This is probably a silly question that has no physical meaning, but I want to be sure:

Assume we have the potential barrier $$ V(x) = \begin{cases} V_0 & x>0 \\ 0 & x<0 \end{cases} $$ Now assume the particle is coming from the right. Is that possible? Physically I think not, but mathematically what is making it not possible?

Although it seemed wrong to continue, I wrote down the equations and tried to solve them:

$$x<0: \ \ \phi_1(x)=Ae^{ik_1x}+Be^{-ik_1x}, k_1=\sqrt{\frac{2mE}{\hbar^2}}$$ $$x>0: \ \ \phi_2(x)=Ce^{k_2x}+De^{-k_2x}, k_2=\sqrt{\frac{2m(V_0-E)}{\hbar^2}}$$
From physical limits, $A=0$ and $C=0$.

And from boundary conditions, $B=D$ and $ik_1=k_2$ (an equation that I think supports the notion that the whole thing is garbage).

My question is, are these really meaningless, non-physical results? Or am I wrong, and there is a physically meaningful interpretation of the results?

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    $\begingroup$ Is the energy of the particle more or less than V0? $\endgroup$ – The Photon Feb 23 '18 at 4:05
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    $\begingroup$ For the model, it doesn't really matter, if the energy is more or less than V0. The particle could fly over the barrier or tunnel out of it form whichever source in the barrier. The source of the particle is simply beyond this model. There's nothing wrong with the math here. Whether you can construct a realistic situation, where this model applies, is a totally different question. $\endgroup$ – engineer Feb 23 '18 at 4:32
  • $\begingroup$ @ThePhoton, the solution I made was for $0<E<V_0$ $\endgroup$ – Mr.OY Feb 23 '18 at 10:52
  • $\begingroup$ @engineer, if that is the case, can you explain (if there is explanation for unrealistic situation such this) the result I got. Maybe in terms of probability. $\endgroup$ – Mr.OY Feb 23 '18 at 10:56

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