The document summarizes the history of solving cubic equations and the discovery of complex numbers. It describes how del Ferro, Tartaglia, and Bombelli contributed to solving cubic equations of the form x3 = bx + c. Bombelli realized that taking the cube root of a negative number results in an "imaginary" number containing the square root of -1. He used this insight to find all three cube roots of a complex number. Modern algebra approaches finding the cube roots of a complex number using polar form and trisecting angles. This allowed solving any cubic equation with three real or complex solutions. The document traces the conceptual breakthroughs that led to the acceptance and understanding of imaginary and complex numbers.

1. 1

2. 2
if iwere not there it would have been
impossible …
A man is like a fraction whose numerator is what he is and whose
denominator is what he thinks of himself. The larger the
denominator the smaller the fraction.
Priyatosh Dutta√-1

3. A Few Puzzles
1. A 25 cubic feet water Tank is 1 ft taller than it is wide, and 1 ft longer than it is
the dimension of the Tank.
~ 24.5 inches wide; x(x+1)(x+2) =25, a simple(!) cubic equation! What is the exact
solution?
2. Find he Number (x) whose triple is 4 plus its square. [1884, Nicolas Chuquet]
X2 + 4 = 3x, x = (3/2) ± √(-7/14)
It’s a solution, but impossible solution!! It can not be placed in the real
number Line.
[(3/2) ± √(-7/14)]2 +4 (definitely) = 3 [(3/2) ± √(-
7/14)]
3. 1803, Lazare Carnot, Geometrie de Position:
How a line segment AB of length a can be divided so that the product of the len
Of the shorter segments is one-half of the square of the original length a?
x (a-x) = a2/2 ; x = a/2 ± a √(-1)/2
Newton: “… the equation has one true affirmative solution, two negative ones
and two
impossible ones”.
3

4. Attempts To Understand √(-m)
Socrates’ Square
Puzzle
To Construct a Square whose area is twice the
Area of
A given Square
1
1
1
1
√2
1
1
√2
√2 
Diagonal;
A sq ft = Area  √A ft = side
Geometrically √ is perceived as equivalent to
length
What if Area (A) < 0 ? Say, A= -m, Side = √A =
√(-m)
Negative Area and Square Root of (Negative): Non Real
Situation
Definition: The Square Root of a positive Number (N) is Just a quantity (S) whose
square (S2)
Is that Number (N). √N = S or N = S2 ; But S2 = Always Positive, so also N.
So a Negative number is not entitled to have a Square Root.
4

5. There Is No More
OR
There Is [Even] More …
Problems versus Good Problems!!!
Category I SOLVE THIS! An Equation, Perhaps; Just Solve It! There is No M
Category II
important: Conceptual Structure of the solution, NOT the Solution Itself.
There is EVEN
MORE.
• Solving Gets You to a Deeper Level of QUESTION
ASKING
• A Letter of Introduction to a Level of Interaction that You Have NOT Achieved Be
• An Invitation to Extend Your
Imagination
Poincare Conjecture (Yet to achieve!):
TRUE or NOT TRUE? It Extends our 3D geometric Intuition
So, √-2 Exists or Not? : What [even] more It HAS? How To Reach What √-2 Really Is
5

6. Time: Past, Present and Future and i =√-1: A Mystical Approach, Abbe Buee (1748-
1826)
If t = future time, -t= past time, t 0 = present time, then t0 is composed
of
(t √-1)/2 and (–t √-1)/2;
t0 = c1. (t √-1)/2 + c2 (–t √-1)/2; for c1=c2, t0 =0  present is as mysterious as
ZERO!
c1 ≠ c2, t0 = (c1 – c2). (t √-1)/2 ; t0 contains √-1, so present can not be
placed on the
Real line of past and future. So where to place it?
• √-1 is the sign of perpendicularity. Failed to define multiplication of line
segments.
The Thinker? “We are the bees of the Invisible” (Rainer Maria Rilke)
“… to go to the edge and report
back”
“… having-gone-to-the-very-end in an
experience,
to where one can go no further.”
“The eureka moment in a Story. A moment of restless anticipation in the face
of slowly
emerging act of imagining”.
The story  The history of Mathematics; Act  Not in any single mind
and
Moment a period that may stretch over centuries.
The bees took 300 years to go to the edge and bring back the drop of honey [√-1 or i ]
from the invisible, imaginative world. So: √-1 = Imagination, an imaginary Number.
6

7. •What is Imagination? How does it work? How it differs from that in Poe
•Can one feel the “working” of imagination in mathematics?
•Do these different imaginations are complimentary to each other?
1. Phantasies
(Greek)
Sight’ of something seen before,
absent now.
‘Mone koro ma jachhi
Ghorai Chare…
Visio (Latin)= sight  Imagination (English)
2. A number of abstract ideas compounded into one image [J.
Bentham]
Rakkhas, Khokkas,
Rupkatha
3. Function of connecting mere facts with eternity/infinity [no Poetry otherwise],
Wordsworth
Bar Aaschhe Topar Mathai Diye; Topar  symbol of eternal unity
4. Coleridge: Fancy: Less Daring Sibling of
Imagination
Mode of memory emancipated from
Order of Time and space.
5. Object and Felt Inner Experience “Imagining a Flower” vs. “Fearing an Earthquake”
6. “The Missing mystery of Philosophy and an Unacknowledged Question Mark”, Eva Bro
7. “Horn of Light, Intermediary between Being (Its Narrow End) and Not Being (Its infinitely
Wide End)”, 2oth century Sufi ,Ibn al-Arabi
7

8. How We Imagine and Express That …
1. Visualization vs.
Experiencing
• Mind’s Game Beyond Mind or “Mind’s Mind” [9th
consciousness?]
[familiar picture on mind’s screen]
• One Can Express Difficult to Express or
describe• Gharana (Classify, Articulate and explain On Stage Performance (Nikhil
Banerjee)
2. Phantastikon: A Kind of Elementary Particle of
Imagination;
(Interior Analogue of Sensory Perception). Stoic,
Chrysippus
3. “…better to go down upon your marrow-bones,
for to articulate sweet sound together
is to work harder than all these…” W.B. Yeats in Adam’s Curse
4. “…all the images rose up [to me] as things, with a parallel production of the expression
without any sensation or consciousness of effort…” , Coleridge
Claims to have composed at least 200-300 lines in his SLEEP [of poem
“Kabul Khan”
5. “…unsettling feeling of a “poem to come” –as disturbing as the aura of
Migraine”
Stephen Dobyns (Poet).
Joy Goswami: Kabita in a moving Tram, Ekta Ghorer mato…
8

9. Breakthrough for √-1: Not from Quadratic [x2 +1 =0], rather from Cubic
equations
The Duel: 1535: Venice: M. Fiore and N. TartagliaX3 = X + 1; X3 = b X + c
NO Sridharacharyya, NO Simple formula: del Ferro [1465-1526], Not so
simple!!!
X
=
3
c/2 + c2/4 – b3/27
3
c/2 - c2/4 – b3/27
+
[Gift from a Magician
Genius]
1. The Duel-Challenge Problem: X3 = X + 1, b=1, c=1, d= 1/4 – 1/27 = 23/108
> 0
X = {1/2 + √(23/108)}1/3 + {1/2 - √(23/108)}1/3 = 1.324717957
2. X3 = 3 X -2, b=3, c= -2; d= 4/4 – 27/27 = 0;
X =
3
-
1
3
-1+
How this curious expression would give the two real solutions X = 1 and
X = -2?
Rafael Bombelli [1526-72], L’Algebra
You must have cubed -1? (-1)3 = -1; so -1 is one of the cube roots of -1.
Ever cubed (1 ± √-3)/2? {(1 ± √-3)/2}3 = -1 also. Got 2 other cube roots of -1.
So X = -1 + (-1) X = (1 + √-3)/2 + (1 - √-3)/2X = 1/2 + 1/2 = 1
X = {c/2 + √d}1/3 + {c/2 -√d}1/3 ; where d = c2/4 –
b3/27
9

10. 3. X3 = 15 X + 4, b= 15, c= 4, d= 16/4- 3375/27 = 4 -125 = -121 ; √d= √(-121) =
11i
X = {c/2 + √d}1/3 + {c/2 -√d}1/3 ; X = {2 + 11i}1/3 + {2 – 11i}1/3
X = {D}1/3 + {D*}1/3 ; where D = 2 + 11i and D* = 2 – 11i
How to find the cube roots of D and D* , two complex numbers, conjugate to each
other?
Bombelli Insight Again!
Let z = (2 + i q) is a complex number which is the required cube root of D.
Z = {D}1/3 or Z3 = D i.e. (2 + iq)3 = D or 8 – iq3 + 12i q – 6 q2 = D = 2 + 11i
Or, 8 – 6q2 = 2 or 6q2 = 6 or q = 1 and i(12q – q3)= 11i or q(12-q2) = 11  q=1
So, z = (2 + 1i) is the cube root of D, similarly Z* = (2 – 1i )= {D*}1/3
X= Z + Z* = 2 + 1i + 2 – 1i = 4  one of the roots (X) of the given equation X3 =
15 X + 4
Two other solutions are X = -2
± √3
10

11. 4. X3 = 3 X + 1, b= 3, c= 1, d= c2/4 – b3/27 = 1/4- 1 = -3/4; √d= √(-3/4)= (√3 i)/2
X = {c/2 + √d}1/3 + {c/2 -√d}1/3 ; X = {1/2 + √(3i)/2}1/3 + {1/2 - √(3i)/2}1/3
X = {D}1/3 + {D*}1/3 ; where D = 1/2 + √(3i)/2 and D* = 1/2 - √(3i)/2
How to find the cube roots of D and D* , two complex numbers, conjugate to each
other?
Modern algebra (of complex number)! The Algorithm follows:
• D = a + i b; the polar form of D  (r, ɸ) where r = {a2 + b2}, ɸ = tan-1(b/a)
• The three cube roots of D (say, P, Q and R) will have the same modulus “ r1/3 ” and
• angles Ɵ1 = ɸ/3, Ɵ2 = ɸ/3 + 120, and Ɵ3 = ɸ/3 + 240 [you need only to be able to
trisect an angle (ɸ) and extract the cube root of a real number “r”].
• P = r1/3 {cos Ɵ1 + i sin Ɵ1 ) ; Q = r1/3 {cos Ɵ2 + i sin Ɵ2 ) and R = r1/3 {cos Ɵ3 + i
sin Ɵ3 )
Similarly for D*, the three cube roots are P*, Q*, and R*
P* = r1/3 {cos Ɵ1 - i sin Ɵ1 ) ; Q* = r1/3 {cos Ɵ2 - i sin Ɵ2 ) and R* = r1/3 {cos Ɵ3 - i
sin Ɵ3 )
Now, the three solutions are X1 = P + P* , X2 = Q + Q* , and X3 = R + R*
11

12. Example: X3 = 3 X + 1, b= 3, c= 1, d= c2/4 – b3/27 = 1/4- 1 = -3/4; √d= √(-3/4)=
(√3 i)/2
X = {c/2 + √d}1/3 + {c/2 -√d}1/3 ; X = {1/2 + √(3i)/2}1/3 + {1/2 - √(3i)/2}1/3
X = {D}1/3 + {D*}1/3 ; where D = 1/2 + √(3i)/2 and D* = 1/2 - √(3i)/2
• D = a + i b; a = 1/2 and b= √3/2, r = {(1/2)2 + (√3/2)2}1/2 = (1/4 + 3/4)1/2 = 1,
and
• ɸ = tan-1[ (√3/2)/(1/2) ] = tan-1[ √3]  ɸ = 60;
• angles Ɵ1 = 60/3 = 20, Ɵ2 = ɸ/3 + 120 = 140, and Ɵ3 = ɸ/3 + 240 = 260
• P = 11/3{cos 20 + i sin 20); Q = r1/3{cos 140 + i sin 140) and R = r1/3 {cos 260 +
i sin 260)
• X1 = 2 cos 20 = 1.879385, X2 = 2 cos 140 = -1.532088 and X3 = 2 cos 260= -
0.347296Similarly for X3 = 15 X + 4 we get all the three solutions: X = 4 and
X = -2 ± √3
12

13. X=
3
c/2 + c2/4 – b3/27 3
c/2 - c2/4 – b3/27
+
Unfolding the Gift Pack (from the Magician Genius] X3 = b X +
c
X = {D}1/3 + {D*}1/3How did del Ferro know to do this???
Let X = u + v  X3 –bX = c  (u + v)3 – b(u + v) = c  u3 + v3 + (3uv –b) (u+v) = c ,
3uv –b = 0 and u3 + v3 = c
v = b/(3u) u3 + b3 /(27 u3 ) = c
u6 – c u3 + b3 /27 = 0 a huge step backward? Not
Really!
sixth degree, but quadratic in u3
Magician’s Trick: Splitting into two
parts:
{u3}2 – c u3 + b3 /27 = 0;  u3 = c/2 ± {c2/4 - b3 /27}1/2
u = [ c/2 + {c2/4 - b3 /27}1/2 ]1/3  v3 = c - u3  v = [ c/2 - {c2/4 - b3 /27}1/2 ]1/3
X = {D}1/3 + {D*}1/313

14. Visualizing/ Representing Z = a + i b:
Targ
et
2. The Negative of Z and complex conjugate of Z: Any Graphical
Representation?
1. Is a point good enough to represent Z? What does it
mean?
3. Addition of two complex numbers: Z = Z1 + Z2 : What
does it give?
5. The Polar Form of complex number Z  (r, ɸ).
What it is?6. Representing three cube roots of unity. four 4th roots of unity
8. The three cube roots of Z= 1 + i : An example
and finally9. Visualizing the three solutions of X3 = b X + c; An
example
1. Comprehend the idea of number as
transformation, and2. To work at visualizing these
transformations.
Imagining imaginary number is not a simple act of visualization. Two Steps
are there:
4. Multiply by i: What happens
visually?
7. Raising to power: Zm [Z3 for
example].
MarriagebetweenAlgebraand
Geometry
14

15. Z = x + i y
x
y
Z = x + iy in a complex plane
Z
-Z
The Negative of Z is its reflection through the
origin
Z
Z*
The complex conjugate of Z is its
reflection through the axis.
Z
r
ɸ
The polar form of Z 
(r, ɸ)
Z=x + i y = r. cosɸ + I
.r.sinɸ
15

16. Z1 + Z2
Z1
Z2
Addition of Z1 and Z2 :
Vector Parallelogram Law
ɸ = 3π/4√
2
Z = -1 + i
The complex Number Z = -1 + I in polar form
Z
Z3
(r, ɸ) (√2, 450)
(r3, 3ɸ)
(2√2, 1350)
Z3 and Z in the complex plane; Z
= 1+ i
90o a + i b
-b + i a
Multiplying by i is rotation by 90o
Zm = rm (cos mɸ + i sin mɸ)
Z3 = 2√2(cos 135 + i sin 135)
= -2 + 2 i
16

17. Three Cube Roots of Unity Four 4th Roots of Unity
Three Cube Roots of Z =
1 + i
17

18. PQ
R
Q
R
P a =
0.93693…
X =
1.879386…
a=-0.766044…
X=-1.532088…
a = -0.173648…
X = -
0.347296…
Three Solutions of X3 = 3 X + 1
X = {D}1/3 + {D*}1/3
P, Q, and R  three cube roots of
D
P or Q or R  a + i b; X = 2 a
18

19. Complex Number: An Ant: Carrying Load Much More Than its weight
A pair of frequently used Approximations and
an even more useful Exact Relation
 ex ≈ 1 + x; ln(1 + x) ≈ x, x is small  e±Ɵ = cos Ɵ ± i sin Ɵ
Taylor’s Series: f(x) = f(a) + (x-a) fˊ (a) + {(x-a)2/2!} fˊˊ(a) + {(x-a)3/3!} fˊˊˊ(a) + . . .;
take a=0,
f(x) = f(0) + x fˊ (0) + (x2/2!) fˊˊ(0) + (x3/3!) fˊˊˊ(0)
+ . .
Example 1 ln(x) = (x-1) - (x-1)2/2 + (x-1)3/3 - …
ln(1+x) = x- x2/2 + x3/3 - …
Clearly, for small x, xn ≈ 0 for n ≥ 2; ln(1 + x) ≈ x
Example 2 sin(x) = x - x3/3! + x5/5! – x7/7! + x9/9! - …
Example 3 cos(x) = 1 - x2/2! + x4/4! – x6/6! + x8/8!
- …Example 4 ex = 1 + x + x2/2! + x3/3! + x4/4! + x5/5! + x6/6! + x7/7! + …
Clearly, for small x, xn ≈ 0 for n ≥ 2; ex ≈ 1 +
x
Now comes His Majesty complex number Z = x + i y
19

20. ez = 1 + z + z2/2! + z3/3! + z4/4! + z5/5! + z6/6! + z7/7! + …
Not Happy? Well, may take a purely imaginary number, say z = i y
eiy = 1 + (iy) + (iy)2/2! + (iy)3/3! + (iy)4/4! + (iy)5/5! + (iy)6/6! + (iy)7/7! + …
=1 + iy - y2/2! - iy3/3! + y4/4! + iy5/5! - y6/6! - iy7/7! + …
= (1 - y2/2! + y4/4! - y6/6! + …) + i(y - y3/3! + y5/5! - y7/7! + …)
eiy = cos (y) + i sin(y) similarly e-iy = cos (y) - i sin(y)
Take y = π, eiπ = cos (π) + i sin(π) = 1 eiπ + 1 = 0, The
most Beautiful
(e, π), (i), (0,1), complex  real
combination
π from i using eiπ + 1 = 0 : The beauty of the beautiful [Euler]
eiπ + 1 = 0 Now ln(1+ i) – ln(1- i) = 2 i (1 – 1/3 +1/5 -1/7
+1/9 - …)
eiπ = -1 = i2 Expand ln{(1+i)/(1-i)} = ln i = 2 i (1 – 1/3 +1/5 -1/7 +1/9 - …)
ln eiπ = ln i2 ln(1±z), i π/2 = 2 i (1 – 1/3 +1/5 -1/7 +1/9 - …)
iπ = 2 ln i take z=i π/4 = 1 – 1/3 +1/5 -1/7 +1/9 - …20

21. Form DOES Matter: A few examples
1. Raising Z to power m Z = x + i y then Zm = ?
Z = r cos ɸ + i r sin ɸ = r(cos ɸ + i sin ɸ) = reiɸ
Zm = (reiɸ )m = rm eimɸ = rm [cos (mɸ) + sin
(mɸ)]
2. Excellent Trigonometric
IdentitieseiA = cos A + i sin A, eiB = cos B + i sin B
eiA.eiB =(cos A + i sin A)(cos B + i sin B) =(cos A cos B – sinA sinB) + i (sin A
cos B +cos A sin B)
But eiA.eiB = ei(A+B) = cos (A+B) + i sin (A+B)
cos (A+B) = cos A cos B – sinA sinB sin (A+B) = sin A cos B +
cos A sin B
Special Case: A = B,
cos 2A = cos2 A – sin2 A sin 2A = 2 sin A cos
A21

22. 3. A famous formula in the history of π
Z1 = r1 eiɸ1 , Z2 = r2 eiɸ2 ; and Z3 = r3 eiɸ3 ;
Take Z3 = Z1 Z2  r3 eiɸ3 = (r1 eiɸ1) ( r2 eiɸ2 ) = (r1 r2) ei(ɸ1+ ɸ2) So, r3 = r1r2, ɸ3 =
ɸ1 + ɸ2
Z1 = 2 + i, ɸ1 = tan-1 (1/2), and Z2 = 3 + i, ɸ2 = tan-1 (1/3)
Z3 = Z1 Z2 = (2+i) (3+i) =5 + 5i,  ɸ3 = tan-1 (5/5) = tan-1 1 = π/4;
from ɸ1 ,ɸ2 and ɸ3 we have tan-1 (1/2) + tan-1 (1/3) = π/4
4. A common saying may not be true in general: 1x = 1 for any x?
Obviously its true for x being ± integers including zero, or fraction. What
about 1π ?
1  (1eiɸ), ɸ= tan-1 (0/1) = 2nπ, n=0,1,2,…, so 1= (e2inπ ) , or 1π = (e2inπ )π [Lamber
Or, 1π = cos(2nπ2) + i sin(2nπ2) ; clearly for n=0, 1π =1.
Now, 2nπ2 = π(2nπ) and for n ≠ 0, 2πn is never an integer [ π is irrational] and
Hence sin(2nπ2) will never vanish. So, 1π will always have a nonzero imaginary p
Similarly 1e [ e is irrational] also has an infinite distinct complex values. [Euler 1
22

23. Touch of Genius: Euler and Poisson
I1 = ∞∫0 sin(s)2 ds = I2 = ∞∫0cos(s)2 ds = (1/2) √(π/2)
1743: He Could find convergent infinite series for numerical calculation only of I1
and I2.And he took another nearly 40 years to evaluate them exactly: April 30,
1781.
And he did it using Complex quantities.
That “HE” is none other than all time great Leonhard Euler.
6. I = 1∫-1 dx/x = ??? [Poisson (1781-1840) in 1815]
From Symmetry Argument: I = 0 as 1/x is an odd function.
There are infinite negative area from -1 to 0 and infinite positive area from
0 to 1.
But ∞ + (-∞) can be anything, not just zero.
Integrating along the real axis, at the origin x=0, the integrand blows up.
Poisson’s Trick to Avoid Integrand Explosion: Using complex number: x =-eiƟ , 0 ≤ Ɵ≤
1∫-1 dx/x = π∫0 (-i eiƟ /-eiƟ )dƟ = π∫0 i dƟ = i π
Integration path swings around the fatal point x=0, along a smicircular arc in the
upper half of the complex plane.
5. Using The Complex To Do The Real :Euler
23

24. Imagining yourself As Erwin Schrodinger in 1926
[How √-1 appears in a real-world physical problem?]
1900: Planck: BBR (E = hν) 1905: A. Einstein: P.E.E. (E = hν)
1924: de Broglie: Matter Wave (p=h/λ)
1926: Schrodinger, In Search of the matter wave equation!
A few features of the Expected Equation of matter wave
 The Equation can not be derived from other principle
It itself constitutes a fundamental law of nature
 Fix the Solution first, then trace back to the Equation that gives the Solution.
 The correctness can be judged only by consequent agreement with observed pheno
(a posteriori proof).
The wave equation that motivated Schrodinger for a Clue
Maxwell’s equation for each component of electric and magnetic fields in vacuum
Ψ – (1/c2 ) ∂2 Ψ/∂t2 = 0 where 2 [the Laplacian or “del-
squared” is
= ∂2 /∂x2 + ∂2 /∂y2 + ∂2 /∂z2 ;
you want to have an analogous equation for the de Broglie matter
wave
2
2
Let us consider a wave motion propagating in X-direction. At x and t, the form of the
might be represented by a product of periodic functions, one periodic in space (x) an
the one is periodic in time (t )such as
24

25. Ψ(x) = f(2 π x/λ) and Ψ(t) = g(2πνt).
f repeats its value for x increases by one wavelength λ.
ν represents the number of cycles of the wave per unit time.
Taking into account both x- and t- dependence, consider the wave Ψ(x,t) as
Ψ(x,t) = Ψ(x) Ψ(t) = f(2 π x/λ). g(2πνt).
f and g might be sinusoidal functions like sin Ɵ, cos Ɵ, or some linear
combination of them.
Scrodinger’s choice: Derivatives of exponentials are simpler than those of sines or
cosines.
Is there any suitable relation that bind exponentials and sinusoidals together?
You are Lucky and should be grateful to Euler. Complex exponential gives that
expression: e±iƟ = cos Ɵ ± i sin Ɵ;
So, f (2 π x/λ) = ei 2πx/λ and g (2πνt) = e±2πiνt
Ψ(x,t) = Ψ(x) Ψ(t) = f(2 π x/λ). g(2πνt) = ei 2πx/λ . e-2πiνt = Exp[2
πi(x/λ - νt)]
(-) sign indicates that the wave is travelling from left to right.
So far, so Good! It has ν and λ i.e. you’ve got the wave form.
But where is the particle? How is it a matter wave?
Nowhere the particle, neither its mass (m) nor its energy(E) or
momentum(p).
Simpler” at the cost of a “complex”? Grammatical? Simple and Complex get
synonymous.
You won’t mind really as you’ve not demanded so far that Ψ(x,t) has to be real!
25

26. Ψ(x,t) = Exp[2πi(x/λ - νt)]
Planck + de Broglie  Scrodinger
Planck: E = hν or ν = E/h de Broglie: p=h/λ or 1/λ = p/h
Ψ(x,t) = Exp[2πi(px/h – Et/h)]
Ψ(x,t) = Exp[ ipx/ħ ] Exp[ -iEt/ħ] ħ  “aitch-bar”, Dirac
At last the wave like nature of a particle (p, E) got its expression.
It’s supposed to be solution of an equation. But where is the Equation?
We need a “backward journey” or “Reverse Engineering” to get that …
What is that? How can you do that? The Reverse Engineering?
From child (solution) to its mother (parent equation that produces that
solution)???
Ψ = Ψ(x,t): Your character (variation) is your identity! Why don’t you see the
variation of the man [Ψ(x,t)]. Yes, rightly guessed, its all about differentiation o
w.r.to its variable x and t. Do it!26

27. Ψ(x,t) = Exp[ i(px –Et)/ħ]
∂Ψ(x,t)/∂t = (∂/∂t) Exp[ i(px –Et)/ħ]
= (-iE/ħ) Exp[ i(px –Et)/ħ]
= (-iE/ħ) Ψ(x,t), or
Ψ(x,t) = Exp[ i(px – Et)/ħ]
∂Ψ(x,t)/∂x = (∂/∂x) Exp[ i(px – Et)/ħ]
= (ip/ħ) Exp[ i(px – Et)/ħ]
= (ip/ħ) Ψ(x,t), or
Tempted for a 2nd derivative? Well,
Do it!
-ħ2 (∂2Ψ/∂x2 ) = p2 Ψ
p2 /2m = E or p2 =2mE
-ħ2 (∂2Ψ/∂x2 ) = (2mE) Ψ
-(ħ2/2m)(∂2Ψ/∂x2 ) = E Ψ = (iħ)(∂/∂t)Ψ
(iħ)(∂/∂t)Ψ = -(ħ2/2m)(∂2Ψ/∂x2 )
(iħ)(∂/∂t)Ψ = -(ħ2/2m)(∂2Ψ/∂x2 ) + V(x)Ψ,
[for a particle with P.E = V(x,t)].
(iħ) (∂/∂t)Ψ = E Ψ,
-ħ2 (∂2Ψ/∂x2 ) = p2 Ψ
You’ve got two derivatives involving E & p.
Are you going to get two equations then?
Never mind. Try to merge them into one…
How you can do that?
For waves in 3 dimension:
(iħ)(∂/∂t)Ψ(r,t) = -(ħ2/2m) 2 + V(x)] Ψ(r,t)
(iħ)(∂Ψ/∂t) = HΨ GOT it? The Baby Got
Adult.27
(iħ) (∂/∂t)Ψ = E
Ψ
(-iħ) (∂/∂x)Ψ = p
Ψ

28. HΨ = (iħ)(∂Ψ/∂t) , where Ψ = Ψ(r,t) = Ψ(r) Ψ(t) = Ψ(r) Exp [ (-i E t) /ħ]
Now consider a conservative [E ≠ f(t)] system i.e. V(x) and H ≠ f(t) . What follows?
H Ψ(r) Ψ(t) = (iħ)(∂[Ψ(r) Ψ(t)]/∂t) , Ψ(t) = Exp [ (-i E t) /ħ]
Ψ(t) [H Ψ(r)] = (iħ) Ψ(r) (∂Ψ(t)/∂t) (∂Ψ(t)/∂t) = (-i E /ħ) Ψ(t)
= (iħ) Ψ(r) (-i E /ħ) Ψ(t)
= E Ψ(r) Ψ(t) or
H Ψ(r) = E Ψ(r)  Time independent Schrodinger
equation.
Haven't You witnessed the Birth of Modern Quantum Mechanics?!
Rewind a little bit: (-iħ) (∂/∂x)Ψ = p Ψ: If you like the colour-match, you can
equate them
(-iħ) (∂/∂x) = p or p= (-iħ) (∂/∂x)= (ħ/i) (∂/∂x)  you may accept it as a justified
Postulate.
28

29. if i were not there: A Postscript on the Wave function Ψ(r,t)
Ψ(r,t) is complex (in general). You don’t insist it to be real either. What the hell is it then
Just a mere mathematical abstraction! But abstraction of what reality?
Truly, Ψ(r,t) has no “physical significance” in the sense that it does not correspond to
any
”observable” or physically measurable quantity. But for Schrodinger’s sake, please
hold your
tongue and let me LOVE (this non-real one)!
z = (a + i b), z* = (a – i b), Amplitude of z = z = (a2 + b2 )1/2 (REAL)
Amplitude Ψ i.e. (Ψ* Ψ) or Ψ 2 is a real quantity. Now You can assign it
the duty of representing something real and meaningful:
In EMR (Maxwell’s), Amplitude of the wave is intensity of field. For matter
wave?
Max Born: (Ψ* Ψ) = probability density.As a Result: Constraints on its behavior, to be a “well Behaved” one.
As a Result: Quantization, Naturally arising quantum numbers.29

30. if i were not there: A Postscript on What matters and matters
not …
1. p= (-iħ) (∂/∂r) operator (linear momentum)
2. (iħ)(∂Ψ/∂t) = HΨ time dependent Schrodinger
equation
3. Ψ(r,t) = Ψ(r) Exp [ (-i E t) /ħ] The matter wave
4. [x, px ] = i ħ [commutation]
5. (d/dt) <Q> = (i /ħ) <[H, Q]> [commutation and conservation]
6. Uncertainty: Δx Δpx = ΔEΔt = ħ/2 Δx Δpx ≥ (1/2) |˂ [x, px ]˃| and [x, px ]
= i ħ
7. [H, px ] = (i ħ) (d/dx) V(x)  (d/dt)<px > = -<(d/dx)V(x)>= force
 Newton’s
[H, x ] = (ħ/I ) (p/m)  m (d/dt)< x >= < px >= momentum  Law
A bridge (one way) between Classical and Quantum: Ehrenfest’s Theorem.“Classical” deals with Average values, “quantum” deals with underlying details.
The minute details of your Mind is subject to Quantum which, on average manife
The classical YOU.30

31. References
1. An imaginary Tale: The Story of √-1, Paul J Nahin
2. Imagining Numbers, Bary Mazur
3. GAMMA: Exploring Euler’s Constant
4. What Is Mathematics, R. courant and H. Robbins, Revised by Ian Stew
5. Mathematical Methods for Scientist and Engineers, Donald A. McQua
6. Molecular Quantum mechanics, Atkins and Friedman
7. Introduction To Quantum Mechanics, S.M. Blinder
8. A Life of Schrodinger, W.J. Moore
31