Interpreting the probability distribution of the potential stepby Identity Tags: distribution, interpreting, potential, probability, step 

#1
May310, 10:24 AM

P: 150

Say you have a potential step problem where the potential steps up from [tex]V=0[/tex] to [tex]V=V_0[/tex] at [tex]x=0[/tex]. If the incident particle has energy [tex]E <V_0[/tex], you get a nonnormalisable solution for the wavefunction.
How can you interpret [tex]\psi^2[/tex] for this nonnormalisable solution? Is it still the probability density or does it mean something else? Thanks 



#2
May310, 11:14 AM

P: 842

You get two solutions, one is normalizable and one is not. We throw out the nonnormalizable 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
May410, 04:16 AM

P: 150

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_1k_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] Hence, [tex]\psi(x)^2 = \begin{cases} A^2\left[1+\left(\frac{k_1k_2}{k_1+k_2}\right)^2+2\frac{k_1k_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? 



#4
May410, 06:24 AM

Sci Advisor
P: 1,395

Interpreting the probability distribution of the potential stepIf you want to make your life a little harder, you can try solving the timedependent 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 overbarrier 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 1D problem (and many others), are taught in intro courses. 



#5
May410, 10:31 AM

P: 842

I don't see how you got those solutions for E less than V. For X > 0 you should have
Cexp[sqrt(2m(VE))x/hbar] 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
May410, 11:45 AM

Mentor
P: 6,037

For [itex]x>0[/itex], [itex]\psi \left( x \right)[/itex] is okay if [itex]k_2[/itex] is imaginary, but [itex]\left \psi \left( x \right) \right^2[/itex] is incorrect because then [itex]\left \exp \left( i k_2 x \right) \right \ne 1[/itex]. 



#7
May410, 11:53 AM

P: 842





#8
May410, 01:27 PM

Sci Advisor
P: 1,395




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