# States with defined energy and time evolution

Consider the following simple problem: We have a step potential: $$V=V_0\Theta (x)$$ so the Hamiltonian is: $$H=\frac{p^2}{2m}+V_0\Theta(x)$$ and we want to find the eigenfunctions of the Hamiltonian $$\psi _E$$, so the states with defined energy for our system. This is easy enough to understand and solve.

Since $$H$$ is not a function of $$t$$ the time evolution operator $$S(t,t_0)$$ acting on an eigenfunction of $$H$$ with energy $$E$$ has the following structure: $$S(t,t_0)=\exp{\left[\frac{1}{i\hbar}E(t-t_0)\right]}$$ Wonderful! Now we can use the time independent Schrodinger equation: $$H|\psi _E\rangle = E |\psi _E \rangle \ \Rightarrow \ \langle x |H|\psi _E\rangle = E \psi _E (x)$$ and so we get two different differential equations: one for $$x<0$$ and one for $$x \geq 0$$. Form here, with some calculation, we get plane waves on both sides if $$E>V_0$$ and plane waves on the left side and decreasing exponential in the right side if $$E. At last if we like we can specify the time evolution, given by $$S$$, to write $$\psi(x,t)$$ instead of $$\psi(x)$$: $$\psi _E(x,t)=\psi _E(x)\exp{\left[\frac{1}{i\hbar}E(t-t_0)\right]}$$

Wonderful. I have no problem with this.

But now suppose that our objective changes: we now want to find, for the same system, what happens when a plane wave (so a particle with wave function that is a plane wave) coming from the left side impacts our system.

I feel like I don't know how to solve this problem: this problem seems like a scattering problem, or a time evolution problem where the function is not an eigenfunction of the system.

But I have seen that the way to solve this problem is practically identical to the way we solved the first question: to solve this last question we simply find the eigenfunction of the system, as if the objective was to find them, and then, in the case of $$E>V_0$$, we simply remove, the plane wave on the right side with $$k<0$$, and this should represent the fact that the wave comes from the left side.

I know how to solve this last question, but I don't understand why this procedure work! For example: why the particle coming from the left, after impacting the barrier, stays into an eigenstate with defined energy? What theorem ensures this? Couldn't the impact put the particle into a superposition of energy? Remember that the particle has wavefunction that is eigenfunction of $$H$$ on the left side, but not in all space.

Even worse: the procedure that I cited seem to work almost for all potentials: step potential, well potential, Dirac's Delta potential, and so on. So there must be something essential about quantum particles "dynamics" that I am missing here.

TL;DR: What do the eigenfunction of $$H$$ have to do with a scattering problem for a plane wave?

(Bonus question: what if the particle impacting from the left side has wavefunction that is not a plane wave? What if, for example, we have a gaussian wave packet coming from the left and impacting our potential?)

• Any function can be written as a superposition of eigenfunctions. Therefore, if you know eigenfunctions with proper asymptotic you know what happens to an arbitrary function Commented Jun 3, 2021 at 16:36
• "the procedure that I cited seem to work almost for all potentials [...]" - all the potentials you list are very specific potentials - they are constant and only jump discontinuously to other constant values. Those certainly are not "all potentials". Have you tried this with e.g. a well with smooth sides? Commented Jun 3, 2021 at 17:44
• @ACuriousMind You are right, intuitively I get that maybe it will not work for other kinds of potentials, but why exactly? Can you expand on this in an answer? Commented Jun 3, 2021 at 19:22
• Actually, after reading your question again, I'm not sure what procedure exactly you mean in the paragraph starting with "But I have seen". Can you give a reference where this "removal" of one part of the plane waves is claimed to answer the question of what happens when a plane wave moves into the system from one side? What does it even mean for a plane wave to "come from the left"? Is the initial wavefunction supposed to be a plane wave at the left and zero after some point? Commented Jun 3, 2021 at 19:37
• Mathematically correctly posed problem includes not only the equation, but also the boundary conditions. Thus eigenvalue problem and scattering problem are not the same: the boundary conditions are different. Most QM textbooks containa chapter on scattering problems, but much after they deal with the potential step. You may want to check this answer as well: physics.stackexchange.com/a/597816/247642 Commented Jun 4, 2021 at 9:59

In $$1D$$, when you are looking for the stationary states (ie the eigenstates of the Hamiltonian), you are trying to solve the second order linear differential equation : $$\frac{-\hbar ^2}{2m}\frac{\text d^2\psi}{\text dx^2}+ V(x)\psi(x) = E\psi(x)$$ (and to find solutions which can be properly normalized).
If the potential function $$V(x)$$ satisfies : $$V(x) = \left\{ \begin{array}{cl} V_- & \text{for } x\text{ near } - \infty \\ V_+ & \text{for } x \text{ near } + \infty \end{array}\right.$$
and you are looking for solutions with $$E\geq \max (V_-, V_+)$$, then you see that the solution is given, near $$\pm\infty$$ by : $$\psi(x) = A_\pm e^{ik_\pm x} + B_{\pm}e^{-ik_\pm x}$$ where : $$k_{\pm} = \sqrt{\frac{2m(E - V_\pm)}{\hbar^2}}$$
More precisely, what this means is that, by setting the constant $$A_-$$ and $$B_-$$ (ie the form of the solution near $$-\infty$$), you can extend it up to $$+\infty$$ (as this is a second order linear differential equation) where it will have also a simple expression. Put in another way, of the $$4$$ constants $$A_\pm$$, $$B_\pm$$, only two are free parameters for the solution.
One of this state corresponds physically to a situation where a plane wave with amplitude $$A_-$$ and wavevector $$k_-$$ is incident from $$-\infty$$ and a plane wave with amplitude $$B_+$$ and wavevector $$-k_+$$ is incident from $$+\infty$$. If you want to find the state which corresponds to a plane wave with amplitude $$1$$ incident from $$-\infty$$ (implying that no wave is incident from $$+\infty$$), you have to find the solution with $$A_- = 1$$ and $$B_+ = 0$$.