Write the spatial factors of the stationary wavefunction as

ψ (x, t) = e^{- i Et /ℏ} ⎩ ⎨ ⎧ e^{ik x} + R e^{- ik x}, C e^{k x} + D e^{- k x}, T e^{ik x}, x < - a, - a < x < a, x > a,

where

k = \frac{2 m E}{ℏ} .

Since

U_{0} - E = E

, the decay constant inside the potential barrier is also

k

. Continuity of the wavefunction and its first derivative at

x = - a

and

x = a

gives four linear equations. Solving them yields

T = \frac{e ^{- 2 ika}}{c o s h ( 2 ka )} .

Consequently the transmitted wave is

ψ_{tr} (x, t) = \frac{exp ( i [ k ( x - 2 a ) - Et /ℏ ] )}{cosh ( 2 ka )}, x > a,

ρ_{tr} (x, t) = ∣ ψ_{tr} ∣^{2} = sech^{2} (2 ka) .

ii

Solution

A e^{i (k x - Et /ℏ)}

J = \frac{ℏ k}{m} ∣ A ∣^{2} .

The incident amplitude is one and the transmitted amplitude is

T

, so

\frac{J _{tr}}{J _{in}} = ∣ T ∣^{2} = sech^{2} (2 ka) .

\partial ρ / \partial t = 0

, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

\frac{∣ J _{refl} ∣}{J _{in}} = 1 - sech^{2} (2 ka) = tanh^{2} (2 ka) .

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