Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
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!

share|improve this answer

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 $t\to\infty$. Here unitarity encodes conservation of probabilities over time. On the other hand, Griffiths' book, 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 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 an 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,

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

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

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

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

or equivalently,

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

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

  1. $\psi(x)$ is a solution to the time-independent Schrödinger equation $$ \hat{H} \psi(x) ~=~ E \psi(x), \qquad \hat{H}~:=~\frac{\hat{p}^2}{2m}+V(x),\qquad \hat{p}~:=~\frac{\hbar}{i}\frac{\partial}{\partial x},\qquad\qquad (6) $$ 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 $\pm k$, $$k ~:=~\frac{\sqrt{2mE}}{\hbar} ~\geq~ 0,\qquad\qquad (7) $$ 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 $k$.

  4. Moreover, it follows that there exists a bijective linear map $$ \begin{pmatrix} A(k) \\ B(k) \end{pmatrix} ~\mapsto~ \begin{pmatrix} F(k) \\ G(k) \end{pmatrix}.\qquad\qquad (8) $$ In the problems of Chapter 2, the transfer matrix $M(k)$ is defined as the corresponding matrix $$ \begin{pmatrix} F(k) \\ G(k) \end{pmatrix}~=~ M(k) \begin{pmatrix} A(k) \\ B(k) \end{pmatrix}.\qquad\qquad (9) $$ 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 $x$, $$ \frac{dW(\psi,\psi^*)(x)}{dx} ~=~\psi(x)\psi^{\prime\prime}(x)^*-\psi^{\prime\prime}(x)\psi(x)^* ~\stackrel{(6)}{=}~0.\qquad\qquad (10)$$ Equation (5) is equivalent to the statement that $$\left. W(\psi,\psi^*)\right|_{I}~=~\left. W(\psi,\psi^*) \right|_{III}.\qquad\qquad (11)$$ Drazin and Johnson, Solitons: An Introduction, mention that eq. (10) encodes conservation of energy in the scattering.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.