One-dimensional scattering

One-dimensional scattering#

So far, beyond the free particle, we have focused on bound state problems. However, an important class of experiments we need to address are scattering experiments, in which we probe the structure of some system – a metal, an atomic nucleus, and so forth – by throwing another particle at it and seeing what comes out.

Here we will consider an extremely simplistic setup in which the background probed is nondynamical (so scattering is elastic) and represented by a fixed potential of finite range. More specifically we will focus on a finite, rectangular potential barrier:

V(x)={0x<0V0>00xa0x>

The question we want to ask is the following. Consider a particle with fixed energy E which is approaching the barrier from large negative x. Its initial energy is kinetic, E=12mv2. What happens after it hits the barrier?

Classical scattering#

Classically, this is a well defined question. For 0<E<V0, the particle will reflect off of the barrier and travel towards x=.

For E>V0, the particle will pass over the barrier, although it will slow down as it passes over the top of the barrier. This latter phenomenon will lead to a time delay for the particle to reach a point past the barrier, relative to the time it would take absent any such potential. We can work out this time delay in the latter case. If the initial velocity is v0=2E/m it would take time t0=a/v=am2E to cross the barrier. For a barrier of height V0E, the velocity above the barrier is v1=2(EV0)m, so the time to cross the top is t1=am2(EV0). The time difference is thus

δt=t1t0=av1av0=ma22[1EV01E]

Note that if V0<0 the same basic story applies and the above formulae apply, but the particle always passes through the region of nonzero potential, time delay will be negative.

For E=V0 exactly, the velocity will become zero as soon as the particle reaches the barrier, and it will stick to the left edge at x=0.

Quantum scattering#

Quantum-mechanically the story is different in interesting ways!

The first thing we should note is that this problem (as all scattering problems) has a different character than the bound state problems we have been describing. Since the potential is short range, far from the potential barrier the particle is just the free particle. There is a state for every energy E>0 (once we properly specify the boundary conditions). The nontrivial part of the problem is specifying the boundary conditions, and finding the wavfunction as a function of the energy and those boundary conditions.

Ideally one would prepare a localized wave packet and study its evolution. In general this will both propagate and spread and the interaction with the potential barrier will be somewhat complicated. Instead we will consider the following idealization. To the left or the right of the barrier, the solutions are of the form e±ikx where E=2k22m. Recall that the probability ρdx=|ψ(x)|2dx that a single particle is found in an interval of length dx about position x, where ρ is the probability density. Recall from Conservation of probability that this density satisfies a conservation law tρ+xJ where

(307)#J=2im(ψxψxψψ)

If the wavefunction is Ae±ikx, then J=|A|2km. Here k is the momentum of the particle, k/m behaves as a velocity. Nowfor |ψ|2 to be a good probability density, A has units of (length)1/2, so |A|2v has units of inverse time. We can think of it as the probability per unit time that the particle crosses x from left to right (if k and thus v is positive).

More generally, we can imagine sending a flux of particles, all prepared in the same state. Then in the limit that the flux is large, ρ becomes to good approximation the particle density, and J the number of particles per second crossing x.

We assume that any flux of particles coming from x=pm must be prescribed. We we will consider an experiment for which the particles are injected from x=. They can then pass over or be scattered from the potential barrier and travel to x=±. I claim that a good model for this is to consider wavefunctions for x<0,x>a which take the form

(308)#ψ(x)={Ainceikx+Arefleikxx<0Atranseikxx>a

where k=12mE. We will choose a normalization for the wavefunctions such that Jinc=|Ainc|2km is the incoming flux of particles. Jrefl=|Arefl|2km is then the flux of particles that are reflected from the barrier, and Jtrans=|Atrans|2km is the flux of particles which are transmitted over or through the barrier. We impose as a boundary condition that the wavfunction has no combinent with momentum k for x>a/2, meaning we are not injecting particles from x. The other boundary condition is the prescription of Ainc. These completely specify the solution to the time-independent Schroedinger equation. Arefl and Atrans (and the associated fluxes) are completely determined from Ainc (and its associated flux).

Because we are looking at stationary states – Eq. scattering_psi is an energy eigenstate – ρ is time-independent. We can integrate xJ over the barrier; it is a total derivative so that the flow on one side of the barrier is equal to the flux on the other side of the barrier: that is,

|Ainc|2|Arefl|2=|Atrans|2

The detailed formulae arise from matching the above to the wavefunction in the barrier region. This is well described in many textbooks and we won’t dig into the details. We take our notation (and the next figure) from [Schiff, 1955]. The result for general E is

AreflAinc=(k2α2)(1e2iaα)(k+α)2(kα)2e2iaαAtransAinc=4kαei(αk)a(k+α)2(kα)2e2iaα

where α=2m(EV0)2. Note that this formula holds for E<V0; we can then write α=iκ.

To get a sense of what is happening, let us plot the transmission coefficient, that is, the ratio of transmitted to incident flux. This is

(309)#T=|Atrans|2|Ainc|2=11+V02sin2αa4E(EV0)

See the plot below, for mV02a22=8.

Transmission coefficient

For E>V0, note that the transmission becomes perfect whenever ααa=0, or αa=nπ; this occurs when the barrier is an integer number of half-wavelengths. Note that at V=E0, αsinEV0, so that sin2αa/(EV0) goes to a constant. The full coefficient is:

T(E=V0)=11+mV0a222

which is nonzero, unlike the classical case.

Note further that the transmission coefficient is nonzero even when E<V0. Here sinαa=isinhκa; the (i)2 term in trans_coeff cancels the change of sign in EV0. The result is:

T(E<V0)=11+V02sinh2κa4E(V0E)

To build some intuition, take κa1, which can happen when a or V0 gets large at fixed E. In this case sinhκa12eκa becomes exponentially large and dominates the denominator in T. The resulting expression is

T16E(V0E)V02e2κa

this exponential suppression with the width (and square root of the height) of the barrier is characteristic of tunneling/below-barrier transmission. Note that κ=2m(V0E) so the exponential has a non-analytic dependence on if we scale 0.