# Time-dependent Quantum step potential: Short-time tunneling possible?

Consider the Schrödinger equation

$$i \partial_t \psi(x,t) = -\frac{\hbar^2}{2m} \Delta \psi(x,t) + V(x) \psi(x,t) + W(t) \psi(x,t)$$

where $$\psi(x,t)$$ is the wave function, $$m$$ the particle mass and $$V(x)$$ a time-Independent potential that is Zero if $$x<0$$ and has constant value $$V_0$$ on all other Points. Furthermore, we have a time-dependent, uniform driving potential

$$W(t)=W \sin (\Omega t)$$

that oscillates with Amplitude $$W$$ and frequency $$\Omega$$. This equation I want to study nonperturbatively. If the particle has average energy of

$$\frac{\hbar^2 k^2}{2m}+W < V_0$$,

while moving with velocity $$v=\frac{\hbar k}{m}$$

then time-Independent calculations (driving potential = const.) will give me an exponentially decreasing wave function in the Region $$x \geq 0$$. The particle will almost not tunnel over the step barrier. But when I have a time-dependent excitation, I will have some energy uncertainty

$$\Delta E \ge \frac{\hbar}{2 \Delta t}$$

with time uncertainty $$\Delta t$$ of order $$\Omega^{-1}$$. Will there be a short-time Tunneling over the step, more precisely: Will I observe a wave function that is not exponential decreasing with distance? Will the energy uncertainty give the particle a temporary push that we have wave Solutions for $$x \geq 0$$ over a short time?

Your function $$W(t)=W\sin(\Omega t)$$ added to $$V(x)$$ can be thought of as a gauge transformation with the gauge function $$\Lambda = W\frac{\cos(\Omega t)}{\Omega}$$. The solutions for the Schrödinger equation are then just the solutions without $$W(t)$$ multiplied by $$e^{-i\Lambda(t)}$$. So there will be no change in the tunneling.
Alternatively, you can just separate variables with your form. You have the same energy eigenfunctions, and the solution of the separated time equation will be $$e^{-i\Lambda(t)-E_nt}$$ again showing that this changes all states by an overall phase which has no physical significance.