XI-25
1
Delta Function Potential
A particle of mass m is in a region of potential V (x) = −Vo Lδ(x). First, use the variational principle to get
an upper bound on the particle’s ground state energy. Then, solve for the ground state energy exactly and
show that is is less than the determined bound.
y and momentum of a photon are E = hν
c formulas for the energy and momentum
2 Gravitational Perturbations
nd p = γmv. (Hint: Consider a mass m
at this decay in both the rest frame of the
Consider
a particle
is in a one-dimensional harmonic potential V (x) = 12 mw2 x2 . Suppose an
ss moves at speed
v. You’ll
needoftomass
usemthe
gravitational field is applied, so that the potential is shifted by an amount V 0 (x) = −mgx. Find the lowest
non-vanishing correction to the energy levels. Also, solve the system exactly and show that the resulting
energies are consitent with the perturbation results.
Compton
They collide 3at an
angle θ andScattering
create a
A photon of energy Eγ is incident on a particle of mass m, at rest in the lab frame. Find Eγ 0 , the energy of
the photon after it has scattered as a function of the scattering angle. Also, find the maximum energy that
a back-scattered photon can have. (This maximum will be independent of Eγ . If you’re clever, you can do
andthissticks
to aor small
is 5 lines
less.) mass m,
v, collides
ss of the resulting object? (Work in the
λ
(charged) particle of mass m. If the photon
1.15), show that the resulting wavelength,
wavelength, λ, by
h
(1 − cos θ).
mc
m
λ'
θ
(11.78)
m
Figure 11.15
= hν = hc/λ.)
E collides (elastically) head-on with a stathat the final energy of mass m is
+ E(M 2 + m2 )
.
M 2 + m2 + 2EM
2M
(11.79)
messy, but you can save yourself a lot of
must be a root of an equation you get for
rbitrarily large. Find (approximately) the
1