# 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?

• – yohBS Mar 5 '12 at 19:26
• Scattering matrices are unitary in order to conserve probability. – nibot Mar 18 '12 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.

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,

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

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

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 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 $x$, $$\frac{\mathrm dW(\psi,\psi^*)(x)}{\mathrm dx} ~=~\psi(x)\psi^{\prime\prime}(x)^{\ast}-\psi^{\prime\prime}(x)\psi(x)^{\ast} ~\stackrel{(6)}{=}~0.\tag{12}$$ 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. – an offer can't refuse Oct 15 '14 at 18:04