In Griffith's QM book, he introduces scattering matrices as an end-of-the-chapter Problem 2.52.

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?

  • 1
    $\begingroup$ Scattering matrices are unitary in order to conserve probability. $\endgroup$
    – nibot
    Mar 18, 2012 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!

  • $\begingroup$ In the book on Nuclera Physics by Weisskopf, there is a statement, that $\sum \limits_{\beta}S^{*}_{\alpha \beta}S_{\beta \gamma} = \delta_{\alpha \gamma}$. Is this the same to what you derived. I guess, that you are talking about $SS^{\dagger}=\mathbb{1}$, which in matrix form reads $\sum \limits_{\gamma}S^{*}_{\alpha \gamma}S_{\beta \gamma} = \delta_{\alpha \beta}$ $\endgroup$ Feb 20 at 14:13
  • $\begingroup$ @AlexanderCska: It is the same as $S^*S=1$, not $SS^*=1$. $\endgroup$ Feb 21 at 10:43
  • $\begingroup$ So basically $S^{\dagger}S=\mathbb{1}$. Could you recommend a good book to learn more about the properties of the scattering matrix. In particular, the time reversal property (detailed balance). I am interested in how it works in nuclear physics. I tried to read Weisskopf, but didn't like it very much. $\endgroup$ Feb 21 at 15:08
  • $\begingroup$ @AlexanderCska: Begin with en.wikipedia.org/wiki/S-matrix - For a deep understanding, I recommend the book by R.G. Newton, Scattering Theory of Waves and Particles (2nd ed., Springer, New York 1982). This excludes QFT, where there is no book I can really recommend. $\endgroup$ Feb 22 at 13:47
  • $\begingroup$ But Chapter 3 of Weinberg's vol.1 on QFT gives a sensible account, including detailed balance. $\endgroup$ Feb 22 at 13:50

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, Ref. 1, 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. Ref. 1 defines 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 $\mp k$ with $k>0$) are related to two asymptotic outgoing (left- and right-moving) waves. In formulas,

$$\begin{align}\left. \psi(x) \right|_{I}~=~& \underbrace{A(k)e^{ikx}}_{\text{incoming right-mover}} + \underbrace{B(k)e^{-ikx}}_{\text{outgoing left-mover}}, \tag{1} \cr \left. \psi(x)\right|_{III}~=~& \underbrace{F(k)e^{ikx}}_{\text{outgoing right-mover}} + \underbrace{G(k)e^{-ikx}}_{\text{incoming left-mover}}, \tag{2}\cr k~>~&0,\end{align} $$

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

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

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

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) $$\begin{align} \hat{H} \psi(x) ~=~& E \psi(x), \cr \hat{H}~:=~&\frac{\hat{p}^2}{2m}+V(x),\cr \hat{p}~:=~&\frac{\hbar}{i}\frac{\partial}{\partial x},\end{align}\tag{6}$$ 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 Ref. 2, the transfer matrix $M(k)$ is defined as the corresponding matrix $$ \begin{pmatrix} F(k) \\ G(k) \end{pmatrix}~=~ M(k) \begin{pmatrix} A(k) \\ B(k) \end{pmatrix}.\tag9$$ 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 $$ W(\psi,\psi^{\ast})(x)~=~\psi(x)\psi^{\prime}(x)^{\ast}-\psi^{\prime}(x)\psi(x)^{\ast},\tag{10}$$ or equivalently the probability current $$ J(x)~=~\frac{i\hbar}{2m} W(\psi,\psi^{\ast})(x),\tag{11}$$ does not depend on the position $x$, $$\begin{align} \frac{\mathrm dW(\psi,\psi^*)(x)}{\mathrm dx} ~=~&\psi(x)\psi^{\prime\prime}(x)^{\ast}-\psi^{\prime\prime}(x)\psi(x)^{\ast}\cr ~\stackrel{(6)}{=}~&0.\end{align}\tag{12}$$ Unitarity (5) is equivalent to the statement that $$\left. W(\psi,\psi^*)\right|_{I}~=~\left. W(\psi,\psi^*) \right|_{III}.\tag{13}$$ Ref. 3 mentions that eq. $(12)$ encodes conservation of energy in the scattering.


  1. D.J. Griffiths, Introduction to Quantum Mechanics; Section 2.7 in 1st edition from 1994 and Problem 2.52 in 2nd edition from 1999.

  2. D.J. Griffiths, Introduction to Quantum Mechanics; Problem 2.49 in 1st edition from 1994 and Problem 2.53 in 2nd edition from 1999.

  3. P.G. Drazin & R.S. Johnson, Solitons: An Introduction, 2nd edition, 1989; Section 3.2.

  • $\begingroup$ In this argument, have you used the time-reversal symmetry? I wonder equation 10 is a result of this. $\endgroup$ Oct 15, 2014 at 18:04
  • $\begingroup$ Hi @an offer can't refuse: Can you elaborate on your reasoning? $\endgroup$
    – Qmechanic
    Oct 15, 2021 at 7:16
  • $\begingroup$ Notes for later: 1. $\hat{H}$ Hermitian $\Rightarrow$ $V$ real $\Rightarrow$ time-reversal symmetry $\Rightarrow$ $S=S^T$ symmetric, cf. Wikipedia. 2. Optical theorem: $\quad S=\sigma_x + iT$; $\quad T=\begin{pmatrix} r & t \cr t & s \end{pmatrix}$; $\quad S^{\ast}S={\bf 1}_{2\times 2} \quad \Rightarrow \quad s=r^{\ast}\frac{1+it}{1-it^{\ast}} \quad \wedge \quad 2{\rm Im}(t)=|r|^2+|t|^2$. $\endgroup$
    – Qmechanic
    Nov 20, 2021 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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