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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.

Roger V.
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Tachyon209
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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.

Another tip: for calculating the scattering matrix, it is convenient to consider separately the waves incident from the left and from the right.

Roger V.
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  • 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. – Tachyon209 Apr 08 '20 at 08:02
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    @Tachyon209 I added more details. – Roger V. Apr 08 '20 at 08:25
  • I understand your calculations completely but I am not able to exactly interpret the line $$J_{right} = j(x_R,t) = j(x_L,t) = J_{left}.$$ Since $x_L$ and $x_R$ are the boundaries of the potential (essentially $-a/2$ and $+a/2$), I understand that I obtain $j(-a/2,t) = j(+a/2,t) $. But how are these terms equal to $J_{right}$ and $J_{left}.$? – Tachyon209 Apr 08 '20 at 08:32
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    These are the currents/fluxes in the two sides, I merely change to your notation. – Roger V. Apr 08 '20 at 08:37
  • Okay. One more thing. You wrote that since the problem is time-independent, you can take the derivative of probability density to be zero. Are you sure that this problem is time-independent? Because as soon as I latch on the rotating phases (with time) $e^{-i\omega t}$, the $ |\psi(x,t)|^2 $ isn't gonna be time-independent anymore. – Tachyon209 Apr 08 '20 at 09:29
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    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. – Roger V. Apr 08 '20 at 09:34
  • Oh okay okay. I made a mistake in my algebra. Thanks a lot for your help. – Tachyon209 Apr 08 '20 at 09:41
  • Also, does this also work for wave packets instead of stationary states? – Tachyon209 Apr 08 '20 at 09:42
  • Strictly speaking, it should not work for the wave packets. You will probably have to expand the packet in plane waves, and reassemble it after scattering. – Roger V. Apr 08 '20 at 09:46
  • I will try to do the math. If I get stuck, can I ask it to you here itself? – Tachyon209 Apr 08 '20 at 09:47
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    I think it would be better, if you post it as a new question - in case I have no time and/or somebody else can also help. – Roger V. Apr 08 '20 at 09:48