A1  ^?? Show that in the well the wavefunction $\Psi_\text{in}$ is
\[ \Psi_\text{in} = A e^{i q x} + B e^{-iqx}. \]Express $q$ in terms of $k$, $m$ and $U_0$.

Substituting the wave function $Ae^{\pm iqx}$ into the Schrödinger equation gives the relation
\[ \frac{\hbar^2 q^2}{2ma} -U_0 = E = \frac{\hbar^2 k^2}{2ma},\]which is satisfied by $A e^{\pm iqx}$.

A2  ^?? Suggest the problem about wave optics, which is completely analogous to the one currently being considered.

We found that the wave vector inside the well is greater than outside, which corresponds to the refractive index
\[ n= \sqrt{1+ \frac{2maU_0}{\hbar^2k^2}}.\]Then the problem under consideration is equivalent to the problem of interference in a thin film with thickness $a$ and refractive index $n$, i.e., a Fabry-Perot resonator.

A3  ^?? Find $|t|^2$ — the probability that the particle transmits through the well.

Using an analogy with wave optics, we can solve the problem by considering the refraction and reflection of de Broglie waves at the boundaries of the interface (as is done for a Fabry-Perot resonator).

Instead, we will solve the problem straightforwardly by writing down a system of equations based on the boundary conditions:
\[\Psi_-(0) = \Psi_\mathrm{in}(0), \quad \Psi_-‘(0) = \Psi_\mathrm{in}’(0)\]
\[\Psi_+(a) = \Psi_\mathrm{in}(a), \quad \Psi_+(a) = \Psi_\mathrm{in}'(a)\]
In terms of $r$, $t$, $A$ and $B$, we get the following system:
\[\begin{cases} 1+r = A+B \\ k(-1+r) = q(A-B) \\ te^{-ika} = A e^{iqa} + Be^{-iqa} \\ -k te^{ika} = q(Ae^ {iqa} - B e^{-iqa}) \end{cases}\]

It can be solved as follows

1. Eliminate $r$ from the first two equations to obtain the relation \[2=A(1-q/k) + B(1+q/k)\]
2. Eliminate $t$ from the last two equations and obtain the relation \[Ae^{iqa}(1+q/k)+Be^{-iqa}(1-q/k) =0\]
3. We find \[A=\frac{2}{(1-q/k) - e^{2iqa} \frac{(1+q/k)^2}{1-q/k}}\]
4. Let us express $t$ in terms of $A$ and obtain \[t = -\frac{4qk}{(k-q)^2 - (k+q)^2 e^{2iqa}} e^{-ika}\]

Let's calculate $|t|^2$:

\[\begin{split}|t|^2 = t t^* = \frac{16k^2 q^2}{(k-q)^4 + (k+q)^4 - 2(k-q)^2(k+q)^2\cos 2qa } =\\= \frac{4k^2q^2}{4k^2q^2+ (k^2-q^2)^2 \sin^2 qa }\end{split}\]

A4  ^?? Sketch the graph of $|t|^2$ vs $k$.

To scetch the graph, first note that when $qa=\pi n$, the particle completely passes through the well. When $qa = \pi/2 + \pi n$,
\[|t|^2_\text{min} = \frac{4}{4+\frac{(k^2-q^2)^2}{k^2q^2}}=\frac{4}{4+\frac{C}{k^2(k^2+C)}},\]where $C=2maU_0/\hbar^2$. Thus, $|t|^2$ from $k$ is an oscillating function enclosed between the envelope $|t|^2_\text{min}$ and unity.