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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?

Qmechanic
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countunique
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2 Answers2

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$S^{-1}=S^*$ 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!

Arnold Neumaier
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  • 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 $\sum \limits_{\gamma}S^{}{\alpha \gamma}S{\beta \gamma} = \delta_{\alpha \beta}$ – Alexander Cska Feb 20 '23 at 14:13
  • @AlexanderCska: It is the same as $S^S=1$, not $SS^=1$. – Arnold Neumaier Feb 21 '23 at 10:43
  • 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. – Alexander Cska Feb 21 '23 at 15:08
  • @AlexanderCska: Begin with https://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. – Arnold Neumaier Feb 22 '23 at 13:47
  • But Chapter 3 of Weinberg's vol.1 on QFT gives a sensible account, including detailed balance. – Arnold Neumaier Feb 22 '23 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 $\langle \phi ^{} | H | \psi ^{}\rangle$ – Alexander Cska Feb 22 '23 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. – Arnold Neumaier Feb 24 '23 at 12:39
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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, 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.

Qmechanic
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  • In this argument, have you used the time-reversal symmetry? I wonder equation 10 is a result of this. – an offer can't refuse Oct 15 '14 at 18:04
  • Hi @an offer can't refuse: Can you elaborate on your reasoning? – Qmechanic Oct 15 '21 at 07:16
  • Notes for later: 1. $\hat{H}$ Hermitian $\Rightarrow$ $V$ real $\Rightarrow$ time-reversal symmetry $\Rightarrow$ $S=S^T$ symmetric, cf. Wikipedia. 2. Optical theorem: $\quad S=\sigma_x + iT$; $\quad T=\begin{pmatrix} r & t \cr t & s \end{pmatrix}$; $\quad S^{\ast}S={\bf 1}_{2\times 2} \quad \Rightarrow \quad s=r^{\ast}\frac{1+it}{1-it^{\ast}} \quad \wedge \quad 2{\rm Im}(t)=|r|^2+|t|^2$. – Qmechanic Nov 20 '21 at 11:41