Infinite well particle subject to additional time dep. potential

Question

I am asked to find the wavefunction of the particle in a well subject to an additional potential $$V(x,t)=\frac{\pi x \hbar}{L}\delta(t).$$ I have already solved that $$\psi(x,t)=\exp\left(\frac{-i}{\hbar}\int^t_0H(t')dt'\right)\psi(x,0).$$ I understand this equation, but I am unsure of how to treat the delta potential because 0 is not included. I was thinking of integrating from -$\epsilon$ to $ \epsilon$. And we assume that it is in ground state for t<0. I assume this means $$\psi(x,0)=\sqrt{2/L}\sin{(\pi x/L}).$$ Any input is appreciated! (I saw some post regarding perturbation theory, but we have not covered anything like that)

Answer 1

Well, without the delta potential the wave function is

$$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$

where

$$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$

Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to "half" of the effect if we wrongly choose to start at $t=0$). In other words, we only know (1) is true for strictly negative times $t<0$. If $\epsilon>0$ denotes an infinitesimal small positive quantity, then

$$ \tag{3} \psi(x,-\epsilon)~=~ \phi(x).$$

Therefore

$$\tag{4} \psi(x,t)~=~T\exp\left[ -\frac{i}{\hbar}\int_{t_0}^t\!dt^{\prime}\hat{H}(t^{\prime}) \right]\psi_0(x,t_0) $$

for $t_0<0$, where $T$ denotes time ordering. However it is not so easy to work directly in terms of eq. (4). It is easier to integrate the Schrödinger equation from $t=-\epsilon$ to $t=\epsilon$, as OP suggests:

$$ \tag{5} i\hbar(\psi(x,\epsilon)-\psi(x,-\epsilon))~=~\int_{-\epsilon}^{\epsilon}\!dt^{\prime} \hat{H}(t^{\prime})\psi(x,t^{\prime})~=~\frac{\pi\hbar x}{L}\psi(x,0). $$

So the wave function has a discontinuity at $t=0$. Next we assume that the value at $t=0$ is the average of the limits from right and left: $$\tag{6} \psi(x,0)~=~\frac{\psi(x,\epsilon)+\psi(x,-\epsilon)}{2}.$$

From eqs. (3), (5) and (6), we deduce that

$$\tag{7} \psi(x,\epsilon) ~=~\frac{1-\frac{i\pi x}{2L}}{1+\frac{i\pi x}{2L}}\phi(x) .$$

What remains is to find the wave function $\psi(x,t)$ for finite $t>0$. We leave that to OP.

Answer 2

You may want to try a Laplace or Fourier transform in time on the Schrodinger equation. Then you will get what is effectively an eigenfunction equation like $\hat{H} \Psi = E \Psi$, but with the frequencies (i.e. the actual "energy") $\omega$ determined by the boundary conditions on $\Psi$. Given that you know the initial condition, I would say that you should use a Laplace transform since that will give you the time dependence of the new state.

As for how to handle the delta function at zero, think of it like this: \begin{equation} F(0) = \lim_{\epsilon \rightarrow 0} \int_{-\epsilon}^{\epsilon} \delta(x) F(x) = \lim_{\epsilon \rightarrow 0} \int_{-\epsilon}^{0} \delta(x) F(x) + \int_{0}^{\epsilon} \delta(x) F(x) \end{equation} and $\delta(x)$ is assumed an even function, so $\delta(-x) = \delta(x)$. So the usual rule is to take the integral over "half" a delta function as half the function value at that point.