Download Lecture: P1_Wk1_L6 The Most General Inter

Lecture: P1_Wk1_L6
The Most General Inter-Molecular Force:
The London Dispersion Force
Ron Reifenberger
Birck Nanotechnology Center
Purdue University
2012
1
This Lecture: Intermolecular Interactions Between
Non-polar Molecules
(5)
fixed angle
interacts
with
polar molecule
angle averaged
Keesom
polar molecule
interacts
with
non-polar molecule
polarization
(7)
Debye
This Lecture
non-polar molecule
interacts
with
P1_Wk1_L6
(6)
non-polar molecule
induced
dipole
fluctuating
induced
dipoles
(8)
London
2
Motivation
Tip approximated
by uncharged
sphere, radius R
R
d
z
How does Uinter
depend on z?
Uncharged Insulating plate
3
8. Interaction potential energy between two
non-polar molecules – Dispersion Forces
non-polar
molecule 1
z
δ-(t)
δ+(t)
δ-
δ+
α1
non-polar
molecule 2
α2
• the London or Dispersion Force acts between ALL
atoms/molecules even if they have zero permanent dipole moments
• fundamentally quantum mechanical in nature, arising from a
“fluctuating dipole-induced dipole” interaction
• requires a time-correlation between fluctuating dipoles
• develop simple model for a fluctuating dipole system:
A typical fluctuation that occurs at some time:
z1, v1
+q
P1_Wk1_L6
m
fixed
k
non-polar
molecule 1
z2, v2
m
-q
z
+q
m
fixed
k
non-polar
molecule 2
m
-q
4
Assume for the moment a small fluctuation in molecule 1 will induce a
fluctuation in molecule 2. Model the fluctuation as a particle trapped
in a parabolic well. The confining potential can be described as
U=
1 ( z)
1 2
=
k z1 mωo2 z12
2
and U=
2 ( z)
1 2
=
k z2 mωo2 z22 .
2
Because the two oscillators are uncoupled, Schrödinger’s equation
reduces to two separate equations
 2 ∂ 2 
1
2 2
−
Ψ
+
m
ω
z1 Ψ1 = E1Ψ1
o
1


2
2
 2m ∂z1 
 2 ∂ 2 
1
−
Ψ
+
mωo2 z22 Ψ 2 = E2 Ψ 2
2

2
2
 2m ∂z2 
Under these circumstances, it is well known that the allowed energy
eignevalues are
1

E1 =
 n +  ωo
2

P1_Wk1_L6
and
1
k

E2 =
0,1, 2,...; ωo =
 n '+  ωo ; n, n ' =
2
m

5
When the fluctuation in the molecule involves a charge, the situation
changes because there is now an electrostatic interaction term given
by

q2
q2
q2
1  q2
U
( z=
)
−
+
 −

electr
4πε o  z z + z2 z − z1 ( z − z1 ) + z2 
The consequences of this interaction term is fully described in
Appendix A where we show the energy of the 1-dimensional system is
lowered in energy by an amount
  1 q 2 2  1
1
 6
∆U ( z ) =
− ω o  

  4πε o k   z
2


This result can be further developed (see Appendix A) for two
similar molecules in 3-dimensions, embedded in a dielectric with
dielectric constant κ:
CL
3 αo2 I 1
−
=
− 6
U London ( z ) =
2
6
2 (4πκε o ) z
z
where I is the ionization energy of the atom/molecule and αo is
the polarizability of the atom/molecule under consideration.
P1_Wk1_L6
6
For dissimilar molecules with ionization energies I1 and
I2 and polarizabilities αo,1 and αo,2
C 'L
3 α o1α o 2  I1 I 2  1
−
=
−
U London ( z ) =


2 (4πκε o ) 2  I1 + I 2  z 6
z6
Source: http://en.wikipedia.org/wiki/Ionization_energy
P1_Wk1_L6
7
Dispersion Forces: State of the Art
Electronic-structure Calculation
Full QM treatment of many-electron, non-covalent vdW interactions
Two main approaches: Hartree-Fock (HF) and Kohn-Sham density-functional theory (DFT)
Ar-Ar (Z=18)
Kr-Kr (Z=36)
Limitations: HF - accurate calculations only feasible for ~50 light atoms
DFT – system with “thousands“ of atoms (SIESTA)
P1_Wk1_L6
Source: A. Tkatchenko, et al., MRS Bulletin 35, 435 (2010).
8
The above discussion focuses mainly on atoms &
molecules. Any modifications required for solids?
Lifshitz Theory – treats solids as continuous materials with
BULK properties – macroscopic (condensed matter) treatment
1
3
2
 ionization energy I replaced by frequency dependent dielectric
constant ε(iω)
 polarizability α replaced by static dielectric constant ε
More Later: see P1_Wk2_L1
P1_Wk1_L6
9
8. Dispersion Forces - comments
 The ability of fluctuating molecular dipoles to attract one
another depends strongly on the frequency of the fluctuations in
electron density - two transiently induced dipoles will attract
each other only if their frequencies nearly match (are equal or
multiples of one another). Quantities that vary with frequency
are said to exhibit dispersion – hence the name Dispersion
Force, Dispersion Energy = vdW Force, vdW Energy.
 Extensions to frequency dependent polarizations by McLachlan
(1963)
 As z increase (>100nm), the time for fluctuating dipole electric
fields to reach a second atom/molecule must be taken into
account. This is known as the Cashimer interaction and a proper
accounting of the retardation effects gives a power law
dependence that varies as z-7 .
 Casimir-Polder force is a generalization to include finite
conductivity
P1_Wk1_L6
10
Summary - van der Waals Forces
The van der Waals force is the sum of three different components
of the electrostatic interaction between molecules: orientation,
induction, and dispersion. Each electrostatic interaction produces a
potential energy that varies as 1/z6, where z is the separation
– Orientation or Keesom Force is the angle-averaged dipole-dipole
interaction between two polar molecules.
– Debye Force is the angle averaged dipole-induced dipole
interaction between a polar and non-polar molecule
– The London or Dispersion force acts between all molecules with
non-zero polarizability
UvdW ( z ) = UKeesom ( z ) + UDebye ( z ) + ULondon ( z )
2
2 C
2
2
p
p
+
α
2 1 p1 p2
1
3 α oCV
o ,1 1
1 o ,2
2 αV
,1α 0,2  I1 I 2  1
=
−
−
−
 CV 
z 6 2 (4πκε 0 )2  I1 + I2  z 6
3 kBT ( 4πκε 0 )2 z 6
(4πκε 0 )2
P1_Wk1_L6
CKeesom CDebye CLondon
=
−
−
−
z6
z6
z6
C
= − vdW
z6
11
What it means
molecule 1
α1
δ+(t)
molecule 2
z
δ+
-
δ (t)
zeq
UvdW ( z ) = −
U(z) or
F(z)
CvdW
z6
δ-
α2
Positive U(z)
Repulsive force
z
U (z)
Umin
F (z) = −
dU ( z )
dz
Negative U(z)
Attractive force
Flocal max
12
Simple Molecules
CKeesom
CDebye
CLondon
Source: http://www.ntmdt.com/spm-basics/view/intermolecular-vdv-force
P1_Wk1_L6
13
Comparing theory and experiment
Source: Physics and Chemistry of Interfaces, H.-J. Butt, K. Graf, M. Kappl, Wiley-VCH (2003).
P1_Wk1_L6
14
Summary of Coulombic Intermolecular Forces
System of interacting
atoms and/or molecules
and/or ions
Is H bonded to
O, N or F
(permanent dipoles)?
NO
NO
Ions
Involved?
YES
εo ?
or
κεo ?
Electrically
neutral atoms or
molecules ?
NO
YES
YES
Dipole – dipole
Interactions
(hydrogen bonding)
NO
Do the atoms or
molecules have a
permanent dipole?
YES
Dipoles
fixed
Induced dipoleinduced dipole
Angleaveraged
Dipoles
London Dispersion
Keesom
P1 Wk1 L6
Dipole
rotating
Polarization
Force
Debye
Ion – dipole
Interaction
Dipole
fixed
Classical
ion-ion
Interaction
van der Waals
Forces
15
Next up, the interaction between a tip and a substrate
R
Tip approximated
by uncharged
sphere, radius R
do
d
z
Uncharged Insulating plate
P1_Wk1_L6
16
Appendix A: Dispersion Forces - a simple model
Consider the diagram below which shows two small masses vibrating in a parabolic well that is
modeled by an effective spring with spring constant k. The two fixed (non-vibrating) atoms are
separated by a distance z as shown.
z1, v1
+q
m
fixed
k
z2, v2
m
-q
z
+q
m
fixed
k
m
-q
As indicated, the instantaneous displacement of the moveable mass m from equilibrium is
specified by z1 and z2, respectively. Assume for the moment that the masses have zero
charge. Schrödinger’s equation for the two uncoupled oscillators requires a specification of
the interaction potential energy U for each oscillator. For a parabolic well, U is specified as
U=
1 ( z)
1 2
=
kz1 mωo2 z12
2
and U =
2 ( z)
1 2
=
kz2 mωo2 z22 .
2
Because U is uncoupled, Schrödinger’s equation reduces to two separate
equations with solutions that are given in many quantum textbooks:
P1_Wk1_L6
 2 ∂ 2 
1
Ψ
+
mωo2 z12 Ψ= E1 Ψ
−

2
2
 2m ∂z1 
 2 ∂ 2 
1
Ψ
+
mωo2 z22 Ψ= E2 Ψ
−

2
2
 2m ∂z2 
17
The known solutions of these two equations tells us that the quantum mechanical
eigenvalues E1 and E2 for the oscillator are quantized, with the lowest energy
solution (the ground state energy) given by
Etot = E1 + E2 =
where

1
( ωo + ωo ) = ωo
2
is Planck’s constant divided by 2π and
ωo =
k
m
.
If each spring system acquires a charge ±q as shown in the diagram above, then
each spring will acquire an instantaneous, fluctuating dipole moment p1=qz1 and
p2=qz2. What is the resulting electrostatic potential energy Uelectr that results
when these two dipoles interact? The answer can be written down by inspection:

1  q2
q2
q2
q2
U
( z=
)
−
−
+


electr
4πε o  z z + z2 z − z1 ( z − z1 ) + z2 
If z>>z1 and z>>z2, you can derive a simple expression for Uelectr by using the binomial
expansion as follows:
P1_Wk1_L6
18

q2
q2
q2
1  q2
U electr
=
−
+
 −

4πε o  z z + z2 z − z1 ( z − z1 ) + z2 




1 q
1
1
1
=
−
+
1 −

z2
z1
z2 − z1 
4πε o z 
1+
1−
1+
z
z
z 

2
  z2 −1  z1 −1  z2 − z1 −1 
1 − 1 +  − 1 −  + 1 +

z 
z
z  


 
n ( n − 1) 2
n ( n − 1) 2
n
n
ε + .....
ε + .....
(1 + ε )  1 + nε +
(1 + (−ε ) )  1 − nε +
2!
2!
we have :
1 q2
=
4πε o z
−1
−1
z1  z1 
  + z1 
−
+
+   + ...
1
1



z
z


 z
2
z2  z2 
 z2 
+
−
+   + ...
1
1



z
z


 z 
−1
2
z2 − z1 
z2 − z1  z2 − z1 

1
+
1
−
+


 + ...
z 
z

 z 
2
2

 
  z − z ( z22 − 2 z2 z1 + z12 )
z1 ( z1 )
1 q 2   z2 ( z2 )
2
1
1 − 1 − + 2 + .....  − 1 + + 2 + .....  + 1 −
+
+ .....  
U electr
=

 
 
z
z
z
4πε o z  
z
z
z2

 
 
 
2
2
2
z2 ( z2 )
z1 ( z1 )
z2 z1 z22 2 z2 z1 z12 
q2 
1 − 1 + − 2 − 1 − − 2 + 1 − + + 2 − 2 + 2 

z
z
z
z z
4πε o z 
z
z
z
z 


2
2
1 q  2z z 
1 q
=  − 22 1  =
−
z2 z1
4πε o z  z 
2πε o z 3
1
P1_Wk1_L6
19
This gives a new expression for the total potential energy of the system, Utot:
U tot = U1 + U 2 + U elect
1 2 1 2
1 q2
z z
= kz1 + kz2 −
3 2 1
2
2
2πε o z
When this expression is used in Schrödinger’s equation, the solutions are not so simple
because the electrostatic potential energy term now contains both z1 and z2. Not
surprisingly, the vibrational frequencies for each spring will be altered because of this
electrostatic interaction. The question is by how much will they change?
The cross-term in the expression for Utot prevents a simple answer to this question.
However, if we could somehow rewrite the expression for Utot to have a separable form,
something like
U tot=
1
1
2
2
ks ( z1 + z2 ) + ka ( z1 − z2 )
2
2
then we could define new coordinate variables zs=(z1+z2)/√2 and za=(z1-z2) /√2, with
ks and ka defined as some “equivalent” spring constants. Schrödinger’s equation could
then be rewritten in the same form as the uncoupled case above:
P1_Wk1_L6
 2 ∂ 2 
1
−
Ψ
+
mωs2 zs2 Ψ= Es Ψ

2 
2
 2m ∂zs 
 2 ∂ 2 
1
2 2
Ψ
+
m
ω
za Ψ= Ea Ψ
−

a
2
2
 2m ∂za 
20
This can be accomplished by setting
1 2 1 2
1 q2
1 ( z1 + z2 )
1 ( z1 − z2 )
kz1 + kz2 −
=
z
z
k
+
ka
,
s
3 2 1
2
2
2πε o z
2
2
2
2
2
2
The algebra is given below:
1 2 1 2
1 q2
1 ( z1 + z2 )
1 ( z1 − z2 )
kz1 + kz2 −
z
z
k
ka
=
+
s
2 1
2
2
2πε o z 3
2
2
2
2
2
Let
2
1 q2
1
k (z + z ) −
z
z
=
( ks + ka ) ( z12 + z22 ) + z1 z2 ( ks − ka )
3 2 1
2
πε o z
2
1
2
2
evidently, 2k =
k s + ka
and
1 q2
substituting gives −
=
πε o z 3
1 q2
−
=
( k s − ka )
3
πε o z
( 2k − k a − k a ) =
2(k − ka )

1 q2
1 q2 
−
=k − ka ⇒ ka = k +
3 
2πε o z 3
2
πε
z
o



1 q2  
1 q2 
k s = 2k − k a = 2k −  k +
= k −

3 
2
2πε o z 3 
πε
z
o

 
P1_Wk1_L6
21
It follows that for the electrostatically coupled system, the new eigenfrequencies are now given by
=
ωs
1 q2
k−
2πε o z 3
=
and ωa
m
1 q2
k+
2πε o z 3
m
What is the change in the quantum ground state energy, ΔU, from the uncoupled
case and how does it vary with z? Define ΔU by
=
∆U
1
1
( ω a + ω s ) − 2 × ω o
2
2
If ΔU is a positive number, the coupled system will have a higher potential energy
and the two charged springs will experience a repulsive force. If ΔU is a negative
number, the coupled system will have a lower potential energy and the two charged
springs will experience an attractive force.
The details of the calculation for ΔU are provided on the next page. The result is
  1 q 2 2  1
1

∆U ( z ) =
− ω o  
  4πε o k   z 6
2


Since ΔU is a negative number, the coupled system will have a lower potential
energy and the two springs will experience an attractive potential that varies
as z-6.
P1_Wk1_L6
22
1
1
( ωa + ωs ) − 2ωo =
( ω a + ω s ) − ω o
2
2

1 q2
1 q2 
k
−
 k+

2πε o z 3
2πε o z 3 

k
=
+
−
=


2
m
m
m






1

2
1 1 q2 
1
 k 
k 
=
+
 1+

 1−
3 



2 m 
2πε o k z 
2m 
2πε o

∆U =
1 1 q2
 k
1+
+

2  m
2πε o k z 3
1q 

k z 3 
2
1
2
1 1 q2 
k
k
1−
−
3 
2πε o k z 
m
m

− k

m

1
1


2  2
2  2

1 1q
1 1q
 k 
=
 1+
 +  1−
 − 2
3 


2 m  
2πε o k z 3 
2
k
πε
z
o




n ( n − 1) n − 2 2
n
expanding using ( a + ε )  a n + na n −1ε +
a ε + .....
2!
 1  1 1 q 2  1 1  1   1 1 q 2 2

1 + 

+
⋅
⋅
−
+
1
....


3 
3 

 k  2  2πε o k z  2 2  2   2πε o k z 



2
2 m

1
1 1 q2  1 1  1  
1 1 q2 
+ 1+ −
+ ⋅ ⋅  − 1  −
+ ... − 2 

3 
3 
2  2πε o k z  2 2  2   2πε o k z 


2
2

1  1 1 q2  1  1 1 q2 
 k  1  1 1 q2  1  1 1 q2 

....
1
...
2
=1 + 
−
+
+
−
−
+
−








2 m  2  2πε o k z 3  8  2πε o k z 3 
2  2πε o k z 3  8  2πε o k z 3 


2
 k  1  1 1 q2  
− 
=
 
2 m  4  2πε o k z 3  


2
  1 1 q2  
  1 q 2 2  1
1
1
 = − ω o  

= − ωo  
  2πε o k z 3  
  4πε o k   z 6
8
2




P1_Wk1_L6
23
Note that the final answer contains q and k, parameters that are easy to define
but difficult to know. These two parameters can be eliminated by realizing that an
electric field causes the charge separation by exerting a force which must be
balanced by the “effective” spring that tethers the separated charges together.
The spring will “stretch” until the restoring force of the spring matches the
electrostatic force produced by the electric field.
E
+q
mm
E
-q
+q
fixed
m
fixed
z2
k
m
-q
In equilibrium, this implies that
q E = kz2
In equilibrium, the electric field will induce a dipole moment pind=qz2 =αE. This gives
q 
pind ≡ q z=
q
2
 E=
 αE
k 
q2
evidently
=α
k
P1_Wk1_L6
24
This finally allows us to write
  1 q 2 2  1
  α 2  1
1
1
1 α 2 ωo 1
∆U ( z ) =
− ωo  
− ωo  
−
  6 =
  6 =


2
4
k
2
4
2 ( 4πε o )2 z 6
πε
πε
z
z
o
o








The parameter ωo represents a characteristic “motional” frequency of the
charge in a polarized atom/molecule and is often replaced by the ionization
energy I, which is a well known and easy-to-measure quantity. This gives rise to
the result often quoted in the literature
1 α2 I 1
∆U ( z ) =
−
2 ( 4πε o )2 z 6
The equivalent result in 3-dimensions is
3 α2 I 1
∆U ( z ) =
−
4 ( 4πε o )2 z 6
P1_Wk1_L6
25