4More results in one dimensions

IB Quantum Mechanics



4.3 Potential step
Consider the time-independent Schr¨odinger equation for a step potential
V (x) =
(
0 x 0
U x > 0
,
where
U >
0 is a constant. The Schr¨odinger equation is, in case you’ve forgotten,
~
2
2m
ψ
00
+ V (x)ψ = Eψ.
x
V (x)
We require ψ and ψ
0
to be continuous at x = 0.
We can consider two different cases:
(i)
0
< E < U
: We apply the standard method, introducing constants
k, κ >
0
such that
E =
~
2
k
2
2m
, U E =
~
2
κ
2
2m
.
Then the Schr¨odinger equation becomes
(
ψ
00
+ k
2
ψ = 0 x < 0
ψ
00
κ
2
ψ = 0 x > 0
The solutions are
ψ
=
Ie
ikx
+
Re
ikx
for
x <
0, and
ψ
=
Ce
κx
for
x >
0
(since ψ has to be bounded).
Since ψ and ψ
0
are continuous at x = 0, we have the equations
(
I + R = C
ikI ikR = κC
.
So we have
R =
k
k +
I, C =
2k
k +
I.
If
x <
0,
ψ
(
x
) is a superposition of beams (momentum eigenstates) with
|I|
2
particles per unit length in the incident part, and
|R|
2
particles per
unit length in the reflected part, with p = ±~k. The current is
j = j
inc
+ j
ref
= |I|
2
~k
m
|R|
2
~k
m
,
The probability of reflection is
P
ref
=
|j
ref
|
|j
inc
|
=
|R|
2
|I|
2
= 1,
which makes sense.
On the right hand side, we have
j
= 0. So
P
tr
= 0. However,
|ψ
(
x
)
|
2
6
= 0
in this classically forbidden region.
(ii) E > U: This time, we set
E =
~
2
k
2
2m
, E U =
~
2
κ
2
2m
,
with k, κ > 0. Then the Schr¨odinger equation becomes
(
ψ
00
+ k
2
ψ = 0 x < 0
ψ
00
+ κ
2
ψ = 0 x > 0
Then we find
ψ
=
Ie
ikx
+
R
ikx
on
x <
0, with
ψ
=
T e
ikx
on
x >
0. Note
that it is in principle possible to get an
e
ikx
term on
x >
0, but this
would correspond to sending in a particle from the right. We, by choice,
assume there is no such term.
We now match ψ and ψ
0
at x = 0. Then we get the equations
(
I + R = T
ikI ikR = ikT.
We can solve these to obtain
R =
k κ
k + κ
I, T =
2k
k + κ
I.
Our flux on the left is now
j = j
inc
+ j
ref
= |I|
2
~k
m
|R|
2
~k
m
,
while the flux on the right is
j = j
tr
|T |
2
~κ
m
.
The probability of reflection is
P
ref
=
|j
ref
|
|j
inc
|
=
|R|
2
|I|
2
=
k κ
k + κ
2
,
while the probability of transmission is
P
tr
=
|j
tr
|
|j
inc
|
=
|T |
2
κ
|I|
2
k
=
4kκ
(k + κ)
2
.
Note that P
ref
+ P
tr
= 1.
Classically, we would expect all particles to be transmitted, since they all
have sufficient energy. However, quantum mechanically, there is still a
probability of being reflected.