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In an exercise, I consider a particle moving from $x=-\infty$ towards a potential step, where $V(x)=0$ for $x\leq 0$ and $V(x)=V_0$ for $x>0$.

If we consider the case of $0<E<V_0$, we have;

$$\psi_1(x)=\begin{cases} A \text{e}^{ikx} + B \text{e}^{-ikx} & x<0, k=\frac{\sqrt{2mE}}{\hbar} \\ F\text{e}^{-\kappa x} & x>0, \kappa= \frac{2m(V_0-E)}{\hbar} \end{cases} $$

This makes sense because we have an evanescent wave when the particle approaches the potential step for $0<E<V_0$. Now if $E>V_0$, we have;

$$\psi_1(x)=\begin{cases} A \text{e}^{ikx} + B \text{e}^{-ikx} & x<0, k=\frac{\sqrt{2mE}}{\hbar} \\ F\text{e}^{i q x} & x>0, q= \frac{2m(E-V_0)}{\hbar} \end{cases} $$

In this case the energy of the particle is larger than the potential well for both $x>0$ and $x<0$, so my question is; Why is the wavefunction $F\text{e}^{iqx}$ and not $F\text{e}^{iqx}+G\text{e}^{-iqx}$ in the case $x>0$? Why do we 'exclude' the part '$G\text{e}^{-iqx}$' - do we not have have the same superposition of waves for $x>0$ as we did for the case $x<0$, of course with different eigenenergies?

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In principle we have always two linearly-independent solutions, since we deal with the second order differential equations (in one dimension). There are various ways to choose the two solutions, the most common of which is by considering the wave flux propagating from left-to-right and from right-to-left. See this answer for more details (the question is technically not a duplicate, but the core issue is the same. See also Deriving Unitarity of S -matrix in 1D Quantum Mechanics.

Remarks:

  • For $E<V_0$ we could have also obtained two linearly independent solutions, but there one of the solutions (growing in the infinity) is dropped for physical considerations.
  • To spell it more explicitly: if we proceed in a purely mathematical way, we write the solution as $$\psi(x)=\begin{cases} A \text{e}^{ikx} + B \text{e}^{-ikx} \text{ for } x<0,\\ F\text{e}^{i q x} + G^{-iqx} \text{ for } x>0. \end{cases} $$ This solution contains four constants: $A,B,F,G$. The boundary conditions for the continuity of the wave function and its derivative at $x=0$ fix the two of these constants (readily allowing derivation of the scattering matrix.) The remaining two constants are the amplitudes of the two linearly independent solutions added together. A possible choice is taking one solution with $G=0$ and the other with $F=0$ which correspond to the wave. In this case the flux in the region $x>0$ will be correspondingly $+\frac{\hbar q}{m}|F|^2$ or $-\frac{\hbar q}{m}|G|^2$ - i.e., flowing to the right/left. The usual interpretation is that the wave is incident from the left with amplitude $A$, so that $B$ and $F$ are the amplitudes of the reflected and the transmitted waves. This interpretation can be somewhat justified by adiabatic switching of the potential in more advanced treatments of the scattering matrix, but it is ultimately a choice of convenience. Following this interpretation one also often considers solution with the wave incident from the right with amplitude $G$ and transmitted/reflected with amplitudes $B$ and $F$ - in this case we set $A=0$, but the solution is a linear superposition of the previously chosen states with $F=0$ and $G=0$. Finally, if we incorporate time reversal, we can also swap $F$ and $G$.
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    $\begingroup$ Okay, thank you for the good links you provided. I understand that we can determine the two solutions to TISE by for instance considering the wave flux propagating from left-to-right (in this problem specifically). Now, if we do consider this derivation-method for the two solutions, I still cannot see how the term $G\text{e}^{-iqx}$ is discarded. Although the links you provided give good insight, I am not quite sure how they answer this question in particular. $\endgroup$ Jul 18, 2023 at 10:25
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    $\begingroup$ @RasmusAndersen term $Fe^{iqx}$ corresponds to the current flowing to the right, whereas term $G^{-iqx}$ is the current flowing to the left - you can easily calculate the current for a plane wave using usual prescription. The solution in the OP is for the (incoming) wave incident from the left. If you consider a wave incident from the right or if you decide to work with outgoing waves, you will have $G^{-iqx}$. $\endgroup$
    – Roger V.
    Jul 18, 2023 at 10:59
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    $\begingroup$ @RasmusAndersen see also the remark that I added to the answer. $\endgroup$
    – Roger V.
    Jul 18, 2023 at 11:14
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    $\begingroup$ Okay, thank you. This clears up some of the confusion, but I will sit down and study the details at a later point. $\endgroup$ Jul 19, 2023 at 8:13

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