Physical Chemistry -Two Electrons - Excited States_lecture26

5.61 Physical Chemistry Lecture #26 page 1



TWO ELECTRONS: EXCITED STATES
In the last lecture, we learned that the independent particle model gives a reasonable
description of the ground state energy of the Helium atom. Before moving on to talk
about many­electron atoms, it is important to point out that we can describe many
more properties of the system using the same type of approximation. By using the
same independent particle prescription we can come up with wavefunctions for
excited states and determine their energies, their dependence on electron spin, etc.
by examining the wavefunctions themselves. That is to say, there is much we can
determine from simply looking at Ψ without doing any significant computation.

We will use the excited state 1s2s configuration of Helium as an example. For the
ground state we had:
Ψ space (r1,r2 ) × Ψspin (σ1,σ 2 )

ψ1s (r1 )ψ1s (r2 ) 1 ⎛α σ β σ − β σ α σ ⎞
1s ⇒ ⎜ ( 1) ( 2 ) ( 1 ) ( 2 ) ⎟⎠
2⎝
In constructing excited states it is useful to extend the stick diagrams we have used
before to describe electronic configurations. Then there are four different
configurations we can come up with:
Ψ space (r1,r2 ) × Ψspin (σ1,σ 2 )
2s
? ?
1s

2s
? ?
1s

2s
? ?
1s

2s
? ?
1s

, 5.61 Physical Chemistry Lecture #26 page 2



Where the question marks indicate that we need to determine the space and spin
wavefunctions that correspond to these stick diagrams. It is fairly easy to come up
with a reasonable guess for each configuration. For example, in the first case we
might write down a wavefunction like:
ψ1sα (r1σ1 )ψ 2sα (r2σ 2 ) =ψ1s (r1 )ψ 2s (r2 ) α (σ1 ) α (σ 2 )
However, we immediately note that this wavefunction is not antisymmetric. We can
perform the same trick as before to make an antisymmetric wavefunction out this:
1 ⎛ψ
⇒ (r1 )ψ 2s (r2 )α (σ1 )α (σ 2 ) −ψ1s (r2 )ψ 2s (r1 )α (σ 2 )α (σ1 ) ⎟⎞⎠
2 ⎜⎝ 1s

⇒ 12 ⎛⎜ψ1s (r1 )ψ 2s (r2 ) −ψ1s (r2 )ψ 2s (r1 ) ⎟⎞ α (σ1 ) α (σ 2 )
⎝ ⎠

Ψ space Ψ spin
Applying the same principle to the 1s↑2s↓ configuration gives us a bit of trouble:
1 ⎛ψ
⇒ 2 ⎜⎝ 1s (r1 )ψ 2s (r2 )α (σ1 ) β (σ 2 ) −ψ1s (r2 )ψ 2s (r1 )α (σ 2 ) β (σ1 ) ⎟⎠⎞ ≠ ΨspaceΨspin
Hence, the pure ↑↓ configuration can’t be separated in terms of a space part and a
spin part. We find a similar result for 1s↓2s↑:
1 ⎛ψ
⇒ 2 ⎜⎝ 1s (r1 )ψ 2s (r2 ) β (σ1 )α (σ 2 ) −ψ1s (r2 )ψ 2s (r1 ) β (σ 2 )α (σ1 ) ⎟⎠⎞ ≠ ΨspaceΨspin
Since we know the wavefunction should separate, we have a problem. The solution
comes in realizing that for an open shell configuration like this one, the 1s↑2s↓ and
1s↓2s↑ states are degenerate eigenstates and so we can make any linear combinations
of them we like and we’ll still obtain an eigenstate. If we make the “+” and “­“
combinations of 1s↑2s↓ and 1s↓2s↑ we obtain:

( 1s (r1 )ψ 2s (r2 ) α (σ1 ) β (σ 2 ) −ψ1s (r2 )ψ 2s (r1 ) α (σ 2 ) β (σ1 )) ±
⇒ ψ

(ψ1s (r1 )ψ 2s (r2 ) β (σ1 ) α (σ 2 ) − ψ1s (r2 )ψ 2s (r1 ) β (σ 2 ) α (σ1 ))
⇒ (ψ (r )ψ (r ) � ψ (r )ψ (r )) (α (σ ) β (σ ) ± β (σ ) α (σ ) )
1s 1 2s 2 1s 2 2s 1 1 2 2 1

Ψ space Ψ spin

which separates nicely. Performing similar manipulations for the ↓↓ configuration and
taking care to make sure that all our spatial and spin wavefunctions are individually
normalized allows us to complete the table we set out for the 1s2s excited states: