VAN DER WAALS FORCES

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1 VAN DER WAALS FORCES JAN-LOUIS MÖNNING The Van der Waals Force is an attracting force between neutral atoms with R the distance of the atoms. Our aim will be to show that such a force exists, which means we need to show that the energy would decrease if we move the atoms closer together. We will try to show the following inequality e(r) e 1 + e 2 C R 6 for some constant C > 0, where e(r) is the total energy and e 1, e 2 the energy of each atom. We will restrict us to the case of two hydrogen atoms (one electron and one nucleon each.) First we have to find the Hamiltonian H for our system, which is given by: H i i 1 (i 1, 2) x i 1 V x 1 x 2 + R 1 x 2 + R 1 x 1 + R + 1 R H H 1 + H 2 + V Where H i is the Hamiltonian of each of the atoms and V the interaction between the two atoms, x i is always the distance from electron to its corresponding nucleon and R ist the vector pointing from the position of nucleon one to the position of nucleon two. Theorem 1: There exists a function ψ s.t. ψ, Hψ 2e C(R) + b( R ), where C > 0, e is the energy given by the infimum of the spectrum of a hydrogen atom, b( R ) is exponentially decreasing and C( R ) decreases like R 6 for R big enough. Proof: Step 1: Construction of the Trial function ψ. Because we are just looking at the hydrogen atom we know its ground state wavefunction φ 0, namely: φ i 0 ce x /2 with c > 0 and it holds true that H i φ i 0 eφ i 0. Next we try to construct a cut-off function, so that the two wavefunctions of the 1

2 2 JAN-LOUIS MÖNNING atoms do not overlap with each other. Let φ i (x i ) φ i 0(x i )f(x i ) with f(x i ) 1 for x i R 2 1 f(x i ) 0 for x i R 2, f is smooth, i f, i f < K with K independent of R and we assume φ i 1 (else normalize φ). Now it holds that: H i φ i i φ i 0(x i )f(x i ) 1 x i φi 0(x i )f(x i ) i (φ i 0(x i ))f(x i ) 1 x i φi 0(x i )f(x i ) (φ i 0(x i ) i f(x i ) + 2 i φ i 0(x i ) i f(x i )) eφ i 0 (φ i 0(x i ) i f(x i ) + 2 i φ i 0(x i ) i f(x i )) This yields us φ i, H i φ i e(φ i ) 2 dx i φ i ( i φ i i f + i φ i i f + φ i 0 i f)dx i (φ i 0 e if)) e + (φ i 0(x i ) i f(x i ) i (φ i 0)fφ i 0 i f φ i 0 i (f)φ i 0 i f)dx i e + φ i 0 2 i f 2 dx i } {{ } b i 1 Where b i 1 is exponentially decreasing, because φ i 0 ce x /2, i f(x i ) 0 for all x i, x 2 i R and 2 i f(x i ) < K. Next we define the polarized atoms as the following: ψm i : m i φ i for m R and m 1. Then ψm i and φ i are orthogonal, because: ψm, i φ i m i (φ i )φ i dx i m i (φ i )φ i dx i ψ i m, φ i 0 In addition to that let ψ i m, ψ i m m i φ i 2 dx i : r i, where r i does not depend on m because φ is spherically symmetric.

3 Finally we can define our trial function: VAN DER WAALS FORCES 3 (1) ψ φ 1 φ 2 +λ ψm 1 ψn 2 φ ψ Step 2: Calculate the expected value of H i. To do that we calculate every possible combination of, H i with φ and ψ. First Next φ, H i φ φ i, H i φ i φ j, φ j e + b i 1. 1 φ, H i ψ φ i, H i ψm i φ j, ψn j 0. 0 And for the last let P i m i, then (P i ) P i, then ψ, H i ψ ψ i m, H i ψ i m ψ j n 2 1/2 φ i, ([P i, [H i, P i ]] + H i (P i ) 2 + (P i ) 2 H i )φ i r j where [P, H] P H HP. Now we split up the sum again and get for 1/2 φ i, (H i (P i ) 2 + (P i ) 2 H i )φ i 1/2( φ i, (P i ) 2 H i )φ i + φ i, (H i (P i ) 2 )φ i ) H i is symmetric 1/2( (P i ) 2 φ i, 2H i )φ i + H i φ i, (P i ) 2 φ i ) the functions are just inr 3 (P i ) 2 φ i, H i )φ i eφ(x i )(P i ) 2 φ(x i ) (P i ) 2 φ(x i )(φ i 0(x i ) i f(x i ) + 2 i φ i 0(x i ) i f(x i ))dx i e(p i φ(x i )) 2 dx i + (P i ) 2 φ(x i )(φ i 0(x i ) i f(x i ) 2φ i 0(x i ) i f(x i ))dx i } {{ } b i 2 er 1 + b i 2 where b i 2 is exponentially decreasing, because φ i 0 ce x /2, (P i ) 2 φ f(x i ), i f 0 for all x i R 2 1, x i R 2 and i f(x i ) < K. x 0

4 4 JAN-LOUIS MÖNNING For the remaining part we get [H i, P i ] P i ( 1 x i ) ( 1 x i )P i P i ( 1 x i ) [P i, P i ( 1 x i )] P i P i ( 1 x i ) P i 1 x i P i (P i ) 2 1 x i Now we average m over the orthogonal base (1, 0, 0), (0, 1, 0), (0, 0, 1), which works because 1 x i is spherical symmetric so (P i ) 2 1 x i can t depend on m, so we get (P i ) 2 1 x i 1/3 1 x i 4π 3 δ(x i) Where δ is the Dirac Delta distribiution i.e. f(x)δ(x)dx f(0) [3, chapter 1.15, p.83]. Together this yields us that ψ, H i ψ er 1 r 2 + b i 2r j + 2π 3 φ, δ(x i) 1 x i φ r j e ψ + (b i 2 + Q i )r j Q i Step 3: Calculate the expected value of V For this we need Newton s Theorem [1, chapter 10, p.249]: φ(x) dx 1 x y y φ(x)dx if φ is spherical symmetric and y supp{φ}. We use this Theorem to calculate every possible combination of, V with φ and ψ. First 1 φ, V φ ( x 1 x 2 + R 1 x 2 + R 1 x 1 + R + 1 R )( φ(x 1) 2 φ(x 2 ) 2 )dx 1 dx 2 Newton ( 1 R + 1 R ) φ(x 1 ) 2 φ(x 2 2 )dx 1 dx 2 0

5 VAN DER WAALS FORCES 5 Secondly φ, V ψ φ 1 m 1 φ 1 φ 2 n 2 φ 2 V dx 1 dx m 1(φ 1 ) n 2(φ 2 ) 2 V dx 1 dx (φ1 ) (φ2 ) 2 m 1 n 2 V dx 1 dx R 1/4 ( φ(x 3 1 ) 2 φ(x 2 ) 2 )dx 1 dx R 3 Where averaged over this orthogonal base ( 1 0, 1 0 ), ( 1 0, 0 1 ), ( ( 0 1, 1 0 ), ( 0 1, 0 1 ), ( ( 0 0, 1 0 ), ( 0 0, 0 1 ), ( and then used Newton Theorem. Now the last, 0 0 ), 1, 0 0 ), 1, ) < ψ, V ψ > average over the same base Newton Theorem 0 (m 1 φ 1 ) 2 (n 2 φ 2 ) 2 V dx 1 dx 2 ( ( 1 9 ( i,j d dx i 1 3 i,j 3 i,j φ 1 ) 2 ( d φ 2 ) 2 V dx dx i 1 dx 2 2 d dx i 1 d dx i 1 φ 1 ) 2 ( d φ 2 ) 2 ( 1 dx i 2 R + 1 R )dx 1dx 2 φ 1 ) 2 ( d φ 2 ) 2 dx dx i 1 dx 2 ( 1 2 R + 1 R )

6 6 JAN-LOUIS MÖNNING Step 4: Finding a good λ recall formula(1): From all steps before it follows that ψ, Hψ e + e + b b λ 1 9 R + 3 r1 e ψ + r 2 e ψ + λ 2 ((b Q 1 )r 2 + (b Q 2 )r 1 ) Q 2e ψ + b b λ 1 9 R + 3 λ2 Q Now choose for λ 1, then ψ This yields us then 18 R 3 Q 18 2 R 6 (Q ) 2 ψ, Hψ 2e ψ + b b R 6 Q R 6 Q ψ 2 ψ (2e + b1 1 + b 2 1 ) 1 ψ 18 2 Q ( R 6 + r1 r Q ) 1 Because b i 1, b i 2 are exponentially decreasing this yields our result if we simply normalize ψ. References [1] Elliott H. Lieb, Michael Loss, Analysis, Graduate Studies Mathematics, American Mathematical Society, Volume 14, 2001 [2] Elliott H. Lieb, Walter E. Thirring, The Stability of Matter: From Atoms to Stars, Universal nature of Van der Waals forces for Coulomb systems, Springer Berlin Heidelberg, Volume 22, , 1986 [3] George B. Arfken, Hans J. Weber Mathematical Methods for Physicists, Elsevier Academic Press sixth edtion, 2005

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