**Problem:**

A one-dimensional potential well is given in the form of a delta function at x =
0,

U(x) = Cδ(x), C < 0.

(a) A non-relativistic particle of mass m and energy E is incident from one
side of the well.

Derive an expression for the coefficient of transmission T(E).

(b) Since a bound state can exist with the attractive potential, find the
binding energy of the ground state of the system.

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:

U(x) = 0 everywhere except at x = 0. - Details of the calculation:

(a) E > 0.

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx) for x < 0. k^{2}= 2mE/ħ^{2}.

Φ_{2}(x) = A_{2}exp(ikx) for x > 0.

Φ is continuous at x = 0. Φ_{1}(0) = Φ_{2}(0). A_{1 }+ A_{1}’ = A_{2}.

∂Φ/∂x has a finite discontinuity at x = 0.

∂^{2}Φ(x)/∂x^{2}+ (2m(E - U)/ħ^{2})Φ(x) = 0.

Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.

∂Φ_{2ε}(x_{1 }+ ε)/∂x - ∂Φ_{1ε}(x_{1 }- ε)/∂x = -(2m/ħ^{2})∫_{x1-ε}^{x1+ε }(E - Cδ(x)) Φ(x) dx

= (2mC/ħ^{2})Φ(0).

iA_{1}k – iA_{1}’k = iA_{2}k - A_{2}2mC/ħ^{2}, A_{1}– A_{1}’ = A_{2}- A_{2}2mC/(ikħ^{2}) = [1 - 2mC/(ikħ^{2})]A_{2}.

Eliminate A_{1}’:

2A_{1}= [2 - 2mC/(ikħ^{2}) ]A_{2}, A_{2}/A_{1}= 1/[1 - mC/(ikħ^{2})] = ikħ^{2}/(ikħ^{2}- mC).

T(E) = (k|A_{2}|^{2})/(k|A_{1}|^{2}) = ħ^{4}k^{2}/(ħ^{4}k^{2}+ m^{2}C^{2}) = E/(E + mC^{2}/(2ħ^{2})).

(b) E < 0.

Φ_{1}(x) = A_{1}exp(ρx) + A_{1}'exp(-ρx) for x < 0. ρ^{2}= -2mE/ħ^{2}.

Φ_{2}(x) = A_{2}exp(ρx) + A_{2}'exp(-ρx) for x > 0.

Φ is finite at infinity. A_{1}’ = A_{2 }= 0. Φ is continuous at x = 0. Φ_{1}(0) = Φ_{2}(0). A_{1 }= A_{2}’.

∂Φ_{2}/∂x|_{x=0}= ∂Φ_{1}/∂x|_{x=0}+ (2mC/ħ^{2})Φ(0).

-ρA_{2}' - (2mC/ħ^{2})A_{2}' = ρA_{2}', ρ = -mC/ħ^{2}.

m^{2}C^{2}/ħ^{4}= -2mE/ħ^{2}, E = -mC^{2}/(2ħ^{2}).

Only one bound state exists.

**Problem:**

Consider the non-relativistic motion in one dimension of a particle outside an infinite barrier at x ≤ 0 with an additional delta function potential at x = a, i.e. U(x) = ∞ for x ≤ 0, U(x) = Fδ(x - a) for x > 0, where F is a positive constant. Derive an analytical expression for the phase shift δ(k) for a particle approaching the origin from x = +∞ with momentum ħk.

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:

U(x) = 0 for x > 0 except at x = a. U(x) = ∞ for x ≤ 0. - Details of the calculation:

The most general solution of the" time-independent" Schroedinger equation in region 1 is Φ_{1}(x) = A sin(kx) because Φ_{1}(x) = 0 due to the boundary condition at x = 0.

The most general solution in region 2 is Φ_{2}(x) = B sin(kx + δ(k)). The boundary conditions at x = a are Φ_{1}(a) = Φ_{2}(a), ∂Φ_{1}/∂x|_{a}= ∂Φ_{2}/∂x|_{a}- (2mF/ħ^{2})Φ(a).

A sin(ka) = B sin(ka + δ(k)), kA cos(ka) = kB cos(ka + δ(k)) - (2mF/ħ^{2})A sin(ka).

cot(ka) + (2mF/ħ^{2}) = cot(ka + δ(k)),

δ(k) = cot^{-1}(cot(ka) + (2mF/ħ^{2})) - ka.