# Does the Schrodinger Equation for a Bound Particle Feeling an Impulse have an Exact Solution?

I'm interested in solving the case of a particle bound in a 1-D infinite potential well feeling an impulsive force, so: $$V(x,t) = \left\{ \begin{array} A-Ax\delta(t-t_0), & \text{if } 0 I specify that the particle begins in the ground state and that it satisfies the normal boundary and normalization conditions: $$\psi(x,t=0)=\sqrt{\frac{2}{L}}\sin(\frac{{\pi}x}{L})$$ $$\psi(0,t)=\psi(L,t)=0$$ $$\int_{0}^{L}|\psi(x,t)|^{2}dx=1\tag{2}$$ And then wish to solve the Schrodinger equation for the particle in the box at any given time: $$i\hbar\frac{\partial{\psi(x,t)}}{\partial{t}}=-\frac{\hbar^2}{2m}\frac{\partial^2{\psi(x,t)}}{\partial{x^2}}-Ax\delta(t-t_0)\psi(x,t)\tag{3}$$

My physical intuition (supported by the discussion here) says that the exact solution to this problem should be something to the effect of : $$\left\{ \begin{array} A\theta(t-t_0)e^{-\frac{iAx}{\hbar}}\psi(x,0), & \text{if } A=\frac{\hbar\pi}{L}(n-1), n\in\mathbb{Z}\\ {\psi(x,0)}, & \text{otherwise } \end{array} \right\}\tag{4}$$

Where $$\theta(t-t_0)$$ is the Heaviside step function. In other words, if the impulsive potential is such that it imparts a force which increases the kinetic energy of the particle to exactly the next quantized energy level of the box, it does so at $$t=t_0$$, but if it does not have that threshold energy, nothing happens. Essentially a simplified model of quantized absorption.

The other Stack Exchange discussion which I linked above uses an approximation method to obtain a solution similar to this (without the piecewise dependence), but I wasn't convinced by the approximation used there and I wanted to try to solve it for myself. Ideally, I wanted to find the exact solution.

I tried using a Laplace transform $$\Psi(x,s)\equiv\mathscr{L}(\psi(x,t))$$. The Laplace transform changes equation (3) into: $$-i\hbar(s\Psi(x,s)-\psi(x,t=0))=\frac{\hbar^2}{2m}\frac{d^2\Psi(x,s)}{dx^2}+Axe^{-st_0}$$ $$\text{Plugging in boundary condition}: \psi(x,t=0)=\sqrt{\frac{2}{L}}\sin(\frac{{\pi}x}{L})$$ $$-i\hbar(s\Psi(x,s)-\sqrt{\frac{2}{L}}\sin(\frac{{\pi}x}{L}))=\frac{\hbar^2}{2m}\frac{d^2\Psi(x,s)}{dx^2}+Axe^{-st_0}$$ $$\Rightarrow\frac{d^2\Psi}{dx^2}+is\alpha\Psi={\gamma}\sin(\frac{{\pi}x}{L})+x{\beta}e^{-st_0}$$ $$\alpha\equiv\frac{2m}{\hbar}$$ $$\beta\equiv-\frac{2mA}{\hbar^2}$$ $$\gamma\equiv-\sqrt{\frac{8m^2}{\hbar^{4}L}}\tag{5}$$

Plugging this equation into Maple, it says that the solution is:

$$\Psi(x,s)=Ae^{k(s)x}cos(k(s)x)+Be^{-k(s)x}\sin(k(s)x)+\frac{1}{g(s)}[s{\lambda}\sin(\frac{{\pi}x}{L})+({\kappa}x+\epsilon)e^{-st_0}]\tag{6}$$

Where:

$$k(s)\equiv\sqrt{\frac{ms}{\hbar}}$$ $$g(s)\equiv-\frac{i4L^2m^2}{\hbar}s^2+2m{\pi^2}s$$ $$\lambda\equiv\sqrt{\frac{32m^3L^3}{\hbar^4}}$$ $$\kappa\equiv\frac{4AL^2m^2}{\hbar^2}$$ $$\epsilon{\equiv}\frac{i2{\pi^2}}{\hbar}$$

Apologies for how cumbersome those equations are, I cleaned them up as best as I could. Now, under normal circumstances, I would take the inverse Laplace transform $$\mathscr{L}^{-1}(\Psi(x,s))$$ to find the solution for my original $$\psi(x,t)$$. But, as proven here, you cannot take the inverse Laplace transform of a periodic function. And indeed, Maple can't do the inverse Laplace of (6). This means I've either done a step in my calculations wrong, or that there is no analytic solution to (3) whatsoever. It is not clear to me where I would have gone wrong, but it also doesn't seem right to me that (3) has no exact solution, given how seemingly "obvious" the physical solution is which I've "guessed" in equation (4).

Could someone point out where I'm going wrong here and tell me, does equation (3) have an exact solution that I should continue working to find? Or is it a lost cause? If it has no exact solution, could you please make an effort to explain why?

• I made a small Mathjax edit for readability. I hope you find it ok. Jun 20 at 19:56

The most common definition of a delta function in time potential used in physics is that the resulting important part of the differential equation around the delta function $$\frac{\partial}{\partial t} f(t) = -i \delta(t) f(t)$$ can be rewritten as $$\begin{equation} \frac{\partial}{\partial t} \ln[f(t)] = -i\delta(t) \end{equation}$$ with solution $$\ln[f(0^+)]-\ln[f(0^-)] = -i$$, where the superscript $$\pm$$ indicates, as usual, an infinitesimal time slightly greater or slightly smaller. This same analysis shows you that your wave function at time $$t=t_0^+$$ is $$\begin{equation} \psi(x,t_0^+) = e^{-\frac{i}{\hbar} A x} \psi(x,t_0^-) \,. \end{equation}$$ where $$\psi(x,t_0^-)$$ is your ground state multiplied by the $$e^{-\frac{i}{\hbar} E_1 t_0}$$ phase factor, with $$E_n = \frac{\hbar^2 n^2 \pi^2}{2mL^2}$$.
For $$t>t_0$$ the solution is simply this $$\psi(x,t_0^+)$$ "initial'' solution for the square well propagated forward in time. To the extent you consider the expansion in eigenstates of the original problem an exact solution, then this is an exact solution. The integrals $$\begin{equation} I_{n1} = \langle n |e^{-\frac{i}{\hbar}A x}|0\rangle = \frac{2}{L} \int_0^L dx \sin\left (\frac{n\pi x}{L}\right) e^{-\frac{i}{\hbar}A x} \sin\left (\frac{\pi x}{L}\right) \end{equation}$$ are elementary, and give the transition amplitudes to the excited states. Notice all values of $$A$$ can give transitions. Taking $$t_0 = 0$$ for convenience $$\begin{equation} \psi(x,t>0) = \sum_{n=1}^\infty I_{n1} \sqrt{\frac{2}{L}}\sin\left (\frac{n\pi x}{L}\right) e^{-\frac{i}{\hbar} E_n t} \,. \end{equation}$$
• Thank you for your answer! I have one follow-up question. That transition probability you calculate -- which is valid for all values of $A$, as you point out -- becomes zero if the phase factor is such that it pushes the function it's acting upon into the next stationary state, right? So for the special case where $A$ satisfies my condition in equation (4), $e^{-iAx/{\hbar}}|0\rangle=|1\rangle$ and the transition probability to any stationary state higher than $|1\rangle$ goes to zero in that case. Am I correct? Jun 21 at 21:15
• If $e^{-iAx/\hbar}|0\rangle$ was equal to $|1\rangle$ then what you say would be correct. However, if I take $A = \hbar\pi/L$, then $\psi(x,t=0^+) = [\cos(\pi x/L)+i\sin(\pi x/L)]\sqrt{2/L}\sin(\pi x/L)$. The first term is 1/2 times the first excited state, but the second is a linear combination of the odd $n$ states with $I_{n1} = -i8/(\pi n(n^2-4))$. Jun 21 at 21:51