Chapter2–!IntrotoMathTechniquesfor!Quantum…scienide2.uwaterloo.ca/~nooijen/Chem356/Chem+356+pdf/Ch_2.pdf ·...

Fall 2014 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

2 5 6 0d y dy ydxdx

− + = x∀ (1)

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

Then xdy edx

λλ=

2

22

xd y edx

λλ=

n

n xn

d y edx

λλ=

Substitute into differential equation (1)

2 5 6 0x x xe e eλ λ λλ λ− + = x∀ Or

2 5 6 0λ λ− + = ( 3)( 2) 0λ λ− − =

Solutions: 3λ = , 2λ =

3xe : 3 3 39 15 6 0x x xe e e− + =

2 xe : 4e2x −10e2x + 6e2x = 0

But: any linear combination of solutions is also solution


Chapter 2 – Intro to Math Techniques for Quantum Mechanics 23

3 21 2( ) x xy x c e c e= +

9c1e

3x + 4c2e2x ( d 2 y

dx2 )

−15c1e

3x −10c2e2x (−5 dy

dx)

6c1e3x + 6c2e

2x (6y)

0 0

Let us try another one

2

2 0d y ydx

+ = x∀

2 0x xe eλ λλ + = 2 1 0λ + = iλ = ±

General Solution: 2ix 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

2 cos ( sin ) cosd dx x xdxdx

= − = −

2

2 sin (cos ) sind dx x xdxdx

= = −

2

2 0d y ydx

+ = , as expected

Type of solutions xeλ , xe λ− , cos xλ , sin xλ , real, >0λ λ



When does this work?

2 3

1 2 32 30 .....Bdy d y d yc y c c cdx dx dx

= + + + +

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

→ not: 2

22 0d y x y

dx+ =

(2) Linear in function y

→ not: 2

2 0d y dyydxdx

+ =

(3) Homogeneous equation

→ not: c2

∂2 y∂x2 + c1

∂y∂x

+ c3 y = f (x)

For inhomogeneous differential equation (last case 3): → 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

2 5 0d y dy ydxdx

− + =

3 21 2( ) x xy x c e c e= +

Now impose further conditions. Eg: (0) 0y = 1 2 0c c+ =

0

1x

dydx =

⎛ ⎞ =⎜ ⎟⎝ ⎠ 1 23 2 1c c+ =

2 1c c= −

→ 3c1 − 2c1 = 1 1 1c = → c2 = −1

3 2( ) x xy x e e= − Satisfies both 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 eλ= 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

Differential Equation with partial derivatives:

2 2

2 2 2

1( , ) ( , )u ux t x tx 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

∂u∂x

⎛⎝⎜

⎞⎠⎟ t

In math we typically do not write is kept constant (in contrast to 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 xdx 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= . It is clear that this 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! I.e. The same constant Let us call this constant, the separation constant 2k− (for later simplicity)

2

22 ( )d X k X x

dx= − (0) ( ) 0X X a= =

2

22 2

1 ( )d T k T tv dt

= − no boundary conditions

Now we can use techniques discussed before ( k is constant) Try ( ) xX x eλ=

2 2x xe k eλ λλ = − ikλ = , ikλ = − 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 2ikx 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(x) = 0 ? 0, , 2 , 3 ...x π π π= ± ± ± ±

x → ka = nπ nkaπ=

General solution: ( ) sinnn xX x daπ⎛ ⎞= ⎜ ⎟⎝ ⎠

2 2

22n

nkaπ− = −



This is solution for x at particular value for nnkaπ=

Now consider corresponding solution for ( )T t

22

2 2

1 ( )d T n T tav dtπ⎛ ⎞= −⎜ ⎟⎝ ⎠

d 2Tdt2 = −ω n

2T (t) ω n =

nπva

Similar equation as before:

T (t) = c1 sinω n(t)+ c2 cos(ω nt) If we combine this with X we get

u(x,t) = An sin

nπ xa

⎛⎝⎜

⎞⎠⎟

sinnπvt

a⎛⎝⎜

⎞⎠⎟+ Bn sin

nπ xa

⎛⎝⎜

⎞⎠⎟

cosnπvt

a⎛⎝⎜

⎞⎠⎟

This is a solution for any value of nA , nB and any value of 0, 1, 2, 3n = ± ± ±

Most general solution (we can restrict to non-‐negative n):

u(x,t) = sin

nπ xa

⎛⎝⎜

⎞⎠⎟n=0,1,2,3...

∑ An sin(ω nt)+ Bn cos(ω nt)⎡⎣ ⎤⎦

You can verify that this indeed satisfies PDE (quite some work).

How can we interpret this?

sin n xaπ

:



sin sinn x n xa aπ π−⎛ ⎞ = −⎜ ⎟⎝ ⎠

:

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.

sin n xaπ

ω n =

nπva

= nω0

All multiples of fundamental frequency πva

=ω0

This defines the pitch of the sound ω0

The other modes are called overtones

Meaning of coefficients nA , nB ?

0t = Bn cos

nπva

t⎛⎝⎜

⎞⎠⎟ t=0

= Bn → u(x,t = 0) = Bn

n∑ sin nπ x

a⎛⎝⎜

⎞⎠⎟

The initial shape of function



dudt

→ An cosnπvt

a⎛⎝⎜

⎞⎠⎟ t=0

= An ;du(x,t)

dt t=0

= An sin(nπ x

an∑ )

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 → ω0 the general harmonic

Period sin(ωt) = sinω (t +T )

sin(ωt + 2π )→ sin(ωt)

2π =ωT

T = 2π

ω ;

ω = 2π

T angular frequency

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.

1 1010 2 980



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

1000 1

6000 6000 6i

in

P≈ = =

In the limit that we throw a very large (infinite) times, we get closer and closer to 16

ii

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 iitot

n aN ∑ i i

iPa∑

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 iiP a Aσ = −∑ 0≥

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

ia =1, 2, 3, 4, 5 , 15iP =

3 995 4 1025 5 1030 6 960

Total 6000



A =3

2 2 2 2 2 21[(1 3) (2 3) (3 3) (4 3) (5 3) ]5Aσ = − + − + − + − + −

1[4 1 0 1 4] 25

= + + + + =

Standard Deviation 2Aσ =



I can write the variance differently as

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

P a A P a a A A− = − +∑ ∑

2 22i i i i ii i iPa A Pa A P= − +∑ ∑ ∑

2 2A AA AA= − +

22 2

AA A σ= − ≡

Let us check for the 5 face dice:

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

A = + + + +

1 (1 4 9 16 25)5

= + + + +

1 (55) 115

= =

2 3 3 9A = ⋅ = 22 22 AA A σ− = = (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.

ρ(x)dx = dm

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

ρ(x)dx = M

−∞

∞

∫ total mass

Also ρ(x)dx =

a

b

∫ mass between points a and b

If we normalize

P(x)dx = 1

Mρ(x)dx

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

We can define average position



x = xP(x)dx−∞

∞

∫

x2 = x2P(x)dx−∞

∞

∫

σ x

2 = x2 − x2

Example: take a box between 0 and a

Uniform 1( )P xa

= 0 x a< <

0= elsewhere

1)

P(x)dx = 1a

dx0

a

∫ = xa⎤

⎦⎥0

a

= 1−∞

∞

∫

“normalized”

2)

xP(x)dx = 1a

x dx0

a

∫ = x2

2a⎤

⎦⎥

0

a

= a2−∞

∞

∫

3)

x2P(x)dx = 1a

x2 dx0

a

∫ = x3

3a⎤

⎦⎥

0

a

= 13

a2

−∞

∞

∫

4) σ x

2 = x2 − x2= 1

3a2 − 1

2a

⎛⎝⎜

⎞⎠⎟

2

= a2

12 0≥ Always

More complicated distributions are possible of course Famous is the Gaussian distribution

P(x) = ce− x2

2a2 ‘a ’ parameter (will be xσ )

c : normalization constant

P(x)−∞

∞

∫ = 1 c = 1

2πa2

Then

x = 1a 2π

xe− x2

2a2

−∞

∞

∫ = 0

Odd function ( ) ( )f x f x− = −



x2 = 1a 2π

x2e− x2

2a2

−∞

∞

∫ dx

2 2

21 2 222

a a aa

ππ

= ⋅ ⋅ =

x aσ = as advertised

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

See book for discussion integrals

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