Wavepacket incident on a potential step

[id00022]

Hello,
I am writing a Fortran 95 program to model the scattering of a wavepacket by a potential step of height V₀ at x=0. My wavepacket is formed by the superposition of numerous travellling waves of different k values. The wavepacket has the dispersion relation ω(k)=k². I want the wavepacket to be in its undispersed state at t=0 at a start position x₀. Therefore each component wave is composed of an incident wave, reflected wave, and transmitted wave.

At x≤0 :
ψ(x,t)=A(e^{i(k(x-x₀)-ωt)} +[itex]\frac{k-k'}{k+k'}[/itex]e^{-i(k(x+x₀)-ωt)})

At x>0 :
ψ(x,t)=A[itex]\frac{2k}{k+k'}[/itex]e^{i(k'(x-[itex]\frac{k}{k'}[/itex]x₀)-ωt)}

The factor of [itex]\frac{k}{k'}[/itex] in the bottom equation was found analytically to ensure continuity in the wavefunctions at the boundary. Well, at least that's what I thought: It works as long as E>V₀ otherwise there is discontinuity. Can anyone help me as to why? I don't think this is a Fortran programming problem, but more of a physics one...

 

[sweet springs]

Hello,
The factor of [itex]\frac{k}{k'}[/itex] in the bottom equation was found analytically to ensure continuity in the wavefunctions at the boundary. Well, at least that's what I thought: It works as long as E>V₀ otherwise there is discontinuity.

Exactly. Aren't you satisfied yet?

 

[id00022]

Exactly. Aren't you satisfied yet?

Well not really: when I work through the maths, this factor should work fine in all situations, but in practice it only works when E>V₀. If you have any idea why this might be then I would be very grateful. I'm fairly certain its not a problem with my code as I have applied it in many different ways all with the same result.

 

[sweet springs]

Hi

for E<V0, ψ for x>0 is no longer a plane wave but an exponentially dumping function as e^-Kx where k'=iK pure imaginary number.

 

[id00022]

Hi

for E<V0 [tex]\psi[/tex] for x>0 is no longer plane wave but exponentially dumping function as e^-Kx

Hi,
Thanks, but this is built into my program already: k' becomes complex reducing ψ=e^ik'x to e^-αx the dumping function you describe.

 

[sweet springs]

Hi.

In both of the cases, e^ik'x = e^-Kx = 1 for x=0. So continuity conditions should be the same in the both cases.
What is the discontinuity you are worrying about?

 

[id00022]

Because if I do not offset by x₀, then the wavepacket impacts the step at exactly t=0. Therefore to watch it scatter you must set the time range to be from -t to t. In the period -t to 0 the wavepacket starts off slightly dispersed, becoming less dispersed until it hits the barrier. This is unphysical so I have offset the position of the wavepacket at t=0 to x=x₀. Although e^ikx and e^-αx both equal 1 at x=0, e^ik(x-x₀) and e^-α(x-x₀) are not equal there. So the factor [itex]\frac{k}{k'}[/itex] has to be introduced as above. This works fine on paper but not in practice.

 

[sweet springs]

Hi.

Although e^ikx and e^-αx both equal 1 at x=0, e^ik(x-x₀) and e^-α(x-x₀) are not equal there.

The dumping depends on the distance from the wall, x-0, not the distance from x0, x-x0.
The wave function for x>0 is e^-αx+ikx₀.