Second Trial Wavefunction
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Ψ = exp −α ⋅ r1 ⋅ exp −α ⋅ r2 + exp −α ⋅ r1 ⋅ exp −β ⋅ r2 + exp −β ⋅ r1 ⋅ exp −α ⋅ r2 + exp −β ⋅ r1 ⋅ exp −β ⋅ r2
When the wavefunction shown above is used in a variational method calculation for the ground state energy for two-electron atoms or ions the two-parameter equation shown below for the energy is obtained. This equation is then minimized simultaneously with respect to the adjustable parameters, α and β. Nuclear charge:
Z := 2
α := Z
Seed values for scale factors:
β := Z + 1
Variational energy expression: 1.5 1.5 α 2 + β 2 8⋅ α ⋅ β α ⋅ β − Z⋅ ( α + β ) − ⋅Z − 2 2 α + β ( ) α + β ... E( α , β ) := 1.5 1.5 8⋅ α ⋅ β 1+ 3 (α + β ) 2 2 8 ⋅ α 2.5⋅ β 1.5⋅ 11⋅ α 2 + 8 ⋅ α ⋅ β + β 2 5 2⋅ α ⋅ β ⋅ α + 3⋅ α ⋅ β + β ... ⋅(α + β) + + 4⋅ 8 (α + β ) 3 ( α + β ) 2⋅ ( 3 ⋅ α + β ) 3 8 ⋅ α 1.5⋅ β 2.5⋅ 11⋅ β 2 + 8 ⋅ α ⋅ β + α 2 ... + 2 3 ( ) ( ) α + β ⋅ 3⋅ β + α 3 3 + 20⋅ α ⋅ β ( α + β ) 5 +
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8 ⋅ α 1.5⋅ β 1.5 4 ⋅ 1 + 3 (α + β ) α := Minimize( E , α , β ) β Experimental ground state energy:
Calculate error in calculation:
α β
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2
1.2141 2.1603
E( α , β ) = −2.8603
Eexp := −2.9037
Error :=
Eexp − E( α , β )
Error = 1.4931 %
Eexp
Fill in the table and answer the questions below:
Ψ α β E atom Eatom( exp) %Error
H
He
Li
0.3703
1.2141
2.0969
1.0001
2.1603
3.2778
−0.487 −2.8603 −7.235 −0.5277 −2.9037 −7.2838 7.72
1.49
0.670
2.9993 4.3756 −13.6098 −13.6640 0.397 Be
Fill in the table below and explain why this trial wave function gives better results than the first trial wave function.
α 2 + β 2 8 ⋅ α 1.5⋅ β 1.5 α ⋅ β + ⋅ 2 2 α + β ( ) α + β T( α , β ) := 1.5 1.5 8⋅ α ⋅ β 1+ 3 (α + β )
1.5 1.5 8⋅ α ⋅ β −Z⋅ ( α + β ) − ⋅Z 2 (α + β) ( ) Vne α , β := 1.5 1.5 8⋅ α ⋅ β 1+ 3 (α + β )
T( α , β ) = 2.8603
Vne( α , β ) = −6.7488
Vee( α , β ) := E( α , β ) − T( α , β ) − Vne ( α , β )
Vee( α , β ) = 1.0281
WF2 H He Li Be
E
T
Vne
−0.4870
0.4870
−1.3705
−2.8603
2.8603
−6.7488
−7.2350
7.2350
−16.1243
−13.6098 13.6098 −29.4995
0.3965 1.0281 1.6544 2.2799 Vee
Demonstrate that the virial theorem is satisfied. E( α , β ) = −2.8603
−T( α , β ) = −2.8603
Vne ( α , β ) + Vee( α , β ) 2
= −2.8603
Add the results for this wave function to your summary table for all wave functions. E T Vne Vee H WF1 −0.4727 0.4727 −1.375 0.4297 WF2 −0.4870 0.4870 −1.3705 0.3965
E T Vne Vee He WF1 −2.8477 2.8477 −6.7500 1.0547 WF2 −2.8603 2.8603 −6.7488 1.0281
E T Vne Vee Li WF1 −7.2227 7.2227 −16.1250 1.6797 WF2 −7.2350 7.2350 −16.1243 1.6544
E T Vne Vee Be WF1 −13.5977 13.5977 −29.5000 2.3047 WF2 −13.6098 13.6098 −29.4995 2.2799
These tables show that the improved agreement with experimental results (the lower total energy), is due to a reduction in electron-electron repulsion.
