Problem 1:

(a)  Calculate the transmission coefficient for a particle with mass m and kinetic energy E < U0 passing through the rectangular potential barrier
U(x) = 0 for x < 0, U(x) = U0 > 0 for 0 < x < a, and U(x) = 0 for x > a.
(b)  Show that for E << U0 and  2mU0a22 >> 1 the transmission coefficient can be written as  T ≈ (16E/U0)exp[-2(2mU02)1/2a].
(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 < R0,  U(r) = U0R0/r for r > R0,
where R0 is the radius of the nucleus and U0 is the barrier height for r0 = R.  The energy E of the alpha particle can be assumed to be much smaller than U0.  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 << U0, we have T ≈ exp[-2∫R1R2 dr (2m(U(r) - E)/ħ2)1/2].
The integration limits R1 and R2 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 < U0.
The most general solutions in regions 1, 2, and 3 are
Φ1(x) = A1exp(ikx) + A1'exp(-ikx),
Φ2(x) = B2exp(ρx) + B2'exp(-ρx),
Φ3(x) = A3exp(ikx).
Here k2 = (2m/ħ2)E and ρ2 = (2m/ħ2)(U0 - E).
The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.
x = 0:  A1 + A1' = B2 + B2', ikA1 - ikA1' = ρB2 - ρB2'.
x = a:  B2exp(ρa) + B2'exp(-ρa) = A3exp(ika),
ρB2exp(ik2a) - ρB2'exp(-ρa) = ikiA3exp(ika).
We need to solve these equations for A3 and A1' in terms of A1.

(i)  Solve for B2 and B2' in terms of A3.
B2 = ½exp((ik - ρ)a)(1 + ik/ρ )A3 = C A3.
B2' = ½exp((ik + ρ)a)(1 - ik/ρ)A3 = C' A3.

(ii)  Now solve for A1 in terms of A3 to find T.
2A1 = B2 + B2' + (ρ/ik)(B2 - B2') = (C + C' + (ρ/ik) (C - C'))A3.
A1 = ([(k2 - ρ2)/(2ikρ)]sinhρa + coshρa)exp(ika)A3.
T = |A3/A1|2 = 4k2ρ2/[(k2 - ρ2)2sinh2ρa + 4k2ρ2cosh2ρa]
= 4k2ρ2/[(k2 + ρ2)2sinh2ρa + 4k2ρ2].
[cosh2x - sinh2x = 1.]
T = 4E(U0 - E)/[U02sinh2[(2m(U0 - E)/ħ2)1/2a] + 4E(U0 - E)].

(b)   E << U0 and  2mU0a22 >> 1.
sinh(x) = ½(ex - e-x) ≈ ½ex, if x >> 1.
Therefore
T ≈ 4E(U0 - E)/[¼U02exp[2(2m(U0 - E)/ħ2)1/2a] + 4E(U0 - E)]
≈ 4EU0/[¼U02exp[2(2mU02)1/2a] + 4EU0)]
≈ (16E/U0)exp[-2(2mU02)1/2a].

(c)  Assume that T ≈ exp[-2∫R1R2 dr (2m(U(r) - E)/ħ2)1/2].

R1 = R0,  R2 = U0R0/E,  R2 >> R1.
R1R2 dr (U(r) - E)1/2 = ∫R1R2 dr (U0R0/r - U0R0/R2)1/2
= (U0R0)1/2R1R2 dr (1/r - 1/R2)1/2.
R1R2 dr (1/r - 1/R2)1/2 = ∫R1R2 dr [(R2 - r)/(rR2)]1/2
= √R2[(R1/R2 - (R1/R2)2)1/2 + π/2  - tan-1(R1/(R2 - R1))1/2
-->  √R2 π/2  if R2 >> R1.
T ≈ exp[-√(2m/E) πU0R0/ħ) = e-λ',  λ' = √(2m/E) πU0R0
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/R0.
escape probability ≈  √(2m/E)/R0 * 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) = ½mω2x2 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 ∫0xmax 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
0xmax (2m(E - ½mω2x2)1/2dx = (n - ¼)πħ.
xmax = (2E/(mω2))1/2.
0xmax (2m(E - ½mω2x2)1/2dx = mω∫0xmax (2E/(mω2) - x2)1/2dx
= ½mω(2E/(mω2))(sin-11 - sin-10) = (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, ... .