Fall$2014$ Chem%356:%Introductory%Quantum%Mechanicsscienide2.uwaterloo.ca/~nooijen/Chem356/Chem+356+pdf/Ch_3.pdf · Fall$2014$Chem%356:%Introductory%Quantum%Mechanics$ Chapter$3$–$Schrodinger$Equation,$Particlein$aBox$38$

Fall 2014 Chem 356: Introductory Quantum Mechanics

34

Chapter 3 – Schrodinger Equation, Particle in a Box .................................................................... 34

Introduction to the Schrodinger Equation ................................................................................ 34

Linear Operators ....................................................................................................................... 36

Quantization of energy ............................................................................................................. 39

Interpretation of Wave Function .............................................................................................. 40

Determination of Constant C .................................................................................................... 42

Useful integrals for particle in the box ..................................................................................... 44

Demonstration of Uncertainty Principle ................................................................................... 44

Particle in a 3 dimensional box ................................................................................................. 46

Chapter 3 – Schrodinger Equation, Particle in a Box

Introduction to the Schrodinger Equation

De Broglie suggested one can associate a wave with a particle and take hpλ

=

ikxe 2kπλ =

p = h

2πk = !k

Generalization to 3 dimensional wave

ei!k i!x

!p = "!k

In chapter 2 we saw that waves in general satisfy a wave equation.

Try to postulate a wave equation for “electron-‐waves” (a guess) Provide some rational for Schrodinger equation:

Wave equation

∂2u∂x2 = 1

V2

∂2u∂t2

Choose solution with particular ω = ν

2π

u(x,t) =ψ (x)cos(ωt)


Chapter 3 – Schrodinger Equation, Particle in a Box 35

d 2ψdx2 + ω 2

V2 ( ) 0xψ =

ω = 2πv νλ = V , ω

V= 2πν

νλ= 2π

λ ν (nu) frequency ; V

velocity

d 2ψdx2 + 4π 2

λ 2 ψ (x) = 0

hp

λ =

4π 2

λ 2 = 4π 2

h2 ⋅ p2 = p!

⎛⎝⎜

⎞⎠⎟

2

!2 ∂2ψ

dx2 + p2ψ (x) = 0

Now substitute 2p :

Let V = V(x) indicate potential:

p2

2m+V = E

!2

2m∂2ψ∂x2 + p2

2mψ (x) = 0

!2

2m∂2ψ∂x2 + (E −V )ψ (x) = 0

Or Eψ (x) = − !

2

2m∂2ψ∂x2 +V (x)ψ (x)

Hψ (x)

We obtain a differential equation for function ( )xψ

Hψ (x) = Eψ (x) E is a constant, the energy H is “operator” that acts on a function. Summarizing:

1) p2ψ (x) = −!2 ∂2ψ

∂x2 (using de Broglie + classical wave equation)



2) Substitute 2 2 ( ( ))p m E V x= − 2 2

2 ( ) ( ) ( )2

V x x E xm x

ψ ψ ψ∂− + =∂

h

ˆ ( ) ( )H x E xψ ψ=

H = p2

2m+ V (x) ‘energy operator’ (see later)

We need to discuss 2 mathematical items

a) Operators p , H , 2p ….? b) Eigenvalue equations

H Eψ ψ= E, !p : numbers

pψ = !pψ Operators will be indicated by “^” hat or carot

Linear Operators (in 1 dimension first)

ˆ ( ) ( )Af x g x=

Acting with an operator on a function yields a new function.

A ( )f x ˆ ( ) ( )Af x g x= 2

2

ddx

2x 0

2

2 2 3d ddxdx

⎛ ⎞+ +⎜ ⎟

⎝ ⎠ 3x 2 36 6 3x x x+ +

dxdx

2x x

ddx

(x2 ) = x ⋅2x = 2x2

d xdx

2x

ddx

(x ⋅ x2 ) = ddx

(x3) = 3x2

−i!

ddx

ikxe !keikx

− !

2

2md 2

dx2 +V (x)⎛⎝⎜

⎞⎠⎟ cos( )kx

!2k 2

2m+V (x)

⎛⎝⎜

⎞⎠⎟

cos(kx)



The operators we consider are linear operators:

1 1 2 2ˆ( ( ) ( ))A c f x c f x+

= ( ) ( )1 1 2 2ˆ ˆ( ) ( )c Af x c Af x+

Where 1c , 2c are (complex) constants

Example of operator that is not linear: SQR( ( ))f x 2( ( ))f x

2 2SQR( ( ) ( )) ( ( )) ( ( )) 2 ( ) ( )f x g x f x g x f x g x+ = + + =SQR( ( )) SQR( ( )) 2 ( ) ( )f x g x f x g x+ +

Not linear therefore We can act with operators in sequence ˆ ˆˆ ˆ( ) ( ( ))ABf x A Bf x= In general: ˆ ˆˆ ˆ( ) ( )ABf x BAf x≠

Example A x= , ˆ dBdx

=

( )d dfx f x xdx dx

⎛ ⎞ =⎜ ⎟⎝ ⎠

( ) ( ( )) ( )d d dfx f x xf x f x xdx dx dx

⎛ ⎞ = = +⎜ ⎟⎝ ⎠

If ˆ ˆˆ ˆ( ) ( )ABf x BAf x= , for any ( )f x we write

ˆ ˆˆ ˆ 0AB BA− =

ˆ ˆ[ , ] 0A B = A and B commute, the order does not matter. This will play an important role later on.

Eigenvalue equations (by example)

Aψ (x) = aψ (x)

Acting with A on a function yields the same function multiplied by a constant. Example:

−i!

∂∂x

eikx = (−i!)(ik)eikx

= !keikx



22

ikxh eππ λ

⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

interpretation: ikx ikxh e peλ

= =

2i xikxe eπλ= periodic with period λ

We say px = −i!

∂∂x

ˆ ikx ikxx xp e p e=

Number The wave function ikxe is an eigenfunction of operator ˆ xp

px = −i!

ddx

, with eigenvalue

!k = h

λ

ˆ ( ) ( )p x p xψ ψ=

A particle with definite momentum xp is described by eigenfunction of operator ˆ xp

Consider kinetic energy operator

p2

2m=

−i!∂∂x

⎛⎝⎜

⎞⎠⎟

2

2m= − !

2

2md 2

dx2

Eigenfunctions of Kinetic energy:

− !

2

2md 2

dx2 eax = − !2a2

2m 0 !!< (if a is real)

Not physical

− !

2

2md 2

dx2 sin(ax) = + !2

2ma2 sin(ax)

Constant Eigenvalue Or

− !

2

2md 2

dx2 cos(ax) = !2

2ma2 cos(ax)

Also

eiax

!2

2ma2eiax



Or Hamiltonian operator:

2ˆˆ ( )

2pH V xm

= + = − !

2

2m∂2

∂x2 +V (x)

ˆ ( ) ( )H x E xψ ψ=

: particle described by eigenfunction ( )xψ has the definite energy E , (to be discussed in more detail in chapter 4)

Quantization of energy

We saw that a fundamental feature of ‘new’ quantum mechanics was that energy cannot take on any value, but only certain values. Why is that? Let us consider a particle in a box problem:

( ) 0V x = 0 x a< < ( )V x = ∞ elsewhere

We wish to solve

− !

2

2md 2

dx2ψ (x)+V (x)ψ (x) = Eψ (x)

E is a Constant Outside the box ( )V x →∞we want finite values of E , the only possibility is ( ) 0xψ = outside the box. We also wish ( )xψ to be continuous:

Inside the box we have 0V =



− !

2

2mdψ 2

dx2 = Eψ (x)

Boundary Condition: (0) ( ) 0aψ ψ= =

We considered before this equation

General Solution: sin( ) cos( )c kx b kx+

0x = 0ψ = 0b = E = !

2k 2

2m

x a= sin( ) 0c ka = , 1,2,3nk naπ= =

Any c , c not equal to 0

( ) sin n xx caπψ ⎛ ⎞= ⎜ ⎟⎝ ⎠

E = !

2n2π 2

2ma2 = h2n2

8ma2 1,2,3.....n =

-‐ Quantization: Combination of wave equation + Boundary conditions -‐ 1, 2, 3n = − − − also possible, but yields “same” solutions

sin sinn x n xc ca aπ π⎛ ⎞− = −⎜ ⎟⎝ ⎠

-‐ c can be anything (still)

For any operator A , with eigenfunction ψ (x) ˆ ˆ( ) ( )Ac x cA xψ ψ= ( )ca xψ= ( ( ))a c xψ=

If ( )xψ is an eigenfunction of operator A then also ( )c xψ is eigenfunction. (c is constant)

Interpretation of Wave Function In Mathchapter B we discussed probability distribution ( )p x dx :



( ) 0p x ≥ x∀

( ) 1p x dx∞

−∞

=∫

( )x xp x dx∞

−∞

= ∫ etc.

The absolute square of the wave function 2 *( ) ( ) ( )x x xψ ψ ψ= is to be interpreted like a

probability distribution.

2( ) ( )p x dx x dxψ=

Probability to find particle between x and x dx+ ( ) ( ) ( )x f x ig xψ = + complex ( )f x , ( )g x real

* ( ) ( ) ( )x f x ig xψ = − * ( ) ( ) [ ( ) ( )][ ( ) ( )]x x f x ig x f x ig xψ ψ = − +

2 2( ) ( ) [ ( ) ( ) ( ) ( )]f x g x i f x g x g x f x= + + −

= f (x)2+ g(x)

2

(real always)

Also 2( )xψ 0> everywhere

Probability distribution

Moreover (we should impose): 2( ) 1x dxψ∞

−∞

≡∫

Normalization Multiply ( )xψ by constant c , choose c such that ( ) ( )newc x xψ ψ= is normalized

Particle in the box (later)

2( ) sinn

n xxa a

πψ ⎛ ⎞= ⎜ ⎟⎝ ⎠

Further Interpretation

*( ) ( )high

low

x

x

x x dxψ ψ∫

Probability to find particle between lowx and highx

And



x = xψ *(x)ψ (x)dx−∞

∞

∫ = ψ *(x)xψ (x)dx−∞

∞

∫

Determination of Constant C We will impose that the wave functions are normalized

* ( ) ( ) 1x x dxψ ψ∞

−∞

≡∫ For reasons discussed before

* ( )xψ : complex conjugate of functions ( ) ( ) ( )x f x ig xψ = + ( )f x , ( )g x real

* ( ) ( ) ( )x f x ig xψ = −

* ( ) ( ) [ ( ) ( )][ ( ) ( )]x x f x ig x f x ig xψ ψ = − +

2 2[ ( )] [ ( )] [ ( ) ( ) ( ) ( )]f x g x i f x g x g x f x= + + −

2 2( ) ( )f x g x= +

2( )xψ= 0≥ everywhere

If ( )xψ is real then 2 2( ) ( )x xψ ψ=

Consider particle in the box wave functions:

( )n xψ = Cn sin nπ x

a 0 ≤ x ≤ a

0, elsewhere

2

2 2

0

( ) sina

nn xx dx C dxaπψ

∞

−∞

⎛ ⎞= ⎜ ⎟⎝ ⎠∫ ∫

2 12naC= =

Choose Cn =

2a

Simplest,

2a

eiθ would work too.

We can always choose the function ( )xψ to be normalized (for meaningful wave

functions) A physically meaningful wave function would be normalized

If ˆ ( ) ( )A x a xψ ψ= eigenfunction of A , eigenvalue a



Then ( )* ˆ( ) ( )x A xψ ψ

* ( ) ( )x a xψ ψ= * ( ) ( )a x xψ ψ=

And: ( )* ˆ( ) ( )x A x dxψ ψ∞

−∞∫

* ( ) ( )a x x dxψ ψ∞

−∞

= ∫

1a= ⋅ IF ( )xψ is normalized

We define:

A = ψ *(x) Aψ (x)dx−∞

∞

∫

Called the expectation value of operation A , depending on ( )xψ , also called the

average value of A

If ( )xψ is normalized, then A would be the average value measured for quantity A

If ( )xψ is an eigenfunction of A , then one would always measure a , and the

average value A a= IF ( )xψ is normalized

If ( )xψ is not an eigenfunction of A , then many values could be obtained if A is

measured. The average value would be A (much more discussion later)

One more definition:

( )2ˆ Â A− : The standard deviation from the average. The spread of the

measured values

( )( )ˆ ˆ ˆ Â A A A− −

22ˆ ˆ ˆ ˆ2A A A A= − +

22ˆ ˆ ˆ ˆ2A A A A= − +

22ˆ Â A= −

2Aσ= Depends on wave function ( )xψ



Using definition:

A2 = ψ *(x) A2ψ (x)dx−∞

∞

∫

Useful integrals for particle in the box

sin2 bx dx = x

2− sin2bx

4b∫

xsin2 bx dx = x2

4− xsin2bx

4b∫ − cos2bx8b2

x2 sin2 bx dx = x3

6− x2

4b− 1

8b3

⎛⎝⎜

⎞⎠⎟∫ sin2bx − x

cos2bx4b2

Definite Integrals (Most important). Use b = nπ

a; bx

x=a= nπ

aa = nπ

2

0

sin2

a n x adxaπ =∫

2

2

0

sin4

a n x ax dxaπ =∫

3 3

2 22 2

0

sin6 4

a n x a ax dxa nπ

π= −∫

sinnπ x

a0

a

∫ cosmπ x

adx = 0, ∀n,m integers

Demonstration of Uncertainty Principle Using the above integrals, we can calculate the following

a) Normalize sinn nn xCaπψ =

22 2

0

sin 12

a

n nn x aC dx Caπ⎛ ⎞ = ≡⎜ ⎟⎝ ⎠∫ 2

nC Ca

= =

Normalized particle in the box eigen states: 2 sin n xa a

π⎛ ⎞⎜ ⎟⎝ ⎠

b) Calculate x for normalized ( )n xψ :



0

2 sin sina n x n xx x dx

a a aπ π= ⋅ ⋅∫

224 2a a

a= ⋅ = center of the box

c) Calculate 2x

2 2

0

2 sin sina n x n xx x dx

a a aπ π= ⋅ ⋅∫

3 3 2 2

2 2 2 2

26 34 2a a a a

a n nπ π⎛ ⎞

= ⋅ − = −⎜ ⎟⎝ ⎠

d) Standard deviation in x : 22 2

x x xσ = −

2 22 2 2 2 2 2

2 2 2 2 23 2 12 2 32 2a a a a a a n

nn nπ

ππ π⎡ ⎤⎛ ⎞ ⎛ ⎞= − − = − = −⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦

e)

Px = 2a

sin nπ xa

−i!ddx

⎛⎝⎜

⎞⎠⎟

sin nπ xa

dx0

a

∫

= 2

a−i!

nπa

⎛⎝⎜

⎞⎠⎟

sinnπ x

acos

nπ xa

dx = 00

a

∫

f)

Px2 = 2

asin nπ x

a0

a

∫ −!2 d 2

dx2

⎛⎝⎜

⎞⎠⎟

sin nπ xa

= 2

a!2 ⋅ n2π 2

a2 sin nπ xa

⎛⎝⎜

⎞⎠⎟

2

0

a

∫ dx 2a

= !

2n2π 2

a2 = h2n2

4a2 ( 2 nmE= , of course!)

( )2xhnPa

σ =

We can test the Heisenberg Uncertainty Principle

12 2 2

22 3 2x pa n hnn a

πσ σπ

⎡ ⎤= ⋅ − ⋅⎢ ⎥

⎣ ⎦

= !

2π 2n2

3− 2

⎡

⎣⎢

⎤

⎦⎥

12

> !

2



Note 1: 12xaσ → as n→∞ is the same as uncertainty in uniform distribution:

2

0

12 2

ax ax

a= =

2 3 2

0

1 1 13 3

a

x x aa

= =

2 2 2

3 4 12x uniform

a a aσ = − =

xP

σ grows with n. Why?

Pn = ± (2mEn ) = ± n2π 2!2

a2

Spiked distribution

Large Uncertainty This represents the classical limit of particle of energy nE bouncing back and forth in the box

Note 2: x , 2x , xP , 2xP Can be calculated for any wave function

for example: ( ) ( )x Cx a xψ = − also satisfies the boundary conditions

Particle in a 3 dimensional box

Consider rectangular box of length , ,a b c



3D Schrodinger Equation:

− !

2

2m∂2

dx2 +∂2

dy2 +∂2

dz2

⎛⎝⎜

⎞⎠⎟ψ (x, y, z) ( , , )E x y zψ=

Boundary Conditions: (0, , ) ( , , ) 0y z a y zψ ψ= = ,y z∀ ( ,0, ) ( , , ) 0x z x b zψ ψ= = ,x z∀ ( , ,0) ( , , ) 0x y x y cψ ψ= = ,x y∀

“The wave function at the faces of sides of a box is zero” Technique to solve: Separation of variables.

Try ( , , ) ( ) ( ) ( )x y z X x Y x Z zψ = Substitute in Schrodinger equation and divide by ( , , )x y zψ (as we did for vibrating

strings)

− !

2

2m1

X (x)d 2 Xdx2 − !

2

2m1

Y ( y)d 2Ydy2 − !

2

2m1

Z(z)d 2Zdz2 = E

This can only be true if each term itself is constant: , ,x y zE E E

We get 3 equations

a) 2 2

2 ( )2 xd X E X x

m dx− =h (0) ( ) 0X X a= =

b) 2 2

2 ( )2 yd Y E Y y

m dy− =h

(0) ( ) 0Y Y b= =

c) 2 2

2 ( )2 zd Z E Z z

m dz− =h (0) ( ) 0Z Z c= =

x y zE E E E+ + =

This is just 3 times the 1D particle in the box equation! We know the (normalized) solution:

2( ) sin k xX xa a

π⎛ ⎞= ⎜ ⎟⎝ ⎠

Ex =

h2

8mk 2

a2

⎛⎝⎜

⎞⎠⎟

2( ) sin l yY yb b

π⎛ ⎞= ⎜ ⎟⎝ ⎠

Ey =

h2

8ml2

b2

⎛⎝⎜

⎞⎠⎟

2( ) sin n zZ zc c

π⎛ ⎞= ⎜ ⎟⎝ ⎠

2 2

28yh nEm c⎛ ⎞

= ⎜ ⎟⎝ ⎠

Or



8 sin sin sinx y z

y yx x z zn n n

nn nabc a b c

ππ πψ⎛ ⎞⎛ ⎞ ⎛ ⎞= ⋅ ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

22 22

2 2 28yx znn nhE

m a b c⎛ ⎞

= + +⎜ ⎟⎜ ⎟⎝ ⎠ , , 1, 2,3....x y zn n n =

Degeneracies for Cubic box Consider the special case of a Cubic box a b c= = . Then the energy takes the form

( )2

2 2 228 x y z

hE n n nma

= + +

For each triplet , ,x y zn n n we get a different wave function, but different values of , ,x y zn n n may

yield the same energy. Such energy levels are called degenerate. Eg.for atoms we know there are 1 s-‐orbital, 3 p-‐orbitals, 5 d-‐orbitals. Table of energies

2

28hEma

= ( ), ,x y zn n n Degeneracy

14 (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) 6 12 (2,2,2) 1 11 (1,1,3), (1,3,1), (3,1,1) 3 9 (2,2,1), (2,1,2)(1,2,2) 3 6 (1,1,2), (1,2,1), (2,1,1) 3 3 (1,1,1) 1

Documents

Fall$2014$ Chem%356:%Introductory%Quantum%Mechanicsscienide2.uwaterloo.ca/~nooijen/Chem356/Chem+356+pdf/Ch_3.pdf · Fall$2014$Chem%356:%Introductory%Quantum%Mechanics$ Chapter$3$–$Schrodinger$Equation,$Particlein$aBox$38$