# B.C. for time-decaying delta barrier inside an infinite well

So let's say I have an infinite well with walls up at $$x=-L$$ and $$x=L$$. Suppose that inside the well, there is a time-dependent potential

$$V(x,t)= \alpha_0\delta{(x)}f(t)$$

where $$f(t)$$ is a monotonically decreasing function. Then the time-dependent Schrodinger equation in the region $$x \, \in \,[-L,L]$$ is then

$$-\frac{\hbar^2}{2m}\partial_{x}^2\Psi(x,t)+\alpha_0\delta{(x)}f(t)\Psi(x,t) = i\hbar \partial_t \Psi(x,t).$$

Am I right in assuming that these are the boundary conditions? Is there anything I'm missing?

1. $$\Psi(-L,t)=0$$ and $$\Psi(L,t)=0$$
2. At any $$t=\textrm{constant}$$, $$\Delta(\partial_x \Psi(x,t)) = \partial_x \Psi(x,t) \vert_{+q} - \partial_x \Psi(x,t) \vert_{-q} = \frac{-2m\alpha_0}{\hbar^2}\Psi(x=0,t).$$
3. As $$t\rightarrow \infty$$, $$\Psi(x,t) \rightarrow \Psi_{ISQ}(x,t)$$. Where ISQ stands for the infinite square well

If you have a time dependent potential, you need to solve the time dependent Schroedinger equation -- and the solution is unlikely to become a solution of the time-independent infinite square well problem as $$t\to \infty$$.

• So that removes condition 3? How about the first 2 conditions?
– jboy
Sep 24, 2020 at 11:51
• They look Ok. WSavfunction has to be continuous at $x=0$ also Sep 24, 2020 at 16:15