Step Potential Transmission Probability as t->inf

[elanier]

At t=0 we have a particle incident from the left on a step potential (where V(x) = V₀ for x ≥ 0 and V(x) = 0 otherwise). The particle has energy of 5/4 V₀ and the question asks to find the probability that it will be on the right (ie. x > 0) as t→∞.

I understand how to solve this problem and that I'm looking for the transmission coefficient T, but what's throwing me off is the time evolution thing. I know that the wavefunction can be evolved by simply multiplying by e^-iEt/ħ (ie. ψ(x,t) = e^-iEt/ħφ₀(x) where φ₀(x) was found using the TDSE and relevant boundary conditions).

But...does this even matter when it comes to finding T? T depends on the wavenumber from either side of the barrier (k and q, if it were), which are related solely to the position part of the wavefunction, not the time part. Am I right in this? Since the potential isn't changing, wouldn't T be the same for all time?

Thanks in advance for any help! This has been bugging me all day and none of my books explicitly address this in their treatment of the step potential.

 

[mark726]

Hello,

You are correct that the time evolution of the wavefunction does not affect the transmission coefficient in this case. The transmission coefficient, as you mentioned, depends on the wavenumbers from either side of the barrier, which are determined by the position part of the wavefunction. Since the potential is not changing, the transmission coefficient will remain the same for all time.

However, it is still important to consider the time evolution of the wavefunction in order to find the probability of the particle being on the right side of the barrier as t→∞. This can be done by finding the time-independent solution (φ0(x)) and then multiplying it by the time evolution factor e-iEt/ħ. This will give you the time-dependent wavefunction, which can then be used to find the probability of the particle being on the right side of the barrier at any given time.

I hope this helps clarify things for you. Let me know if you have any other questions. Good luck with your problem!