6.1: The Potential Step
6.2: The Potential Barrier
We’ll continue 6.2 in the next class
Plane wave solutions
 Right-moving: Ψ 𝑥𝑥, 𝑡𝑡
= 𝐴𝐴𝐴𝐴 +𝑖𝑖(𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡) =𝐴𝐴𝐴𝐴 +𝑖𝑖𝑖𝑖𝑖𝑖−𝑖𝑖𝜔𝜔𝑡𝑡
 Left-moving: Ψ 𝑥𝑥, 𝑡𝑡
= 𝐴𝐴𝐴𝐴 −𝑖𝑖𝑖𝑖𝑖𝑖−𝑖𝑖𝜔𝜔𝑡𝑡
−𝑖𝑖𝜔𝜔𝑡𝑡
 Time solution is the same 𝜙𝜙 𝑡𝑡 = 𝑒𝑒
We’ll focus on the space parts:
+𝑖𝑖𝑖𝑖𝑖𝑖
 Right-moving: 𝜓𝜓 𝑥𝑥 = 𝐴𝐴𝐴𝐴
−𝑖𝑖𝑖𝑖𝑖𝑖
 Left-moving: 𝜓𝜓 𝑥𝑥 = 𝐴𝐴𝐴𝐴

In general A may be complex

We are really considering a beam of traveling
particles. So the amplitude A*A represents the
number of particles per unit distance.
Review
ℏ2 𝜕𝜕2 𝜓𝜓 𝑥𝑥
 Time indep. Schrodinger: −
2𝑚𝑚 𝜕𝜕𝜕𝜕 2
ℏ2 𝜕𝜕2 𝜓𝜓 𝑥𝑥
 When U = 0: −
= 𝐸𝐸𝜓𝜓 𝑥𝑥
2𝑚𝑚 𝜕𝜕𝜕𝜕 2
2
𝜕𝜕 𝜓𝜓 𝑥𝑥
 Rearrange:
= −𝑘𝑘 2 𝜓𝜓 𝑥𝑥 where
2
𝜕𝜕𝜕𝜕
+ 𝑈𝑈 𝑥𝑥 𝜓𝜓 𝑥𝑥 = 𝐸𝐸𝜓𝜓 𝑥𝑥
𝑘𝑘 2 =
2𝑚𝑚𝑚𝑚
ℏ2
All four functions--sin (kx), cos (kx), e+ikx and e-ikx -are solutions. For bound states we used sines
and cosines to describe standing waves. We
could have used complex exponential functions,
but that would have been an unnecessary
complication.
 Now we need to describe waves travelling to
the right or left and we must use complex
exponential functions.

 Potential
is zero to the left (x
< 0).
 Particle is free in this region.
 The free particle encounters
a potential step (at x = 0)—
steep increase in potential
energy due to a charged,
narrow capacitor.
 The potential energy is a
nonzero constant U0 for x > 0.



𝑑𝑑𝑑𝑑
Remember 𝐹𝐹 = − , a force to
𝑑𝑑𝑑𝑑
the left.
If E > U0, according to classical
physics, the particle continues
moving rightward with a less
kinetic energy (moves slower).
However if the potential jumps
abruptly in an region that is small
compared to the wavelength of
the particle, then we must model
the particle as a wave and apply
quantum mechanics.
According to quantum
mechanics, when a right moving
wave encounters the potential
step, it may be transmitted, but it
also may be reflected. (much
like light encountering a
boundary between two media.)
 It does NOT matter if there is a
step up or a step down.
 Let’s find probability of
reflection and transmission. (be
patient)

•
For x <
𝜕𝜕2 𝜓𝜓 𝑥𝑥
0:
𝜕𝜕𝜕𝜕 2
2
2
= −𝑘𝑘 𝜓𝜓 𝑥𝑥 where 𝑘𝑘 =
2𝑚𝑚𝑚𝑚
ℏ2
−𝑖𝑖𝑖𝑖𝑖𝑖
And the solution is 𝜓𝜓𝑥𝑥<0 𝑥𝑥 = 𝐴𝐴𝐴𝐴 +𝑖𝑖𝑖𝑖𝑖𝑖 +𝐵𝐵𝐵𝐵
ℏ2 𝜕𝜕2 𝜓𝜓 𝑥𝑥
• For x >0: −
+ 𝑈𝑈0 𝜓𝜓 𝑥𝑥 = 𝐸𝐸𝜓𝜓 𝑥𝑥
2𝑚𝑚 𝜕𝜕𝜕𝜕 2
𝜕𝜕2 𝜓𝜓 𝑥𝑥
2𝑚𝑚(𝐸𝐸−𝑈𝑈0 )
2
′
2
′
•
= −𝑘𝑘 𝜓𝜓 𝑥𝑥 where 𝑘𝑘 =
𝜕𝜕𝜕𝜕 2
ℏ2
• Mathematically the solution is the sum of
•
rightward and leftward travelling waves, but there
is nothing to cause the wave to travel leftward in
this region. So we reject the leftward solution.
′
+𝑖𝑖𝑖𝑖
The physical solution is 𝜓𝜓𝑥𝑥>0 𝑥𝑥 = 𝐶𝐶𝑒𝑒 𝑥𝑥

𝜓𝜓𝑥𝑥<0 𝑥𝑥 = 𝐴𝐴𝐴𝐴 +𝑖𝑖𝑖𝑖𝑖𝑖 +𝐵𝐵𝐵𝐵 −𝑖𝑖𝑖𝑖𝑖𝑖
Reflected
Incident
 𝜓𝜓𝑥𝑥>0
𝑥𝑥 =
′
+𝑖𝑖𝑖𝑖
𝐶𝐶𝐶𝐶 𝑥𝑥
Transmitted
Number of incident particles (per unit distance):
𝜓𝜓 2𝑖𝑖𝑖𝑖𝑖𝑖 = 𝐴𝐴∗ 𝐴𝐴
 Number of reflected particles (per unit distance):
𝜓𝜓 2𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐵𝐵 ∗ 𝐵𝐵
 Number of transmitted particles (per unit
distance): 𝜓𝜓 2𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐶𝐶 ∗ 𝐶𝐶

 Apply
and
boundary conditions:
2𝑘𝑘
𝐶𝐶 =
𝐴𝐴
𝑘𝑘 + 𝑘𝑘𝑘
𝑘𝑘 − 𝑘𝑘𝑘
𝐵𝐵 =
𝐴𝐴
𝑘𝑘 + 𝑘𝑘𝑘
Transmission Probability:
𝐸𝐸(𝐸𝐸 − 𝑈𝑈0 )
4𝑘𝑘𝑘𝑘𝑘
𝑇𝑇 =
=4
′
2
(𝑘𝑘 + 𝑘𝑘 )
[ 𝐸𝐸 + (𝐸𝐸 − 𝑈𝑈0 )]2
And reflection probability
𝑘𝑘 − 𝑘𝑘𝑘
𝑅𝑅 =
𝑘𝑘 + 𝑘𝑘𝑘
2
=
𝐸𝐸 − (𝐸𝐸 − 𝑈𝑈0 )
𝐸𝐸 + (𝐸𝐸 − 𝑈𝑈0 )
2
 Ch
6: 14, 15
Reflected
Incident
 𝜓𝜓𝑥𝑥<0
•
For x
•
𝑥𝑥 = 𝐴𝐴𝐴𝐴 +𝑖𝑖𝑖𝑖𝑖𝑖 +𝐵𝐵𝐵𝐵 −𝑖𝑖𝑖𝑖𝑖𝑖
𝜕𝜕2 𝜓𝜓 𝑥𝑥
>0:
𝜕𝜕𝜕𝜕 2
= +𝛼𝛼 2 𝜓𝜓 𝑥𝑥
2
where 𝛼𝛼 =
2𝑚𝑚(𝑈𝑈0 −𝐸𝐸)
ℏ2
𝑥𝑥 = 𝐶𝐶𝐶𝐶 −𝛼𝛼𝑥𝑥 +𝐷𝐷𝑒𝑒 +𝛼𝛼𝛼𝛼
 But D = 0 otherwise wave function diverges.
 𝜓𝜓𝑥𝑥>0
 Apply
And
And
boundary conditions:
𝐴𝐴 + 𝐵𝐵 = 𝐶𝐶
𝑖𝑖𝑘𝑘 𝐴𝐴 − 𝐵𝐵 = −𝛼𝛼𝛼𝛼
𝛼𝛼 + 𝑖𝑖𝑖𝑖
𝐵𝐵 = −
𝐴𝐴
𝛼𝛼 − 𝑖𝑖𝑖𝑖
|B| = |A|
 Reflection
probability: R = 1.
 So transmission probability is zero.
 Wave reflects completely, but there is some
nonzero probability it is to the right of the
step.
 Penetration depth is the same as before:
1
ℏ
𝛿𝛿 ≡ =
𝛼𝛼
2𝑚𝑚(𝑈𝑈0 − 𝐸𝐸)
 Ch
 So
6: 17
Ch 6: 14, 15 and 17
are due on Tuesday
24NOV (just before Tday break)
𝜓𝜓𝑥𝑥<0 𝑥𝑥 = 𝐴𝐴𝐴𝐴 +𝑖𝑖𝑖𝑖𝑖𝑖 +𝐵𝐵𝐵𝐵 −𝑖𝑖𝑖𝑖𝑖𝑖
𝜓𝜓0<𝑥𝑥<𝐿𝐿 𝑥𝑥 = 𝐶𝐶𝑒𝑒 +𝑖𝑖𝑖𝑖′𝑥𝑥 +𝐷𝐷𝑒𝑒 −𝑖𝑖𝑖𝑖′𝑥𝑥
𝜓𝜓𝑥𝑥>𝐿𝐿 𝑥𝑥 = 𝐹𝐹𝑒𝑒 +𝑖𝑖𝑖𝑖𝑖𝑖
 Reflection
 Resonant
Or
and transmission probabilities:
Transmission, when R = 0:
2𝑚𝑚(𝐸𝐸 − 𝑈𝑈0 )
𝐿𝐿 = 𝑛𝑛𝜋𝜋
ℏ
1 𝑛𝑛𝑛𝑛𝑛
𝐸𝐸 = 𝑈𝑈0 +
2𝑚𝑚 𝐿𝐿
2