r 2 dr h2 α = 8m2 q 4 Substituting we find that variational estimate for the energy is m e q 4 E G = 4

Transcription

1 Variational calculations for Hydrogen and Helium Recall the variational principle See Chapter 16 of the textbook The variational theorem states that for a Hermitian operator H with the smallest eigenvalue E, any normalized ψ satisfies E ψ H ψ Please prove this now without opening the text Given a Hamiltonian the method consists in starting with a clever and tractable guess for the so-called trial wave function with one or more free variational parameters Normalize the wave function and evaluate ψ H ψ This yields a function of the variational parameters Find the value of the parameters that minimizes this function and this yields the variational estimate for the ground state energy Gaussian trial wave function for the hydrogen atom: Try a Gaussian wave function since it is used often in quantum chemistry Start from the normalized Gaussian: 2 3/4 ψr = α 3/4 e αr2 π The kinetic energy can be evaluated to be r 2 dr h2 2m ψr ψr = h2 2m 3α Observe that since the dimensions of α is L 2 we can guess the dependence on dimensional considerations only The potential energy is given by Minimize the energy with respect to α e2 r 2 dr ψ 2 r 1 2 ɛ r = 2 π q2 α 1/2 α = 8m2 q 4 9π h 4 Substituting we find that variational estimate for the energy is E G = 4 3π m e q 4 The exact energy has a factor of 1/2 in front and the error is 151% If one uses the sum of two Gaussians Ae αx2 + Be βx2 one can do better but this is tedious to implement The Hamiltonian is given by H = h2 2m 2 1 h2 2m 2 2 r 1 1 r 2 + q 2 r 1 r 2

2 where the first two terms correspond to the kinetic energies of the two electrons and the last term is the Coulombic repulsion between the two electrons For Helium the nuclear charge Z = 2 As a variational trial wave function we place the two electrons in a 1s-like state with the spatial part being symmetric both electrons are in the same state and the spin part in the antisymmetric spin singlet So we write the spatial part of the wave function as φ r 1 φ r 2 1 We try a form inspired by the 1s state of the hydrogen atom which is exponentially decaying: φr = 1 2α 3/2 e αr This is normalized and has one variational parameter α which controls the spatial extent of the wave function The angular part is a constant, ie, l = We need to evaluate Ψ H Ψ The first term we need is d 3 r 2 φ r 1 φ r 2 h2 2m 2 1 φ r 1 φ r 2 Since the operator only involves the first coordinate r 1 the integral over d 3 r 2 yields 1 since φ r 2 is normalized We are left with 2m [ φr] 2 and this is the same calculation as the kinetic energy of the electron in the hydrogen atom and we find h2 α 2 ; this must be multiplied by a factor of 2 for the kinetic energy of the 2m second electron which involves identical integrals 2 Now for the potential energies For the attractive energy of the first electron and the nucleus we have d 3 r 2 φ r 1 φ r 2 r 1 Again the d 3 r 2 integral yields 1 and we are left with φ r 1 r 1 φ r 1 φ r 1 φ r 2 which can be done since this is spherically symmetric and yields αzq 2 ; this should also be multiplied by 2 for the contribution from the other electron 1 Note that the actual wave function will not be of this form and can depend upon r 1 r 2 which cannot be written as a product of a function of r 1 and r 2 separately However, we choose the product form for calculational convenience If we want a better approximation to the ground state energy we can use other forms of the wave function 2 Observe that when the Hamiltonian involves a single-particle term, ie, depends on the coordinates of one of the particles only, the expectation value involves a three-dimensional integral over the coordinates of that particle only 2

3 The interaction energy is much harder to calculate Please see the end of this note if you are mathematically curious and yields 5q 2 α/8 Together we have Eα = h2 α 2 m q2 α 2Z 5 8 Minimizing with respect to α we obtain 3 α = mq2 Z 5 16 The minimum energy is given by Eα = α and is E g = mq4 Z Recall that the ground state of the hydrogen atom is given by E = mq4 2 Thus we write our estimate of the ground state energy is E g = 2E Z If we had just naively used the 1s hydrogen atom energy except with nuclear charge Z, each electron would have an energy E Z 2 and so for two electrons we would have 2E Z 2 excluding the repulsive energy Instead the nuclear charge is reduced from Z = 2 to a lower value thanks to the shielding effect of the other electron Evaluating we find an energy of 28477Ry per electron while the experimental value is 294Ry per electron for a 2% error which is rather good The variational estimate for the ground state energy is 775eV in contrast to the experimental value of 79eV S Chandrasekhar and G, Herzberg used 1 variational parameters and then 18 Later T Kinoshita used 39 and later 8 parameters If we had set Z = 1 this calculation corresponds to the variational estimate for H : note that our calculation for Z = 1 yields 121/128 Ry which is higher than that of the hydrogen atom and so this makes the prediction that H is unstable with respect to the hydrogen atom and the electron far away However, experimentally, H is stable Bethe was one of the first to ask this question Superior trial wave functions have been tried For 3 2 α m q2 2Z 5 8 α=α = 3

4 example, multiply by 1+W r 12 where r 12 = r 1 r 2 4 Using two variational parameters we obtain an energy that yields a stable ion with an energy of Ry Calculation of the interaction energy for the Helium atom with the simplest variational ansatz The trial wave function is given by u r = 1 2α 3/2 e αr We need to evaluate the six-dimensional inetgral I d 3 r 2 u r 1 2 q 2 r 1 r 2 u r 2 2 There are many ways of doing this integral Here is one: We use the fact that the three-dimensional Fourier transform of the Coulomb potential is /k 2, a very important result 5 : 1 r 1 r 2 = d 3 k 2π 3 k 2 ei k r 1 r 2 Substituting this in the integral and changing the orders of integration I = d 3 k q 2 2π 3 k 2 u r 1 2 e i k r 1 d 3 r 2 u r 2 2 e i k r 2 Clearly, we need we need the Fourier transform of u r 2 which occurs twice: 6 we have Using e i k r 1 α 3 dω e i k r 1 π e 2αr 1 = e i k r 1 α 3 π dθ sin θ π e 2αr 1 2π dφ e i k r 1 = sinkr 1 kr 1 dr r 2 4α 3 sinkr 1 kr 1 e 2αr 1 = 16α 4 4α 2 + k This is a non-trivial calculation done most simply by using spherical coordinates for r 1 and cylindrical coordinates for r 2 with the z axis aligned along r 1 and ρ = r Even with this one has to work hard to evaluate the different terms 5 One way of convincing yourself of this is to Fourier transform the equation 1 2 = δ r r 6 Since we can change r 2 r 2 the sign of the argument in the exponential does not matter You have done essentially this integral in an earlier homework when you considered the momentum distribution of the hydrogen atom ground state 4

5 So we finally need the integral 7 d 3 k q 2 2π 3 k 2 16α 4 4α 2 + k = 5 8 q2 α 7 This is not difficult to do or you can be use Maple or Mathematica 5