Few doubts in the solutions of step potential in Quantum mechanics Assume we have a step potential
$$
V(x)=\left\{\begin{array}{ll}
0, & x<0 \\
V_{0}, & x \geq 0
\end{array}\right.
$$ and we fire particles from a distance $s$ towards the barrier from the left. Using the Schrödinger equation we get the solutions in the two regions as:
\begin{align}
\varphi_{1} & =A e^{i p_{1} x / \hbar}+B e^{-i p_{1} x / \hbar}
\\
\text{for} \quad 
p_{1} & =\sqrt{2 m E},
\\
\text{and} \quad 
\varphi_{2} & =c e^{i p_{2} \frac{x}{\hbar}}+D e^{-i p_{2} x}
\\
\text{for} \quad 
p_{2} & =\sqrt{2 m\left(E-V_{0}\right)}
.
\end{align}
My questions:
It's said that the term $B e^{-i p_{1} x / \hbar}$ represents a wave going to the left and is related to the particles being reflected from the barrier.

*

*How can we say that this wave is related to the particles being reflected from the barrier? Or how the presence of this wave means that particles are traveling back after being reflected? Isn't it a logical leap?


*Now if this is so then let's say I fire particles light years away from the barrier but from the above solution I'll find a wave going back from where I fired them since the equations remain the same! But this violates the speed-of-light limit.
 A: 

*

*How can we say that this wave is related to the particles being reflected from the barrier? Or the presence of this wave means that particles are traveling back after being reflected?


The latter. There is a nonzero probability of reflection, and the amplitude of this reflected wave is given by this term.



*Now if this is so then let's say I fire particles light years away from the barrier but from the above solution I'll find a wave going back from where I fired them since the equations remain the same! But this violates the speed-of-light limit.


In this formalism you are solving the time-independent Schrödinger equation, which implies that you are only solving for static configurations, and any transient behaviour has died out. If your particle source is far off at $x<0$, then that transient behaviour will take correspondingly longer to die out $-$ by which time the speed-of-light constraint is long gone.
If you want to model a particle source 'firing' individual particles from far away in this way, then you need to consider the time-dependent Schrödinger equation, with an initial condition $\psi(x,0)$ equal to a wavepacket localized at large negative $x$ and moving to the right. This wavepacket can be decomposed as a superposition of the eigenstates that you've found already, but then the Schrödinger equation ensures that, until the bulk of the population actually hits the step, the reflected-wave components cancel out.
A: *

*It is an incomplete analogy for pedagogical purposes. That not saying it is not useful. As your notation already suggests, if you were to compute the momentum of each term in $\varphi_1$ you will either get $+p_1$ or $-p_1$ so the wave vector of the wave or alternatively its momentum is pointing forwards or backwards. Remember that in the end you are looking to solve the time-independent Schrödinger equation, so no statements about time can be made directly.


*This fails because of the classical idea of trajectories. You are still thinking in classical terms where particles are points and follow a route. But here one uses an idealization, namely a plane wave $e^{ipx}$ which is everywhere in space. So there are actually no trajectories in this problem. Nobody is actually shooting things.
P.D. If you want to consider shooting things you have to start putting time into the problem and study perhaps scattering theory. But that is the next step in complexity probably.
A: You have only given solutions for particles with a precise initial momentum $p_1$. These have a wave function spread over all space, contradicting your notion that you can

fire particles light years away from the barrier.

If you want to do that, you must formulate an initial wave function for a wave packet which starts light years away from the barrier, using the FT. Then you will use the inverse FT to plot the evolution of the wave packet. In this case you will find that the wave packet travels until it hits the barrier, and it then splits into a transmitted wave and a reflected wave, in accordance with the description.
In practice, when we talk of the physical realisation plane waves, or particles with definite momentum, we are really talking of wave packets with a narrow range of momenta. It would be tedious to make this explicit in every description, and you will find that descriptions leave the analysis of wave packets implicit. It makes little conceptual change, and the writer expects you to fill in the detail yourself.
A: 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.
