# Deriving Unitarity of $S$-matrix in 1D Quantum Mechanics

I was studying about scattering across a one-dimensional unknown potential ( pretty elementary Quantum Mechanics) and how, if we know the $$S$$-matrix of such a system, we can deduce an awful lot of information about the potential. Also, the $$S$$-matrix satisfies some properties. First, for the sake of notational clarity, let me define it. Suppose there exists a potential $$V(x)$$ such that it is zero everywhere but some other arbitrary function between $$-a/2$$ and $$+a/2$$.

Now, by treating this time independently such that plane waves hit the potential and are reflected or transmitted accordingly, I can write the wave function as follows:

$$\psi(x)= \begin{cases} Ae^{ikx} + Be^{-ikx},& \text{for } x\leq -a/2\\ Ce^{ikx} + De^{-ikx},& \text{for} x\geq +a/2\\ \end{cases}$$

Now we create a 2x2 matrix called the S-matrix which relates incoming amplitudes $$A,D$$ to outgoing amplitudes $$B,C$$ such that

$$\begin{pmatrix} B \\ C \\ \end{pmatrix}= \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \\ \end{pmatrix} \begin{pmatrix} A \\ D \\ \end{pmatrix}$$

Now to prove that this matrix is unitary, many sources including Wikipedia use the fact that since the integral of probability density $$\int_{-\infty}^{\infty}|\psi(x,t)|^2=1$$ is time-independent, $$J_{left}=J_{right}$$ where $$J_{left}$$ and $$J_{right}$$ are probability currents to left and right of the potential, which implies that $$|A|^2-|B|^2=|C|^2-|D|^2$$ which can then further be used to prove unitarity. My main question is how did everyone deduce that $$J_{left}=J_{right}$$ and that current inside the potential region is 0? How do I know there exists no probability for the particle to stay inside that region? And even if I know that, how can the above result be derived? Any sort of help would be really appreciated.

The reason for the equality of the currents is particle conservation. You can start with the continuity equation $$\partial_t |\psi(x,t)|^2+ \partial_x j(x,t)=0;$$ integrating it over the central region we get $$\int_{x_L}^{x_R}dx \partial_t |\psi(x,t)|^2 + \int_{x_L}^{x_R}dx \partial_x j(x,t) = \frac{dQ(t)}{dt} +j(x_R,t) - j(x_L,t) = 0,$$ where we assumed that the potential is limited to the interval $$[x_L, x_R]$$ and $$Q(t)$$ is the charge in this region. Since we are dealing with a time-independent problem, $$\partial_t |\psi(x,t)|^2 =0$$, i.e. $$J_{right} = j(x_R,t) = j(x_L,t) = J_{left}.$$ Note that the continuity equation is directly derivable from the Schrodinger's equation.
• Using the continuity equation, I get that the total net current over the entire real axis would be equal to 0. But how would that make $J_{left}=J_{right}$ ? I understand the continuity equation but I'm not able to get that result. Apr 8, 2020 at 8:02
• You solve the stationary wave equation for scattering: all your solutions correspond to the same energy/frequency. Factor $e^{i\omega t}$ will not affect neither the value of $|\psi(x,t)|^2=\psi^*(x,t)\psi(x,t)$, nor that of the current $j(x,t) = \frac{-i\hbar}{2m}\left[\psi^*\partial_x\psi - \partial_x\psi^*\psi\right]$ - it disappears in either expression. Though I should have probably used the term stationary rather than time-independent. Apr 8, 2020 at 9:34