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