First consider the free particle, which has Hamiltonian $H=\frac{1}{2m}P^2 = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$. The "eigenstates" of this Hamiltonian are the non-normalizable plane waves of the form $\phi_k(x) = e^{ikx}$, which possess energy $E_k=\frac{\hbar^2 k^2}{2m}$ and momentum $p_k=\hbar k$. Let's also define the frequency $\omega_k \equiv E_k/\hbar$.
Because these states are not normalizable, they are not allowed quantum states. However, we can combine them in an integral to obtain a physical state of the form
$$\psi(x) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty A(k)\phi_k (x)dk= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty A(k) e^{ikx} dk$$
this can be inverted to give $A(k)$ in terms of $\psi(x)$:
$$A(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \psi(x)e^{-ikx} dx$$
This is called a wave packet in scattering theory. The time evolution of $\psi$ is given by applying the operator $e^{iHt/\hbar}$. Since $H\phi_k=\hbar\omega_k \phi_k$, it follows that $e^{-iHt/\hbar}\phi_k = e^{-i\omega_k t}\phi_k$, and so the result is
$$\psi(x,t) = e^{-iHt/\hbar}\psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty A(k) e^{i(kx-\omega_k t)}$$
So the recipe here goes as follows:
- First, we obtain the (non-normalizable) eigenstates of the Hamiltonian.
- Next, we write an arbitrary solution as a superposition of these eigenstates.
- From there, we can obtain $A(k)$ by inverting the superposition in the last step
- Finally, we can calculate the time evolution of the system by appending $e^{-i\omega_k t}$ to each plane wave inside the integral.
The procedure for the step potential is more interesting because the eigenstates are not given by $\phi_k(x) = e^{ikx}$, but by the more complicated expression
$$\phi_k(x) = \cases{e^{ikx} + R_k e^{-ikx} & $x<0$\\T_ke^{iq_k x} & $x>0$} $$
where $R_k$ is the reflected amplitude, $T_k$ is the transmitted amplitude, and $q_k \equiv \sqrt{\frac{2m(E-V_0)}{\hbar^2}}$. In writing this, I have already assumed that there is no wave incident on the step from the right, and so $k\geq 0$.
Despite the added complexity, the procedure remains the same. A generic state can be written
$$\psi(x) = \frac{1}{\sqrt{2\pi}}\int_{0}^\infty A(k) \phi_k(x) dk $$
$$= \cases{\left(\frac{1}{\sqrt{2\pi}}\int A(k) e^{ikx} dk\right) +R_k\left(\frac{1}{\sqrt{2\pi}} \int A(k)e^{-ikx} dk\right) & $x<0$ \\
T_k\left(\frac{1}{\sqrt{2\pi}}\int A(k) e^{iq_k x}dk \right)& $x>0$}$$
This can be inverted as before, though you have to get a bit creative to do so. Once you've obtained your $A(k)$, then you can push this wavefunction forward in time as before:
$$\psi(x,t) = \cases{\left(\frac{1}{\sqrt{2\pi}}\int A(k) e^{i(kx-\omega_kt)} dk\right) +R_k\left(\frac{1}{\sqrt{2\pi}} \int A(k)e^{-i(kx+\omega_kt)} dk\right) & $x<0$ \\
T_k\left(\frac{1}{\sqrt{2\pi}}\int A(k) e^{i(q_k x-\omega_kt)}dk \right)& $x>0$}$$
There's a reason this is not usually covered in a first pass through scattering theory - even in the simplest cases, it's a mess. If you assume that $A(k)$ is sharply peaked around some $k_0$, then you can make some reasonable approximations and end up with a not-so-horrible result - I've done so, and plotted the result below as a GIF (you may need to click on it, depending on your browser).
The other reason this level of detail is rarely covered in a first course is that it simply isn't necessary for quite a few problems of interest. In particular, finding the non-normalizable "eigenstates" $\phi_k(x)$ is enough to tell you how much reflection and how much transmission you can expect at a given $k$. In higher dimensions, these eigenstates tells you what how much scattering to expect at various scattering angles.
All of this information holds for (unphysical) particles with exact momentum $k$, but it extends approximately to particles with momentum which is peaked at $k$, and often times this is all the information you need. The only thing I gained from the work I did, for example, is a nice little movie of a bump moving along a line.
