Quantum Mechanics I: Physics 6210: Wheeler
Assignment 6
Note the last paragraph on page 100:
"It is assumed that the reader of this book has some experience
solving the time-dependent and time-independent wave equations."
While I’m sure this is true, I’m also sure that a little more practice wouldn’t
hurt. So let’s take some time to go back and solve the Schrödinger equation for
the standard cases.
A good reference for wave equation solutions is Merzbacher. Check out
his solution for the Linear Harmonic Oscillator, pp. 53 - 79. He solves the
problem in Sections 1 & 2 and gives properties of the Hermite polynomials
in section 3. Note especially the similarity between Merzbacher’s expression
for the Hermite polynomials in terms of a generating function, eq. 5.19, and
Sakurai’s expression for < xin >, eq.(2.3.32). In Sections 4, Merzbacher writes
the matrix representation using the number basis, |n >, for XOP , and leaves
POP as a simple exercise. In Sections 5 and 6, he solves the problem of the
double oscillator.
PROBLEMS: Our main focus is to review solutions to the Schrödinger equation. The standard problems are piecewise constant potentials, barrier problems, the central force problem and hydrogen atom. We’re doing hydrogen
in class; below are some 1-dimensional problems. Continue in Sakurai with
some selected problems in Chapter 2 that let you get more practice solving the
Schrödinger equation: 21, 22, 23, 24, 26, 27.
• W.1: Solve the Schrödinger equation for a free particle.
• W.2: Solve the Schrödinger equation for a square well:

x<0
 ∞
0 0≤x≤L
V =

∞
x>L
and find the allowed energies. Observe the symmetry of the even and odd
solutions.
• W.3: Solve the Schrödinger equation for a plane wave incident from the
left on a potential barrier:

x<0
 0
V0 0 ≤ x ≤ L
V =

0
x>L
The solution will be composed of a wave in the region x < - a travelling
to the right with amplitude A, a reflected wave in the same region with
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amplitude B, some stuff going on inside the barrier, and a transmitted
wave moving to the right in the region x > a with amplitude C. You
won’t be able to normalize this form of the solution, but you can find the
transmission coefficient when 0 < E < V o. The transmission coefficient
is the ratio of the absolute squares of the amplitudes of the outgoing
transmitted wave (x > L) to the incoming wave
T =
|ψ (L)|
2
2
|ψ (0)|
This represents the probability that a particle defies the classical expectation and makes it past the barrier. See my notes below for boundary
conditions.
• S.2.20: This just involves putting a boundary condition on the solutions
for the harmonic oscillator.
• S.2.21: You’ve already written down the wave function for this problem
in problem S.1.20. It’s a nice simple one to practice time evolution on.
• S.2.22: This is an odd potential, but a nice problem. The thing you
have to do is to match boundary conditions at the point where the δfunction diverges. First, assume that E < 0 and solve the Schrödinger
equation outside the well. This gives solutions which decay exponentially
away from the origin. Now try and match those solutions at the origin,
using boundary conditions derived in the usual way from the Schrödinger
equation.
• S.2.23: The time dependence of problem 22.
SUPPLEMENT: Boundary conditions for the Schrödinger equation.
We already know that the wave function ψ must be continuous, or the probability wouldn’t be continuous, and that would strange at the very least. To find
the condition on ψ 0 , we begin with
−
~
ψ” + V ψ = Eψ
2m
and integrate across the boundary. Take the boundary to be at the origin, x = 0.
Then integrating from −ε to +ε:
−
~
(ψ 0 (ε) − ψ 0 (−ε)) + 2V εψ (0) = 2εEψ (0)
2m
where the first term on the left integrates immediately and the other two may
be approximated for sufficiently small ε by
ˆ ε
f (x) dx ≈ f (0) 2ε
−ε
2
This becomes exact in the limit as ε → 0. Taking this limit, the second two
integrals vanish provided E and V remain finite. Defining
ψ−
=
ψ+
=
lim ψ (−ε)
ε→0
lim ψ (+ε)
ε→0
we are left with the continuity of ψ 0 across the boundary,
ψ− = ψ+
This condition is altered for infinite potentials. See if you can justify the boundary conditions used for the infinite square well.
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