Consider the Hamiltonian $H$ on functions on the line with \begin{eqnarray} H=H_0+V,\\ H_0=-\frac{1}{2m}\frac{d^2}{dx^2} \end{eqnarray} where $V$ is a potential vanishing outside of a bounded interval and $m>0$. To avoid discrete spectrum of $H$ one may assume $V\geq 0$. One even may assume that $V(x)=V_0>0$ for $x\in [0,a]$ and $V(x)=0$ otherwise.

How to write down explicitly the $S$-matrix for this Hamiltonian? Namely how does $S$ act on $e^{ipx}$?

Remark. Since $S$ commutes with $H_0$ then necessarily $$S(e^{ipx})=A(p)e^{ipx}+B(p)e^{-ipx}.$$

My question is how to write down $A(p),B(p)$ explicitly.

ADDED. Let me ask a more precise question. The equation $H\psi=\frac{p^2}{2m}\psi$ has a solution $\psi_p$ such that \begin{eqnarray} \psi_p(x)=\left\{\begin{array}{cc} e^{ipx}+\tilde B(p)e^{-ipx},&x<\inf supp(V)\\ \tilde A(p)e^{ipx},&x>\sup supp(V) \end{array}\right. \end{eqnarray} where $\tilde A(p),\tilde B(p)$ are appropriate constants.

Is it true that for $p>0$ one has $\tilde A(p)=A(p)$ and $\tilde B(p)=B(p)$?

(If my understanding is correct, $\psi_p$ is the IN state corrresponding to plane wave $e^{ipx}$ provided $p>0$.)

I believe this should be a basic example in the scattering theory, so a reference will be most helpful.

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    $\begingroup$ Is this not essentially a duplicate of your earlier question? $\endgroup$ Nov 12 '19 at 17:37
  • $\begingroup$ No, but of course it is closely related. Now my question is more precise and more general at the same time (see the ADDED part for general compactly supported $V$). $\endgroup$
    – makt
    Nov 12 '19 at 17:44

Paraphrasing L.I.Schiff, "Quantum Mechanics", the $S$-matrix $S=\langle \beta | \alpha^{+} \rangle $ is the amplitude of the final asymptotic state $\beta $ contained in what became of an initial asymptotic state $\alpha $ after the scattering has taken place. So you should construct this for a complete set of states $\alpha $ and $\beta $. Evidently, you have decided to use bases of energy eigenstates. In your case, this infinite-dimensional matrix decomposes into $2\times 2$ blocks on the diagonal, associated with given energy $p^2 /2m$, which indeed contain your amplitudes $\tilde{A} $ and $\tilde{B} $, namely, $$ S(p^2 )=\left( \begin{array}{cc} \tilde{B} (p) & \tilde{A} (p) \\ \tilde{A} (p) & -\tilde{B}^{*} (p) \tilde{A} (p) / \tilde{A}^{*} (p) \end{array} \right) $$ (the first column is directly read off your wave function solution, the second column follows from unitarity and time reversal invariance). Your first question and the "Remark" following it are phrased a bit too sloppily. You have to distinguish between the asymptotic states for $x\rightarrow \pm \infty $. The incoming states are $e^{ipx} $ for $x\rightarrow -\infty $ and $e^{-ipx} $ for $x\rightarrow \infty $. The outgoing states are $e^{-ipx} $ for $x\rightarrow -\infty $ and $e^{ipx} $ for $x\rightarrow \infty $ (in that specific order, for the concrete matrix representation given).

  • $\begingroup$ Many thanks. Do you have an explanation why the first column of $S(p^2)$ looks as you wrote? I am looking not for an intuitive explanation like in Schiff's book, but rather more formal one based on the definition of $S$-matrix as composition of Moller operators etc. (And actually I think $\tilde A$ and $\tilde B$ should be interchanged, but this is less important.) $\endgroup$
    – makt
    Nov 13 '19 at 6:55
  • $\begingroup$ The less important comment first: The ordering of the rows in $S(p^2)$ of course simply corresponds to a choice of ordering the basis of outgoing states. The choice I specify in the last sentence of my answer is often adopted because it leads to $S^T =S$. You're completely free to reorder the basis, swapping the two rows in my $S(p^2 )$. Given the ordering I chose, I simply read off the first column of $S$ from your wave function; $\tilde{B} $ comes with the outgoing state $e^{-ipx} $ for $x\rightarrow -\infty $, $\tilde{A} $ comes with the outgoing state $e^{ipx} $ for $x\rightarrow \infty $. $\endgroup$ Nov 13 '19 at 17:57
  • $\begingroup$ Now, about using the Moller machinery (which I'm sure you've thought about much more than I, so I'm only taking a bird's eye view): I don't see how using the energy eigenstates that you invoke in your question is compatible with that. By virtue of being energy eigenstates, time evolution will never make the interacting and free states congruent, regardless of how far you evolve back into the past. So I don't see how it makes sense to ask, "what is $S(e^{ipx} ) $ ?". (continued in next comment ...) $\endgroup$ Nov 13 '19 at 18:10
  • $\begingroup$ ... Using the Moller machinery, you force yourself to use a different basis, consisting of superpositions of energy eigenstates, and you have to modify your notion of completeness. But let's say that all works out (otherwise, you're stuck anyway) - then, all you've really done is combine the $S$ I gave with changes of basis. You can decompose your bona fide Moller ingoing state into energy eigenstates, use the $S$ I gave, supply the proper time evolution phases, and reassemble your Moller outgoing state. (continued ...) $\endgroup$ Nov 13 '19 at 18:26
  • $\begingroup$ The physical scattering information is all in the $S$ I gave, and you're just putting transformations around it to more properly describe asymptotic states and time evolution. $\endgroup$ Nov 13 '19 at 18:32

construction of the scattering states, $\psi_p(x)$ incident from the left, $\psi'_p(x)$ incident from the right (for $p>0$) \begin{eqnarray} \psi_p(x)=\left\{\begin{array}{cc} e^{ipx}+r(p)e^{-ipx},&x\rightarrow-\infty\\ te^{ipx},&x\rightarrow+\infty \end{array}\right. \end{eqnarray} \begin{eqnarray} \psi'_p(x)=\left\{\begin{array}{cc} e^{-ipx}+r'(p)e^{ipx},&x\rightarrow+\infty\\ t'(p)e^{-ipx},&x\rightarrow -\infty \end{array}\right. \end{eqnarray} I write $x\rightarrow\pm\infty$, since if the line has a nonzero extension in the transverse $y$-direction it is not enough to take $x$ outside of the support of $V$, because of evanescent waves: waves that decay exponentially into the region where $V=0$, but have not yet decayed to zero.

the reflection coefficients $r(p),r'(p)$ and transmission coefficients $t(p),t'(p)$ define the scattering matrix:

$$S(p)=\begin{pmatrix} r(p)&t'(p) \\ t(p)&r'(p) \end{pmatrix}$$ unitarity: $S(p)S^\dagger(p)=\mathbb{1}$, time-reversal symmetry: $S^t(p)=S(p)$, so $t(p)=t'(p)$. These constraints may be incorporated in the polar decomposition $$S=\begin{pmatrix} e^{2i\phi}\sqrt{1-T}&e^{i\phi+i\phi'}\sqrt{T} \\ e^{i\phi+i\phi'}\sqrt{T}&-e^{2i\phi'}\sqrt{1-T} \end{pmatrix},$$ with $\phi(p),\phi'(p)\in[0,2\pi)$ and $T(p)\in[0,1]$.

There is no simple closed-form expression for $\phi,\phi',T$ for arbitrary $V(x)$, this will typically require a numerical solution. For an overview of approximate methods, you might take a look at Scattering by one-dimensional smooth potentials: between WKB and Born approximation.

  • 2
    $\begingroup$ If one assumes, as the OP does, that $V\geq 0$, there are no evanescent waves. Once $V=0$, the solutions to the Schr\"odinger equation are strictly $e^{\pm ipx} $. $\endgroup$ Nov 13 '19 at 0:03
  • $\begingroup$ indeed, thanks for correcting me; I was thinking of the more general case where the "line" has a nonzero extension in the transverse direction, then there appear evanscent (decaying) modes even when $V=0$. $\endgroup$ Nov 13 '19 at 7:36
  • $\begingroup$ Oh yes, all kinds of fun stuff happens in more general cases; and physically relevant potentials usually don't have bounded support anyway. $\endgroup$ Nov 13 '19 at 17:48

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