Origin of the atttractive part of Lennard-Jones potential
Masahiro Yamamoto
Department of Energy and Hydrocarbon Chemistry,Graduate School of Engineering, Kyoto University, Kyoto 606-8501, Japan(Dated: March 4, 2008)

Now we condsider two hydrogen atoms. If the two atoms are located far away, the Hamiltonian H0 and thewavefunction Ψ of the system is given by two independent H atoms,h̄2e2e2(∇21 + ∇22 ) −−2m4πϵ0 r1
4πϵ0 r1Ψ = ψ1 (r1 )ψ2 (r2 ) = Rnl (r1 )Ylm (r̂1 )Rnl (r2 )Ylm (r̂2 ) = |nlm〉|nlm〉

H0 = −(1)(2)

The ground state Ψ0 is given by |100〉|100〉. If the two H atoms get closer, they interact each other and the potentialer

e1

r

2＋＋zr

FIG. 1: van der Waals interaction between two hydrogen atom

V (perturbation) is given byV =e2
e2
1e2
1e2
1−−+4πϵ0 r 4πϵ0 | − r1 + r| 4πϵ0 |r + r2 | 4πϵ0 | − r1 + r + r2 |(3)

If we assume r >> r1 , r2 , and vector r is in the z direction. r = (0, 0, z). If we consider the order up to r−1 (x21 /r2 )because the (1/r)(z1 /r) order terms are cancelled out. If we use (1 + x)−1/2 ≅ 1 − x/2 + 3x2 /8...
[ 2]−1/2z − 2zz1 + x21 + y12 + z12|r − r1 |−1 = [(x − x1 )2 + (y − y1 )2 + (z − z1 )2 ]−1/2 = r−1r2[]−1/2−2zz1 + x21 + y12 + z12−2zz1 + x21 + y12 + z12
3z 2= r−1 1 +≅ r−1 (1 −+ 12 )22r2r2r22222zz2 + x2 + y2 + z23z|r + r2 |−1 ≅ r−1 (1 −+ 22 )2r22r222222−2x−2zz+2zz+x3 (z2 − z1 )2
1 x2 + x2 + y1 − 2y1 y2 + y2 + z1 − 2z1 z2 + z2
121|r − r1 + r2 |−1 ≅ r−1 [1 −+]2r22r2−2zz1 + x21 + y12 + z123z 22zz2 + x22 + y22 + z22
3z 2e2(1 − 1 +− 1 −1+− 2V ≅224πϵ0 r
2r22r2−2zz1 + 2zz2 + x21 − 2x1 x2 + x22 + y12 − 2y1 y2 + y22 + z12 − 2z1 z2 + z22
3 z12 + z22 − 2z1 z2+1 −+)2r22r2e2e2=(x1 x2 + y1 y2 + z1 z2 − 3z1 z2 ) =(x1 x2 + y1 y2 − 2z1 z2 )(4)3
4πϵ0 r4πϵ0 r3
This is the instantaneous dipole-dipole interaction 1/(4πϵ0 r3 )[⃗µ1 ·⃗µ2 −3(⃗µ1 ·r̂)(⃗µ2 ·r̂)]. Here µ⃗ 1 = e(−r1 ), µ⃗ 2 = e(−r2 ).
Using perturbation theory the energy change is given by(2)En = En(0) + ∆(1)n + ∆n + ...

= 〈n|V |n〉 = Vnn∑|Vnk |2∆(2)
=n(0)
(0)
k̸=n En − Ek∆(1)
n(5)
(6)
(7)

2(0)

For ground state n = 0, E0= ϵ0 (atom1) + ϵ0 (atom2) ande2(x1 x2 + y1 y2 − 2z1 z2 )|0〉
4πϵ0 r3e2=〈1, 0, 0(atom1)|〈1, 0, 0(atom2)|(x1 x2 + y1 y2 − 2z1 z2 )|1, 0, 0(atom1)〉|1, 0, 0(atom2)〉
4πϵ0 r3e2=〈1, 0, 0(atom1)|x1 |1, 0, 0(atom1)〉〈1, 0, 0(atom2)|x2 |1, 0, 0(atom2)〉 + ...
4πϵ0 r3= 0∆(1)= 〈0|n(8)In the last equation the integral of the product of even, odd, even functions becomes zero. and the second order(2)∆0=∑k̸=0=|V0k |2ϵ0 (atom1) + ϵ0 (atom2) − ϵk (atom1) + ϵk (atom2)∑ 〈0|x1 x2 + y1 y2 − 2z1 z2 |k〉〈k|x1 x2 + y1 y2 − 2z1 z2 |0〉e4(4πϵ0 )2 r6
ϵ0 (atom1) + ϵ0 (atom2) − ϵk (atom1) + ϵk (atom2)k̸=0(9)∑∑Using the relationk |k〉〈k| = 1 = |0〉〈0| +k̸=0 |k〉〈k|, and we assume that the denominator is constant andϵ0 (atom1) + ϵ0 (atom2) − ϵk (atom1) + ϵk (atom2) ≅ −(I1 + I2 ), here I1 , I2 is the ionization energy of atom1 and 2.
(2)

∆0e4
1
〈0|(x1 x2 + y1 y2 − 2z1 z2 )2 |0〉(4πϵ0 )2 r6 I1 + I2
e4
1
= −[〈0(atom1)|x21 |0(atom1)〉〈0(atom2)|x22 |0(atom2)〉26(4πϵ0 ) r I1 + I2
= −+〈0(atom1)|y12 |0(atom1)〉〈0(atom2)|y22 |0(atom2)〉 + 4〈0(atom1)|z12 |0(atom1)〉〈0(atom2)|z22 |0(atom2)〉]
In the same perturbation theory the atomic polarizability is given by
α = −2e2∑ 〈0|z|k〉〈k|z|0〉(0)k̸=0(0)E0 − Ek≅= −2e2〈0|z 2 |0〉I(10)Here the electric ﬁeld is in z-direction. Then we can get 〈0|z 2 |0〉 = 〈0|x2 |0〉 = 〈0|y 2 |0〉 = −(αI)/(2e2 ) With thisrelation[]e41
α1 I1 α2 I2
α1 I1 α2 I2α1 I1 α2 I2(2)∆0 = −++4(4πϵ0 )2 r6 I1 + I2 2e2 2e2
2e2 2e22e2 2e23 α1 α2I1 I2(11)= −262 (4πϵ0 ) r I1 + I2
This equation was derived by F. London (Z. Phys. 63, 245-279 ,1930). Since the vdW interaction can not be estimatedfrom usual density functional theory( because we should consider excitation and dynamical dipole-dipole interaction),some semiempirical correction based on Eq.(11) is reported. (e.g. F. Ortmann et al. PRB, 73, 205101, 2006)
The ﬁnal result give the attractive part of the Lennard-Jones potential, φ(r) = 4ϵ[(σ/r)12 − (σ/r)6 ]. In the mixedsystem one can use the following convention(Lorentz-Berthelot mixing rule)
σi + σj2√ϵij = ϵi ϵjσij =(12)(13)

In the case of σij we can understand the rule from the view point of hard-sphere. In the case of ϵij we should consider
Eq.(11)
3 αj2 Ij
−4ϵj σj6 = −2 (4πϵ0 )2 2
√Ii Ij3 αi αj√ϵij = ϵi ϵj =3216 (4πϵ0 ) σi σj3−4ϵi σi6 = −3 αi2 Ii,
2 (4πϵ0 )2 2(14)3From the ﬁnal equation the approximation of Eq.(13) may be not bad. If one consider the deviation form the standard
mixing rule, the modiﬁed Lorentz-Berthelot mixing rule can be used. ξ is the deviation parameter.
√ϵij = ξ ϵi ϵj ,σij =σ i + σj2(15)FIG. 2: from http://socrates.berkeley.edu/ peattiea/research main.html

The Gecko (やもり）can stick to the wall. (both on the hydrophilic and hydrophobic surace and even dead gackocan stick!!) They have 1 billion contacting point per foot and interact via van der Waals force!!. The total
adhesive stress from Autumn’s force measurements is 576 kPa (5.68 atm).
[ Kellar Autumn, Yiching A. Liang, S. Tonia Hsieh, Wolfgang Zesch, Wai Pang Chan, Thomas W. Lenny, Ronald
Fearing and Robert J. Full (2000) ”Adhesive force of a single gecko foot-hair”, Nature, 405(6787). 681-685, and Proc
Natl Acad Sci U S A. 2002 99, 12252-6. ”Evidence for van der Waals adhesion in gecko setae.” Autumn K, Sitti M,Liang YA, Peattie AM, Hansen WR, Sponberg S, Kenny TW, Fearing R, Israelachvili JN, Full RJ.]
Recently Huber et al. showed that the evidence for capillary force contribution to gecko adhesion from single
spatula AFM force measurment. (PNAS, 102, 16293-16296, 2005) The pull-oﬀ force is increased lineary (from 7 nN
to 12 nN) with the humidity for hydrophilic ﬂat surface and the the increase is not so much (from 6 nN to 9 nN) forhydrophobic surface. At the humidity around 90 % the monolayer of water is adsorbed on the hydrophilic surface.

FIG. 3: capillary force