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

1 Answer 1


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 $t\to \infty$.

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

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