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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.

Qmechanic
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EM_1
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1 Answers1

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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 $$R+T=1$$

Cuntista
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