231
3.3. Diffraction
leys represent destructive interference. It is evident from the figure
that the two rays have a path length difference given by d sin θ,
where d is the separation distance of the two slits. Constructive
interference occurs when this path length difference is equal to an
integral number of wavelengths λ. Equivalently,
d sin θ = n λ,
n = 0, ±1, ±2, ±3, . . . .
(3.337)
Destructive interference occurs when the path length difference is
equal to a half-odd number of wavelengths. Equivalently,
d sin θ = n +
1
2
λ,
n = 0, ±1, ±2, ±3, . . . . (3.338)
Next we consider the specific viewing angle θ for which the intensity distribution I(x) has its first minimum. This is shown in the
figure, where the path length difference between the two rays is
λ/2, and n = +1 in (3.338). We assume the particle has momentum p, which is related to its de Broglie wavelength λ by
h
p = ,
(3.339)
λ
where h is Planck’s constant. This momentum has a transverse
component Δp, which satisfies
Δp = sin θ.
(3.340)
p
Since Δp represents the half-width of the first interference fringe
with n = 0, we ascribe an uncertainty Δp to the transverse momentum of the particle.
Separately, one has no knowledge about which of the two slits
the particle passed through. Following Feynman [30, Chapter 1,
Volume 3], any attempt to measure which slit the particle passed
through would perturb the wave function, thereby irreparably destroying the interference. Consequently, we ascribe an uncertainty
Δx = d/2 to the transverse position of the particle. It is left as a
brief exercise to show (3.338, 3.339, 3.340) that
h
Δx Δp = .
(3.341)
4
3.3. Diffraction
leys represent destructive interference. It is evident from the figure
that the two rays have a path length difference given by d sin θ,
where d is the separation distance of the two slits. Constructive
interference occurs when this path length difference is equal to an
integral number of wavelengths λ. Equivalently,
d sin θ = n λ,
n = 0, ±1, ±2, ±3, . . . .
(3.337)
Destructive interference occurs when the path length difference is
equal to a half-odd number of wavelengths. Equivalently,
d sin θ = n +
1
2
λ,
n = 0, ±1, ±2, ±3, . . . . (3.338)
Next we consider the specific viewing angle θ for which the intensity distribution I(x) has its first minimum. This is shown in the
figure, where the path length difference between the two rays is
λ/2, and n = +1 in (3.338). We assume the particle has momentum p, which is related to its de Broglie wavelength λ by
h
p = ,
(3.339)
λ
where h is Planck’s constant. This momentum has a transverse
component Δp, which satisfies
Δp = sin θ.
(3.340)
p
Since Δp represents the half-width of the first interference fringe
with n = 0, we ascribe an uncertainty Δp to the transverse momentum of the particle.
Separately, one has no knowledge about which of the two slits
the particle passed through. Following Feynman [30, Chapter 1,
Volume 3], any attempt to measure which slit the particle passed
through would perturb the wave function, thereby irreparably destroying the interference. Consequently, we ascribe an uncertainty
Δx = d/2 to the transverse position of the particle. It is left as a
brief exercise to show (3.338, 3.339, 3.340) that
h
Δx Δp = .
(3.341)
4