Scattering problem with a time-dependent rectangular potential

Question

I was wondering if it makes sense to define transmission/reflection (T/R) coefficients for Hamiltonian of the form

$$ \mathcal{H} = \frac{p^2}{2m} + V(x)f(t) $$ where $ V(x) = V_0 \left[\theta(x) - \theta(x-x_0) \right]$ is a rectangular potential and $f(t)$ is some function of time.

In the time-independent case ($f(t) = 1$), it is well known that the wave function can be written as $$ \Psi(x,t) = \psi(x)\phi(t)$$ with $\phi(t)=e^{-iEt/\hbar}$ where $E$ is the solution of the eigenvalue equation $\mathcal{H}\psi = E \psi$. $\psi(x)$ are plane-wave or exponential functions and can be written as $\psi(x) = A e^{ikx} + B e^{-ikx}$. Then the T/R coefficients, defined as the ratio of the transmitted/reflected current to the incident current, are obtain by imposing continuity of the wave function and its derivative at $x=0, x_0$. This quantities are interesting because we can predict the T/R of an arbitrary energy-superposition wave packet.

Now what happens with a $f(t)$ that really depends on time ? I don't think we can use $\Psi(x,t) = \psi(x)\phi(t)$ anymore, even outside the potential barrier, because the T/R coefficients would be time-independent in this case.

To make my questions perfectly clear : is it still possible to use plane wave solutions outside the barrier to obtain T/R coefficients ? Do these coefficients still have a well-defined meaning in this time-dependent framework ? Can we still use them afterwards to obtain a wave packet T/R ?

Answer 1

Plane waves are still the solutions to the Schrodinger equation in the $x>x_0$ and $x<0$ regions, so it is still easiest to decompose whatever's going on there into plane waves in order to find the full time-dependent solution. However as you observed, the transmitted and reflected waves will not necessarily have the simple time dependence $\phi(t)$. Instead, they will be a superposition of many waves with many different $\omega$ and $k$.

In the time-dependent case, the value of the wave function at the boundary $x=x_0$ can oscillate at other frequencies besides the original incident frequency, and therefore new frequencies can appear that could not appear in the time-independent problem. Physically, this corresponds to the time-dependent potential adding energy to the system, so the output wave's oscillation frequency is no longer constrained by conservation of energy. But because you can still decompose solutions in the $x>x_0$ and $x<0$ regions in terms of plane-waves, the $T$ and $R$ coefficients still parameterize the wave function. However there are in a sense an infinite number of $T$ and $R$ coefficients, one for each output frequency. The wave function in the $x>x_0$ region can be written as

$$\psi(x>x_0,t) = \int_{-\infty}^\infty d\omega A(\omega) e^{-ik(\omega)x - i\omega t}$$

and similarly for the reflected wave (and of course $k(\omega) = \sqrt{2m\omega/\hbar}$ as usual for plane waves). We can then use these $A$s to define transmission coefficients as usual.

The hard work is in finding $A(\omega)$ and likewise for the reflected wave. Once you have that, you can solve the problem for all possible input waves and then compute the solution for an arbitrary input wavepacket using the superposition principle. One odd thing you'll find is that in general the potential well will 'radiate' even in the absence of an incidence wave.