Consider spinless non-relativistic free particles of mass M
moving in a two dimensional square box. Find an expression for the density of
states ρ(E).

Remember ρ(E)dE is
defined as the number of energy levels per unit volume between E and E + dE.

Solution:

- Concepts:

The two-dimensional, infinite square well, the density of states - Reasoning:

We are asked to find the density of states for a two-dimensional, infinite square well - Details of the calculation:

The eigenfunctions and eigenvalues of the 2D infinite square well are

Φ_{nm}(x,y) = (2/L)sin(k_{x}x)sin(k_{y}y), with k_{x}= πn/L, k_{y}= πm/L, n, m = 1, 2, ... .

The associated eigenvalues are

E_{nm}= (n^{2}+ m^{2})π^{2}ħ^{2}/(2ML^{2}) = (k_{x}^{2}+ k_{y}^{2})ħ^{2}/(2M) = k^{2}ħ^{2}/(2M)

k_{x}= πn/L, k_{y}= πm/L, Δk_{x}= Δk_{y}= π/L.

Let N(k') be the number of states with |**k**| ≤ k'.

The area of a circle of radius k' in 2D k-space is A = πk^{'2}.

The number of states in the circle is N(k') = (A/4)/(π/L)^{2}= k^{'2}L^{2}/(4π).

Only positive values of k_{x}and k_{y}are allowed.

The density of states in k-space is n(k') = dN(k')/dk' = k'L^{2}/(2π).

E = k^{2}ħ^{2}/(2M). dE/dk = kħ^{2}/M.

n(E) = dN/dE = (dN/dk)(dk/dE) = ML^{2}/(2πħ^{2}).

Consider a particle in the ground state of a one-dimensional square well of
with a and depth U_{0}.

Assume that the well is very deep and

k/k_{0} = (2m(E + U_{0})/(2mU_{0}))^{½} << 1

for the ground state, so that the ground state wave function is nearly identical
to that of the infinite square well.
At t = 0, a time dependent perturbation W(t) = Wcosωt
is turned on.

(a) What is the minimum frequency necessary to free the particle
from the well?

(b) For frequencies greater than this minimum frequency, use
perturbation theory to find the transition rate.

You can assume that the free particles will be in a box of size L >> a.

Solution:

- Concepts:

Fermi's golden rule, the infinite square well - Reasoning:

The initial wave function of the particle look very much like the wave function of a particle in an infinite square well. The final state of the particle is a continuum state. - Details of the calculation:

(a) Let H_{0}be the Hamiltonian of the infinite square well.

Eigenfunctions of H_{0}: Φ_{n}(x) = (2/a)^{½}sin(nπx/a) with eigenvalues E_{n}= n^{2}π^{2}ħ^{2}/(2ma^{2}).

The ground state energy is E_{1}= π^{2}ħ^{2}/(2ma^{2}).

In a very deep well the ground state energy is approximately equal to -V_{0}+ E_{1}~ -V_{0}.

The minimum frequency required to free the particle from the well therefore is

ω_{min}= V_{0}/ħ.

(b) To find the transition rate we use Fermi's golden rule.

If W(t) = Wcos(ωt), then the transition probability per unit time is given by

w(i,E) = (π/(2ħ))ρ(E)|W_{Ei}|^{2}δ_{E-Ei,ħω}, where W_{Ei}= <Φ_{E}|W|Φ_{i}>.

Assume that the ejected particle is a free particle confined to a region of width L >> a. Use periodic boundary conditions. Then

Φ_{k}(x) = L^{-½}exp(ikx), with k = 2πn/L, n = 0, ±1, ±2, ... . k^{2}= 2mE/ħ^{2}= 2m(ω - ω_{min})/ħ.

The number of states with wave vectors whose magnitudes lie between k and k + dk is

dN = 2 dk/(2π/L) = Ldk/π. (The particle can move towards the left or to the right.)

dN/dk = L/π.

The density of states is dN/dE = (dN/dk)(dk/dE).

With E = ħ^{2}k^{2}/(2m) we have ρ(E) = dN/dE = Lm/(πkħ^{2}).

W_{Ei}= (2/La)^{½}W∫_{0}^{a}dx e^{ikx}sin(πx/a)

= (2/La)^{½}W(e^{ika}+1)πa/(π^{2}- k^{2}a^{2})

|W_{Ei}| = W^{2}(2/La)cos^{2}(ka/2)4π^{2}a^{2}/(π^{2}- k^{2}a^{2})^{2}.

w(i,E) = (π/(2ħ))Lm/(πkħ^{2})W^{2}(2/La)cos^{2}(ka/2)4π^{2}a^{2}/(π^{2}- k^{2}a^{2})^{2}

= W^{2}cos^{2}(ka/2)4π^{2}am/[kħ^{3}(π^{2}- k^{2}a^{2})^{2}].

This is the transition probability per unit time.

A hypothetical particle of mass 2m can decay into two different final states
under the action of a perturbation H'.

(a) 2 m --> m + γ,

(b) 2 m --> 1.5 m + γ.

Here m and 1.5 m represent particles of mass m and 1.5 m respectively. If the
matrix elements of H' are equal, i.e. <2 m|H'|m> = <2 m|H'|1.5 m>, what
will be the relative likelihood of decay to m as opposed to 1.5 m? You may
assume the recoil fragment (m or 1.5 m) receives negligible kinetic energy.

Solution:

- Concepts:

Fermi's golden rule, - Reasoning:

The transition probability per unit time is

w(i,βE) = (2π/ħ)ρ(β,E)|W_{Ei}|^{2}δ_{E-Ei,ħω};

ρ(β,E)dE is the number of final states in the interval dE characterized by some discrete index b.

Here ρ(β,E) is the density of photon states for photons of energy ħω = (m_{initial}- m_{final})c^{2}. - Details of the calculation:

The number of photon states ρ(β,E)dΩdE, i.e. the number of photon states in an energy interval between E and E + dE with wave vector**k**in the solid angle dΩ is proportional to d^{3}**k**.

ρ(E)dΩdE ∝ d^{3}**k**∝ p^{2}dpdΩ.

Therefore ρ(E) ∝ p^{2}dp/dE.

p = E/c, ρ(E) ∝ E^{2}/c^{3}∝ ω^{2}/c^{3}.

So if the matrix elements are equal, the likelihood of decay is proportional to (m_{initial}- m_{final})^{2}.

P_{(a)}/P_{(b)}= (m/0.5 m)^{2}= 4.

A uniform periodic electric field acts upon a hydrogen atom, which at t = 0 is in its ground state. Determine the minimum frequency of the field necessary to ionize the atom and use perturbation theory to evaluate the probability for ionization per unit time. For the sake of simplicity, assume the electron in the final state to be free.

Solution:

- Concepts:

Time dependent perturbation theory, Fermi's golden rule - Reasoning:

Fermi's golden rule gives the probability per unit time for a transition from a state of the discrete spectrum to a state corresponding to an infinitesimal interval in the continuous spectrum. - Details of the calculation:

w(i,βE) = (π/2ħ)ρ(β,E)|W_{Ei}|^{2}d_{E-Ei,ħω}for W(t) = Wcos(ωt).

ρ(β,E)dE is the number of final states in the interval dE characterized by some discrete index β.

Here W(t) = W_{DE}= e**r∙E**= eE_{0}zcos(ωt).

The minimum frequency necessary to ionize the hydrogen is ω_{0}= E_{I}/ħ = μe^{4}/(2ħ^{3}).

Assume the ejected electron is free.

The eigenfunctions and eigenvalues of a free electron confined to a cubical 3D infinite square well with periodic boundary conditions are

ψ_{nml}(x,y,z) = (1/L)^{3/2}exp(i**k∙r**)= (1/L)^{3/2}exp(ik_{x}x)exp(ik_{y}y)exp(ik_{z}z),

with k_{x}= 2πn/L, k_{y}= 2πq/L, k_{z}= 2πl/L, n, q, l = ±1, ±2, ... .

The associated eigenvalues are

E_{nql}= (n^{2}+ q^{2}+ l^{2})4π^{2}ħ^{2}/(2mL^{2}) = (k_{x}^{2}+ k_{y}^{2}+ k_{z}^{2})ħ^{2}/(2m) = ħ^{2}k^{2}/(2m).

[H = p_{x}^{2}/(2m) + p_{y}^{2}/(2m) + p_{z}^{2}/(2m), if x, y , z < L.

ψ(x,y,z) = Φ(x)χ(y)γ(z).

(-ħ^{2}/(2m))(∂^{2}/∂x^{2})Φ(x) = E_{x}Φ(x).

Φ(x) = (1/L)^{½}exp(ik_{x}x), Φ(x) = Φ(x + L) --> k_{x}= 2πn/L.

E_{x}= ħ^{2}k_{x}^{2}/(2m) = 4π^{2}n^{2}ħ^{2}/(2mL^{2}) .]

Δk_{x}= Δk_{y}= Δk_{z }= 2π/L.

To each allowed

**k**_{nml}there corresponds a wave function ψ_{nml}(x,y,z).

The tips of the vectors**k**_{nml}divide k-space into elementary cubes of edge length 2π/L. We have one vector per (2π/L)^{3}volume of k-space.

The number of vectors in a volume d^{3}k of k-space therefore is dN = d^{3}k/(2π/L)^{3}.

d^{3}k = k^{2}dkdΩ,

dN = k^{2}dkdΩ/(2π/L)^{3}= k^{2}dkVdΩ/(2π)^{3}= number of state with a wave vector of magnitude between k and k + dk in a solid angle dΩ.

dN/dk = k^{2}dΩ/(2π/L)^{3}= k^{2}VdΩ/(2π)^{3}. (V = L^{3}.)

The density of states for the electron is dN/dE = (dN/dk)(dk/dE), with E = ħ^{2}k^{2}/(2m).

dN/dE = ρ(β,E) = 2^{½}m^{3/2}VE^{½}dΩ/(2πħ)^{3}.

(We do not multiply by 2 since the interaction cannot flip the spin.)

The ground state wave function of hydrogen is (πa_{0}^{2})^{-½}exp(-r/a_{0}).

We therefore have W_{E1}= (Vπa_{0}^{2})^{-½}∫exp(-i**k∙r**-r/a_{0})(-e**r∙E**)d^{3}r.

To evaluate this integral let**k**point into the z-direction and use spherical coordinates.**r**=**r**(r,θ,φ),**E**=**E**(r_{0},θ_{0},φ_{0}).

Then**E∙r**= E_{0}r[cosθcosθ_{0}+ sinθsinθ_{0}cos(φ-φ_{0})] and

W_{E1}= -(Vπa_{0}^{2})^{-½}eE_{0}∫∫∫exp(-i**k∙r**-r/a_{0})[cosθcosθ_{0}+ sinθsinθ_{0}cos(φ-φ_{0})]r^{3}drsinθdθdφ

= 2πcosθ_{0}(Vπa_{0}^{2})^{-½}eE_{0}∫_{-1}^{1}dcosθ[∫_{0}^{∞}exp(-ikrcosθ - r/a_{0})r^{3}dr]cosθ.

W_{E1}= -2πcosθ_{0}(Vπa_{0}^{2})^{-½}eE_{0}∫_{-1}^{1}dx (3! x)/(a_{0}^{-1}+ ikx)^{4}= -i2πcosθ_{0}(Vπa_{0}^{2})^{-½}eE_{0 }16ka_{0}^{5}/(1 + k^{2}a_{0}^{2})^{3}.

Notation:

We have E_{I}= ħω_{0}= μe^{4}/(2ħ^{2}), a_{0}= ħ^{2}/(me^{2}), so we may write a_{0}^{2}= ħ^{4}/(m^{2}e^{4}) = ħ/(2mω_{0}).

Energy conservation requires that E = ħω - ħω_{0}.

Therefore k^{2 }= 2mE/ħ^{2}= 2m(ω - ω_{0})/ħ, k^{2}a_{0}^{2}= (ω - ω_{0})/ω_{0}, 1 + k^{2}a_{0}^{2}= ω/ω_{0}.

The probability that an electron is ejected into a solid angle dΩ with energy E per unit time therefore is

w(1,dΩ E) = ( π/2ħ)2^{½}m^{3/2}VE^{½}dΩ/(2πħ))^{3}|-i2πcosθ_{0}(Vπa_{0}^{2})^{-½}eE_{0 }16ka_{0}^{5}/(ω/ω_{0})^{3}|^{2 }= E_{0}^{2}(cosθ_{0})^{2}(ω/ω_{0})^{6}[(ω - ω_{0})/ω_{0}]^{3/2}a_{0}^{3}dΩ 2^{8}/(4πħ).

To find the probability per unit time that an electron is ejected with energy E we must integrate over all possible angles between the direction of the field and the direction of**k**.

∫_{0}^{2π}∫_{-1}^{1}(cosθ_{0})^{2}dcosθ_{0}dφ_{0}= 4π/3.

w(1,dΩ E) = (256/3)(a_{0}^{3}/ħ) E_{0}^{2}(ω/ω_{0})^{6}[(ω - ω_{0})/ω_{0}]^{3/2}.

Near the threshold for ionization the probability increases from zero as (ω - ω_{0})^{3/2}. If ω >> ω_{0}it decreases as ω^{-9/2}. It has a maximum for ω = 4ω_{0}/3.

Nuclei sometimes
decay from excited states to the ground state by internal conversion, a
process in which an atomic electron is emitted instead of a
photon. Let the initial and final nuclear states have wave
functions Ψ_{i}(**r**_{1},**r**_{2},..,**r**_{z})
and Ψ_{f}(**r**_{1},**r**_{2},..,**r**_{z}),
respectively,
where **r**_{i} describes the protons.
The perturbation giving rise to the transition is the proton-electron
interaction,

W = -∑_{i=1}^{z} e^{2}/|**r** -
**r**_{i}|,

where **r **is the electron coordinate.

(a) Write
down the matrix element for the process in lowest-order perturbation
theory, assuming that the electron is initially in a state characterized
by the quantum numbers nlm, and that its energy, after it has
been emitted, is large enough so that its final state may be described
by a plane wave. Neglect spin.

(b) Write
down an expression for the internal conversion rate.

(c) For
light nuclei, the nuclear radius is much smaller than the Bohr radius
for the given Z, and we can use the expansion

1/|**r** - **r**_{i}| ≈ 1/r +
**r**∙**r**_{i}/r^{3}.

Use this approximation to express the transition rate in terms of the dipole matrix element

d = <Ψ_{f}|∑_{i=1}^{z}
**r**_{i}|Ψ_{i}>.

Solution:

- Concepts:

Fermi's golden rule, a constant perturbation - Reasoning:

We find the internal conversion rate using Fermi's golden rule,

w(0-->1) = (2π/ħ)ρ(E)|W_{10}|^{2}δ_{E1,E0},

because the final state of the electron is a continuum state. - Details of the calculation:

(a) The initial state is Φ_{0}= ψ_{lmn}(**r**)Ψ_{i}(**r**_{1},**r**_{2},..,**r**_{z}),

the final state is Φ_{1}= V^{-½}exp(i**k∙r**) Ψ_{f}(**r**_{1},**r**_{2},..,**r**_{z}).

The final state of the electron is a plane wave. We confine the plane wave to a volume V = L^{3}and use periodic boundary conditions. The matrix element needed to calculate the internal conversion rate is

<Φ_{1}|W|Φ_{0}> = V^{-½}∫d^{3}r exp(i**k∙r**)<Ψ_{f}(**r**_{i})|∑_{i}e^{2}/|**r**-**r**_{i}||Ψ_{i}(**r**_{i})>ψ_{lmn}(**r**).

The energy eigenstates of an electron confined to a cubical box with periodic boundary conditions are

Φ_{nx,ny,nz}(x,y,z) = L^{-3}exp(i2π(n_{x}x+n_{y}y+n_{z}z)/L),

with n_{x}, n_{y}, n_{z}= 0, ±1, ±2, ... .

We have k_{x}= 2πn_{x}/L, k_{y}= 2πn_{y}/L, k_{z}= 2πn_{z}/L.

If k is large, then the number of states with wave vectors whose magnitudes lie between k and k + dk is

dN = 4πk^{2}dk/(2π/L)^{3}= 4πVk^{2}dk/(2π)^{3}.

dN/dk = 4πVk^{2}/(2π)^{3}.

The density of states is dN/dE = (dN/dk)(dk/dE).

With E = ħ^{2}k^{2}/(2m) we have dN/dE = 2mVk/(4π^{2}ħ^{2}) = (2m)^{3/2}VE^{½}/(4π^{2}ħ^{3}).

We are instructed to neglect spin. If we do not neglect spin we multiply the density by 2.

We therefore have for the internal conversion rate

w(0-->1) = (½πħ)(2m/ħ^{2})^{3/2}E^{½}|∫d^{3}r exp(i**k∙r**)<Ψ_{f}|∑_{i}e^{2}/|**r**-**r**_{i}||Ψ_{i}>ψ_{lmn}(**r**)|^{2}δ(E_{1}- E_{0}),

(c) We use 1/|**r**-**r**_{i}| ≈ 1/r +**r**∙**r**_{i}/r^{3}.

<Ψ_{f}|∑_{i}e^{2}/|**r**-**r**_{i}||Ψ_{i}> = (e^{2}/r^{3})**r∙**<Ψ_{f}|∑**r**_{i}||Ψ_{i}> = (e^{2}/r^{3})**r∙d**, since Ψ_{i}and Ψ_{f}depend only on the nuclear coordinates**r**_{i }and not on the electron coordinate**r**.

**d**is independent of**r**. We can therefore write

w(0-->1) = (e^{4}/2πħ)(2m/ħ^{2})^{3/2}E^{½}|**d∙**∫d^{3}r exp(i**k∙r**) (**r**/r^{3}) ψ_{lmn}(**r**)|^{2}δ_{E1,E0},