While reading Griffiths Introduction to Quantum mechanics and using MIT 8.04 QP-1 lectures by Adam Allans as a supplementary source to understand the topic of scattering of particles for step potential, I came across a problem which Griffiths interestingly mentions for the Delta potential barrier but it's resolution isn't very clear to me.

When we do the analysis of a particle with energy $E$ greater than the step potential barrier $V$ i.e. $E > V$ and assuming the barrier is positioned at $x=0$, then the energy eigenfunctions on solving the energy eigenvalue equation (time-independent Schrodinger Equation) comes out to be $$ \psi(x)=Ae^{ik_1x} +Be^{-ik_1x} \qquad x<0\ \\ \psi(x)=Ce^{ik_2x} +De^{-ik_2x} \qquad x>0\ $$ Now since these solutions aren't normalizable, we should construct a wave packet of the form $$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} e^{ikx}\phi(k)dk \qquad $$ (not very sure about the limits of the wave number $k$) and then use these to satisfy the boundary conditions of continuity and differentiablility at $x=0$. But after looking at many sources, the authors mostly use these non-normalizable solutions for further analysis of transmission and reflection probablilities of the wave function.

It would be very nice if someone could provide a reasoning as to why these functions can be used and if not, then how would the wave packet be used the satisfy the boundary conditions at $x=0$.

  • 1
    $\begingroup$ This post (v2) seems to be essentially some version of the question Why can we treat quantum scattering problems as time-independent? $\endgroup$
    – Qmechanic
    Dec 9 '19 at 18:44
  • $\begingroup$ Well, the question is quite similiar but I didn't find any answer understandable for my level of knowledge. None of the answers explain how if we apply the boundary conditions to the incoming, reflected and transmitted wave superpositions at the step would still result in a similiar result to the analysis. In brief, I am stuck as to how to do the similiar excercise for normalizable solutions even if they give the same result. $\endgroup$
    – Tachyon209
    Dec 9 '19 at 20:27
  • $\begingroup$ Which normalizable solutions? You mean time-dependent wave-packets? $\endgroup$
    – Qmechanic
    Dec 9 '19 at 20:35
  • $\begingroup$ @Qmechanic Yes. So, everyone in their analysis uses the the forward and backward moving waves ( A times the exp + B times the exp) as the wave function to satisfy the bou dary conditions. What I feel is that we should use wave packets (time dependent or independent because we are just satisfying boundary conditions at any arbitrary time, say t=0). That's what I mean by using the normalizable solutions. $\endgroup$
    – Tachyon209
    Dec 9 '19 at 20:39
  • $\begingroup$ There are no normalizable time-independent solutions. $\endgroup$
    – Qmechanic
    Dec 9 '19 at 21:09

In the comments on the question I mentioned that the plane-wave solution "gets the right answers". The reason for this is related to the notion of time-independence.

Any experiment where you shoot a single particle at a barrier/scattering-site and then measure where the outgoing particle lands in intrinsically a time-dependent experiment and it corresponds to the wave-packet treatment of the problem.

But if we want to handle the problem in the context of the TISE, we need an experiment that is time-independent, and that means a continuous beam experiment. A class-room example is a diffraction experiment in which we point a laser through a slit/system-of-slits/etalon/etc and observe the intensity pattern that develops on a screen. Any such demonstration lasts for a Saganesque number of periods, and can be reasonably treated as eternal.

But in the case of a (ideally eternal) beam we don't expect the wavefunction to be normalized to one because there isn't just one particle. There are (in the ideallized version) an infinite number of beam particles, so having $$ \int_\text{space} \psi^* \psi \,\mathrm{d}x $$ diverge is not only OK but expected (in the sense that physicist-math just waves away annoying technical details). I suppose there ought to be some requirement that the infinity that comes from that sum remains constant (whatever that would mean).

For this to work we do rather need the individual beam particles to not interact with one another, but that's more or less automatic in the case of a low power laser.

  • $\begingroup$ I would not prefer to go into a complex situation like time independent or dependent situations and whether the probability would diverge or not in such a system. My question is much more simpler, why does this analysis with plane waves and with wave packets give the same answer? I hope you understand my point. $\endgroup$
    – Tachyon209
    Dec 10 '19 at 6:30
  • $\begingroup$ Do you understand the reason that the normalization requirement exists in the first place? $\endgroup$ Dec 10 '19 at 13:22
  • $\begingroup$ Well, to be honest, I don't exactly know the reason for normalization. I have read about the fact that the particle's gotta be somewhere and since $$ | \psi |^2 $$ is the probability density which implies that $$ \int_{-\infty}^{\infty} | \psi |^2 dx =1$$ This is what I know about the normalization for a single particle. $\endgroup$
    – Tachyon209
    Dec 10 '19 at 13:34
  • $\begingroup$ Right. Normalization is about enforcing the representation of a single particle or system. But if you say "I want to solve a version of the problem with an unbounded number of particles" (and you don't have to worry about inter-particle interactions, and there is a physical system that is reasonably modeled that way) you can drop that requirement and do the easier version of the math problem. You get the same probabilities because you are still solvong the same differential equation, you just get to use a simplier function. $\endgroup$ Dec 10 '19 at 16:16
  • $\begingroup$ So you wanna say that if we assume the system to be an arbitrary number of particles system, then we can drop the normalization condition and solve the differential equation for a wave that is "not of one particle" but " a many particle wave function"? Something on that lines, huh? $\endgroup$
    – Tachyon209
    Dec 10 '19 at 18:05

In this problem we don't have to consider any time dependence. It is as if the particle is present in the vicinity of the step potential and 'frozen' in time. Again, we cannot pin point the position of particle in the Energy versus Position graph since that will violate the Uncertainty principle and that's why we have to consider/solve for the entire range of space (i.e. to the left and right of the step potential) at the same instant of time. It is not like a motion picture of classical world where a particle travels from infinity, strikes the potential and flies off; where all the three events happen at different instants of time.

You can crudely picture this like there's some sort of a one way channel, one going to left and other to right with weight factors A and B. The particle is only allowed to make use of this pre-existing tracks. By no means we see a particle travelling in space as a function of time. When we solve for $|{\frac{B}{A}}|^2$ or $|\frac{C}{A}|^2$, we are determining the likelihood of particle ending up on the left or right of $x=0$ when we, say observe it.

You also asked about using a wave packet and then applying the boundary conditions. I think it may be a difficult task since a wave packet (particle) moving in space starts to decay the moment we introduce time in the equation. It is as if the wave tends to elongate at both the extremes making it difficult to normalize and apply boundary conditions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.