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

fixed angle (5) polar molecule interacts with angle averaged (6) Keesom polar molecule

interacts non-polar molecule polarization with (7) Debye induced dipole This Lecture

fluctuating non-polar molecule non-polar molecule induced (8) interacts dipoles with London

2 P1_Wk1_L6 Motivation

Tip approximated by uncharged sphere, radius R R

How does Uinter depend on z? z d

Uncharged Insulating plate

3 8. Interaction potential energy between two non-polar molecules – Dispersion Forces

- z non-polar non-polar + δ (t) - δ (t) δ+ δ molecule 1 molecule 2 α1 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 z2, v2

-q -q m m m m +q fixed +q fixed k z k

non-polar non-polar 4 P1_Wk1_L6 molecule 1 molecule 2 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

11 U( z) = k z2 = mωω 22 z and U( z) = k z2 = m 22 z . 122 11oo 2 22

Because the two oscillators are uncoupled, Schrödinger’s equation reduces to two separate equations

22∂ 1 − Ψ+ω 22 Ψ= Ψ 2 1 mzo 1 1 E 11 22mz∂ 1 22∂ 1 − Ψ+ω 22 Ψ= Ψ 2 2 mzo 22 E 22 22mz∂ 2

Under these circumstances, it is well known that the allowed energy eignevalues are

11  k E12=+=+== nωo and E  n' ωω oo ;n , n ' 0,1,2,...; 22  m

5 P1_Wk1_L6 When the fluctuation in the molecule involves a charge, the situation changes because there is now an electrostatic interaction term given by 1 qq22 q 2 q 2 Uz()= −−+ electr  4πε o z zz+2 zz − 1( zz −+ 12) z 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 2 11q2 1 ∆=−Uz() ω  24o πε k z6 o

This result can be further developed (see Appendix A) for two similar molecules in 3-dimensions, embedded in a dielectric with dielectric constant κ: 31α 2 I C =−=o −L UzLondon () 26 6 2 (4πκε o ) zz

where I is the ionization energy of the atom/molecule and αo is the polarizability of the atom/molecule under consideration.

6 P1_Wk1_L6 For dissimilar molecules with ionization energies I1 and I2 and polarizabilities αo,1 and αo,2 31αα II C' =−=oo12 12 −L UzLondon () 2 66 2 (4πκε o ) IIz12+ z

Source: http://en.wikipedia.org/wiki/Ionization_energy 7 P1_Wk1_L6 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) 8 P1_Wk1_L6 Source: A. Tkatchenko, et al., MRS Bulletin 35, 435 (2010). 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

9 P1_Wk1_L6 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

10 P1_Wk1_L6 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 () zU=Keesom () zU ++Debye () zULondon () z 22 22C 21p p 1 pp1αoo ,2+ 2 α ,1 1 3 αα oCV,1 0,2 II 1 =−1 2 −−V CV12 3 kT 2 zz6 (4πκε )26 2 (4πκε )2II+ z 6 B (4πκε 0 ) 0 0 12 CCC =−−−Keesom Debye London zzz6 66 C = − vdW 6 z 11 P1_Wk1_L6 What it means

molecule 2 molecule 1 z - - + δ δ+(t) δ (t) δ α2 α1

zeq C = − vdW UzvdW () 6 z Positive U(z) U(z) or Repulsive force F(z) z

dU() z U (z) Fz()= − dz Negative U(z) Attractive force

Umin Flocal max

12 Simple Molecules

CKeesom CDebye CLondon

Source: http://www.ntmdt.com/spm-basics/view/intermolecular-vdv-force

13 P1_Wk1_L6 Comparing theory and experiment

Source: Physics and Chemistry of Interfaces, H.-J. Butt, K. Graf, M. Kappl, Wiley-VCH (2003). 14 P1_Wk1_L6 Summary of Coulombic Intermolecular Forces

System of interacting εo ? atoms and/or molecules or and/or ions κεo ?

Is H bonded to NO YES Ions Electrically O, N or F Involved? neutral atoms or (permanent dipoles)? molecules ?

NO YES YES NO

Dipole – dipole NO Do the atoms or Interactions molecules have a (hydrogen bonding) permanent dipole?

YES

Dipoles Dipole Ion – dipole Dipole Classical fixed rotating Interaction fixed ion-ion Interaction

Angle- Induced dipole- averaged Polarization van der Waals induced dipole Dipoles Force Forces

London Dispersion Keesom Debye 15 P1Wk1L6 Next up, the interaction between a tip and a substrate

Tip approximated R by uncharged sphere, radius R

z

do d

Uncharged Insulating plate

16 P1_Wk1_L6 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 z2, v2 -q -q m m m m +q fixed +q fixed k z k

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 11 U( z) = kz2 = mωω 22 z and U( z) = kz2 = m 22 z . 122 11oo 2 22 Because U is uncoupled, Schrödinger’s equation reduces to two separate equations with solutions that are given in many quantum textbooks: 22∂ 1 − Ψ+ω 22 Ψ= Ψ 2 mzo 11 E 22m ∂z1 22∂ 1 − Ψ+ω 22 Ψ= Ψ 2 mzo 22 E 22m ∂z2 17 P1_Wk1_L6 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 1 E=+= EE (ωω +) =  ω tot 122 o o o k where is Planck’s constant divided by 2π and ω = .  o 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 qq22 q 2 q 2 Uz()= −−+ electr  4πε o z zz+2 zz − 1( zz −+ 12) z

If z>>z1 and z>>z2, you can derive a simple expression for Uelectr by using the binomial expansion as follows:

18 P1_Wk1_L6 1 qq22 q 2 q 2 Uelectr = −−+ 4πε o z zz+2 zz − 1( zz −+ 12) z  1q2  11 1 =1 −−+ 4πε z z z zz− o 111+−+2 1 21 zz z −−11 − 1 1 q2 z  z  zz−  =11 −+2 −− 11 ++ 1 21 πε     4 o zz  z  z 

nnnn( −−11) nn( ) (1+εεε)  1 +nn +22 +.....( 1 +− ( εεε )) 1 − + + ..... 2! 2! we have : −−1 2 12 z2  zz 2  2  + z1  zz 11  1+  1 −+  +...  1 − 1 ++  +... z  zz  z  zz  −12 zz21−  zz 21 −−  zz 21 1+  1 −+  +... z  zz 

2  22 22 1 q zzz( zz21) ( ) − z ( z2−+2 zz 21 z 1) U =1 −− 1 212 + +..... −+ 1 + +..... +− 1 1 ++..... electr 4πε zz zz22 z z z2 o   

2 22 22 1 q z( zz21) z( ) zzz2 zzz  11−+2 − −− 11 − +− 1 2 + 1 + 2 − 21 + 1 4πε z z2 z2 zz 222 o z z zzz 11qq222zz = −=−21 23zz21 42πε ooz zzπε 19 P1_Wk1_L6 This gives a new expression for the total potential energy of the system, Utot:

11 1q2 =++ =22 + − Utot U1 U 2 Uelect kz1 kz 2 3 z21 z 222πε 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 1122 U= kz( ++ z) kz( − z) tot22 s 12 a 12

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: 22∂ 1 − Ψ+ω 22 Ψ= Ψ 2 mzss E s 22m ∂zs 22  ∂ 1 22 − Ψ+mzωaa Ψ= E a Ψ ∂ 2 22m za 20 P1_Wk1_L6 This can be accomplished by setting 22 11 1q2 1( zz+−) 1( zz) 22+− =12 + 12 kz1 kz 2 3 z21 z ksak , 222πε o z 2 2 2 2 The algebra is given below:

22 11 1q2 1( zz+−) 1( zz) 22+− =12 + 12 Let kz1 kz 2 3 z21 z ksak 222πε o z 2 2 2 2 11q2 22+− = +22 ++ − kzz( 12) 3 zzkkzz21 ( sa)( 12) zzkk 12( sa) πε o z 2 1 q2 =+−=− evidently,2 k ksa k and 3 (ksa k ) πε o z 1 q2 − = −− = − substituting gives 3 (2k kaa k) 2( k k a ) πε o z 1 q2 1 q2 − =−⇒= + 3 kkaa k k 3 2πε o z 2πε o z 11qq22   = −= −+ =− ksa22 kk k k 33 k  22πε oozz πε 

21 P1_Wk1_L6 It follows that for the electrostatically coupled system, the new eigen- frequencies are now given by

11qq22 kk−+ 22πε zz33πε ωω= ooand = samm 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 11

∆U =(ωωas +) −×2  ω o 22

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

2 11q2 1 ∆=−Uz() ω  24o πε k 6 o z 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. 22 P1_Wk1_L6 11 ∆=+−U (ωωas) 2  ω o =(  ωω as +−)  ω o 22 11qq22 kk+− 22πε zz33πε k k11 qk2211 q k =oo + −=11 + + − − 2m m m22 mπε kmzz33 2πε k m oo   11 2222  k11 qk 11 q  k =11 +  +−  − 33   2mk 2πε oozz 22 mkπε m    11 22  k 11qq22 11  = 1+  +− 12  − 33   2 mk22πε oozzπε k   

n nn( −1) expanding using( a+εε)  ann ++ na −−1 an22 ε +..... 2! 2 1 11qq22 111  11 1++⋅⋅ −1 +.... πε 33πε  k 22ookkzz 22 2  2  2 2 m 1 11q2 11 1 11q2 ++ − + ⋅ ⋅ − − +− 1 3 1 3 ... 2 22πε o k z 222 2πε o k z 22 22 22  k1 11 qq 1 11 1  11 qq 1 11 =+−+1 ....+ 1−−+ ...− 2 2mkk 22πε 33 82πε 22πε kk33 82πε oozz oo zz 2 2  kq1 11 = −  2mk 42πε 3 o z 22 1 11qq22   1  1  1 = − ωω  = −    8 oo2πε kk36  24πε  oozz     23 P1_Wk1_L6 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 E z -q 2 -q mm m m +q +q fixed fixed k

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 ≡= qz2 q E =α E k q2 evidently = α k

24 P1_Wk1_L6 This finally allows us to write

22 2   2 1 1q  11 α 11αω o 1 ∆=−Uz() ωωoo  =−  =− 24πε k z6 24 πε zz6 2πε 26 oo     (4 o )

The parameter  ω o 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

11α 2 I ∆=−Uz() 2 26 (4πε o ) z

The equivalent result in 3-dimensions is

31α 2 I ∆=−Uz() 4 26 (4πε o ) z

25 P1_Wk1_L6