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May4-10, 06:24 AM
Sci Advisor
P: 1,395
Interpreting the probability distribution of the potential step

Quote Quote by Identity View Post
I'm not sure if you understood what I meant, sorry

The general solution is of the form:

[tex]\psi(x) = \begin{cases} Ae^{ik_1x}+A\frac{k_1-k_2}{k_1+k_2}e^{-ik_1x} \ \ \ \ \ \ x < 0\\\\ A\frac{2k_1}{k_1+k_2}e^{ik_2x}\ \ \ \ \ \ x \geq 0\end{cases}[/tex]


[tex]|\psi(x)|^2 = \begin{cases} A^2\left[1+\left(\frac{k_1-k_2}{k_1+k_2}\right)^2+2\frac{k_1-k_2}{k_1+k_2}\cos{2k_1x}\right] \ \ \ \ \ \ x < 0 \\\\ A^2\frac{4k_1^2}{(k_1+k_2)^2}\ \ \ \ \ \ x \geq 0\end{cases}[/tex]

For [tex]x<0[/tex] we have a cosine wave which can't be normalised, since it has a vertical translation

For [tex]x>0[/tex] we have a constant, which is certainly not normalisable.

The are continuous and differentiable at [tex]x=0[/tex], and in all regions satisfy the schrodinger equation for the potential step

However, since they are NOT normalisable, how do you interpret them?
The solutions you are talking about are the plane wave solutions, and all of the problems you raise are also problems for a plane wave. Mathematically, plane waves are the momentum eigenstates, so you could interpret your results as showing the intereference between the incoming and reflected plane waves in the x<0 region, and as showing the transmitted plane wave in the x>0 region. However, there is a chicken and egg problem with this analysis, because it is time-independent and distributed over all space. Thus, as you say, it is a mathematical solution with little direct physical significance.

If you want to make your life a little harder, you can try solving the time-dependent version of this problems, where you start with a (normalized) wavepacket incident on the barrier from one side or the other. You can then propagate the wavepacket and see what happens when it encounters the barrier. The math is significantly more difficult, but the results are more physically significant. However, the qualitative insights about quantum phenomena are the same; the wavepacket splits into a reflected part and a transmitted part, and the reflected part interferes with the incoming wavepacket on its way back out (for a finite time).

Thus the static plane wave picture still gets you the quantum weirdness (barrier penentration for E<V and over-barrier reflection for E>V), but you have to work a lot less hard on the math. I think that is the main reason that the plane wave solutions for this 1-D problem (and many others), are taught in intro courses.