Let us consider the behavior of tan energy eigenfunction Φ(x) at a potential
step.

Φ(x) is a solution of the differential equation

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

(a) Assume E > U_{0}.

The most general solutions in regions 1, and 2 are

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

Φ_{2}(x) = A_{2}exp(ik_{2}x) + A_{2}'exp(-ik_{2}x),

Here k_{1}^{2} = (2m/ħ^{2})E and k_{2}^{2}
= (2m/ħ^{2})(E - U_{0}).

Φ is continuous at x = 0.
Therefore A_{1 }+ A_{1}' = A_{2 }+ A_{2}' .

(∂/∂x)Φ(x) is continuous at x = 0.
Therefore ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2
}- ik_{2}A_{2}' .

We have two equations and four unknowns. No unique solution exists.

Let us limit
ourselves to particles approaching from the left being transmitted or reflected at the
barrier. Assume A_{2}' = 0.

We can then solve for A_{1}'/A_{1} and A_{2}/A_{1}.
We find

A_{1}'/A_{1} = (k_{1} - k_{2})/(k_{1}
+ k_{2}), A_{2}/A_{1} = 2k_{1}/(k_{1}
+ k_{2}) .

The ratios are real and positive, there is no
phase shift upon reflection and transmission.

**Probability of Reflection (R) and Transmission (T):**

Our solutions in regions 1 and 2 are plane wave solutions. They are not square-integrable,
and a proper normalization is not possible. Plane wave solutions are useful in describing
steady streams of particles. To evaluate the probability of reflection and transmission
for such beams, we compare the reflected and transmitted
flux with the incident flux.

Let P(x) be the probability density at x, let x_{0} be some
point in region 1, and let x_{0}^{’} be some point in region
2. We then have

incident flux ∝ v_{1}P_{i}(x_{0})

reflected flux ∝ v_{1}P_{r}(x_{0})

transmitted flux
∝ v_{2}P_{t}(x_{0}')

The **reflectance** R is R = (v_{1}P_{r}(x_{0}))/(v_{1}P_{i}(x_{0}))
= |A_{1}'/A_{1}|^{2}_{,
}and the **transmittance** T is T = (v_{2}P_{t}(x_{0}'))/(v_{1}P_{i}(x_{0}))
= (k_{2}/k_{1})|A_{2}/A_{1}|^{2}_{,
}We have R = 1 - 4k1k_{2}/(k_{1} + k_{2})^{2},
T = 4k1k_{2}/(k_{1} + k_{2})^{2}, R + T =
1.

Classically the probability of reflection is zero.

In Quantum Mechanics
R --> 0, T --> 1 if E >> U_{0} and k_{2 }
--> k_{1}.

(b) Assume E < U_{0}.

Let k_{1}^{2}
= (2m/ħ^{2})E and ρ_{2}^{2}
= (2m/ħ^{2})(U - E).

Then Φ_{1}(x)
= A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x)

and Φ_{2}(x)
= B_{2}exp(ρ_{2}x) + B_{2}'exp(-ρ_{2}x)

are the most general solutions. We need B_{2 }= 0 for the solution to
remain finite at x --> ∞.

Φ is continuous at x = 0.
Therefore A_{1 }+ A_{1}' = B_{2}' .

(∂/∂x)Φ(x) is continuous at x = 0.
Therefore ik_{1}A_{1 }- ik_{1}A_{1}' =
_{ }-ρ_{2}B_{2}'.
We find

A_{1}'/A_{1} = (k_{1} -
iρ_{2})/(k_{1}
+ iρ_{2}),
B_{2}/A_{1} = 2k_{1}/(k_{1} +
iρ_{2}) .

The ratios are complex, a
phase shift appears upon reflection.

We compute R by comparing the
incident and the reflected flux, R = |A_{1}'/A_{1}|^{2}
= 1.

Therefore T = 0, there is no transmitted flux.

Note: As U_{0} --> ∞,
ρ_{2 }--> ∞. Then A_{1}'/A_{1
}--> -1, B_{2}'/A_{1 }--> 0,
we have a phase shift of 180^{0}.

If U_{0} = ∞ then A_{1}' = -A_{1},
B'_{2 }= 0, Φ_{2}(x) = 0, Φ(x) is
continuous at x = 0 since Φ(0) = 0, but
∂Φ/∂x is not continuous.

We have Φ_{1}(x)
= 2iA_{1}sin(k_{1}x), ∂Φ_{1}/∂x
= 2iA_{1}k_{1}cos(k_{1}x), ∂Φ_{1}/∂x|_{x=0}
≠ 0.