SME
Quantum Mechanics
Wavefunctions and operators
Use this for “An electron is in the ground state of a 2 nm box. What is the chance of finding it between 0.5 nm and 1 nm from the left wall?”
Diagram
Eₙ
=
n
2
h
2
8mL²
:
the
levels
grow
as
n
2
,
and
ψ
=
0
at
both
walls.
The shaded area of |ψ₁|² is P = 0.8183; a classical particle would give 0.5.
This is an example. Type your own values to start a new problem.
∫
2
L
sin²
πx
L
dx
=
P
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)
Assumptions
•
ψₙ = √(2/L) sin(nπx/L) inside the box and 0 outside; it is normalized, since ∫|ψ|² dx over the whole box is 1.
•
|ψ|² is a probability per length, so the chance between x₁ and x₂ is the area under |ψ|² there.
•
That area is ∫ from x₁ to x₂ of (2/L)sin²(nπx/L) dx; the sines in its closed form are of radians.
Step-by-step
WE KNOW
n = 1, L = 1 nm, x₁ = 0.25 nm, x₂ = 0.75 nm
FIND
probability (P)
Step 1 · Find probability (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
)
=
(
1
L
)(1
−
cos(2nπx/
L
)):
the
stretch’s
share
of
the
width,
less
the
sine
terms.
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
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)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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