5

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 $x=0$.

Qmechanic
  • 220,844
Tachyon209
  • 813

2 Answers2

4

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.

dmckee --- ex-moderator kitten
  • 87,238
1

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.

Sarthak Hajirnis
  • 11