Variational Wavefunction for the Helium Atom

Transcript

1 Technische Universität Graz Institut für Festkörperphysik Student project Variational Wavefunction for the Helium Atom Molecular and Solid State Physics 53. submitted on: 3. November 9 by: Markus Krammer

2 Contents Problem 3 Optimization of a given wavefunction 3 3 Calculating the energy E α 3. x Ĥ ψ ψ Ĥ ψ ψ Ĥ ψ ψ ψ ψ Ĥ ψ Transforming to spherical coordinates Determinant of the Jacobi matrix One more transformation Solving the integral Optimizing the Energy

3 Problem The Hamitonian for the helium atom is Ĥ = m + e 4πɛ r e 4πɛ r + e 4πɛ r where r is the distance between the two electrons. There is no analytical solution for the Schrödinger equation for this problem. So in a first step we neglect the electron electron interaction term 5 in equation to get an idea of how the real wavefunction could look like. Solving the Schrödinger equation for this simplified problem leads to the following wavefunction for the ground state: ψ r, r = a 3 exp r a This approximation is, of course, not very good. But it gives us an idea of how a better approximation could look like. So we try a variational wavefunction: ψ r, r, α = a 3 exp α r 3 a Now, as we have made a decision of how our approximate wavefunction for the ground state should look like, we can optimize it. Optimization of a given wavefunction To optimize a wavefunction, we have to calculate the energy E α = ψ Ĥ ψ ψ ψ 4 and find the lowest energy depending on our parameter. If we would have more than one Parameter e.g. coefficients of a polynomial that we multiply with our wavefunction we would have to find the global minimum of our energy. But since we have just one parameter, finding the minimum energy is pretty easy. de α dα = 5 Equation 5 gives us the optimized parameter α Energy must be the global minimum for the best α, so we have to check whether this energy is the global minimum or not. 3 Calculating the energy E α 3. x Ĥ ψ The Laplace operator in spherical coordinates looks like this: i = ri ri + r i r i ri sin θ sin θ i + i θ i θ i ri sin θ i ϕ i 6 Now we let the Laplace operator sink on the wavefunction: i ψ = ri r i r i r i a 3 exp α r a i ψ = α a 4 r i ri exp α r r i a i ψ = α a 4 r i r i exp α r α ri exp α r a a a 3

4 i ψ = α a 4 α a r i exp α r a 7 Putting this into x Ĥ ψ returns: with x Ĥ ψ = i= α α e + e m a a r i 4πɛ r i 4πɛ r a 3 exp α r a ma = this leads to e 4πɛ x Ĥ ψ = ma 4 i= α α + exp α r a r i r a ψ Ĥ ψ With equation 9 it is not difficult to write down ψ Ĥ ψ : R6 ψ Ĥ ψ = dx dy dz dx dy dz α α + ma a r i r i= a 6 exp r a The first four terms in this integral are no problem to solve. Only the last term with the r is a bit harder to evaluate. So we split it up into two parts: R6 ψ Ĥ ψ = dx dy dz dx dy dz α α exp r a r i a ma 7 i= and ψ Ĥ ψ = R6 dx dy dz dx dy dz ma 7 exp r r a 3.3 ψ Ĥ ψ The integrals R6 dx dy dz dx dy dz ma 7 α α exp r a r a and R6 dx dy dz dx dy dz ma 7 α α exp r a r a are exactly the same, so we can write R6 ψ Ĥ ψ = dx dy dz dx dy dz ma 7 α α exp r a r a 3 Now we tansform it into spherical coordinates and with the Jacobi determinant of the transformation we get: dx dy dz dx dy dz = r sin θ r sin θ dr dϕ dθ dr dϕ dθ 4 Our integral does not depend on ϕ i and θ i, so we can integrate these 4 variables seperately: π π π π dϕ dθ dϕ dθ sin θ sin θ = 4π 5 4

5 ψ Ĥ ψ = 4π ma 7 This integrals are all of the following form: So the result is: dx x exp λx = d dλ dx x exp λx = d dλ ψ Ĥ ψ = 4π ma 7 dr α r α r exp r dr r exp r a a a dx exp λx = d dλ dx exp λx = d dλ α a a a = λ λ 3 6 = λ λ 7 a ψ Ĥ ψ = π 4 α ma α ψ ψ For calculating the energy see equation 4 we need to know ψ ψ. ψ ψ = dx dy dz dx dy dz ψ ψ 9 R 6 ψ ψ = dx dy dz dx dy dz R a 6 exp r 6 a There is again no ϕ i and θ i dependence in this term, so with equation 4 and 5 we get ψ ψ = 4π a 6 and with equation 6 ψ ψ = 4π a 6 dr r exp r dr r exp r a a a ψ ψ = π α ψ Ĥ ψ As we see in equation this integral is not that easy. The term to transform the integral clever. r makes it a bit more difficult. So we need 3.5. Transforming to spherical coordinates In a first step we transform the integral into spherical coordinates. But not like we did it in equation 4, because this would not make the integral easier. So we choose a different transformation. We transform x, y and z like before: x = r cos ϕ sin θ y = r sin ϕ sin θ z = r cos θ 3 For the other coordinates we choose a different transformation. We use the r direction as z axis and the x axis is in the z z plane. So the ỹ axis is in the xy plane see figure. In this coordinates we now transform x, ỹ 5

6 Figure : Transformation, the z axis has the same direction as r and z as we are used to. x = r cos ϕ sin θ 4 ỹ = r sin ϕ sin θ 5 z = r cos θ 6 So now we need to know the relation between x, y, z and x, ỹ, z. For this we write down the basis vectors: ê z = r r ê z = cos ϕ sin θ sin ϕ sin θ cos θ 7 êỹ = ê z ê z ê z ê z êỹ = ê z ê z cos ϕ sin θ sin ϕ sin θ cos θ êỹ = ê z ê z sin ϕ sin θ cos ϕ sin θ 6

7 ê z ê z = sin ϕ sin θ cos ϕ sin θ sin ϕ sin θ cos ϕ sin θ ê z ê z = sin ϕ sin θ + cos ϕ sin θ ê z ê z = sin θ êỹ = sin ϕ cos ϕ 8 ê x = êỹ ê z ê x = sin ϕ cos ϕ cos ϕ sin θ sin ϕ sin θ cos θ ê x = cos ϕ cos θ sin ϕ cos θ sin θ 9 With x ê x + y ê y + z ê z = x ê x + ỹ êỹ + z ê z 3 and the equations 7, 8 and 9 we get x cos ϕ cos θ sin ϕ cos ϕ sin θ y = sin ϕ cos θ cos ϕ sin ϕ sin θ z sin θ cos θ x ỹ z 3 Now our transformation is complete. Equations to 6 and 3 are all we wanted to know Determinant of the Jacobi matrix For the integral tranformation we need the Jacobi determinant. The Jacobi matrix has the form J = cos ϕ sin θ r sin ϕ sin θ r cos ϕ cos θ M sin ϕ sin θ r cos ϕ sin θ r sin ϕ cos θ M cos θ r sin θ M3 M 7

8 where M, M, M3 and M are the more complicate parts of the Jacobi matrix with the transformation to r, ϕ and θ. r cos ϕ sin θ r sin ϕ cos θ M r sin θ M3 J = cos ϕ sin θ M r sin ϕ sin θ r cos ϕ cos θ M r sin θ M3 sin ϕ sin θ M r sin ϕ sin θ r cos ϕ cos θ M r cos ϕ sin θ r sin ϕ cos θ M + cos θ M J = r cos ϕ sin θ r sin ϕ sin θ r sin ϕ sin θ cos θ r cos ϕ sin θ cos θ r sin θ M3 M r sin θ M3 M r sin ϕ cos θ M M r cos ϕ cos θ M M J = r sin 3 θ M r sin ϕ sin θ cos θ M r cos ϕ sin θ cos θ M For the integral transformation we need the absolute value of the determinant of the Jacobi matrix. J = r sin θ M 3 So now we just need to calculate the absolute value of the determinant of M: x M = y r ϕ θ z M = cos ϕ cos θ sin ϕ cos ϕ sin θ sin ϕ cos θ cos ϕ sin ϕ sin θ sin θ cos θ x ỹ z r ϕ θ As the transformation matrix has no dependence on r, ϕ and θ we can easily get the determinante: cos ϕ cos θ sin ϕ cos ϕ sin θ x M = sin ϕ cos θ cos ϕ sin ϕ sin θ ỹ r ϕ θ sin θ cos θ z 8

9 M = cos ϕ cos θ sin ϕ cos ϕ sin θ sin ϕ cos θ cos ϕ sin ϕ sin θ sin θ cos θ x ỹ z r ϕ θ The determinante of the second part is the same as we were calculating many times bevor: x ỹ r ϕ θ z = r sin θ And the first part is also not really difficult: cos ϕ cos θ sin ϕ cos ϕ sin θ sin ϕ cos θ cos ϕ sin ϕ sin θ sin θ cos θ = cos ϕ cos θ +sin ϕ sin θ +cos ϕ sin θ +sin ϕ cos θ = So the absolute value of the determinante of the Jacobi matrix is: J = r r sin θ sin θ One more transformation With the transformation done before the integral is still not easy to solve. So we need one more transformation. We transform θ to r : r = r r r cos θ The Jacobi matrix for this transformation is: J = r r r r cos θ r r cos θ r r r sin θ r J = r r sin θ 34 We need to take a look at the integration limits. In spherical coordinates they were the same as we are used to. But now we have to get the correct limits for r. The limits for θ were and π, so r now needs to go from r r to r Solving the integral With all these transformations equations 33 and 34 the integral finally has a form that is easy to solve: ψ Ĥ ψ = r+r π π π dr dr dr dϕ dϕ r r ma r a 6 exp r a r+r = ma 7 π dr dr = ma 7 π = ma 7 π r r dr r r exp r r r+r dr dr dr + dr r r r r dr dr r + dr r r r exp r dθ r r sin θ sin θ r r r sin θ a r+r r r dr r a r r exp r a 9

10 r dr r exp λr = r λ dr exp λr = λ exp λr + = λ λ λ exp λr r λ + λ + λ 3 r = exp λr r λ + λ r λ λ 3 + λ 3 = exp λr r λ + r λ + λ 3 + λ 3 r dr r exp λr = λ r dr exp λr = λ exp λr λ r = exp λr λ + λ ψ Ĥ ψ = ma 7 + ma 7 = ma 7 4π dr exp 4π dr r exp 4π dr r a + = a 4π ma 7 r r a a dr With equations 6 and 7 this leads to: ψ Ĥ ψ = a 4π ma 7 = a ma 7 So finally we solved the integral: 4α a 5 4π r r a exp 4α r a a + r a + + a a + a + r exp exp a r + r exp 4α r a a + a 4α a + a a r exp r a r a r + r exp r a r a ψ Ĥ ψ = ma π 5 α Optimizing the Energy Now we calculated all terms to get the energy see equation 4: E α = ψ Ĥ ψ ψ ψ = ψ Ĥ ψ + ψ Ĥ ψ ψ ψ With equation 8, and 35 we get the result: E α = π 4 α ma α 5 π α 6 + π 5 ma α 5 8 E α = 7 ma 8 α α 36

11 As already explained in equation 5 the first derivative of the energy must be zero de α dα = = d 7 dα ma 8 α α = 7 ma 8 = 7 8 So the optimized value for α is: α = As we can easily see from the energy E α this is the global minimum. The corresponding energy is with equation 36: 7 E opt = E 6 = ma = 79 ma E opt = 79 ma The hartree E h is the atomic unit of energy see also equation 8: E h = ma = e 4πɛ a 4 So in hartree the energy would be E opt = hartree =.8477hartree 4 4 Putting the values = Js m = kg a = m e = C into equation 39 results in: E opt =.45 7 J = eV 47 48