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Slater Determinants for 2 electrons in 2 orbitals (Read 4968 times)
Gerrit-Jan Linker
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Slater Determinants for 2 electrons in 2 orbitals
21.08.12 at 13:20:27
 
Slater Determinants for 2 electrons in 2 orbitals
 
There are 6 possible Slater determinants for 2 electrons and 2 orbitals. We use the following notation:
a,b : spatial orbitals
α, : spin functions alpha and beta
 
The product of a spatial and a spin function is a spin orbital, e.g. aα , b.
 
The 6 different Slater determinants we can build are listed below. We indicate whether the Slater determinants are eigenfunctions of Ŝ2 and Ŝz. The results of the working of these operators on one or two electron spin functions are taken from http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1297282533/0#1.
 
|aαbα| : two alpha electrons each in its own orbital
|aαbα|=aαbα-bαaα=a(1)α(1)b(2)α(2)-b(1)α(1)a(2)α(2)=[a(1)b(2)-b(1)a(2)] α(1)α(2)
 
Ŝz|aαbα| = Ŝz [a(1)b(2)-b(1)a(2)] α(1)α(2)  
  = [a(1)b(2)-b(1)a(2)] Ŝz α(1)α(2)
  = [a(1)b(2)-b(1)a(2)] α(1)α(2)
  = |aαbα|
Ŝ2|aαbα| = Ŝ2 [a(1)b(2)-b(1)a(2)] α(1)α(2) =
  = [a(1)b(2)-b(1)a(2)] Ŝ2 α(1)α(2)
  = [a(1)b(2)-b(1)a(2)] 2 α(1)α(2)
  = 2 |aαbα|
 
 
|ab| : two beta electrons each in its own orbital
|ab|=ab-ba=a(1)(1)b(2)(2)-b(1)(1)a(2)(2)=[a(1)b(2)-b(1)a(2)] (1)(2)
 
Ŝz|ab| = Ŝz [a(1)b(2)-b(1)a(2)] (1)(2)  
  = [a(1)b(2)-b(1)a(2)] Ŝz (1)(2)
  = [a(1)b(2)-b(1)a(2)] (-1) (1)(2)
  = - |ab|
Ŝ2|ab| = Ŝ2 [a(1)b(2)-b(1)a(2)] (1)(2) =
  = [a(1)b(2)-b(1)a(2)] Ŝ2 (1)(2)
  = [a(1)b(2)-b(1)a(2)] 2 (1)(2)
  = 2 |ab|
 
 
|aαb| : two electrons with opposite spin each in its own orbital
|aαb|=aαb-baα=a(1)α(1)b(2)(2)-b(1)(1)a(2)α(2)=a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)
 
Ŝz|aαb| = Ŝz a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)  
  = Ŝz a(1)b(2)α(1)(2) - Ŝz b(1)a(2)(1)α(2)
  = a(1)b(2) Ŝz α(1)(2) - b(1)a(2) Ŝz (1)α(2)
  = a(1)b(2) Ŝz α(1)(2) - b(1)a(2) Ŝz (1)α(2)
  = 0 - 0 = 0
Ŝ2|aαb| = Ŝ2 a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)  
  = Ŝ2 a(1)b(2)α(1)(2) - Ŝ2 b(1)a(2)(1)α(2)  
  = a(1)b(2) Ŝ2 α(1)(2) - b(1)a(2) Ŝ2 (1)α(2)
  = a(1)b(2) [α(1))(2) + (1)α(2)] - b(1)a(2) [α(1))(2) + (1)α(2)]
  = [a(1)b(2) - b(1)a(2)] [α(1))(2) + (1)α(2)]
  => |aαb| is not an eigenfunction of Ŝ2
 
 
|abα| : two electrons with opposite spin each in its own orbital  
|abα| = aαb-baα = a(1)α(1)b(2)(2)-b(1)(1)a(2)α(2) = a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)
 
Ŝz |abα| = Ŝz [a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)]
  = a(1)b(2) Ŝz α(1)(2) - b(1)a(2) Ŝz (1)α(2)
  = a(1)b(2) Ŝz α(1)(2) - b(1)a(2) Ŝz (1)α(2)
  = 0 - 0 = 0
Ŝ2 |abα| = Ŝ2 [a(1)b(2)α(1)(2)-b(1)a(2)(1)α(2)]
  = a(1)b(2) Ŝ2 α(1)(2) - b(1)a(2) Ŝ2 (1)α(2)
  = a(1)b(2) [α(1))(2) + (1)α(2) ] - b(1)a(2) [α(1))(2) + (1)α(2) ]
  = [a(1)b(2) - b(1)a(2)] [α(1))(2) + (1)α(2) ]
  => |abα| is not an eigenfunction of Ŝ2  
 
 
|aαa| : two paired electrons in the same orbital
|aαa| = aαa - aaα = a(1)α(1)a(2)(2) - a(1)(1)a(2)α(2) = a(1)a(2)α(1)(2) - a(1)a(2)(1)α(2)
  = a(1)a(2) [α(1)(2) - (1)α(2)]
 
Ŝz |aαa| = Ŝz a(1)a(2) [α(1)(2) - (1)α(2)]
  = a(1)a(2) [Ŝz α(1)(2) - Ŝz (1)α(2)]
  = a(1)a(2) [0 - 0]
  = 0
Ŝ2 |aαa| = Ŝ2 a(1)a(2) [α(1)(2) - (1)α(2)]
  = a(1)a(2) [Ŝ2 α(1)(2) - Ŝ2 (1)α(2)]
  = a(1)a(2) [[α(1))(2) + (1)α(2)] - [α(1))(2) + (1)α(2)] ]
  = a(1)a(2) [ 0 ]
  = 0
 
 
|bbα| : two paired electrons in the same orbital
|bbα| = bbα - bαb = b(1)(1)b(2)α(2) - b(1)α(1)b(2)(2) = b(1)b(2)(1)α(2) - b(1)b(2)α(1)(2)
  = b(1)b(2) [(1)α(2) - α(1)(2)]
 
Ŝz |bbα| = Ŝzb(1)b(2) [(1)α(2) - α(1)(2)]
  = b(1)b(2) [Ŝz (1)α(2) - Ŝz α(1)(2)]
  = b(1)b(2) [0 - 0]
  = 0
Ŝ2 |bbα| = Ŝ2 b(1)b(2) [(1)α(2) - α(1)(2)]
  = b(1)b(2) [Ŝ2(1)α(2) - Ŝ2α(1)(2)]
  = b(1)b(2) [0 - 0]
  = 0
 
 
Let's summarise. The eigenvalue of Ŝz is Ms and the eigenvalue of Ŝ2 is S(S+1).
 
|aαbα| Ms=1, S(S+1)=2, S=1
|ab| Ms=-1, S(S+1)=2, S=1
|aαb| Ms=0, not an eigenfunction of Ŝ2
|abα| Ms=0, not an eigenfunction of Ŝ2
|aαa| Ms=0, S(S+1)=0, S=0
|bbα| Ms=0, S(S+1)=0, S=0
 
The Slater determinants |aαb| and |abα| are not eigenfunctions of Ŝ2. We seek solutions of the Schrodinger equation that are simultaneous eigenfunctions of the Hamiltonian Ĥ, Ŝ2 and Ŝz. These determinants are therefore not good solutions of the Schrodinger equation.
 
|ab| and |ab| are not eigenfunctions of Ŝ2 but linear combinations are:
Ŝ2 [|ab| + |ab|] = Ŝ2 [|ab| - |ba|] = Ŝ2|ab| - Ŝ2|ba|  
  = [a(1)b(2) - b(1)a(2)] [α(1))(2) + (1)α(2)] - [b(1)a(2) - a(1)b(2)] [α(1))(2) + (1)α(2)]
  = ([a(1)b(2) - b(1)a(2)] - [b(1)a(2) - a(1)b(2)]) [α(1))(2) + (1)α(2)]
    = 0
 
Ŝ2 [|ab| - |ab|] = Ŝ2 [|ab| + |ba|]= Ŝ2|ab| + Ŝ2|ba|
  = [a(1)b(2) - b(1)a(2)] [α(1))(2) + (1)α(2)] + [b(1)a(2) - a(1)b(2)] [α(1))(2) + (1)α(2)]
  = ([a(1)b(2) - b(1)a(2)] + [b(1)a(2) - a(1)b(2)]) [α(1))(2) + (1)α(2)]
   = 0
 
See also:
Spin operators
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1297282533/0#1.  
Slater determinant
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1217256942
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« Last Edit: 09.09.12 at 21:11:16 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
Linker IT Software
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Gerrit-Jan Linker
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Posts: 75
Slater Determinants: using LCAO's
Reply #1 - 09.09.12 at 16:38:29
 
Slater Determinants: using LCAO's
 
We usually make molecular orbitals as a linear combination of atomic orbitals. When we use for example a basis a, b, we can also use a linear combination of these AO's as a basis, e.g.: (a+b) and (a-b). An example where this would be appropriate is H2 where the two H atoms have interaction. Suppose a is the 1s AO on one H and b is the 1s AO on the other H let's review what the Slater determinants represent. We use the following short hand notation: |ab|=|a(1)alpha(1)b(2)beta(2)|.
 
The six possible Slater determinants are:
Singlet: |aa|,|bb|,|ab|,|ab|
Triplet: |ab|,|ab|
 
|aa|,|bb|: both electrons are on one H only
|ab|,|ab|,|ab|,|ab|: one electron on each atom.
 
When we would use (a+b) and (a-b) as a basis we can write six possible Slater determinants:
|(a+b)(a+b)|, |(a-b)(a-b)|, |(a+b)(a-b)|, |(a+b)(a-b)|, |(a+b)(a-b)|, |(a+b)(a-b)|
 
We can write these determinants out to see what they represent:
|(a+b)(a+b)|= |aa|+|ab|+|ba|+|bb|  
|(a-b)(a-b)|= |aa|-|ab|-|ba|+|bb|  
|(a+b)(a-b)|= |aa|-|ab|+|ba|-|bb|  
|(a+b)(a-b)|= |aa|-|ab|+|ba|-|bb|  
|(a+b)(a-b)|= |aa|-|ab|+|ba|-|bb|  
|(a+b)(a-b)|= |aa|-|ab|+|ba|-|bb|  
 
We get a mixture of ionic configurations with both electrons on one hydrogen and neutral configurations in which there is one electron on each hydrogen.
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« Last Edit: 28.09.12 at 14:55:12 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
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