I have a bit of confusion in regards of interpreting the solutions of the eigenfunctions. Having the potential,

$$ U(x) = \begin{cases} 0 & x < 0 \\ - |U_{0}| & x\geq 0 \end{cases} $$

This is pretty easy to visualize. For $x<0$, i'll call this region I, and for the other, Region II.

So by looking at the case for $E>0$, assume we are coming from the left side $x<0$

For both Regions I and II, we will have traveling waves in the form:

$$ \psi(x) = \begin{cases} Ae^{ikx} + Be^{-ikx} & x < 0 \\ Ce^{ilx} + D e^{-ilx} & x\geq 0 \end{cases} $$

where the values of $k=\frac{\sqrt{2mE}}{\hbar}$ and $l=\frac{\sqrt{2m(E+|U_{0}|)}}{\hbar}$ are solved from Schrodingers equation with boundary conditions.

So by visualizing the traveling wave coming from the left, I have some trouble knowing if it transmits back, and for which coefficients are set to zero (if any are).

After that, I want to look at the $-|U_{0}| < E < 0 $ where the wave travels from $x>0$

I believe the solutions are this (not sure):

$$ \psi(x) = \begin{cases} Ce^{nx} + De^{-nx} & x < 0 \\ Ae^{imx} + Be^{-imx} & x\geq 0 \end{cases} $$

In which $n=\frac{\sqrt{2mE}}{\hbar}$ and $m=\frac{\sqrt{2m(|U_{0}|-E)}}{\hbar}$.

I believe that wave traveling to the left for $x<0$ is bounded, so the wave does not transmit back. Thus $D=0$.

Am i interpreting case two correctly? And how do I go about case one?


1 Answer 1


For case $1$, since the particle is coming from the left ($x<0$), D is set to $0$, since $De^{ilx}$ is a term that says "there is a wave that is coming in from the left ($x<0$) with a wave number $l$ with an amplitude D"; whereas we know that the wave coming from the left is actually $Ae^{ikx}$. There will definitely be some reflection (stick to the word 'Reflection' instead of 'Transmits back' because transmission and reflection are two different things and you dont want to get them mixed up!)To sum it up, since wave is coming from the left ($Ae^{ikx}$), it sees a potential well and a huge part of it gets transmitted ($Ce^{ilx}$) and a small part gets reflected back ($B^{-ikx}$....quantum effects)

For the second case, since the wave is coming in from the right (ie the wave is coming from positive infinity), C is set to $0$ for the same reason as above......... [wave comes in from the right ($Be^{-imx}$), part of it gets transmitted and eventually dies out ($De^{-nx}$) and the rest of it gets reflected ($Ae^{imx}$)]

As for the boundedness of the second case, it is not fully bounded in the sense that the wavefunction is not contained inside 2 well defined boundaries. It is bounded from the left at $x=0$ but thats it. The other boundary if you want to call it one is at $x\rightarrow\nsim$

  • $\begingroup$ So for the first case, If my wave is traveling from the left, in some sense, how can the wave reflect back if their is no component for the wave to hit? If that makes sense. Wouldn't B = 0 as well as D=0 because E>0? $\endgroup$
    – iron2man
    Oct 19, 2016 at 2:27
  • $\begingroup$ Apologies, you might've misinterpreted the first case. The wave is coming from the left ($x<0$). Second case is when the particle comes from the right. $\endgroup$
    – iron2man
    Oct 19, 2016 at 2:28
  • $\begingroup$ Thats the thing about quantum mechanics. Even though there is nothing for the wave to 'hit' classically, it will still have some finite probability of reflecting back!! So B will not equal $0$ $\endgroup$ Oct 19, 2016 at 2:29
  • $\begingroup$ Sorry for the unnecessary confusion. $\endgroup$ Oct 19, 2016 at 2:33

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