Physical Chemistry (II)

Lecture 3

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.01/25/2017

Quantum Mechanical Principles

Quiz 2

10 minutes

Please stop writing when the timer stops !

Information Encoded in a Wavefunction

ψ r,t( ) r : position t : timeWavefunction is the only quantity needed to represent a quantum state

ρ r,t( ) = ψ r,t( ) 2 =ψ * r,t( )ψ r,t( )Probability Density:

Spatial distribution

x

ϕ x( )higher probability

lower probability

How about momentum ?

Wavefunction of a Free ParticleDefinition of a free particle: V = 0 V : potential

−!2

2m∂2ϕ x( )∂x2

+V (x)ϕ x( ) = Eϕ x( )

Time-independent Schrödinger equation:

− !

2

2m∂2ϕ x( )∂x2

= Eϕ x( )Standard solution:

ϕ x( ) = Aeikx + Be− ikx E = k

2!2

2mtotal energy,

but also the kinetic energy

A,B : "arbitrary" constants ψ * r,t( )ψ r,t( )dr∫ = 1 what is k? p = !k

Planewave

free particle wave function: ϕ x( ) = Ae+ ikx + Be− ikx

Ae+ ikx :right-moving

packet2A

real: imaginary:

Be− ikx :left-moving

packetB

k = 2πL

L L

Although A could be different than B, the two planewaves must have the same k.

Eleft = Eright

e+ ikx = coskx + isin kx

Momentum of a Free ParticleIf the free particle is treated by classical mechanics:

Ek = E =p2

2m→ p = 2mEk p : momentum

kinetic energy given by Schrödinger equation:

Ek =

k2!2

2m→ p = k!

ϕ x( ) = Aeikx + Be− ikxthe wavefunction does have the momentum information

Unfortunately, not every wavefunction can be easily expressed as a combination of plane waves

Formal quantum mechanics definition of momentum: ϕ x( ) : arbitrary wavefunction

p̂ = −i! ∂

∂xmomentum

operator p = ϕ* x( )∫ p̂ϕ x( )dx

expectation valueinformation extractor

Operator

function: f (x) = x2 +1 x : indepedent variable

A function has the magic to change the value of an independent variable.

We define a rule, , so thatD̂D̂f (x) = d

dxf (x)

f (x) = x2 +1when

D̂f (x) = ddx(x2 +1) = 2x

if we define g(x) = 2x D̂f (x) = 2x = g(x)An operator is a function of functions.

x = 1→ f (x) = 2

Important Properties of OperatorsAn operator acting on more than one functions:

D̂ f (x, y),g(x, y)( ) = ∂∂x

f (x, y) + g(x, y)( ) + ∂∂y

f (x, y) + g(x, y)( )

f (x, y) = x2 + y2 ,g(x, y) = x + y

D̂( f (x, y),g(x, y)) = 2x + 2y + 2 = h(x, y)

predefined mixture of functions

Does the order of mixing matter ?

ÂB̂ f (x)( ) = B̂Â f (x)( )

?

??

Commutator of Two OperatorsDefinition of Commutator:

[Â, B̂] = ÂB̂ − B̂Âdifference between two mixing orders

If for any function f x( )

[Â, B̂] f (x) = ÂB̂f (x) − B̂Âf (x) = 0The operators of  and B̂ commute.

A simple example:Âf (x) = d

dxf (x), B̂f (x) = d

2

dx2f (x)

[Â, B̂] f (x) = ddx

d 2

dx2f (x)− d

2

dx2ddx

f (x) = d3

dx3f (x)− d

3

dx3f (x) = 0

One More Example

Âf (x) = ddx

f (x), B̂f (x) = xf (x)

ÂB̂f (x) = ddx

B̂f (x)( ) = ddx xf (x)( ) = xf '(x) + f (x)do they commute?

B̂Âf (x) = x Âf (x)( ) = x ddx f (x)⎛⎝⎜

⎞⎠⎟= xf '(x)

A is after B:

B is after A:

[Â, B̂] f (x) = ÂB̂f (x) − B̂Âf (x) = xf '(x) + f (x) − xf '(x) = f (x)

[Â, B̂] = 1 ≠ 0 They DO NOT commute !

Why Does Commutator Matter?momentum operator:

p̂ = −i!

∂∂x

position operator:

x̂ = xcommutator of momentum and position operators:

[ p̂, x̂]Ψ x( ) = p̂x̂Ψ x( ) − x̂p̂Ψ x( ) = −i! ∂

∂x⎛⎝⎜

⎞⎠⎟xΨ x( )( ) − xi! ∂

∂xΨ x( )( )

[ p̂, x̂]Ψ x( ) = −i! ∂

∂x⎛⎝⎜

⎞⎠⎟xΨ x( )( ) − xi! ∂

∂xΨ x( )( ) = −i!Ψ x( )

momentum and position operators do NOT commute !

There is no way to simultaneously determine the momentum and position of a particle !

residue

Heinsberg Uncertainty Principle

Cauchy-Schwartz Inequality:σ A2σ B

2 ≥ÂB̂ − B̂Â

2

2

→σ A2σ B

2 ≥Â, B̂⎡⎣ ⎤⎦2

2

[ p̂, x̂] = −i!For position and momentum,

σ p2σ x

2 ≥p̂, x̂[ ]2

2

→σ p2σ x

2 ≥!2

4→σ pσ x ≥

!2

σ A = A2 − A 2Standard deviation: : average value

For energy and time,

σ Eσ t ≥

!2

It is impossible to exactly measure a particle’s energy at any given time.

Hamiltonian Operator

−!2

2m∂2ϕ x( )∂x2

+V (x)ϕ x( ) = Eϕ x( )Time-independent Schrödinger equation:

− !

2

2m∂2

∂x2+V (x)

⎛⎝⎜

⎞⎠⎟ϕ x( ) = Eϕ x( )Ĥ : Hamiltonian

operator

a system’s total energy

Using the definition of momentum: p̂ = −i! ∂

∂x Êk =

p̂2

2m= − !

2

2m∂2

∂x2

Ĥ = Êk + V̂ x( )kinetic energy

potential energy

Ĥϕ x( ) = Eϕ x( )eigenfunction-eigenvalue

problem

Eigenfunction and EigenvaluesDefinitions:

D̂ f x( )( ) = cf x( )

For a given operator, only some particular functions satisfy the condition.

f x( ) is the eigenfunction of the operator, D̂, and c is the eigenvalue for f (x)

For a given eigenfunction, we only have one eigenvalue.

A simple example:D̂ =

ddx

f (x) = e3x ddx

f (x) = 3e3x = 3 f (x)

f (x) = e2xddx

f (x) = 2e2x = 2 f (x)

Physical Meaning of Eigenfunction and Eigenvalue

Ĥϕ x( ) = Eϕ x( )Time-independent Schrödinger equation:

The eigenfunction of an operator has a constant observable value, which is the eigenvalue.

The eigenfunction has a constant total energy E. ϕ x( )For a free particle: ϕ x( ) = Aeikx

p̂ = −i! ∂

∂xIs it an eigenfunction of momentum ?

p̂ϕ x( ) = −i! ∂∂x

Aeikx( ) = (−i!)(ik) Aeikx( ) = !kϕ x( )A free particle has a well-defined momentum, !k

Is it an eigenfunction of position ? x̂ = xx̂ϕ x( ) = x Aeikx( ) = xϕ x( )

YES

NO

, but not a well-defined position

x ∈ −∞,+∞( )

Expectation Values

A = ϕ * x( )∫ Âϕ x( )dxexpectation valueaverage value of a physical observable, which is represented by an operator

E = ϕ * x( )∫ Ĥϕ x( )dxaverage total energy Ĥ : Hamiltonian operatorp = ϕ * x( )∫ p̂ϕ x( )dxaverage momentum p̂ : momentum operator

average position x = ϕ * x( )∫ x̂ϕ x( )dx x̂ : position operatorIf ̂Aϕ x( ) = aϕ x( ) eigenfunction-eigenvalue

A = ϕ * x( )∫ Âϕ x( )dx = ϕ * x( )∫ aϕ x( )dx = a ϕ * x( )∫ ϕ x( )dx = a= 1

Special case:

Hermitian OperatorsHermiticity:

ϕi*(x)xϕ j x( )∫ dx = ϕi*(x)xϕ j x( )∫ dx( )*⎛⎝ ⎞⎠

*

= ϕi (x)x*ϕ j

* x( )∫ dx( )*

= ϕi (x)xϕ j* x( )∫ dx( )* = ϕ j* x( )xϕi (x)∫ dx( )*

 is a Hermitian operator

If a physical property is observable, its corresponding operator must be Hermitian !

For example, x̂ = x

ϕi*(x)Âϕ j x( )∫ dx = ϕ j*(x)Âϕi x( )∫ dx( )*

Q.E.D.

Ĥ , p̂, x̂ are all Hermitian operators because they are measurable.

a + bi( )* = a − bi

Hermicity of Quantum OperatorsLet us say Âϕ x( ) = aϕ x( ) a : complex number

a = f + gi If  is a Hermitian operator

A = ϕ * x( )∫ Âϕ x( )dx = ϕ * x( )∫ Âϕ x( )dx( )*

ϕ * x( )∫ aϕ x( )dx = ϕ * x( )∫ aϕ x( )dx( )*

a ϕ * x( )∫ ϕ x( )dx = a* ϕ * x( )∫ ϕ x( )dx( )*a ×1= a* × 1( )* a = a* a : real number

The Hermicity of quantum operators is used to ensure real observable values !

Superposition of Hermitian Eigenfunctions

eigenfunction set:

Âϕ x( ) = aϕ x( )ϕ1(x),ϕ2 (x),...,ϕN (x){ }

eigenvalue set: a1,a2,...,aN{ } If  is a Hermitian operator

ϕi* x( )∫ ϕ j x( )dx = δ ij =

1 if i = j0 if i ≠ j

⎧⎨⎪

⎩⎪spatial overlap between wavefunctions

normalization ruleorthogonality

Ψ x( ) = ciϕii=1

N

∑superimposed wavefunction linear combination

A = Ψ* x( )∫ ÂΨ x( )dx

Expectation Value for a Superimposed Wavefunction

A = Ψ* x( )∫ ÂΨ x( )dxΨ x( ) = ciϕi

i=1

N

∑Ψ* x( ) = cj*ϕ j*j=1

N

∑

A = cj*ϕ j

*

j=1

N

∑∫ Â ciϕii=1

N

∑ dx

A = cj*ci

i=1

N

∑j=1

N

∑ ϕ j*Âϕi dx∫ ϕ j*aiϕi dx∫ai ϕ j

*ϕi dx∫ = aiδ ij

A = cj*ci

i=1

N

∑j=1

N

∑ aiδ ij j = i A = cici*aii=1

N

∑ A = ci 2 aii=1

N

∑

An Example

Ψ x( ) = c1ϕ1 + c2ϕ2two-component decomposition:

ϕ1 and ϕ2 are the eigenfunctions of Hamiltonian operator, ĤE1 and E2 are the corresponding eigenvalues

what is the expectation value of energy for Ψ x( )?

E = c12 E1 + c2

2 E2c1

2 and c22 : Projection of Ψ x( ) onto ϕ1 x( ) and ϕ2 x( )

If Ψ x( ) =ϕ1 x( ) c1 2 = 1 and c2 2 = 0 If Ψ x( ) =ϕ2 x( ) c2 2 = 1 and c1 2 = 0In general, 0< c1

2

Review of Homework 27.7 Suppose that the state of the vibrating atom follows wavefunction ψ x( ) = Nxe− x2 /2a2 . What is the most probable location of the particle?

Probability density: P(x) =ψ *(x)ψ x( )

P(x) = Nxe− x2 /2a2( )* Nxe− x2 /2a2 = Nxe− x2 /2a2Nxe− x2 /2a2 = N 2x2e− x2 /a2

most probable location: dP(x)dx

= 0 and d2P(x)dx2

< 0

dP(x)dx

= 0→ N 2 x2e− x2 /a2 −2x

a2⎛⎝⎜

⎞⎠⎟ + 2xe

− x2 /a2⎛⎝⎜

⎞⎠⎟= 0

Review of Homework 2

2N 2e− x2 /a2 x − x

2

a2+1⎛

⎝⎜⎞⎠⎟= 0→

x = 0x = +ax = −a

⎧⎨⎪

⎩⎪another requirement: d

2P(x)dx2

< 0

d 2P(x)dx2

< 0→ 2N 2e− x2 /a2 1− 5x

2

a2+ 2x

4

a4⎛⎝⎜

⎞⎠⎟< 0

when x = 0, d2P x( )dx2

= 2N 2 > 0 it is actually the least probable location !

when x = ±a, d2P x( )dx2

= 2N 2e−1 1− 5 + 2( ) = −4N 2e−1 < 0

both are the most probable locations

Homework 3

Reading assignment: Chapter 7.5, 7.6, 7.7

Homework assignment: Exercises 7.17 Problems 7.18

Homework assignments must be turned in by 5:00 PM, January 25th, Thursday

to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall