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In Griffith's QM book, he introduces scattering matrices as an end-of-the-chapter problem.

For a Dirac-Delta potential $V(x) = \alpha \delta (x - x_0)$, I've derived the scattering matrix and observed that it is unitary $S^{-1} = S^{\dagger}$.

I'm trying to explain why this is intuitively, but I don't really have an intuitive picture of what hermitian conjugation $S^{\dagger}$ is doing here. Thoughts?

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related:physics.stackexchange.com/q/18539 –  yohBS Mar 5 '12 at 19:26
Scattering matrices are unitary in order to conserve probability. –  nibot Mar 18 '12 at 15:47

2 Answers 2

$S^{-1}=S^*$ is just the condition for unitarity. It is usually written as $S^*S=1$ (together with invertibility) and means that $\psi^*\psi$ doesn't change when $\psi$ is replaced by $S\psi$:


Therefore probability is conserved, a must for a good scattering matrix.

In general, unitarity of the S-matrix is a consequence of the fact that the S-matrix is formally defined as a limit of products of unitary matrices, which are themselves unitary, though the analysis of the limit requires some care.

Actually, I noticed that I might have missed the point of your question, as you asked about what the adjoint does in your calculation. The delta of a selfadjoint operator is itself selfadjoint, did you mean that? Otherwise, please clarify your question!

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Most often, the $S$-matrix is defined as an operator between asymptotic initial and final Hilbert spaces for a time-dependent scattering process, i.e. between $t\to-\infty$ and $t\to\infty$. There unitarity encodes conservation of probabilities over time. On the other hand, the book that OP mentions, D.J. Griffiths, Introduction to Quantum Mechanics, talks about a time-independent scattering process. For a discussion of the connection between time-dependent and time-independent scattering, see this Phys.SE question.

In this answer we will only consider time-independent scattering. Griffiths' QM book defines in the beginning of Section 2.7 for a 1D system (divided into three regions $I$, $II$, and $III$, with a localized potential $V(x)$ in the middle region $II$), a $2\times 2$ scattering matrix $S(k)$ as a matrix that tells how two asymptotic incoming (left- and right-moving) waves (of wave number $k$) are related to two asymptotic outgoing (left- and right-moving) waves. In formulas,

$$\left. \psi(x) \right|_{I}~=~ A(k)e^{ikx} + B(k)e^{-ikx}, \tag{1} $$ $$\left. \psi(x)\right|_{III}~=~ F(k)e^{ikx} + G(k)e^{-ikx}, \tag{2}$$

$$ \begin{pmatrix} B(k) \\ F(k) \end{pmatrix}~=~ S(k) \begin{pmatrix} A(k) \\ G(k) \end{pmatrix}.\tag{3}$$

To show that a finite-dimensional matrix $S(k)$ is unitary, it is enough to show that $S(k)$ is an isometry,

$$ S(k)^{\dagger}S(k)~\stackrel{?}{=}~{\bf 1}_{2\times 2} \quad\Leftrightarrow\quad |A(k)|^2+ |G(k)|^2~\stackrel{?}{=}~|B(k)|^2+ |F(k)|^2,\tag{4}$$

or equivalently,

$$ |A(k)|^2-|B(k)|^2 ~\stackrel{?}{=}~|F(k)|^2-|G(k)|^2.\tag{5} $$

Equation (5) can be justified by the following comments and reasoning.

  1. $\psi(x)$ is a solution to the time-independent Schrödinger equation (TISE) $$ \hat{H} \psi(x) ~=~ E \psi(x), \qquad \hat{H}~:=~\frac{\hat{p}^2}{2m}+V(x),\qquad \hat{p}~:=~\frac{\hbar}{i}\frac{\partial}{\partial x},\tag{6}$$ for positive energy $E>0$.

  2. The solution space for the Schrödinger eq. $(6)$, which is a second-order linear ODE, is a two-dimensional vectors space.

  3. It follows from eq. $(6)$ that the wave numbers $\pm k$, $$k ~:=~\frac{\sqrt{2mE}}{\hbar} ~\geq~ 0,\tag{7} $$ must be the same in the two asymptotic regions $I$ and $III$. This will imply that the $M$-matrix (to be defined below) and the $S$-matrix are diagonal in $k$-space.

  4. Moreover, it follows that there exists a bijective linear map $$ \begin{pmatrix} A(k) \\ B(k) \end{pmatrix} ~\mapsto~ \begin{pmatrix} F(k) \\ G(k) \end{pmatrix}.\tag{8} $$ In the problems of Chapter 2 in Griffiths' QM book, the transfer matrix $M(k)$ is defined as the corresponding matrix $$ \begin{pmatrix} F(k) \\ G(k) \end{pmatrix}~=~ M(k) \begin{pmatrix} A(k) \\ B(k) \end{pmatrix}.\tag9$$ The $S$-matrix $(3)$ is a rearrangement of eq. $(9)$.

  5. One may use the Schrödinger eq. $(6)$ (and the reality of $E$ and $V(x)$) to show that the Wronskian $ W(\psi,\psi^*)(x)$ does not depend on the position $x$, $$ \frac{dW(\psi,\psi^*)(x)}{dx} ~=~\psi(x)\psi^{\prime\prime}(x)^*-\psi^{\prime\prime}(x)\psi(x)^* ~\stackrel{(6)}{=}~0.\tag{10}$$ Unitarity (5) is equivalent to the statement that $$\left. W(\psi,\psi^*)\right|_{I}~=~\left. W(\psi,\psi^*) \right|_{III}.\tag{11}$$ The book Drazin and Johnson, Solitons: An Introduction, mentions that eq. $(10)$ encodes conservation of energy in the scattering.

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In this argument, have you used the time-reversal symmetry. I wonder equation 10 is a result of this. –  buzhidao Oct 15 '14 at 18:04

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