SME
Physical Chemistry II: Quantum & Spectroscopy
Variational and perturbation methods
Use this for “Hydrogen’s trial function e−αr² has α = 0.5 a₀⁻². Find its variational energy and how far it sits above −0.5 Eₕ.”
Chart
Local minimum at (≈ 0.2829, ≈ −0.4244)
E(0.2829) ≈ −0.4244
Level true energy = −0.5 never meets the curve
This is an example. Type your own values to start a new problem.
E
=
3
×
α
÷
2
−
2
2
×
÷
π
EeV
Variational energy in electronvolts
worked out (not typed) · eV
error
Error above −0.5 Eₕ
worked out (not typed) · %
Formulas
E
=
3
×
α
2
−
2
×
√(2
×
α
π
)
−0.4244
=
3
×
0.2829
2
−
2
×
√(2
×
0.2829
π
)
EeV
=
27.211
×
E
−11.55 = 27.211 × (−0.4244)
error
=
100
×
E
+ 0.5
0.5
15.12
=
100
×
−0.4244 + 0.5
0.5
Assumptions
•
Atomic units: lengths in Bohr radii a₀ and energies in hartrees Eₕ, with 1 Eₕ = 27.211 eV.
•
The trial function e−αr² gives the energy 3α ÷ 2 − 2√(2α ÷ π) for the hydrogen atom.
•
The variational principle: no trial function gives an energy below the true ground state, −0.5 Eₕ.
•
The best α makes the energy least: α = 8 ÷ 9π = 0.2829, still 15.1% above the true energy.
Step-by-step
WE KNOW
α = 0.2829
FIND
variational energy (hartrees) (E), variational energy in electronvolts (EeV), error above −0.5 Eₕ (error)
Step 1 · Find variational energy (hartrees): E
Use E = 3 × α ÷ 2 − 2 × √(2 × α ÷ π)
The
kinetic
part
3
α
2
grows
with
α
;
the
attraction
to
the
nucleus,
−2√(
2
α
π
),
grows
more
slowly.
E
=
3
×
0.2829
2
−
2
×
√(2
×
0.2829
π
)
3
×
0.2829
2
−
2
×
√(
0.5658
π
)
0.4244
−
2
×
√
0.1801
0.4244 − 0.8488
E
=
−0.4244
→
This
is
the
best
exponent,
8
9π
:
no
Gaussian
does
better,
and
it
is
still
above
−0.5
Eₕ.
Step 2 · Find variational energy in electronvolts (EeV)
Use EeV = 27.211 × E
One hartree is 27.211 eV.
EeV = 27.211 × (−0.4244)
EeV = −11.55 eV
Step 3 · Find error above −0.5 Eₕ (error)
Use error = 100 × (E + 0.5) ÷ 0.5
How far the energy sits above −0.5 Eₕ, as a percent of 0.5 Eₕ.
error
=
100
×
−0.4244 + 0.5
0.5
error
=
100
×
0.0756
0.5
error
=
7.56
0.5
error = 15.12%
Check
Put the numbers back in. Both sides should match.
−0.4244
=
3
×
0.2829
2
−
2
×
√(2
×
0.2829
π
)
✓
−11.55 = 27.211 × (−0.4244) ✓
15.12
=
100
×
−0.4244 + 0.5
0.5
✓
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Microstates of an open subshell
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Second-order perturbation of two levels
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