VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS Link to: physicspages home page. To leave a comment or report an error, please use the auxiliary blog. References: Griffiths, David J. 005), Introduction to Quantum Mechanics, nd Edition; Pearson Education - Problem 7.18. An interesting variant on the variational principle used to calculate an upper bound on the ground state of the hydrogen ion is one proposed in 1944 by Chandrasekhar. It uses two adjustable parameters instead of the one we ve used so far. The exact hamiltonian for the H ion is H = h m 1 + e ) 1 + 1 ] 1 1) r 1 r r 1 r Here we are using two independent spatial coordinates r 1 and r, one for each electron. The terms in the square brackets are the interaction terms between the electrons and the nucleus and between the two electrons. We use as a test function a combination of hydrogen ground states in the form ψ = Aψ 1 r 1 )ψ r ) + ψ 1 r )ψ r 1 )] ) where Zi 3 ψ i r) = πa 3 e Z ir/a 3) is a normalized ground state hydrogen wave function and Z 1, Z are the parameters. The idea is that each of the two electrons experiences a different amount of shielding due to differing distances from the nucleus. Equation is written as a symmetric function of r 1 and r implying that the spin state is antisymmetric in order to give the two electrons an overall antisymmetric function, as required by the Pauli exclusion principle. To find the optimal values of the Z i we can follow the usual procedure in the variational principle, although the calculations are quite tedious. One approach is simply to plug the equations as they stand into software such as Maple and let it grind through the calculations. However, to get the answer in Griffiths s book requires a bit more work. 1
VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS First, we observe that ψ i r j ) is an eigenfunction of H ij h m j e Z i 4) r j with eigenvalue energy) Z i E 1 where E 1 = 13.6 ev is the ground state energy of a hydrogen atom. Thus we can write the original hamiltonian 1 as H = H 11 + H + e Z1 1 + Z ] 1 + e 1 r 1 r r 1 r = H 1 + H 1 + e Z 1 + Z ] 1 1 + e 1 r 1 r r 1 r 5) 6) In calculating Hψ, we can use the fact that H 11 + H )ψ 1 r 1 )ψ r ) = E 1 Z 1 + Z ) ψ1 r 1 )ψ r ) 7) = H 1 + H 1 )ψ 1 r )ψ r 1 ) 8) To calculate H we therefore need to calculate the means of the remaining terms in 5 and 6. First, however, we need to calculate the normalization constant A in. We get A ˆ ψ d 3 r 1 d 3 r = 1 9) Doing the integrals with Maple, we get A Z 1 6 + 15Z 4 1 Z + 84Z 3 Z 3 1 + 6Z 5 Z 1 + 15Z 4 Z 1 + Z 6 + 6Z 5 1 Z Z 6 1 + 6Z 5 1 Z + 15Z 4 1 Z + 0Z 3 Z 3 1 + 15Z 4 Z 1 + 6Z 5 6 Z 1 + Z = 1 10) This rather frightening expression can be simplified by factoring the numerator and denominator. First, the numerator:
VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS 3 Z 6 1 + 15Z 4 1 Z + 84Z 3 Z 3 1 + 6Z 5 Z 1 + 15Z 4 Z 1 + Z 6 + 6Z 5 1 Z =... 11) Z + 6Z 1 Z + Z ) 1 Z 4 + 14Z Z 4 1 + Z ) 1 =... 1) ] ] Z 1 + Z ) + 4Z 1 Z Z 1 + Z ) 4 4Z1 3 Z + 8Z1Z 4Z 1 Z 3 =... 13) ] ] Z 1 + Z ) + 4Z 1 Z Z 1 + Z ) 4 + 16Z1Z 4Z1 3 Z 8Z1Z 4Z 1 Z 3 =... 14) ] Z 1 + Z ) + 4Z 1 Z Z 1 + Z ) 4 + 16Z1Z 4Z 1 Z Z 1 + Z 1 Z + Z) ] =... 15) ] Z 1 + Z ) + 4Z 1 Z Z 1 + Z ) 4 + 16Z1Z 4Z 1 Z Z 1 + Z ) ] =... 16) x + y ) x 4 + y 4 x y ] = x 6 + y 6 17) where we ve defined The denominator comes out to x Z 1 + Z 18) y Z 1 Z 19) Z 1 6 +6Z 1 5 Z +15Z 1 4 Z +0Z 3 Z 1 3 +15Z 4 Z 1 +6Z 5 Z 1 +Z 6 = x 6 so 0) A = x 6 x 6 + y 6 ) 1) Now we need to grind through the calculations for the terms in 5 and 6. First, we ll do the electron-electron interaction term:
VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS 4 V ee = e A ˆ = e A ˆ = e A a ψ 1 r 1 )ψ r ) + ψ 1 r )ψ r 1 )] d 3 r 1 d 3 r r 1 r ) ψ 1 r 1 )ψ r ) + ψ 1 r )ψ r 1 )] d 3 r 1 d 3 r r1 + r r 1r cosθ 3) y 5Z 3 Z 1 + 5Z 3 1 Z + 8Z Z 1 + Z 4 4 1 + Z ) 5 Z1 + 5Z 4 1 Z + 10Z 3 1 Z + 10Z 1 Z 3 + 5Z 4 5 Z 1 + Z ) = E 1 A y 5Z 3 Z 1 + 5Z 1 3 Z + 8Z Z 1 + Z 1 4 + Z 4 ) To reduce the numerator: 4) x 5 5) 5Z 3 Z 1 + 5Z 3 1 Z + 8Z Z 1 + Z 4 1 + Z 4 =... 6) x 4 + Z 3 Z 1 + Z 3 1 Z + Z Z 1 =... 7) x 4 + Z 1 Z Z 1 + Z ) + Z 1 Z + 0Z 1 Z = x 4 + y x + 5y ) 8) 4 Therefore V ee = E 1 A y 4x4 + y x + 5y ) 4x 5 9) = 1 E 1 xy 4x 4 + 5y 4 + x y ) 8 x 6 + y 6 30) Now for the electron-nucleus terms in 5 and 6. ˆ I 1 = e Z1 1 ψ + Z ] 1 ψ 1 r 1 )ψ r )d 3 r 1 d 3 r 31) r 1 r = E 1 A N x 5 3) The numerator is N = Z 7 + 5Z 1 6 Z 84Z 3 Z 1 3 + 11Z 1 Z 5 6Z 1 5 Z + 47Z 1 4 Z 3 Z 1 6 + 11Z 1 5 Z 15Z 1 4 Z Z 6 6Z 5 Z 1 + 47Z 1 3 Z 4 + 5Z 1 Z 6 + Z 1 7 15Z 4 Z 1 33)
VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS 5 We can break this down by picking out the terms where the exponents sum to 6 denoted by N 6 ) and then the terms where they sum to 7 N 7 ). The terms summing to 6 are: N 6 = Z 6 6Z 1 5 Z Z 1 6 15Z 1 4 Z 6Z 5 Z 1 84Z 3 Z 1 3 15Z 4 Z 1 34) = Z + 6Z 1 Z + Z 1 ) Z 4 + 14Z Z 1 + Z 1 4 ) 35) = x 6 y 6 36) using 17. The terms summing to 7 give: N 7 = Z 7 1 + 47Z 3 1 Z 4 + 5Z 1 Z 6 + Z 7 + 11Z 5 1 Z + 47Z 4 1 Z 3 + 11Z 1 Z 5 + 5Z 6 1 Z 37) = Z + Z 1 ) Z ) 6 + 4Z 5 Z 1 + 7Z 4 Z 1 + 40Z 3 Z 3 1 + 7Z 4 1 Z + 4Z 5 6 1 Z + Z 1 38) We can write the sixth degree polynomial as Z 6 + 4Z 5 Z 1 + 7Z 4 Z 1 + 40Z 3 Z 1 3 + 7Z 1 4 Z + 4Z 1 5 Z + Z 1 6 =... 39) Z 1 + Z ) 6 Z 5 1 Z Z 1 Z 5 8Z4 1Z 8Z 1Z 4 + 0Z 3 1 Z3 =... 40) Putting it together: x 6 y Z 4 1 + Z 4 + 4Z1 3 Z + 4Z 1 Z 3 10Z 1Z) = x 6 y x 4 y 4) 41) N 7 = x x 6 y x 4 y 4)] 4) x x 6 y I 1 = E 1 A x 4 y 4)] x 6 y 6 x = E 1 x x 6 y x 5 43) x 4 y 4)] x 6 y 6 x 6 + y 6 44)
VARIATIONAL PRINCIPLE AND THE HYDROGEN ION: TWO PARAMETERS 6 The other integral is ˆ I = e Z 1 ψ + Z ] 1 1 ψ 1 r )ψ r 1 )d 3 r 1 d 3 r 45) r 1 r which turns out to be equal to I 1. So finally H = E 1 Z 1 + Z) + I1 + I + V ee 46) x x = E 1 Z 1 + Z) 6 y x 4 y 4)] x 6 y 6 E1 x x 6 + y 6 1 E 1 xy 4x 4 + 5y 4 + x y ) 8 x 6 + y 6 47) Using Z 1 + Z = x y /, we can put everything over a common denominator and cancel terms to get H = E 1 x 8 x 6 + y 6 + x 7 + 1 x6 y 1 x5 y 18 x3 y 4 + 118 xy6 1 ] y8 48) To find the minimum of this, we can find the maximum of H /E 1 since E 1 < 0) and we can use Maple s Maximize function to do this numerically. We find These values of x min and y min correspond to x min = 1.345 49) y min = 1.08505 50) H min = 1.066E 1 51) Z 1 min = 1.0393 5) Z min = 0.83 53) Thus the upper bound on the energy is slightly lower than E 1 indicating that the H ion is stable.