SME
Physical Chemistry II: Quantum & Spectroscopy · Problem type
Second-order perturbation of two levels
Use this for “Two states at 2 eV and 8 eV are coupled by 0.5 eV. Compare the second-order lower energy with the exact one.”
Diagram
δ: E₁ → exact lower, −0.09902 eV
δ⁽²⁾: E₁ → second order, −0.1 eV
This is an example. Type your own values to start a new problem.
E₁
Lower level before coupling
example · eV
E₂
Upper level before coupling
example · eV
V
Coupling between the states
example · eV
Δ
Gap between the levels
calculated · eV
δ
Exact shift of the lower level
calculated · eV
Elow
Exact lower energy
calculated · eV
δ⁽²⁾
Second-order shift
calculated · eV
E⁽²⁾
Second-order lower energy
calculated · eV
Formulas
Δ
=
E₂
−
E₁
10 = 11 − 1
δ
=
−(
V
2
Δ
÷ 2 +
√
Δ
2
÷ 4 +
V
2
)
−0.09902
=
−(
1
2
10 ÷ 2 +
√
10
2
÷ 4 +
1
2
)
Elow
=
E₁
+
δ
+0.901 = 1 + (−0.09902)
δ
⁽²⁾
=
−(
V
2
Δ
)
−0.1
=
−(
1
2
10
)
E⁽²⁾
=
E₁
+
δ
⁽²⁾
+0.9 = 1 + (−0.1)
Assumptions
•
Two states at E₁ < E₂ before coupling are mixed by the coupling V; the gap is Δ = E₂ − E₁, all in eV.
•
Exact: the lower energy E solves (E₁ − E)(E₂ − E) = V², so it sits below E₁ by V² ÷ (Δ ÷ 2 + √(Δ² ÷ 4 + V²)).
•
Second-order perturbation theory moves the lower level down by V² ÷ Δ; it is close when V is small next to Δ.
•
The upper level moves up by the same amount: the coupling pushes the two levels apart.
Step-by-step
WE KNOW
E₁ = +1 eV, E₂ = +11 eV, V = 1 eV
FIND
gap between the levels (Δ), exact shift of the lower level (δ), exact lower energy (Elow), second-order shift (δ⁽²⁾), second-order lower energy (E⁽²⁾)
Step 1 · Find gap between the levels (Δ)
Use Δ = E₂ − E₁
The gap is the upper level less the lower one.
Δ
=
11
−
1
Δ
=
10
eV
Step 2 · Find exact shift of the lower level (δ)
Use δ = −(V² ÷ (Δ ÷ 2 + √(Δ² ÷ 4 + V²)))
The
lower
root
of
(
E₁
−
E)(
E₂
−
E)
=
V
2
,
written
as
V
2
over
the
sum
so
nothing
cancels.
δ
=
−(
1
2
10 ÷ 2 +
√
10
2
÷ 4 +
1
2
)
δ
=
−(
1
2
10 ÷ 2 +
√
100 ÷ 4 + 1
)
δ
=
−(
1
2
10 ÷ 2 +
√
25 + 1
)
δ
=
−(
1
2
10 ÷ 2 +
√
26
)
δ
=
−(
1
2
10 ÷ 2 + 5.099
)
δ
=
−(
1
2
5 + 5.099
)
δ
=
−(
1
2
10.099
)
δ
=
−(
1
10.099
)
δ
=
−0.09902
eV
Step 3 · Find exact lower energy (Elow)
Use Elow = E₁ + δ
Move the lower level by the exact shift.
Elow = 1 + (−0.09902)
Elow = +0.901 eV
Step 4 · Find second-order shift (δ⁽²⁾)
Use δ⁽²⁾ = −(V² ÷ Δ)
Second order: the coupling squared over the gap, taken off the lower level.
δ
⁽²⁾
=
−(
1
2
10
)
δ
⁽²⁾
=
−(
1
10
)
δ
⁽²⁾
=
−0.1
eV
→
V
Δ
is
0.1,
small:
second
order
overshoots
the
exact
shift
by
only
0.9902%.
Step 5 · Find second-order lower energy (E⁽²⁾)
Use E⁽²⁾ = E₁ + δ⁽²⁾
Move the lower level by the second-order shift.
E⁽²⁾ = 1 + (−0.1)
E⁽²⁾ = +0.9 eV
Check
Put the numbers back in. Both sides should match.
10 = 11 − 1 ✓
−0.09902
=
−(
1
2
10 ÷ 2 +
√
10
2
÷ 4 +
1
2
)
✓
+0.901 = 1 + (−0.09902) ✓
−0.1
=
−(
1
2
10
)
✓
+0.9 = 1 + (−0.1) ✓
Topic check
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