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Second Trial Wavefunction ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Ψ = exp −α ⋅ r1 ⋅ exp −α ⋅ r2 + exp −α ⋅ r1 ⋅...

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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.