Origin of the atttractive part of Lennard-Jones potential
Masahiro Yamamoto
Department of Chemistry, Konan University, Kobe, 658-8501, Japan(Dated: February 7, 2016)

A.

Origin of the attractive part of L-J potential

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,e2h̄2e2(∇21 + ∇22 ) −−2m4π0 r1
4π0 r1Ψ = ψ1 (r1 )ψ2 (r2 ) = Rnl (r1 )Ylm (r̂1 )Rnl (r2 )Ylm (r̂2 ) = |nlmi|nlmi

H0 = −(1)(2)

The ground state Ψ0 is given by |100i|100i. If the two H atoms get closer, they interact each other and the potentialeer1r2＋＋rz

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

V (perturbation) is given bye2
e21e21e2
1−−+4π0 r 4π0 | − r1 + r| 4π0 |r + r2 | 4π0 | − r1 + r + r2 |V =(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...
−1|r − r1 |2 −1/2= [(x − x1 ) + (y − y1 ) + (z − z1 ) ]
22=r

−1

[ 2]−1/2z − 2zz1 + x21 + y12 + z12
r2[]−1/2−2zz1 + x21 + y12 + z12−2zz1 + x21 + y12 + z123z 2= r−1 1 +' r−1 (1 −+ 12 )22r2r2r22223z2zz2 + x2 + y2 + z2+ 22 )|r + r2 |−1 ' r−1 (1 −2r22r222222−2zz+2zz+x−2x3 (z2 − z1 )2
121 x2 + x2 + y1 − 2y1 y2 + y2 + z1 − 2z1 z2 + z2
1|r − r1 + r2 |−1 ' r−1 [1 −+]2r22r2e2−2zz1 + x21 + y12 + z12
3z 22zz2 + x22 + y22 + z223z 2V '(1 − 1 +− 1 −1+− 2224π0 r2r22r23 z12 + z22 − 2z1 z2−2zz1 + 2zz2 + x21 − 2x1 x2 + x22 + y12 − 2y1 y2 + y22 + z12 − 2z1 z2 + z22+)+1 −2r22r2e2e2=(x1 x2 + y1 y2 + z1 z2 − 3z1 z2 ) =(x1 x2 + y1 y2 − 2z1 z2 )(4)34π0 r4π0 r3
This is the instantaneous dipole-dipole interaction 1/(4π0 r3 )[~µ1 ·~µ2 −3(~µ1 ·r̂)(~µ2 ·r̂)]. Here µ~ 1 = e(−r1 ), µ~ 2 = e(−r2 ).
Using perturbation theory the energy change is given by(2)En = En(0) + ∆(1)n + ∆n + ...

∆(1)
n= hn|V |ni = Vnn

(5)
(6)

2∆(2)=n(0)For ground state n = 0, E0∑|Vnk |2(0)(0)
k6=n En − Ek(7)

= 0 (atom1) + 0 (atom2) ande2(x1 x2 + y1 y2 − 2z1 z2 )|0i
4π0 r3e2=
h1, 0, 0(atom1)|h1, 0, 0(atom2)|(x1 x2 + y1 y2 − 2z1 z2 )|1, 0, 0(atom1)i|1, 0, 0(atom2)i4π0 r3e2=
h1, 0, 0(atom1)|x1 |1, 0, 0(atom1)ih1, 0, 0(atom2)|x2 |1, 0, 0(atom2)i + ...
4π0 r3= 0∆(1)= h0|n(8)In the last equation the integral of the product of even, odd, even functions becomes zero. and the second order
(2)

∆0

=

∑
k6=0

=|V0k |2
0 (atom1) + 0 (atom2) − k (atom1) + k (atom2)∑ h0|x1 x2 + y1 y2 − 2z1 z2 |kihk|x1 x2 + y1 y2 − 2z1 z2 |0ie426(4π0 ) r
0 (atom1) + 0 (atom2) − k (atom1) + k (atom2)
k6=0(9)∑∑Using the relationk |kihk| = 1 = |0ih0| +k6=0 |kihk|, and we assume that the denominator is constant and0 (atom1) + 0 (atom2) − k (atom1) + k (atom2) ' −(I1 + I2 ), here I1 , I2 is the ionization energy of atom1 and 2.
(2)

∆0

e4
1h0|(x1 x2 + y1 y2 − 2z1 z2 )2 |0i(4π0 )2 r6 I1 + I2
1e4
[h0(atom1)|x21 |0(atom1)ih0(atom2)|x22 |0(atom2)i= −(4π0 )2 r6 I1 + I2
= −+h0(atom1)|y12 |0(atom1)ih0(atom2)|y22 |0(atom2)i + 4h0(atom1)|z12 |0(atom1)ih0(atom2)|z22 |0(atom2)i]
In the same perturbation theory the atomic polarizability is given by
α = −2e2∑ h0|z|kihk|z|0i
(0)

k6=0

(0)

E0 − Ek'= −2e2h0|z 2 |0iI

(10)

Here the electric ﬁeld is in z-direction. Then we can get h0|z 2 |0i = h0|x2 |0i = h0|y 2 |0i = −(αI)/(2e2 ) With thisrelation[]e41
α1 I1 α2 I2α1 I1 α2 I2α1 I1 α2 I2(2)∆0 = −++4 2(4π0 )2 r6 I1 + I2 2e2 2e22e2 2e22e 2e2
3 α1 α2I1 I2(11)= −2 (4π0 )2 r6 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)
B.

LJ potential

The ﬁnal result give the attractive part of the Lennard-Jones potential, φ(r) = 4[(σ/r)12 − (σ/r)6 ]. In the mixed
system one can use the following convention(Lorentz-Berthelot mixing rule)
σi + σj2√
ij = i j

σij =(12)(13)

3In the case of σij we can understand the rule from the view point of hard-sphere. In the case of ij we should consider
Eq.(11)
3 αj2 Ij
−4j σj6 = −2 (4π0 )2 2
√Ii Ij3 αi αj√ij = i j =3216 (4π0 ) σi σj3−4i σi6 = −3 αi2 Ii
,2 (4π0 )2 2(14)From 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.
√ij = ξ i j ,σij =σ i + σj2(15)FIG. 2: from http://socrates.berkeley.edu/ peattiea/research main.html

The Gecko (￢?√￠￢?√†￢?√?￢Å jcan stick to the wall. (both on the hydrophilic and hydrophobic surace andeven dead gacko can 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.
C.

Kittel’s derivation of the attractive part of LJ potential: dynamical eﬀect

Now we consider the two neutral atom is separated by R. At a certain moment the atoms induce dipole momentin each other and the induced dipole moments cause an attractive interaction between the atoms.
We assume the two atoms are identical linear harmonic oscillators and only electrons move in the x-direction. Whenthe displacement of electrons are x1 and x2 , the force constants are K, and the momenta of the oscillators are p1 andp2 The hamiltonian of the system without the interaction between the induced dipole moments is
H0 =1 2 1
1 2 1
p1 + Kx21 +p + Kx222m
22m 2 2(16)4FIG. 3: capillary forcex2x1-++-R
FIG. 4: Kittel’s explanation on the vdW attractive potential (ISSP book)

Here m is the mass of electron and K = mω02 . The energy of the ground state is the double of the zero point energ
1
E0 = 2 h̄ω0 = h̄ω02(17)The interaction H1 between the dipoles becomes
H1 =e2e2e2e2−+−4π R 4π0 (R + x1 ) 4π0 (R + x1 − x2 ) 4π0 (R − x2 )| {z0 } |{z}|{z}|{z}+++−−−(18)−+Here e is the elementary charge and 0 is the permittivity of free space. If we assume that R  |x1 |, |x2 |.
()[]4π0 H1
11x1x211x1 − x2
(x1 − x2 )21 (x2x2 )'−1−++1−+−
1++e2
R RR2R2RR2R2RR2R212x1 x2=[−x21 + (x1 − x2 )2 − x22 ] = −2R2R3(19)(20)

Here we used the approximation 1/(1 ± x) ' 1 ∓ x + (1/2)x2 + .... The total hamiltonian becomes
H = H0 + H1 =1 2 1
1 2 12x1 x2
p + Kx21 +p + Kx22 −2m 1 22m 2 2
R3(21)

If we use the symmetric and anti-symmetric coordinate
1xs ≡ √ (x1 + x2 ),2
1x1 = √ (xs + xa ),21xa ≡ √ (x1 − x2 )21x2 = √ (xs − xa )
2(22)
(23)

and also use the symmetric and anti-symmetric momenta
1ps ≡ √ (p1 + p2 ),21pa ≡ √ (p1 − p2 )
2(24)

51
p1 = √ (ps + pa ),21
p2 = √ (ps − pa )2(25)The total hamiltonian becomes[() ] [() ]2e22e21 2 1
1 2 12H =ps +K−x+p+K+x2as2m24π0 R32m a 2
4π0 R3(26)The two vibrational frequencies are found for this interacting system[(ω =K±(= ω0 1 ±2e24π0 R3)]1/2/m2e24π0 KR3)1/2[K=m(2e21±4π0 KR3)]1/2

If assume that 1  (2e2 )/(4π0 KR3 ), we can use (1 ± x)1/2 ' 1 ± (1/2)x − (1/8)x2 + ...
][()()212e2
12e2
ω = ω0 1 ±−+ ...
2 4π0 KR38 4π0 KR3(27)(28)(29)

The zero point energy change is[[()()2 ]()()2 ]
11111
1
12e2
2e2
2e22e2−+ h̄ω0 1 −−− 2 h̄ω0∆E = h̄ω0 1 +22 4π0 KR38 4π0 KR322 4π0 KR38 4π0 KR32()
212e2h̄ω0 e4
A= h̄ω0=−=− 6(30)232268 4π0 KR
32π 0 K R
RThe attractive interaction is proportional to R−6 and A include the dynamics of the oscillator through ω0