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In Griffiths' "Introduction to Quantum Mechanics, second edition" section 2.5.2, p. 73, he states: For the delta-function potential, when considering the scattered states (with $E > 0$), we have the general solutions for the time-independent Schrodinger equation:

$$\psi(x) = Ae^{ikx} + Be^{-ikx}\quad\text{for}\quad x<0 \tag{2.131}$$

and

$$\psi(x) = Fe^{ikx} + Ge^{-ikx}\quad\text{for}\quad x>0.\tag{2.132} $$

In a typical scattering experiment, particles are fired in from one direction-let's say, from the left. In that case the amplitude of the wave coming in from the right will be zero: $$G=0\quad(\text{for scattering from the left}).\tag{2.136}$$

Then $A$ is the amplitude of the incident wave, $B$ is the amplitude of the reflected wave and $F$ is the amplitude of the transmitted wave. Now the probability of finding the particle at a specified location is given by $|\Psi|^2,$ so the relative probability that an incident particle will be reflected back is $$R \equiv \frac{|B|^2}{|A|^2},\tag{2.138}$$ where $R$ is called the reflective coefficient.

Question:

How does the definition of $R$ follow? Where exactly does this probability come from?

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  • $\begingroup$ it's identical to the expression for reflection intensity in terms of the electromagnetic field of a beam of light; here the probability amplitude takes the place of the electric field. $\endgroup$ – Peter Diehr Mar 4 '16 at 19:44
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Sketched proof:

  1. The first question the reader should ask him/herself is:

    Why can we use the time-independent Schrödinger equation to describe scattering of an incoming particle against a fixed potential, which naively sounds like a time-dependent process?

    This is answered in e.g. this Phys.SE post. In particular, note that $e^{+ikx}$ and $e^{-ikx}$ are a right- and left-mover, respectively.

  2. Secondly, to conserve probability over time, impose that the $S$-matrix should be unitary. This is e.g. done in my Phys.SE answer here. Unitarity implies with $G=0$ that $$|A|^2~=~|B|^2+|F|^2.$$

  3. Thirdly, interpret $\frac{|B|^2}{|A|^2}$ and $\frac{|F|^2}{|A|^2}$ as probabilities for reflection and transmission, respectively, thereby adding up to 100 %.

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  • $\begingroup$ Thanks for your answer. I showed that $|A|^2 = |B|^2 + |F|^2$ using the continuity of the probability current, is this fine? Secondly, I don't understand why the probability of reflection and transmission are given by $|B|^2$ and $|F|^2$ respectively? $\endgroup$ – Alex Mar 6 '16 at 9:04

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