Physics 5250, Fall 2024
– Problem Set #7 –
This problem set contains 2problems, worth a total of 40 points.
1. (20 points) A single particle in three dimensions is bound in a spherical “square well”,
V(r) = (−V0, r < a;
0, r ≥a, (1)
where V0>0 is the well depth.
(a) (15 points) Consider the ground-state energy E. Suppose the particle is very weakly bound,
i.e. Eis negative and very close to zero. Show that the binding energy Eis related to the
well depth V0and radius athrough the following approximate formula:
2mV0a2
ℏ2=π2
4+ 2κa +c2(κa)2+... (2)
where κ=√−2mE/ℏ. What is the exact value of the purely numerical constant c2in
this expansion? (Hint: for the solutions inside the well, write the wave number kin terms
of κ, so that you can expand in κa consistently.)
(b) (5 points) The deuteron (proton-neutron bound state) has a binding energy of 2.226 MeV.
Suppose we model it as a single particle with reduced mass µ= (mpmn)/(mp+mn), moving
in a spherical square well with radius a= 1.5 fm. Use the result of part (a) to estimate
the depth of the deuteron potential, V0; give your answer in MeV.
2. (20 points) Suppose a spinless particle is in an eigenstate |l, m⟩of orbital angular momentum
{ˆ
L2,ˆ
Lz}. If we apply a rotation to our system, the probability for the rotated state ˆ
D(R)|l, m⟩
to be measured in any eigenstate |l, m′⟩of ˆ
Lzis given by appropriate matrix elements of the
rotation operator, also known as the Wigner D-matrix,
D(l)
m′,m(R) = ⟨l, m′|exp −i(ˆ
L·n)ϕ
ℏ!|l, m⟩.(37)
(See p.196 of Sakurai for a more detailed explanation.)
Instead of acting the rotation ˆ
D(R) on the ket directly (active transformation), we can view
it as a rotation acting on the operators (passive transformation.) As a vector operator, the
angular momentum ˆ
Lin this approach transforms as
ˆ
Li→ˆ
L′
i=X
j
Rij ˆ
Lj,(38)
where Rij is a classical 3x3 rotation matrix.
Let’s consider a rotation around the y-axis by an angle β, and study what happens to
our initial state |l, m⟩. The matrix describing such a rotation classically is
Ry=
cos β0 sin β
0 1 0
−sin β0 cosβ
.(39)
