# Interpreting the probability distribution of the potential step

1. May 3, 2010

### Identity

Say you have a potential step problem where the potential steps up from $$V=0$$ to $$V=V_0$$ at $$x=0$$. If the incident particle has energy $$E <V_0$$, you get a non-normalisable solution for the wavefunction.

How can you interpret $$|\psi|^2$$ for this non-normalisable solution? Is it still the probability density or does it mean something else?

Thanks

2. May 3, 2010

### LostConjugate

You get two solutions, one is normalizable and one is not. We throw out the non-normalizable solution as it is not in Hilbert Space and is not a probability amplitude. It is curious why it is there, I think it stems back to our method of differential equations, rigorous mathematics may prove it is not a solution.

3. May 4, 2010

### Identity

I'm not sure if you understood what I meant, sorry

The general solution is of the form:

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

Hence,

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

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

For $$x>0$$ we have a constant, which is certainly not normalisable.

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

However, since they are NOT normalisable, how do you interpret them?

Last edited: May 4, 2010
4. May 4, 2010

### SpectraCat

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.

5. May 4, 2010

### LostConjugate

I don't see how you got those solutions for E less than V. For X > 0 you should have

Cexp[-sqrt(2m(V-E))x/h-bar]

I use C as a constant because the amplitude changes when the potential changes. Also the exponential is negative, the positive exponential is thrown out.

Also your incoming and reflected waves functions should not have the same amplitude.

6. May 4, 2010

### George Jones

Staff Emeritus

For $x>0$, $\psi \left( x \right)$ is okay if $k_2$ is imaginary, but $\left| \psi \left( x \right) \right|^2$ is incorrect because then $\left| \exp \left( i k_2 x \right) \right| \ne 1$.
I think they are okay. Note that the reflected wave has a constant multiplying $A$.

Last edited: May 4, 2010
7. May 4, 2010

Ahh..

8. May 4, 2010

### SpectraCat

Yes, you are right ... I had just assumed we were talking about E>V ... for which he gave the correct solutions.