SME
Quantum Mechanics · Problem type
Uncertainty in a box: ⟨x⟩, Δx and Δp
Use this for “An electron is in state n = 2 of a 0.5 nm box. Find ⟨x⟩, Δx and Δp, and check the uncertainty principle.”
Diagram
Eₙ
=
n
2
h
2
8mL²
:
the
levels
grow
as
n
2
,
and
ψ
=
0
at
both
walls.
|ψ|²
is
even
about
L
2,
so
⟨x⟩
=
L
2
for
every
n.
This is an example. Type your own values to start a new problem.
Δx
=
L (nm)
1
12
−
1
2π²
n
2
=
Δx (nm)
Δp
=
n
πħ
÷
L (nm)
=
Δp (kg·m/s)
⟨x⟩
Mean position
worked out (not typed) · nm
⟨x²⟩
Mean square position
worked out (not typed) · nm²
ΔxΔp ÷ ħ
Product, in ħ
worked out (not typed)
Formulas
⟨x⟩
=
L
2
0.5
=
1
2
⟨
x
2
⟩
=
L
2
×
(
1
3
−
1
2 ×
n
2
×
π
2
)
0.2827
=
1
2
×
(
1
3
−
1
2 ×
1
2
×
π
2
)
Δx
=
√
⟨
x
2
⟩ − ⟨x⟩²
0.1808
=
√
0.2827 −
0.5
2
Δp
=
n
×
π
×
1.0546 ×
10
-34
L
×
10
-9
3.3131 ×
10
-25
=
1
×
π
×
1.0546 ×
10
-34
1 ×
10
-9
ΔxΔp
ħ
=
Δx
×
10
-9
×
Δp
1.0546 ×
10
-34
0.5679
=
0.1808 ×
10
-9
×
3.3131 ×
10
-25
1.0546 ×
10
-34
Assumptions
•
ψₙ = √(2/L) sin(nπx/L); ⟨f⟩ = ∫ f|ψ|² dx over the box, so ⟨x⟩ = L ÷ 2 and ⟨x²⟩ = L²(1/3 − 1 ÷ (2n²π²)).
•
⟨p⟩ = 0 and ⟨p²⟩ = (nπħ ÷ L)², so Δp = nπħ ÷ L, with ħ = 1.0546 × 10⁻³⁴ J·s.
•
Each spread is a standard deviation: Δx = √(⟨x²⟩ − ⟨x⟩²). Heisenberg says ΔxΔp ≥ ħ ÷ 2.
Step-by-step
WE KNOW
n = 1, L = 1 nm
FIND
mean position (⟨x⟩), mean square position (⟨x²⟩), spread in position (Δx), spread in momentum (Δp), product, in ħ (ΔxΔp ÷ ħ)
Step 1 · Find mean position (⟨x⟩)
Use ⟨x⟩ = L ÷ 2
|ψ|² is the same either side of the middle, so the mean position is half the width.
⟨x⟩
=
1
2
⟨x⟩ = 0.5 nm
Step 2 · Find mean square position (⟨x²⟩)
Use ⟨x²⟩ = L² × (1 ÷ 3 − 1 ÷ (2 × n² × π²))
Integrate
x
2
×
(
2
L
)sin²(nπx/
L
)
across
the
box
(by
parts,
twice).
⟨
x
2
⟩
=
1
2
×
(
1
3
−
1
2 ×
1
2
×
π
2
)
⟨
x
2
⟩
=
1
2
×
(
1
3
−
1
2 × 9.8696
)
⟨
x
2
⟩
=
1
2
×
(
1
3
−
1
19.7392
)
⟨
x
2
⟩
=
1
2
×
(0.33333
−
0.050661)
⟨
x
2
⟩
=
1
2
×
0.28267
⟨
x
2
⟩
=
0.282673
nm²
Step 3 · Find spread in position (Δx)
Use Δx = √(⟨x²⟩ − ⟨x⟩²)
The spread is the root of the mean square less the square of the mean.
Δx
=
√
0.282673 −
0.5
2
Δx
=
√
0.2827 − 0.25
Δx = 0.18076 nm
Step 4 · Find spread in momentum (Δp)
Use Δp = n × π × 1.0546 × 10⁻³⁴ ÷ (L × 10⁻⁹)
The
state
is
two
waves
of
momentum
±
nπħ
L
,
so
⟨p⟩
=
0
and
Δp
is
nπħ
L
;
L
in
meters.
Δp
=
1
×
π
×
1.0546 ×
10
-34
10
-9
Δp
=
1.0546 ×
10
-34
π
10
-9
Δp
=
3.31312 ×
10
-25
kg·m/s
Step 5 · Find product, in ħ (ΔxΔp ÷ ħ)
Use ΔxΔp ÷ ħ = Δx × 10⁻⁹ × Δp ÷ (1.0546 × 10⁻³⁴)
Multiply
the
two
spreads
(Δx
in
meters)
and
divide
by
ħ.
Heisenberg:
this
is
never
below
1
2
.
ΔxΔp
ħ
=
0.18076 ×
10
-9
×
3.31312 ×
10
-25
1.0546 ×
10
-34
ΔxΔp
ħ
=
5.9888 ×
10
-35
1.0546 ×
10
-34
ΔxΔp
ħ
=
0.5679
Check
Put the numbers back in. Both sides should match.
0.5
=
1
2
✓
0.282673
=
1
2
×
(
1
3
−
1
2 ×
1
2
×
π
2
)
✓
0.18076
=
√
0.282673 −
0.5
2
✓
3.31312 ×
10
-25
=
1
×
π
×
1.0546 ×
10
-34
10
-9
✓
0.5679
=
0.18076 ×
10
-9
×
3.31312 ×
10
-25
1.0546 ×
10
-34
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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