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This is problem 2.8.3 from Miller's Quantum Mechanics For Scientists And Engineers. I'm getting stuck when I try to figure out the wave equation on the right-hand side of the barrier. The original problem is:

Graph the (relative) probability density as a function of distance for an electron wave of energy 1.5 eV incident from the left on a barrier of height 1 eV. Continue your graph far enough in distance on both sides of the barrier to show the characteristic behavior of this probability density.

In the text, Miller solves a similar problem where the energy of the incident particle is less than the energy of the barrier. However, in this case, since the particle's energy is greater than the energy of the barrier, I'm using the same form of the wave function on either side of the barrier. Namely,

$$ \psi_{left}(x) = C \; exp(ik_{left}x)+D \; exp(-ik_{left}x) $$ $$ \psi_{right}(x) = G \; exp(ik_{right}x) $$

where the potential is zero for $x \lt 0$ and 1 eV for $x \ge 0.$ Substituting the above into Schrodinger's equation produces $k_{left}=\sqrt{\frac{2m(E-V_{left})}{\hbar^2}}$ and $k_{right}=\sqrt{\frac{2m(E-V_{right})}{\hbar^2}}.$ Using the boundary conditions that $ \psi_{left}(0)=\psi_{right}(0) $ and $\psi_{left}^\prime(0)=\psi_{right}^\prime(0),$ we obtain,

$$ C+D=G $$ $$ ik_{left}C-ik_{left}D=ik_{right}G. $$

From this, we can solve for $D$ and $G$ to obtain,

$$ D=\frac{k_{left}-k_{right}}{k_{left}+k_{right}}C $$ $$ G=\frac{2k_{left}}{k_{left}+k_{right}}C. $$

Substituting back into the wave functions gives,

$$ \psi_{left}(x) = C \; exp(ik_{left}x)+\frac{k_{left}-k_{right}}{k_{left}+k_{right}}C \; exp(-ik_{left}x) $$ $$ \psi_{right}(x) = \frac{2k_{left}}{k_{left}+k_{right}}C \; exp(ik_{right}x). $$

A this point, I feel like I've done something wrong, since $|\psi_{left}|^2$ is a sinusoidal standing wave, but $|\psi_{right}|^2$ is constant. Have I made a mistake? Will an incident electron with energy greater than that of an infinitely long potential barrier have a probability distribution that is sinusoidal before the barrier and constant after the barrier?

Also, although the original problem uses the word "relative", I'm still troubled by the fact that even in a potential well situation (which is different from the statement of the problem) the above solution would lead to a contradiction. Namely, a finite and constant probability distribution outside of the well would not tend to zero at plus infinity. Can someone either confirm this reasoning, or perhaps point out where I made a mistake?

Please note that I've tagged this question as homework, although it does not pertain to any actual homework assignment. I'm just reading on my own.

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2 Answers

I did not check you algebra very carefully. The question is of Interpretation of The constants C,D and G.

C is the amplitude of the incoming wave.

D is the amplitude of the reflected wave.

G is the amplitude of the transmitted wave.

This implies that is one sends $|C|^2$ particles as the incident beam, then one expects $|D|^2$ to be reflected and $|G|^2$ to be transmitted.

So $|C|^2$ = $|D|^2$ + $|G|^2$ so as to conserve the number of particles. Check this. To be certain of your algebra.

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Thanks, that's a good idea. After substituting in, I'm getting $k_{left}=6.2746\times10^{-9}\;m^{-1},$ $k_{right}=3.62264\times10^{-9}\;m^{-1},$ and $|D|^2+|G|^2=1.67949|C|^2,$ so I must have made a mistake somewhere... –  Andrew Oct 6 '12 at 21:28
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Your solution seems ok. You should compute the transmission coefficient $T$ and reflection coefficient $R$. It is the sum of transmission coefficient and reflection coefficient that give one. $$T + R = 1$$ I obtained $k_{left} = 6.2746×10^{9} \text{ m}^{-1}$ and $k_{right} = 3.62264×10^{9} \text { m}^{-1}$ from your solution. When I used them to compute sum of the transmission coefficient and reflection coefficient, it indeed give one. I think you are on the right track.

See a related post Potential step and its transmission / reflection

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