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Physics 249 Lecture 16, Oct 10th 2012
Reading: Chapter 6, Next time 7.1
Exam scheduled Friday Oct 12th. 50 minutes, in class
Covering chapters 3-6 excluding reflection and transmission of waves.
Closed book, 3-5 index card for equations allowed. Integrals and derivatives more
complex than polynomials, trig functions or first order exponentials will be provided.
A calculator is necessary.
Please bring your own scratch paper.
1) Reflection and transmission of waves.
Reflection and transmission of particle waves from a step potential V(x), E>V
V(x)=0
V(x)=V0
x<0
x>0
and general solutions
𝜓1 (𝑥) = 𝐴𝑒 𝑖𝑘1 𝑥 + 𝐵𝑒 −𝑖𝑘1 𝑥
𝜓2 (𝑥) = 𝐶𝑒 𝑖𝑘2 𝑥
𝑘1 =
√2𝑚𝐸
ℏ
𝑘2 =
√2𝑚(𝐸 − 𝑉0 )
ℏ
Recognizing these as right and left going waves the constant D=0 since there is no left
going wave in region 2.
Completely solving the problem is simply a matter of equating the wave function and
first derivative of the wave function at the potential boundary.
More interesting is comparing the probability for a wave to exist as a right moving,
transmitted wave, or left moving, reflected, wave after encountering the barrier relative to
the original right going wave. A (A=1) is a normalization constant that be set to
normalize the total probability of an incoming particle to one.
|𝐵|2
𝑘1 − 𝑘2 2
𝑅 = 𝑅𝑒 ( 2 ) = (
)
|𝐴|
𝑘1 + 𝑘2
𝑘2 |𝐶|2
4𝑘1 𝑘2
𝑇 = 𝑅𝑒 (
)
=
(𝑘1 + 𝑘2 )2
𝑘1 |𝐴|2
Essentially these equations compare the probability currents on either side of the barrier.
Note that the ratio of the wavenumbers accounts for the fact that the particles have
different momentums on the either side of the step changing velocity or the number of
particles per second.
These should be understood within the context of the probabilistic interpretation of the
wave function. At a given energy percentages R and T of the particles will be reflected
or transmitted.
Even though the E>V some particles/waves will be reflected.
Note that these reflection and transmission probabilities are energy dependent. When
considering a wave packet the potential barrier will distort the wave packet.
2) Reflection of waves.
Reflection of particle waves from a step potential V(x), E<V
Consider the same potential now with E<V.
In region two the Schrodinger equation solution becomes
𝜓2 (𝑥) = 𝐶𝑒 −𝛼𝑥
𝛼=
√2𝑚(𝑉0 − 𝐸)
ℏ
Applying the boundary conditions we find |A|2 = |B|2
|𝐵|2
𝑅=
=1
|𝐴|2
The wave is completely reflected. We can assume that T=0.
If we calculate T using the equation above we find that there is an imaginary component
before we take the real part. This represents the exponentially decaying amplitude of the
wave on the right side of the barrier, which doesn’t correspond to a true transition
through the barrier.
You can calculate the probability of finding a particle on the “wrong” side of the barrier
as:
|𝜓2 (𝑥) =|2 = |𝐶|2 𝑒 −2𝛼𝑥 =
4𝑘12
4𝑘12
2 −2𝛼𝑥
|𝐴|
𝑒
=
𝑒 −2𝛼𝑥
(𝑘12 + 𝛼 2 )
(𝑘12 + 𝛼 2 )
Considering the presence of hbar this is a quantity that decays exponentially over a very
short distance beyond the step function.
3) Transmission through a finite step
Reflection and transmission of particle waves from a finite step potential V(x) of length
a, E<V
V(x)=0
V(x)=V0
x<0, x>a
0<x<a
with general solutions
𝜓1 (𝑥) = 𝐴𝑒 𝑖𝑘1 𝑥 + 𝐵𝑒 −𝑖𝑘1 𝑥
𝜓2 (𝑥) = 𝐶𝑒 −𝛼𝑥 + 𝐷𝑒 𝛼𝑥
𝜓3 (𝑥) = 𝐹𝑒 𝑖𝑘1 𝑥
𝑘1 =
𝛼=
√2𝑚𝐸
ℏ
√2𝑚(𝑉0 − 𝐸)
ℏ
Again understanding reflection and transition is just a matter of equating the wave
functions and first derivatives at the interfaces to enforce continuity. A=1. There are
now four equations and four unknowns. We are interested in transmission to region 3
compared to the incoming wave.
−1
𝑇 = 𝑅𝑒 (
|𝐹|2
|𝐴|
2
) = [1 +
2
𝑠𝑖𝑛ℎ 𝛼𝑎
]
𝐸
𝐸
4 𝑉 (1 − 𝑉 )
0
0
There is a finite probability to tunnel through the barrier for energies less than the
potential!
4): Schrodinger equation in three dimensions. Particle cubic box.
𝐸𝜓(𝑥, 𝑦, 𝑧) = −
ℏ2 𝜕 2 𝜓(𝑥, 𝑦, 𝑧) 𝜕 2 𝜓(𝑥, 𝑦, 𝑧) 𝜕 2 𝜓(𝑥, 𝑦, 𝑧)
(
+
+
) + V(x, y, z)𝜓(𝑥, 𝑦, 𝑧)
2𝑚
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
𝑉(𝑥, 𝑦, 𝑧) = 0, 0 < 𝑥, 𝑦, 𝑧 < 𝐿
𝑉(𝑥, 𝑦, 𝑧) = ∞, outside 0-L
The potential has no dependence on combinations of x,y,z. Therefore the problem can be
formally separated into x, y, and z components. Considering the solution to any one of
the 1D problems.
𝜓 = 𝐴𝑠𝑖𝑛(𝑘𝑥)
Then the full solution must be:
𝜓 = 𝐴𝑠𝑖𝑛(𝑘1 𝑥)𝑠𝑖𝑛(𝑘2 𝑦)𝑠𝑖𝑛(𝑘3 𝑧)
Substituting into the Schrodinger equation
𝐸=
ℏ2 2
ℏ2 𝜋 2 2
(𝑘1 + 𝑘22 + 𝑘32 ) =
(𝑛 + 𝑛22 + 𝑛32 ), 𝑛 = 1,2,3 …
2𝑚
2𝑚𝐿2 1
This solution will point the way to several interesting features of the solution to the 3D
1/r potential.
a) There are three independent quantum numbers.
b) There is a degeneracy of the energy solutions. i.e There are multiple solutions in
terms of the three ns that can give the same energy value.
c) Introducing anything that distinguishes one coordinate from another can break
that degeneracy. For instance making one side longer.