5.61 Physical Chemistry Lecture #27 page 1
MANY ELECTRON ATOMS
Thus far, we have learned that the independent particle model (IPM) gives a
qualitatively correct picture of the eigenstates of the helium atom. What about
atoms with more than two electrons, such as lithium or carbon? As it turns out, the
IPM is capable of giving a realistic picture of atomic structure in essentially an
analogous fashion to the helium case. To begin with, we set up our coordinates so that
the nucleus is at the origin and the N electrons are at positions r1, r2, r3, …rN. In terms
of these variables, we can quickly write down the many-electron Hamiltonian (in atomic
units):
1 N 2 Z N N N
1
Hˆ = − ∑∇ j −∑ + ∑∑
2 j=1 j=1 r̂ j i=1 j >i r̂i − r̂ j
Kinetic Energy Electron-Nuclear Electron-Electron
Attraction Repulsion
Thus, the Hamiltonian has the same three sources of energy as in the two electron
case, but the sheer number of electrons makes the algebra more complicated. As
before, we note that we can make the Hamiltonian separable if we neglect the
electron-electron repulsion:
1 N N Z N ⎛ Z ⎟
⎞ N
Ĥ =− 2
∑ ∇ − ∑ ⎜
= ∑ − ∇ − 1 2 ≡ ∑ ĥ
j = 1 ⎝⎜
NI 2 j = 1 j j = 1 r̂ 2 j r̂ ⎟ i
j j ⎠ j =1
where each of the independent Hamiltonians ĥi describes a single electron in the field
of a nucleus of charge +Z. Based on our experience with separable Hamiltonians, we
can immediately write down the eigenstates of this Hamiltonian as products with
energies given as sums of the independent electron energies:
Ψ = ψ k (1)ψ k ( 2 )ψ k ( 3) ...ψ k (N ) E = Ek + Ek + Ek +... + Ek
1 2 3 N 1 2 3 N
Where (1) is a shorthand for (r1,σ1) and ki ≡ {ni ,li , mi , si } specifies all the quantum
numbers for a given hydrogen atom eigenstate. Of course, there is a problem with
these eigenstates: they are not antisymmetric. For the Helium atom, we fixed this by
making an explicitly antisymmetric combination of two degenerate product states:
ψ1sα (1) ψ1sα ( 2 )
1
(ψ (1)ψ1sβ ( 2) −ψ1sα ( 2)ψ1sβ (1)) =
2 1sα
1
2ψ
1sβ (1) ψ1sβ ( 2)
, 5.61 Physical Chemistry Lecture #27 page 2
where on the right we have noted that this antisymmetric product can also be written
as a determinant of a 2x2 matrix. As it turns out, it is straightforward to extend this
idea to generate an N particle antisymetric state by computing an NxN determinant
called a Slater Determinant:
ψ k (1) ψ k (1) ψ k (1) � ψ k (1)
1 2 3 N
ψ k ( 2) ψ k ( 2) ψ k ( 2) � ψ k ( 2)
1 2 3 N
1
Ψ (1,2,..., N ) =
N ! k1 ( )
ψ 3 ψ k ( 3) ψ k ( 3) � ψ k ( 3)
2 3 N
� � � � �
ψk (N ) ψk (N ) ψk (N ) � ψk (N )
1 2 3 N
As you can imagine, the algebra required to compute integrals involving Slater
determinants is extremely difficult. It is therefore most important that you realize
several things about these states so that you can avoid unnecessary algebra:
• A Slater determinant corresponds to a single stick diagram. This is easy to
see by example:
2px 1sα (1) 1s β (1) 2sα (1) 2 pxα (1)
1sα ( 2 ) 1s β ( 2 ) 2sα ( 2 ) 2 pxα ( 2 )
2s ⇒ Ψ (1, 2, 3, 4 ) =
1sα ( 3 ) 1s β ( 3 ) 2sα ( 3 ) 2 pxα ( 3 )
1s 1sα ( 4 ) 1s β ( 4 ) 2sα ( 4 ) 2 pxα ( 4 )
It should be clear that we can extend this idea to associate a determinant with
an arbitrary stick diagram. Further, recall that for the excited states of
helium we had a problem writing certain stick diagrams as a (space)x(spin)
product and had to make linear combinations of certain states to force things
to separate. Because of the direct correspondence of stick diagrams and
Slater determinants, the same pitfall arises here: Slater determinants
sometimes may not be representable as a space)x(spin) product, in which
case a linear combination of Slater determinants must be used instead.
This generally only happens for systems with unpaired electrons, like the 1s↑2s↓
configuration of helium or the …2px↑2py↓ configuration of carbon.
• A Slater determinant is anitsymmetric upon exchange of any two electrons.
We recall that if we take a matrix and interchange two its rows, the
determinant changes sign. Thus, interchanging 1 and 2 above, for example:
MANY ELECTRON ATOMS
Thus far, we have learned that the independent particle model (IPM) gives a
qualitatively correct picture of the eigenstates of the helium atom. What about
atoms with more than two electrons, such as lithium or carbon? As it turns out, the
IPM is capable of giving a realistic picture of atomic structure in essentially an
analogous fashion to the helium case. To begin with, we set up our coordinates so that
the nucleus is at the origin and the N electrons are at positions r1, r2, r3, …rN. In terms
of these variables, we can quickly write down the many-electron Hamiltonian (in atomic
units):
1 N 2 Z N N N
1
Hˆ = − ∑∇ j −∑ + ∑∑
2 j=1 j=1 r̂ j i=1 j >i r̂i − r̂ j
Kinetic Energy Electron-Nuclear Electron-Electron
Attraction Repulsion
Thus, the Hamiltonian has the same three sources of energy as in the two electron
case, but the sheer number of electrons makes the algebra more complicated. As
before, we note that we can make the Hamiltonian separable if we neglect the
electron-electron repulsion:
1 N N Z N ⎛ Z ⎟
⎞ N
Ĥ =− 2
∑ ∇ − ∑ ⎜
= ∑ − ∇ − 1 2 ≡ ∑ ĥ
j = 1 ⎝⎜
NI 2 j = 1 j j = 1 r̂ 2 j r̂ ⎟ i
j j ⎠ j =1
where each of the independent Hamiltonians ĥi describes a single electron in the field
of a nucleus of charge +Z. Based on our experience with separable Hamiltonians, we
can immediately write down the eigenstates of this Hamiltonian as products with
energies given as sums of the independent electron energies:
Ψ = ψ k (1)ψ k ( 2 )ψ k ( 3) ...ψ k (N ) E = Ek + Ek + Ek +... + Ek
1 2 3 N 1 2 3 N
Where (1) is a shorthand for (r1,σ1) and ki ≡ {ni ,li , mi , si } specifies all the quantum
numbers for a given hydrogen atom eigenstate. Of course, there is a problem with
these eigenstates: they are not antisymmetric. For the Helium atom, we fixed this by
making an explicitly antisymmetric combination of two degenerate product states:
ψ1sα (1) ψ1sα ( 2 )
1
(ψ (1)ψ1sβ ( 2) −ψ1sα ( 2)ψ1sβ (1)) =
2 1sα
1
2ψ
1sβ (1) ψ1sβ ( 2)
, 5.61 Physical Chemistry Lecture #27 page 2
where on the right we have noted that this antisymmetric product can also be written
as a determinant of a 2x2 matrix. As it turns out, it is straightforward to extend this
idea to generate an N particle antisymetric state by computing an NxN determinant
called a Slater Determinant:
ψ k (1) ψ k (1) ψ k (1) � ψ k (1)
1 2 3 N
ψ k ( 2) ψ k ( 2) ψ k ( 2) � ψ k ( 2)
1 2 3 N
1
Ψ (1,2,..., N ) =
N ! k1 ( )
ψ 3 ψ k ( 3) ψ k ( 3) � ψ k ( 3)
2 3 N
� � � � �
ψk (N ) ψk (N ) ψk (N ) � ψk (N )
1 2 3 N
As you can imagine, the algebra required to compute integrals involving Slater
determinants is extremely difficult. It is therefore most important that you realize
several things about these states so that you can avoid unnecessary algebra:
• A Slater determinant corresponds to a single stick diagram. This is easy to
see by example:
2px 1sα (1) 1s β (1) 2sα (1) 2 pxα (1)
1sα ( 2 ) 1s β ( 2 ) 2sα ( 2 ) 2 pxα ( 2 )
2s ⇒ Ψ (1, 2, 3, 4 ) =
1sα ( 3 ) 1s β ( 3 ) 2sα ( 3 ) 2 pxα ( 3 )
1s 1sα ( 4 ) 1s β ( 4 ) 2sα ( 4 ) 2 pxα ( 4 )
It should be clear that we can extend this idea to associate a determinant with
an arbitrary stick diagram. Further, recall that for the excited states of
helium we had a problem writing certain stick diagrams as a (space)x(spin)
product and had to make linear combinations of certain states to force things
to separate. Because of the direct correspondence of stick diagrams and
Slater determinants, the same pitfall arises here: Slater determinants
sometimes may not be representable as a space)x(spin) product, in which
case a linear combination of Slater determinants must be used instead.
This generally only happens for systems with unpaired electrons, like the 1s↑2s↓
configuration of helium or the …2px↑2py↓ configuration of carbon.
• A Slater determinant is anitsymmetric upon exchange of any two electrons.
We recall that if we take a matrix and interchange two its rows, the
determinant changes sign. Thus, interchanging 1 and 2 above, for example:
