**Problem 1:**

(a) Calculate the transmission coefficient for a particle with
mass m and kinetic energy E < U_{0} passing through the rectangular potential
barrier

U(x) = 0 for x < 0, U(x) = U_{0 }> 0 for 0 < x < a, and
U(x) = 0 for x > a.

(b) Show that for E << U_{0} and 2mU_{0}a^{2}/ħ^{2}
>> 1 the transmission coefficient can be written as
T ≈ (16E/U_{0})exp[-2(2mU_{0}/ħ^{2})^{1/2}a].

(c) Many heavy nuclei decay by emitting an alpha particle.
In a
simple one-dimensional model, the potential barrier the alpha particles have to penetrate
can be approximated by

U(r) = 0 for r < R_{0}, U(r)
= U_{0}R_{0}/r for r > R_{0},

where R_{0} is the radius of the nucleus and
U_{0} is the
barrier height for r_{0} = R. The energy E of the alpha particle can
be assumed to be much smaller than U_{0}. For a non constant potential
barrier the expression for the transmission coefficient found in part (b) can be used as a
guide. Assume that for E << U_{0}, we have T ≈ exp[-2∫_{R1}^{R2 }dr (2m(U(r) - E)/ħ^{2})^{1/2}].

The integration limits R_{1}
and R_{2} are determined as solutions to the equation U(r) = E.

Calculate the alpha transmission coefficient and the decay constant λ,
i.e. the decay probability per second.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential. We are asked to find the transmittance T. - Details of the calculation:

(a) Divide space into 3 regions;

region 1: x < 0, region 2: 0 < x < a, region 3: x > a.

(a) potential barrier, E < U_{0}.

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx),

Φ_{2}(x) = B_{2}exp(ρx) + B_{2}'exp(-ρx),

Φ_{3}(x) = A_{3}exp(ikx).

Here k^{2}= (2m/ħ^{2})E and ρ^{2}= (2m/ħ^{2})(U_{0}- E).

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A_{1}+ A_{1}' = B_{2 }+ B_{2}', ikA_{1 }- ikA_{1}' = ρB_{2 }- ρB_{2}'.

x = a: B_{2}exp(ρa) + B_{2}'exp(-ρa) = A_{3}exp(ika),

ρB_{2}exp(ik_{2}a) - ρB_{2}'exp(-ρa) = ik_{i}A_{3}exp(ika).

We need to solve these equations for A_{3}and A_{1}' in terms of A_{1}.

(i) Solve for B_{2}and B_{2}' in terms of A_{3}.

B_{2}= ½exp((ik - ρ)a)(1 + ik/ρ )A_{3}= C A_{3}.

B_{2}' = ½exp((ik + ρ)a)(1 - ik/ρ)A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}to find T.

2A_{1}= B_{2}+ B_{2}' + (ρ/ik)(B_{2 }- B_{2}') = (C + C' + (ρ/ik) (C - C'))A_{3}.

A_{1}= ([(k^{2}- ρ^{2})/(2ikρ)]sinhρa + coshρa)exp(ika)A_{3}.

T = |A_{3}/A_{1}|^{2}= 4k^{2}ρ^{2}/[(k^{2}- ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}cosh^{2}ρa]

= 4k^{2}ρ^{2}/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}].

[cosh^{2}x - sinh^{2}x = 1.]

T = 4E(U_{0}- E)/[U_{0}^{2}sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{1/2}a] + 4E(U_{0}- E)].(b) E << U

_{0}and 2mU_{0}a^{2}/ħ^{2}>> 1.

sinh(x) = ½(e^{x}- e^{-x}) ≈ ½e^{x}, if x >> 1.

Therefore

T ≈ 4E(U_{0}- E)/[¼U_{0}^{2}exp[2(2m(U_{0}- E)/ħ^{2})^{1/2}a] + 4E(U_{0}- E)]

≈ 4EU_{0}/[¼U_{0}^{2}exp[2(2mU_{0}/ħ^{2})^{1/2}a] + 4EU_{0})]

≈ (16E/U_{0})exp[-2(2mU_{0}/ħ^{2})^{1/2}a].(c) Assume that T ≈ exp[-2∫

_{R1}^{R2 }dr (2m(U(r) - E)/ħ^{2})^{1/2}].

R

_{1}= R_{0}, R_{2}= U_{0}R_{0}/E, R_{2}>> R_{1}.

∫_{R1}^{R2 }dr (U(r) - E)^{1/2}= ∫_{R1}^{R2 }dr (U_{0}R_{0}/r - U_{0}R_{0}/R_{2})^{1/2}

= (U_{0}R_{0})^{1/2}∫_{R1}^{R2 }dr (1/r - 1/R_{2})^{1/2}.

∫_{R1}^{R2 }dr (1/r - 1/R_{2})^{1/2}= ∫_{R1}^{R2 }dr [(R_{2}- r)/(rR_{2})^{]1/2}

= √R_{2}[(R_{1}/R_{2}- (R_{1}/R_{2})^{2})^{1/2}+ π/2 - tan^{-1}(R_{1}/(R_{2}- R_{1}))^{1/2}

--> √R_{2}π/2 if R_{2}>> R_{1}.

T ≈ exp[-√(2m/E) πU_{0}R_{0}/ħ) = e^{-λ'}, λ' = √(2m/E) πU_{0}R_{0}/ħ

e^{-λ'}is the transmission coefficient.

To calculate the escape probability per second, we have to multiply T by the rate of the alpha particle hitting the barrier, which is approximately v/R_{0}.

escape probability ≈ √(2m/E)/R_{0}* e^{-λ'}= λ = 1/τ.

This is a reasonable order of magnitude approximation as long as the escape probability is very small.

**Problem 2:**

Let U(x) = ∞ for x < 0, U(x) = ½mω^{2}x^{2}
for x > 0. Use the WKB approximation to find the energy levels of a particle of
mass m in this potential. Compare the WKB energies with the exact energies for
this potential.

Solution:

- Concepts:

The WKB approximation - Reasoning:

We are instructed to use the WKB approximation.

For stationary bound states in a potential well with one vertical wall we want ∫_{0}^{xmax}pdx = (n - ¼)πħ, n = 1, 2, 3, ... . - Details of the calculation:

p = (2m(E - U(x))^{1/2}.

For the given potential energy function

∫_{0}^{xmax}(2m(E - ½mω^{2}x^{2})^{1/2}dx = (n - ¼)πħ.

x_{max}= (2E/(mω^{2}))^{1/2}.

∫_{0}^{xmax}(2m(E - ½mω^{2}x^{2})^{1/2}dx = mω∫_{0}^{xmax}(2E/(mω^{2}) - x^{2})^{1/2}dx

= ½mω(2E/(mω^{2}))(sin^{-1}1 - sin^{-1}0) = (E/ω)(π/2).

(E/ω)(π/2) = (n - ¼)πħ, E = (2n - ½)ħω.

These are the exact energy levels for n = 1, 2, 3, ... .