What is the best estimate for the penetration distance in quantum mechanics?

[sachi]

We have a particle of energy E crossing a potential jump at x=0. for x<=0, V=0, for x>=0 V=V1
We get a wavefunction for x>=0 psi(x) = exp(-iEt/hbar)*exp(-Kx)
where K = (2m(V1-E))^0.5/hbar
N.b E<V1 so classically we get no transmission

we are asked to estimate the penetration distance, and I have found a solution which says let the penetration distance equal 1/K. I can't see physically why we would pick this (it just seems like a random number that means that the wavefunction will decrease by a factor 1/e, but I can't see why this is a sensible estimate).

Thanks

 

[Physics Monkey]

Well, 1/\kappa is the only relevant length scale in the problem, so the penetration depth has to be proportional to it. The point is that if you were to plot the wavefunction as a function of \kappa x, it would look the same no matter what the energy or barrier height were. In other words, if \kappa x &lt; .1 nothing much happens and if \kappa x &gt; 10 the wavefunction is essentially zero. Clearly, \kappa determines the length scale over which the action happens. That being said, you have some freedom in your estimate. Maybe you would like to include a factor of \ln{2} so you get the place where the wavefunction is one half its value at the boundary. In some other problem, it might be nice to include a factor of \pi for convenience, for example. The convention is basically that anything within a power of ten of \kappa is pretty much ok, but this isn't any kind of formal rule and people tend to go with the simple estimate. All you are really doing is identifying the length scale.

 

[Jelfish]

Consider what the probability distribution looks like.

\Psi = e^{\frac{-iE}{\hbar}t}e^{-\kappa x}
P=\Psi^* \Psi = e^{-2\kappa x}

Remember that the second part is completely real since V1>E.

As Physics Monkey said, this is a question of length scale. No matter what threshold of probability you choose (say, .1%), you must scale it must be a constant times 1/kappa because kappa can vary depending on what problem you're doing.

Let x=\frac{d}{\kappa} be your chosen penetration distance, where d is just a constant.
P= e^{-2\kappa x}= e^{-2\kappa \frac{d}{\kappa}} = e^{-2d}

Then choose d according to how close you want the probability to be zero. Since this doesn't depend on kappa, it won't matter what E and V1 are in your problem.