In the step potential \begin{equation} V= \begin{cases} 0 &, \text{x<0}\\ V_0 &, \text{x>0} \end{cases} \end{equation} for the scattering states$(E>V_0)$, the states on the right and left of the step are complex exponentials (waves), but the left-moving wave to the right of the step is thrown away: \begin{equation} \psi = Ce^{ikx} + De^{-ikx},\\ D=0, \end{equation} on the grounds that there is only a transmitted wave on the right. But I don't understand why this is. We don't do the same for a free particle, in which case the solutions are superpositions of right and left-going waves. I thought it might have something to do with conservation of momentum, since the momentum seems to commute with the Hamiltonian (or does it?), but the wavefunctions are not normalizable anyway, so they give rise to infinite expectation values. I read the relevant questions and read some references but ultimately it seems that, as said in Shankar's book "We are only interested in solutions for which $D=0$.) Is that the case? What do solutions in which $D \neq 0$ represent? Why do we apply an argument such as "only transmitted wave on the right" and not an argument such as "zero transmission in a bound state of a potential barrier"?

Note: I thought the momentum commutes with the Hamiltonian because the potential is piece-wise constant, but if we write it as a step function as in the first equation, the commutator gives a delta function. I'm not sure what this means.

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    $\begingroup$ This question has nothing to do with the basic formalism (vector spaces, symmetry, commuting variables, etc.) and nothing to do with anything foundational. Instead, the question is always: "What physical situation are we trying to model?" With these scattering problems, we are modeling a situation where we send in stuff from one direction (say from the left) and see what comes out. Since nothing is coming in from the far right (an assumption about what physical system we are modeling!), there should be no left-moving wave on the right side of the barrier. That is all there is to it. $\endgroup$
    – march
    Commented Sep 21, 2023 at 0:31
  • $\begingroup$ Does this answer your question? Understanding the potential step for a particle in 1D $\endgroup$
    – Roger V.
    Commented Sep 21, 2023 at 8:12
  • $\begingroup$ @RogerVadim It seems similar to march's comment above. I didn't think that was all there is to it, but that seems to be the case. Thanks. $\endgroup$
    – EM_1
    Commented Sep 22, 2023 at 13:06

1 Answer 1


The wave component with coefficient D would be regressive due to the minus sign in the exponent and from a physical perspective its presence would be meaningless (regressive transmitted wave). However, in the symmetric system if the incident wave comes from the right $C=0$ and the trasmitted wave would be regressive, as expected (no progressive trasmitted waves).

The transmission coefficient is related to the probability current $J=\frac{\hbar}{2im}(\psi^* \nabla \psi-\psi\nabla \psi^*)=\frac{\hbar}{m}\Im(\psi^* \nabla \psi)$.

Then, if the particle is in a stationary state, due to the continuity equation $\nabla\cdot J+\frac{\partial \rho}{\partial t}=0$ with $\rho=|\psi|^2$ the derivative of the probability density is zero, so $J$=const.

The probability current has to be invariant passing through the surface $$J_{incident}=J_{reflected}+J_{transmitted}$$ $$T=\frac{J_{transmitted}}{J_{incident}}$$ $$R=\frac{J_{reflected}}{J_{incident}}$$ And then the relation between the coefficients $$R+T=1$$

  • $\begingroup$ Then why not argue, from a physical perspective, that the transmission coefficient for the potential barrier is zero if the particle's energy is below the potential? $\endgroup$
    – EM_1
    Commented Sep 20, 2023 at 22:15
  • $\begingroup$ Because of the tunnel effect even if the particle has an energy lower respect to the potential barrier the probability that the particle will go through is not zero. $\endgroup$
    – Cuntista
    Commented Sep 21, 2023 at 5:43

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