Consider a potential step described by
V (x) = V0 θ(x)
(1)
where θ(x) is the step function, equal to 1 for x > 0 and 0 for x < 0. We will first
consider E > V0 and denote the region of negative x by I and positive x by II. The
Schrödinger equation in the two regions can be written as
√
d2 ψI
2mE
= − k 2 ψI where k =
(2)
2
dx
q h̄
2m(E − V0 )
d2 ψII
2
=
−
q
ψ
where
q
=
.
(3)
II
dx2
h̄
The solutions are simple and are linear combinations of oscillating exponentials.1
ψI (x) = A eikx + Be−ikx
ψII (x) = Ceiqx + De−iqx .
Note that k and q are the magnitudes and are real and positive.
Aside: Recall that if ψ = exp(±ikx), the probability current density is given by
±h̄k/m yielding particles moving to the right and and left respectively. Since
j(x, t) ≡
∂ψ
∂ψ ∗
h̄
[ ψ∗
− ψ
] .
2im
∂x
∂x
(4)
With ψ(x) = A eikx we have
∂ψ
∂ψ
= ik A eikx , ψ ∗ (x)
= |A|2 ik .
∂x
∂x
Obtain the second term by complex conjugation and find the result that the probability
current density associated with the wave function Aeikx is |A|2 h̄k
.
m
The term Aeikx is referred to as the incident wave (incident on the scattering potential step from the left) since the probability current density is to the right; Ceiqx is
the transmitted wave and Be−ikx the reflected wave since the associated currents are
in the direction of and opposite to the incident wave direction respectively.
1
Note that since E > 0 and E > V0 , k and q are real and ψ ′′ = negative constant ψ and so the
solutions are oscillatory.
1
There are 4 unknown constants. We will consider describing an experiment in which
where there is no particle incident from the right or region II. So we set D = 0.
The unknown quantities coefficients are determined by using the continuity of ψ
and dψ/dx at x = 0. Matching boundary conditions we have
ψI (x = 0) = ψII (x = 0) ⇒ A + B = C
dψII dψI =
⇒ ikA − ikB = iqC
dx x=0
dx x=0
(5)
(6)
which can be solved to obtain
B
A
C
A
k−q
k+q
2k
.
k+q
=
=
(7)
(8)
We have only determined the ratio of the coefficients; one might have naively
thought that the overall constant is determined by normalization. Alas, we have shown
that the free particle wave function is not normalizable. So we will define experimentally relevant quantities and avoid this mathematical complication.
Note that the net probability current density in region I is the incident current
density jinc , given by |A|2 h̄k/m minus the reflected current density, jref given by
.
(h̄k/m)|B|2 . The transmitted current density is given by |C|2 h̄q
m
The refection and transmission coefficients are defined to be the ratio of the (magnitude of) reflected current density to the incident current density and transmitted
current density to the incident current density respectively:
R ≡
jref
jinc
and T ≡
jtran
jinc
(9)
Therefore we obtain
R =
|B|2
(k − q)2
=
|A|2
(k + q)2
and T =
|C|2q
4kq
=
.
2
|A| k
(k + q)2
(10)
You can check that R + T = 1 or equivalently jinc = jref + jtran i.e., the incident wave is partially reflected and the rest is transmitted showing conservation of
particle number. Note that the reflection coefficient is finite. Classically a particle
is completely transmitted completely with simply a diminution in its velocity. Recall
the probabilistic interpretation of the quantum mechanical result. A different way of
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thinking about it is to imagine an electromagnetic wave (light) incident on an interface
between two media with different indices of refraction. Light is partially reflected and
partially transmitted. The quantum mechanical particle acts like a wave and travels
with different speeds in the two regions just as light does in different media.
For E < V0 the form of the wave function in the region x > 0 is C eκx + D e−κx
2 2
where h̄2mκ = V0 − E. Please be clear about this. Choose D = 0 for the boundary
condition at x = +∞ to be satisfied. Since the wave function is real, the associated
probability current density vanishes and there is no transmitted current. Thus the
reflection coefficient is unity. The transmission coefficient vanishes since there is no
transmitted current.
3