SME
Quantum Mechanics
Perturbation theory
Use this for “An electron in a 2 nm box is in n = 2. A flat 0.02 eV bump sits from x = 0 to 0.5 nm. Estimate the first-order shift and the new energy.”
Diagram
E⁽¹⁾ = V₀ × P, where P = 0.8183 is the shaded area of |ψ₁|² over the bump: a bump where ψ is large shifts E most.
This is an example. Type your own values to start a new problem.
E⁽¹⁾
=
V₀ (eV)
×
P
=
E⁽¹⁾ (eV)
n
State
example
L
Box width
example · nm
x₁
Bump from
example · nm
x₂
Bump to
example · nm
Eₙ
Unperturbed energy
calculated · eV
E
Estimate
calculated · eV
Formulas
P
=
x₂
−
x₁
L
−
(sin(2
×
n
×
π
×
x₂
÷
L
)
−
sin(2
×
n
×
π
×
x₁
÷
L
))
÷
(2
×
n
×
π)
0.8183
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
E
⁽¹⁾
=
V₀
×
P
+0.04092 = 0.05 × 0.8183
Eₙ
=
0.37603
×
n
2
L
2
0.376
=
0.37603
×
1
2
1
2
E
=
Eₙ
+
E
⁽¹⁾
0.4169 = 0.376 + 0.04092
Assumptions
•
First order keeps the unperturbed ψₙ = √(2 ÷ L) sin(nπx ÷ L): E⁽¹⁾ = ⟨ψₙ|V|ψₙ⟩, V₀ times the chance P over the bump.
•
V₀ is small next to the gap to the next level (at most a tenth of it), so ψ barely changes; a negative V₀ is a dip and lowers E.
•
An electron: Eₙ = 0.37603 eV × n² ÷ L² with L in nm. A bump where |ψ|² is large shifts E most.
Step-by-step
WE KNOW
n = 1, L = 1 nm, V₀ = +0.05 eV, x₁ = 0.25 nm, x₂ = 0.75 nm
FIND
chance over the bump (P), first-order shift (E⁽¹⁾), unperturbed energy (Eₙ), estimate (E)
Step 1 · Find chance over the bump (P)
Use P = (x₂ − x₁) ÷ L − (sin(2 × n × π × x₂ ÷ L) − sin(2 × n × π × x₁ ÷ L)) ÷ (2 × n × π)
Integrate
(
2
L
)sin²(
nπx
L
)
from
x₁
to
x₂
:
the
stretch’s
share
of
L
,
less
the
sine
terms
(radians).
P
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
P
=
0.75
−
0.25
−
(sin(6.283
×
0.75)
−
sin(6.283
×
0.25))
÷
(2
×
π)
P
=
0.75
−
0.25
−
sin(4.712) − sin(1.571)
2 × π
P
=
0.75
−
0.25
−
(−1) − 1
6.283
P
=
0.75
−
0.25
−
−2
6.283
P
=
0.75
−
0.25
−
(−0.3183)
P
=
0.8183
Step 2 · Find first-order shift (E⁽¹⁾)
Use E⁽¹⁾ = V₀ × P
⟨ψₙ|V|ψₙ⟩
is
V₀
times
the
chance
of
being
over
the
bump:
V
is
0
everywhere
else.
E
⁽¹⁾
=
0.05
×
0.8183
E
⁽¹⁾
=
+0.04092
eV
Step 3 · Find unperturbed energy (Eₙ)
Use Eₙ = 0.37603 × n² ÷ L²
An
electron
in
a
1
nm
box
has
E₁
=
0.37603
eV:
times
n
2
,
over
L
2
.
Eₙ
=
0.37603
×
1
2
1
2
Eₙ = 0.376 eV
Step 4 · Find estimate (E)
Use E = Eₙ + E⁽¹⁾
Add the shift to the unperturbed level.
E
=
0.376
+
0.04092
E
=
0.4169
eV
Check
Put the numbers back in. Both sides should match.
0.8183
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
✓
+0.04092 = 0.05 × 0.8183 ✓
0.376
=
0.37603
×
1
2
1
2
✓
0.4169 = 0.376 + 0.04092 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.