Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19 The variational Method to estimate the ground- state energy of a helium atom - We will use the variational method to estimate the groundstate energy of a helium atom. - The Hamiltonian operator for a helium atom is 2 2 2 2 e 1 1 e 1 Hˆ ( ) ( ) 2m e 4 0 r1 r2 4 0 r12 2 1 2 2 (1) - The Schrödinger equation cannot be solved exactly for this system because of the term involving r12 . - Equation (1) can be written in the form 2 e 1 Hˆ Hˆ H (1) Hˆ H (2) 4 0 r12 (2) where Hˆ H (1) 2 2e 2 1 2m e 4 0 r1 2 1 and Hˆ H (2) 2 2e 2 1 2m e 4 0 r2 2 2 So, we can write it as 2 2 2 e 1 Hˆ H ( j ) 2m e 4 0 r j 2 j [1] j = 1and 2 (3) Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19 Hˆ H ( j ) is the Hamiltonian operator for a single electron around a helium nucleus. Thus, Hˆ H (1) and Hˆ H (2) satisfy the equation Hˆ H ( j ) H (rj , j , j ) E j H (rj , j , j ) j = 1 or 2 (4) - where H (rj , j , j ) is a hydrogen like wave function with Z = 2 and where the E j are given by mee 4 me e 4 En 2 2 2 8 0 h n 32 2 02 2 n 2 Z 2mee 4 E j 32 2 02 2 n 2j j = 1 or 2 (5) - with Z = 2. If we ignore the inter electronic repulsion term (e2/4πε0r12), then the Hamiltonian operator is separable and the ground – state wave function would be 0 (r1 , r2 ) 1s (r1 ) 1s (r2 ) (6) from (Table in hydrogen lecture)we take 1s value 1/2 Z3 Zr / a 1s (r j ) 3 e j 0 j = 1 or 2 a0 (7) Where a0 = 4πε0ħ2/mee2. We can use Equations (6) and (7) as a trial function using Z as a variational constant. Thus, we must evaluate E ( Z ) 0 (r1 , r2 )Hˆ 0 (r1 , r2 )dr1dr2 [2] (8) Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19 Z 3 Z ( r1 r2 )/a0 0 (r1 , r2 ) 3 e a0 (9) From equation (1). The Hamiltonian operator of the helium atom is Hˆ 2 2 2e 2 2e 2 e2 2m e 2m e 4 0 r1 4 0 r2 4 0 r12 (10) 2 1 2 2 We substitute equation 9 and 10 in equation 8. The result is mee 4 E (z ) 16 2 02 (Z 2 2 27 Z) 8 (11) Equation (11) suggests that it is convenient to express E in unite of mee 4 16 2 02 2 , and so we can write Equation 11 as E (z ) ( Z 2 27 Z ) (12) 8 If we minimize E(z) with respect to Z, we find that Zmin = 27/16. We substitute this result in equation 11 to obtain mee 4 E (z ) 2.8477 16 2 02 2 (13) The experimental value is mee 4 E 2.9037 16 2 02 [3] 2 (14) Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19 Thus, we achieve a fairly good result, considering the simplicity of the trial function. The value of Z that minimize E can be interpreted as an effective charge. That fact that Z comes out to be less than 2 reflects the fact that each electron partially screens the nucleus from the other , so that net effective nuclear charge is reduced from 2 to 27/16 (1.68). [4]