SME
Quantum Mechanics · Problem type
Two levels: exact against second order
Use this for “Levels at 0.2 eV and 0.7 eV are mixed by a coupling of 0.05 eV. Find the exact levels and compare the lower one with second order.” Two equal levels (degenerate) split to E₁ ± V.
Diagram
E₁ = 0.2 eV, V = 0.1 eV, E₂ = 1.2 eV
D
=
E₁E₂
−
V
2
D = 0.2 × 1.2 − 0.1 × 0.1
D = 0.24 − 0.01 = 0.23 eV²
Down the main diagonal, take away the other diagonal
This is an example. Type your own values to start a new problem.
H
=
E₁
V
V
E₂
(eV)
Δ
Exact push
worked out (not typed)
E₋
Exact lower level
worked out (not typed)
E₊
Exact upper level
worked out (not typed)
E₋⁽²⁾
Second-order estimate
worked out (not typed)
δ
Exact less estimate
worked out (not typed)
e
Error of the estimate’s shift
worked out (not typed) · %
Formulas
Δ
=
V
2
÷
(
E₂
−
E₁
2
+
√((
E₂
−
E₁
2
)²
+
V
2
))
0.009902
=
0.1
2
÷
(
1.2 − 0.2
2
+
√((
1.2 − 0.2
2
)²
+
0.1
2
))
E₋
=
E₁
−
Δ
0.1901 = 0.2 − 0.009902
E₊
=
E₂
+
Δ
1.2099 = 1.2 + 0.009902
E₋⁽²⁾
=
E₁
−
V
2
E₂
−
E₁
0.19
=
0.2
−
0.1
2
1.2 − 0.2
δ
=
Δ
2
E₂
−
E₁
9.8049 ×
10
-5
=
0.009902
2
1.2 − 0.2
e
=
100
×
Δ
E₂
−
E₁
0.9902
=
100
×
0.009902
1.2 − 0.2
Assumptions
•
H = [[E₁, V], [V, E₂]] with E₁ ≤ E₂ and V real; its eigenvalues are the exact levels E₊ and E₋.
•
Second order pushes the lower level down by V² ÷ (E₂ − E₁); it works while V ≪ E₂ − E₁.
•
The picture’s determinant E₁E₂ − V² is the product E₊E₋.
Step-by-step
WE KNOW
E₁ = 0.2 eV, E₂ = 1.2 eV, V = 0.1 eV
FIND
exact push (Δ), exact lower level (E₋), exact upper level (E₊), second-order estimate (E₋⁽²⁾), exact less estimate (δ), error of the estimate’s shift (e)
Step 1 · Find exact push (Δ)
Use Δ = V² ÷ ((E₂ − E₁) ÷ 2 + √(((E₂ − E₁) ÷ 2)² + V²))
H’s
eigenvalues
are
E₁
+
E₂
2
∓
√((
E₂
−
E₁
2
)²
+
V
2
).
With
the
root
rationalized,
each
level
moves
by
V
2
over
half
the
gap
plus
the
root,
so
nothing
cancels.
Δ
=
0.1
2
÷
(
1.2 − 0.2
2
+
√((
1.2 − 0.2
2
)²
+
0.1
2
))
Δ
=
0.1
2
÷
(
1.2 − 0.2
2
+
√(
(
1
2
)
2
+
0.1
2
))
Δ
=
0.1
2
(1.2 − 0.2) ÷ 2 +
√
0.5
2
+
0.1
2
Δ
=
0.1
2
1 ÷ 2 +
√
0.25 + 0.01
Δ
=
0.1
2
1 ÷ 2 +
√
0.26
Δ
=
0.1
2
1 ÷ 2 + 0.5099
Δ
=
0.1
2
0.5 + 0.5099
Δ
=
0.1
2
1.0099
Δ
=
0.01
1.0099
Δ
=
0.009902
eV
Step 2 · Find exact lower level (E₋)
Use E₋ = E₁ − Δ
The smaller eigenvalue: the lower level pushed down.
E₋ = 0.2 − 0.009902
E₋ = 0.1901 eV
Step 3 · Find exact upper level (E₊)
Use E₊ = E₂ + Δ
The larger eigenvalue: the upper level is pushed up by the same amount the lower one drops.
E₊ = 1.2 + 0.009902
E₊ = 1.2099 eV
Step 4 · Find second-order estimate (E₋⁽²⁾)
Use E₋⁽²⁾ = E₁ − V² ÷ (E₂ − E₁)
Second
order:
E₁
plus
|⟨2|
V
|1⟩|²
÷
(
E₁
−
E₂
),
a
push
down
by
V
2
over
the
gap.
E₋⁽²⁾
=
0.2
−
0.1
2
1.2 − 0.2
E₋⁽²⁾
=
0.2
−
0.1
2
E₋⁽²⁾ = 0.2 − 0.01
E₋⁽²⁾ = 0.19 eV
Step 5 · Find exact less estimate (δ)
Use δ = Δ² ÷ (E₂ − E₁)
Second
order’s
push
V
2
E₂
−
E₁
less
the
exact
push
Δ
works
out
to
Δ
2
E₂
−
E₁
:
the
estimate
goes
too
low
by
this
much.
δ
=
0.009902
2
1.2 − 0.2
δ
=
0.009902
2
δ
=
9.8049 ×
10
-5
eV
Step 6 · Find error of the estimate’s shift (e)
Use e = 100 × Δ ÷ (E₂ − E₁)
δ
over
the
exact
push
Δ
=
E₁
−
E₋,
as
a
percent;
since
δ
=
Δ
2
E₂
−
E₁
,
that
is
Δ
over
the
gap:
small
while
V
≪
E₂
−
E₁
.
e
=
100
×
0.009902
1.2 − 0.2
e
=
100
×
0.009902
e
=
0.9902%
Check
Put the numbers back in. Both sides should match.
0.009902
=
0.1
2
÷
(
1.2 − 0.2
2
+
√((
1.2 − 0.2
2
)²
+
0.1
2
))
✓
0.1901 = 0.2 − 0.009902 ✓
1.2099 = 1.2 + 0.009902 ✓
0.19
=
0.2
−
0.1
2
1.2 − 0.2
✓
9.8049 ×
10
-5
=
0.009902
2
1.2 − 0.2
✓
0.9902
=
100
×
0.009902
1.2 − 0.2
✓
Topic check
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