Conceptual question about reflection probability in the Dirac potential

Question

The TISE under a potential $V(x)= -a\delta(x)$ solves to yield a wave function $\psi(x) = A\exp(ikx) + B\exp(-ikx)$ for $x < 0$ and $\psi(x) = F\exp(ikx) + G\exp(-ikx).$

Setting $G=0$ and using boundary conditions yields $B = \frac{i\beta}{1-i\beta}A$ and $F=\frac{1}{1-i\beta}A.$

Here $$k = \frac{\sqrt{2mE}}{h}$$ $$\beta = \frac{m \alpha}{\hbar^2 k}.$$

Now, Griffith's states in Chapter 2 of his text on this potential that the reflection probability is proportional to $|B|^2$ and the transmission is proportional to $|F|^2.$ This, to me, seems to ignore the $A$ term in the wave function for $x < 0.$ Is the reflection probability distinct from the probability of finding the particle in the region $x<0?$ I'm not sure what the reflection probability really "means" and how it relates to a particle scattering from $-\infty$. To me, all it seems like is we have calculated the wave function for a particle with energy $E$ under this potential, not a particle coming in from $-\infty.$ And it seems that the only way to rationally interpret "reflection" for a particle with energy $E$ is the probability of finding it to the left of $0$ (how else could we determine it "reflected?").

Answer 1

To me, all it seems like is we have calculated the wave function for a particle with energy $E$ under this potential, not a particle coming in from $-\infty.$

A fine point, that is usually not explained in the early chapters of quantum mechanics, is that there is difference between solving an eigenvalue problem and a scattering problem. In an eigenvalue problem the solution has to satisfy boundary conditions, which fix the possible free constants and the energies. In a scattering problem we set the energy, whereas the boundary conditions are taken to resemble free solutions in infinity. Since TISE in one dimension is a second order differential equation, we have two independent solutions. A usual choice (but not the only possible) $$ \psi_k^L(x)=\begin{cases}Ae^{ikx}+Be^{-ikx}, x<0\\Fe^{ikx}, x>0 \end{cases} \text{(wave incident from the left)}\\ \psi_k^R(x)=\begin{cases}Be^{-ikx}, x<0\\Fe^{ikx}+Ge^{-ikx}, x>0\end{cases} \text{(wave incident from the right)} $$

Now, Griffith's states in Chapter 2 of his text on this potential that the reflection probability is proportional to $|B|^2$ and the transmission is proportional to $|F|^2.$

This is not quite true. If we consider the wave incident from the left, the reflection and transmission probabilities are given by $$ R=\frac{|B|^2}{|A|^2},T=\frac{|F|^2}{|A|^2}, $$ since $B$ here is the amplitude of the reflected wave, whereas F is the amplitude of the reflected wave. In general case however such a distinction is hard to make, so we express the result in terms of scattering matrix: $$ \begin{pmatrix}B\\F\end{pmatrix}=\hat{S}\begin{pmatrix}B\\F\end{pmatrix} $$

This, to me, seems to ignore the $A$ term in the wave function for $x < 0.$ Is the reflection probability distinct from the probability of finding the particle in the region $x<0?$ I'm not sure what the reflection probability really "means" and how it relates to a particle scattering from $-\infty$. [...] And it seems that the only way to rationally interpret "reflection" for a particle with energy $E$ is the probability of finding it to the left of $0$ (how else could we determine it "reflected?").

Reflection probability is not the probability of finding particle in region $x<0$. Without reflection the wave would be simply $Ae^{ikx}$, the reflection introduces wave running in the opposite direction. If we calculate the current/flux carried by the wave, we find that it is proportional to the transmission probability $T$, whereas $R=1-T$ describes the fraction of the current that was reflected. These are literal meanings of transmission and reflection when talking about particles or light.

See also S-matrix in one-dimensional quantum mechanics