ΔxΔp≥ ℏ/2
ψ
∣ψ∣2
ψ∣ψ∣2
E=hν h
λ=h/p
H
^E
ΔV(x)ψ
E=hν h
E=hν
E=hν
h h ≈ 6, 626 × 10 Joule-seconde
−34
E=hν
λ
p λ =
p
h
d=
400 W= 2, 3 eV
ΔxΔp
x
6
7
8
1
/
8
100%
