Chapter 2 Intro to Math Techniques for Quantum Mechanicsscienide2.uwaterloo.ca/~nooijen/Chem356/Chem_356_pdf_backup/Ch_2.pdf · Winter 2013 Chem 356: Introductory Quantum Mechanics

Winter 2013 Chem 356: Introductory Quantum Mechanics

22

Chapter 2 – Intro to Math Techniques for Quantum Mechanics ............................................................... 22

Intro to differential equations ................................................................................................................ 22

Boundary Conditions ............................................................................................................................... 25

Partial differential equations and separation of variables ..................................................................... 25

Introduction to Statistics ......................................................................................................................... 30

Chapter 2 – Intro to Math Techniques for Quantum Mechanics

Intro to differential equations

Function ( )y y x is to satisfy a differential equation

2

25 6 0

d y dyy

dxdx x (1)

For this ‘type’ of Differential equation (more later), try solution xy e

Then xdye

dx

2

2

2

xd ye

dx

n

n x

n

d ye

dx

Substitute into differential equation (1)

2 5 6 0x x xe e e x

Or

2 5 6 0

( 3)( 2) 0

Solutions: 3 , 2

3xe : 3 3 39 15 6 0x x xe e e

2xe : 3 3 34 10 6 0x x xe e e

But: any linear combination of solutions is also solution


Chapter 2 – Intro to Math Techniques for Quantum Mechanics 23

3 2

1 2( ) x xy x c e c e

3 2

1 29 4x xc e c e

3 2

1 215 10x xc e c e

3 2

1 26 6x xc e c e

0 0

Let us try another one

2

20

d yy

dx x

2 0x xe e

2 1 0 i

General Solution: 2

ix ix

ic e c e

Alternative way to write:

cos sinixe x i x

cos sin( ) cos sinixe x i x x i x

1 2 1 2( ) ( )cos ( )siny x c c x i c c x

1 2cos sind x d x

define 1 1 2d c c ,

2 1 2( )d i c c

→ choose 1d ,

2d ‘real’

Verify:

2

2cos ( sin ) cos

d dx x x

dxdx

2

2sin (cos ) sin

d dx x x

dxdx

2

20

d yy

dx , as expected

Type of solutions xe , xe , cos x , sin x , real, >0



When does this work?

2 3

1 2 32 30 .....B

dy d y d yc y c c c

dx dx dx

(1) Constant coefficients in front of y and its derivatives

→ not: 2

2

20

d yx y

dx

(2) Linear in function y

→ not: 2

20

d y dyy

dxdx

(3) Homogeneous equation

→ not: 2

2 1 320

y yc c c y

xx

For inhomogeneous differential equation:

→ Find particular solution ( )y P x add to this the general solution of inhomogeneous

equation.

For more complicated differential equations (ie. Not homogeneous DE with constant

coefficients) solutions are often hard to find

Many tricks of the trade

Use symbolic math program (it knows many of the tricks)

Numerical approaches (often work very easily → picture of solution)



Boundary Conditions

Let us consider our original differential equation.

2

25 0

d y dyy

dxdx

3 2

1 2( ) x xy x c e c e

Now impose further conditions. Eg:

(0) 0y 1 2 0c c

0

1x

dy

dx

1 23 2 1c c

2 1c c

2 1c

1 13 2 1c c

1 1c

3 2( ) x xy x e e statistics DE and boundary conditions

Solution is completely specified if one supplies as many conditions as one has free

coefficients 1c ,

2c , …. in the solution

→ always linear set of equations

So recipe is very simple

Try ( ) xy x e and work it out!

Partial differential equations and separation of variables

Consider problem of vibrating string (eg. guitar, violin)

We want to describe the amplitude ( , )u x t



DE 2 2

2 2 2

1( , ) ( , )

u ux t x t

x v t

v : Velocity of wave propagation in string, related to spring constant, (as sound of the

string)

Boundary Condition:

(0, ) ( , ) 0u t u a t t

( , )u x t : function of 2 variables → use partial derivatives t

u

x

In math we typically do not write is kept constant (compare thermodynamics)

How to solve PDE (partial differential equation)?

Try solution ( , ) ( ) ( )u x t X x T t

Simple product of a function of x and a function of t

Boundary Condition: (0) ( ) 0X X a

Substitute trial function into PDE

2 2

2 2 2

( ) 1( ) ( )

d X x d TT t X x

dx v dt

Divide both sides by ( ) ( )X x T t :

2 2

2 2 2

1 ( ) 1 1

( ) ( )

d X x d T

X x T tdx v dt

only depends on x only depends on t

Like ( ) ( )f x g t should be true for all x , t

0( ) ( )f x g t → ( )f x is constant

1( )g t → ( )f x is constant, should be same

0( ) ( )g t f x → ( )g t is constant

1( )f x

( ) ( )f x g t ,x t

Can only be true if both functions are constant! Ie. The same constant

Let us call this constant, the separation constant 2k (for later simplicity)



2

2

2( )

d Xk X x

dx (0) ( ) 0X X a

2

2

2 2

1( )

d Tk T t

v dt no boundary conditions

Now we can use techniques discussed before ( k is constant)

Try ( ) xX x e

2 2x xe k e

ik , ik

1 2( ) ikx ikxX x c e c e

Note k could be imaginary im 2 2 2( )k im m 0

However we know that the string will oscillate, and hence can anticipate ikxe , with k real

Using what we did before

1 2

ikx ikxc e c e 1 2 1 2( )cos ( )sinc c kx i c c kx

1 2cos sind kx d kx

This is general solution. Now consider boundary conditions.

0x : 1 2cos 0 sin 0d k d k

1 21 0 0d d

1 0d

x a : 2 sin( ) 0d ka

2 0d (flat string possibility) or sin( ) 0ka

When is sin( ) 0ka ? 0, , 2 , 3 ...x

ka n n

ka

General solution: ( ) sinn

n xX x d

a

2 22

2n

nk

a

This is solution for x at particular value for n

nk

a

Now consider corresponding solution for ( )T t



22

2 2

1( )

d T nT t

av dt

2

2

2( )n

d Tw T t

dt

n

n vw

a

Similar equation as before:

1 2( ) sin ( ) cos( )n nT t c w t c w t

If we combine this with X we get

( , ) sin sin sin cosn n

n x n vt n x n vtu x t A B

a a a a

This is a solution for any value ofnA ,

nB and any value of 0, 1, 2, 3n

Most general solution:

0,1,2,3...

( , ) sin sin( ) cos( )n n n n

n

n xu x t A w t B w t

a

You can verify that this indeed satisfies PDE

How can we interpret this?

sinn x

a

:



sin sinn x n x

a a

:

Same solution, restrict 0,1,2,3n

( 0n , nothing extra)

So a string vibrates as a linear combination of modes, each of the modes oscillates in time at a

different frequency.

sinn x

a

n

n vw

a

0nw

All multiples of fundamental frequency 0

vw

a

This defines the pitch of the sound 0w

The other modes are called over tones

Meaning of coefficients nA ,

nB ?

0t cosn

n vB

a

sinn x

a

The initial shape of function

cosn

du n vA

dt a

the initial velocity of the string.

Different instruments, guitars, violins, cello



different nA ,

nB

How you attack the string determines the initial shape/velocity

compare Chinese zither: hit the snare in different spots or twang the string

None of this affects the pitch 0w the general harmonic

Period sin( ) sin ( )wt w t T

sin( 2 ) sin( )wt wt

2 wt

2

Tw

;

2w

T

Introduction to Statistics

We will see that quantum mechanics is essentially a statistical theory. We can predict

the results and their distribution from a large number of repeated experiments only. We cannot

predict (even in principle) the outcome of an individual experiment.

Let us therefore talk about statistics using a simple example: the dice

If you throw the dice, each throw will yield the result 1, 2, 3, 4, 5 or 6.

If you throw the dice many times,

say 6000 times, we might get

For a fair dice, each number has equal chance, and so we say the probability to throw

for example a ‘3’ is 1

6. This is reflected by the actual numbers we got in the example.

1000 1

6000 6000 6

i

i

nP

In the limit that we throw a very large (infinite) times, we get closer and closer to 1

6

1 1010

2 980

3 995

4 1025

5 1030

6 960

Total 6000



i

i

total

nP

N only has meaning for many repeated experiments

1iP

We might call ia the actual outcome of experiment, here

ia = 1, 2, 3, 4, 5, 6

The average is given by

1

i i

itot

n aN

i i

i

Pa

For dice:

1 21 1

(1 2 3 4 5 6) 36 6 2

The average value does not need to be a possible outcome of an individual throw.

We are also interested in the variance of the results. We call the average A or A . (both are used)

Then the variance is given by:

2 2( )A i i

i

P a A 0

Let us take a dice with 5 sides to make the numbers easier

ia 1, 2, 3, 4, 5 ,

1

5iP

A 3

2 2 2 2 2 21[(1 3) (2 3) (3 3) (4 3) (5 3) ]

5A

1

[4 1 0 1 4] 25

Standard Deviation 2A

I can write the variance differently as

2 2 2( ) ( 2 )i i i i i

i

P a A P a a A A

2 22i i i i i

i i i

Pa A Pa A P

2 2A AA AA



22 2

AA A

Let us check for the 5 face dice:

2 2 2 2 2 21(1 2 3 4 5 )

5A

1

(1 4 9 16 25)5

1

(55) 115

23 3 9A

22 22 AA A (as before)

Of course: We proved this is true mathematically!

This concludes (for now) discussion of discrete statistics.

Now consider the case of a continuous distribution. For example a density distribution.

( )x dx dm

the mass between x and x dx (in 1 dimension)

( )x dx M

total mass

Also ( )

b

a

x dx mass between points a and b

If we normalize

1

( ) ( )P x dx x dxM

probability to find a fraction of the total mass between x and x dx

We can define average position

( )x xP x dx

2 2 ( )x x P x dx

22 2

x x x



Example: take a box between 0 and a

Uniform 1

( )P xa

0 x a

0 elsewhere

1) 00

1( ) 1

aax

P x dx dxa a

“normalized”

2) 2

0 0

1( )

2 2

aax a

xP x dx xdxa a

3) 3

2 2 2

0 0

1 1( )

3 3

aax

x P x dx x dx aa a

4)

2 222 2 21 1

3 2 12x

ax x a a

0 Always

More complicated distributions are possible of course

Famous is the Gaussian distribution

2

22( )

x

aP x ce

‘ a ’ parameter (will be x )

c : normalization constant

( ) 1P x

2

1

2c

a

Then

2

221

02

x

ax xea

2

22 2 21

2

x

ax x e dxa

Odd function ( ) ( )f x f x

2 2

21 22

22

a aa

a

x a as advertised

(this was the reason to define the Gaussian like this)

See book for discussion integrals

Chapter 2 Intro to Math Techniques for Quantum Mechanicsscienide2.uwaterloo.ca/~nooijen/Chem356/Chem_356_pdf_backup/Ch_2.pdf · Winter 2013 Chem 356: Introductory Quantum Mechanics

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