# Why are scattering matrices unitary?

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?

• Scattering matrices are unitary in order to conserve probability. Commented Mar 18, 2012 at 15:47

$S^{-1}=S^*$ 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 $S\psi$:

$(S\psi)^*(S\psi)=\psi^*S^*S\psi=\psi^*\psi$

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!

• 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. Commented Feb 21, 2023 at 15:08
• @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. Commented Feb 22, 2023 at 13:47
• But Chapter 3 of Weinberg's vol.1 on QFT gives a sensible account, including detailed balance. Commented Feb 22, 2023 at 13:50
• thanks to your suggestion from Weinberg, I learned that for anti-unit. $A$ if $A|\phi\rangle = |\phi ^{*}\rangle$ then $\langle \phi ^{*} | \psi ^{*}\rangle =\left ( \langle \phi | \psi \rangle \right)^{*}$ holds. Could you give me some hint how this works, if there is an interaction inbetween $\langle \phi ^{*} | H | \psi ^{*}\rangle$ Commented Feb 22, 2023 at 18:36
• @AlexanderCska: The $H$ turns into its conjugate transposed. But your notation is quite irritating and not recommended. One rather interchanges the bras and kets. Commented Feb 24, 2023 at 12:39

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.

References:

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.

• In this argument, have you used the time-reversal symmetry? I wonder equation 10 is a result of this. Commented Oct 15, 2014 at 18:04
• Hi @an offer can't refuse: Can you elaborate on your reasoning? Commented Oct 15, 2021 at 7:16
• Notes for later: 1. $\hat{H}$ Hermitian $\Rightarrow$ $V$ real $\Rightarrow$ time-reversal symmetry $\Rightarrow$ $S=S^T$ symmetric, cf. Wikipedia. 2. Optical theorem: $\quad S=\sigma_x + iT$; $\quad T=\begin{pmatrix} r & t \cr t & s \end{pmatrix}$; $\quad S^{\ast}S={\bf 1}_{2\times 2} \quad \Rightarrow \quad s=r^{\ast}\frac{1+it}{1-it^{\ast}} \quad \wedge \quad 2{\rm Im}(t)=|r|^2+|t|^2$. Commented Nov 20, 2021 at 11:41
• Notes for later: Unitarity: $\quad S^{\dagger}S={\bf 1}$ $\quad\wedge\quad$ $S={\bf 1}+iT$ $\quad\Rightarrow$ $\quad T^{\dagger}T=\frac{T-T^{\dagger}}{i}$. arxiv.org/abs/2306.16488 section 1.3. Ang. momentum $\ell$. Partial waves: $\quad S_{\ell}=e^{i2\delta_{\ell}}\in B(0,1)$ with ${\rm Im}\delta_{\ell}\geq 0$. $\quad\Leftrightarrow$ $\quad if_{\ell}=\frac{S_{\ell}-1}{2}\in B(-\frac{1}{2},\frac{1}{2})$. Elastic scattering: $S_{\ell}=e^{i2\delta_{\ell}}$ pure phase shift with ${\rm Im}\delta_{\ell}=0$. What about unitarity in partial wave basis? Commented Jul 13 at 12:22
• Poisson resummation formula/Dirac comb/sha fct:$\quad III_{2\pi}(\varphi)=\delta(\varphi-2\pi\mathbb{Z})=\frac{1}{2\pi}\sum_{n\in\mathbb{Z}}e^{in\varphi}$. $\quad III_1(\ell)=\delta(\ell-\mathbb{Z})=\sum_{n\in\mathbb{Z}}e^{2\pi in\ell}$. Commented Jul 13 at 14:25