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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 $$\psi(x,t)=\exp(\frac{-i}{\hbar}\int^t_0H(t')dt')\psi(x,0).$$ 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)

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)

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 $$\psi(x,t)=exp(\frac{-i}{\hbar}\int^t_0H(t')dt')\psi(x,0)$$ 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)

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 $$\psi(x,t)=\exp(\frac{-i}{\hbar}\int^t_0H(t')dt')\psi(x,0).$$ 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)