes 52 smote 53 mites 54 emits 55 smite 56 times 57 items 58 cites 59 modest
An anagram is a scrambled or jumbled word. It is a permutation of the letters in a word. There are six permutations of a three letter word such as “the” and are “the”, “teh”, “hte”, “het”, “eth”, and “eht”. The number of permutations of a word consisting of n letters is n!, for example, 4! = 4 * 3 * 2 * 1 = 4 * 3! = 24, 5! = 5 * 4! = 120, etc. I wrote a program to brute force solve anagrams using an English language dictionary consisting of 152,512 words, a hash table, and a permutation generator. The hash table is generated from the English language dictionary by the formula first character integer ASCII value squared + the second character integer ASCII value . The base hash table entry for the word “an” is 65 * 65 + 78 = 4,303. There will be collisions or words with the same hash code. In this case there is a list of words for each hash code value. The application can solve 12 character anagrams in less than one hour on my computer where 12! = 479,001,600. Some anagrams can represent more than one word so a list of potential anagrams is created. Some example executions of the C# application are illustrated below.
I very recently created a multiple precision signed integer package in C++ using the standard library and a base of 10. I then implemented two integer factoring algorithms trial division and Pollard’s Rho method. Trial division uses all the prime numbers <= 10000 and there are 1229 such primes. Due to the choice of language and the exceedingly small base the resulting application is awfully slow when compared to a similar C# application. The multiple precision signed integer package is largely based on translation of the Pascal source code found in “Prime Numbers and Computer Methods for Factorization” by Hans Riesel. The test number is 2 ^ 72 – 1 which is a Mersenne composite integer. The output large integers start with the number of base 10 digits in the number. For comparison I have included the output from my C# Big Integer factorization program as a screen shot.
Menu
0 Exit Application
1 Test of Package
2 Trial Division
3 Pollard Rho
2
n = 4722366482869645213695
Duration (min:sec.mil) = 00:08.784
n is composite, factors:
3 ^ 3 factor has 1 digits
5 factor has 1 digits
7 factor has 1 digits
13 factor has 2 digits
17 factor has 2 digits
19 factor has 2 digits
37 factor has 2 digits
73 factor has 2 digits
109 factor has 3 digits
241 factor has 3 digits
433 factor has 3 digits
Large prime 5 3 8 7 3 7
Menu
0 Exit Application
1 Test of Package
2 Trial Division
3 Pollard Rho
3
n = 4722366482869645213695
n is composite, factors:
2 2 1
1 3
1 3
4 3 5 1 5
2 1 3
4 1 2 4 1
3 2 4 1
3 1 0 9
3 4 3 3
5 3 8 7 3 7
1 3
1 3
1 3
1 5
1 7
2 1 3
2 1 7
2 1 9
2 3 7
2 7 3
3 1 0 9
3 2 4 1
3 4 3 3
5 3 8 7 3 7
Duration (min:sec.mil) = 00:02.197
I started creating computer programs the Summer Semester of 2002 at Auburn University. I took a course named “Hand Held Software Development”. The course was taught by my research advisor Associate Professor Richard O. Chapman. I created a distributed chess playing client-server Internet system for human chess players. The program was built using the Palm Pilot’s Palm OS and its C language compiler. Later in circa 2006 I built a rule based and neural network chess program for a computer to play itself and for a human observer (voyeur). The language was Java. Then in 2015 I translated and enhanced my Java program by rewriting the code in C#. Below are pictures of one game.
The description of this alphabetic letter game is very facile. Given a word make as many other words as possible using the letters of the given initial word. But first before we enumerate the game solution, we need to refresh the reader’s memory of some elementary mathematics.
The binary number system also known as the base 2 number system is used by computers to perform arithmetic. The digits in the binary number system are 0 and 1. The numbers 0 to 15 in binary using only four binary digits are where ^ is the exponentiation operator (raising a number to a power) are:
0 0000
1 0001 2 ^ 0 = 1
2 0010 2 ^ 1 = 2
3 0011 2 ^ 1 + 2 ^ 0 = 2 + 1 = 3
4 0100 2 ^ 2 = 4
5 0101 2 ^ 2 + 2 ^ 0 = 4 + 1 = 5
6 0110 2 ^ 2 + 2 ^ 1 = 4 + 2 = 6
7 0111 2 ^ 2 + 2 ^ 1 + 2 ^ 0 = 4 + 2 + 1 = 7
8 1000 2 ^ 3 = 8
9 1001 2 ^ 3 + 2 ^ 0 = 8 + 1 = 9
10 1010 2 ^ 3 + 2 ^ 1 = 8 + 2 = 10
11 1011 2 ^ 3 + 2 ^ 1 + 2 ^ 0 = 8 + 2 + 1 = 11
12 1100 2 ^ 3 + 2 ^ 2 = 8 + 4 = 12
13 1101 2 ^ 3 + 2 ^ 2 + 2 ^ 0 = 8 + 4 + 1 = 13
14 1110 2 ^ 3 + 2 ^ 2 + 2 ^ 1 = 8 + 4 + 2 = 14
15 1111 2 ^ 3 + 2 ^ 2 + 2 ^ 1 + 2 ^ 0 = 8 + 4 + 2 + 1 = 15
An algorithm to convert a base 10 (decimal) number to base 2 (binary) number is given below:
Input n a base 10 number
Output b[0], b[1], b[2], … a finite binary string representing the decimal number
Integer i = 0
Do
Integer nmod2 = n mod 2
Integer ndiv2 = n / 2
b[i] = nmod2 + ‘0’
i = i + 1
n = ndiv2
While n > 0
The b[i] will be in reverse order. For example, convert 12 from decimal to using four binary digits:
12 mod 2 = 0
12 div 2 = 6
b[0] = ‘0’
i = 1
n = 6
6 mod 2 = 0
6 div 2 = 3
b[1] = ‘0’
i = 2
n = 3
3 mod 2 = 1
3 div 2 = 1
i = 3
b[2] = ‘1’
n = 1
1 mod 2 = 1
1 div 2 = 0
b[3] = ‘1’
n = 0
So, the reversed binary string of digits is “0011”. And after reversing the string we have 12 is represented by the binary digits “1100”.
Next, we need to define a power set and its binary representation. The index power set of 4 objects which has 2 ^ 4 = 16 entries is specified in the following table:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
A permutation of the set of three indices is given by the following list:
012, 021, 102, 120, 201, 210
A permutation of n objects is a list of n! = n * (n – 1) * … * 2 * 1. A permutation of 4 objects has a list of 24 – 4-digit indices list since 4! = 4 * 3 * 2 * 1 = 24 has the table:
0123, 0132, 0213, 0231, 0312, 0321,
1023, 1032, 1203, 1230, 1320, 1302,
2013, 2031, 2103, 2130, 2301, 2310,
3012, 3021, 3102, 3120, 3201, 3210
Suppose our word is “lost” then we first find the power set:
1 0001 t
2 0010 s
3 0011 st ts
4 0100 o
5 0101 ot to
6 0110 os so
7 0111 ost ots sot sto tos tso
8 1000 l
9 1001 lt tl
10 1010 ls sl
11 1011 lst lts slt stl tsl tls
12 1100 lo ol
13 1101 lot lto olt otl tlo tol
14 1110 los lso slo sol osl ols
15 1111 lost lots slot etc.
Using a dictionary of 152,512 English words my program finds 16 hits for the letters of “lost”:
Dictionary Length: 152512
Word: lost
0 l
1 lo
2 lost
3 lot
4 lots
5 ls
6 o
7 s
8 slot
9 so
10 sol
11 sot
12 st
13 t
14 to
15 ts
Total letters and/or words 16
Next, we use “tear” as our word:
Dictionary Length: 152512
Word: tear
0 a
1 are
2 art
3 at
4 ate
5 e
6 ea
7 ear
8 eat
9 era
10 et
11 eta
12 r
13 rat
14 rate
15 re
16 rt
17 rte
18 t
19 tar
20 tare
21 tea
22 tear
23 tr
Total letters and/or words 24
Finally, we use the word “company”:
Dictionary Length: 152512
Word: company
0 a
1 ac
2 am
3 amp
4 an
5 any
6 c
7 ca
8 cam
9 camp
10 campy
11 can
12 canopy
13 cap
14 capo
15 capon
16 cay
17 cm
18 co
19 com
20 coma
21 comp
22 company
23 con
24 cony
25 cop
26 copay
27 copy
28 coy
29 cyan
30 m
31 ma
32 mac
33 man
34 many
35 map
36 may
37 mayo
38 mo
39 moan
40 mop
41 mp
42 my
43 myna
44 n
45 nap
46 nay
47 nm
48 no
49 o
50 om
51 on
52 op
53 p
54 pa
55 pan
56 pay
57 pm
58 pony
59 y
60 ya
61 yam
62 yap
63 yo
64 yon
Total letters and/or words 65
The C++ program's source code is given below:
// WordToWords.cpp : This file contains the 'main' function. Program execution begins and ends there.
//
#include "pch.h"
#include <algorithm>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> dictWords;
bool ReadDictionaryFile()
{
fstream newfile;
newfile.open("C:\\Users\\james\\source\\repos\\WordToWords\\Dictionary.txt", ios::in);
if (newfile.is_open()) {
int index = 0, length = 128;
char cstr[128];
while (newfile.getline(cstr, length)) {
string str;
str.clear();
for (int i = 0; i < (int)strlen(cstr); i++)
str.push_back(cstr[i]);
dictWords.push_back(str);
}
newfile.close();
sort(dictWords.begin(), dictWords.end());
return true;
}
else
return false;
}
string ConvertBase2(char cstr[], int n, int len)
{
int count = 0;
string str, rev;
do
{
int nMod2 = n % 2;
int nDiv2 = n / 2;
str.push_back(nMod2 + '0');
n = nDiv2;
} while (n > 0);
n = str.size();
for (int i = n; i < len; i++)
str.push_back('0');
n = str.size();
for (int i = n - 1; i >= 0; i--)
if (str[i] == '1')
rev.push_back(cstr[i]);
return rev;
}
vector<string> PowerSet(char cstr[], int len)
{
vector<int> index;
vector<string> match;
for (long ps = 0; ps <= pow(2, len); ps++)
{
string str = ConvertBase2(cstr, ps, len);
int psf = 1;
for (int i = 2; i <= len; i++)
psf *= i;
for (int i = 0; i < psf; i++)
{
if (binary_search(dictWords.begin(), dictWords.end(), str))
{
if (!binary_search(match.begin(), match.end(), str))
{
match.push_back(str);
sort(match.begin(), match.end());
}
}
next_permutation(str.begin(), str.end());
}
sort(match.begin(), match.end());
}
return match;
}
int main()
{
bool done = false;
char cstr[128];
int len;
string str;
vector<int> index;
vector<string> match;
if (!ReadDictionaryFile())
return -1;
cout << "Dictionary Length: " << dictWords.size() << endl << endl;
cout << "Word: ";
cin >> cstr;
cout << endl;
len = strlen(cstr);
if (len != 0)
{
vector<string> match = PowerSet(cstr, len);
for (int i = 0; i < match.size(); i++)
{
cout << setprecision(3) << setw(3) << i << "\t";
cout << match[i] << endl;
}
cout << endl;
cout << "Total letters and/or words " << match.size() << endl;
cout << endl;
}
}
Numerical Explorations 