$V(x)=\infty$ at $x=\pm L$ and $V=0$ for $|x|\leq L$ then consider the ground state$$\psi(x)=\sqrt{\frac{1}{L}}\cos\frac{\pi x}{2L}$$ notice the width is $2L$. The energy is $E=\frac{\hbar^2\pi^2}{8mL^2}$
I am unsure what does the instantaneous removal does to the particle. It becomes a wave packet but how?
I read a sentence from other post (1D Infinite Square Well: Box Suddenly Increases in Size. How treat this?) "wavefunction is just an initial condition", then I guess I also going to treat the infinite well ground state as a wave packet? Which means $$\varphi(k)=\sqrt{\frac{1}{2\pi L}}\int^{\infty}_{-\infty}\cos\frac{\pi x}{2L}e^{-ikx}\,dx=\sqrt{\frac{\pi}{2L}}\left[\delta(\frac{\pi}{2L}-k)+\delta(\frac{\pi}{2L}+k)\right]$$ and the time evolution would be $$\Psi(x,t)=\sqrt{\frac{1}{4L}}\int^{\infty}_{-\infty}\left[\delta(\frac{\pi}{2L}-k)+\delta(\frac{\pi}{2L}+k)\right]e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}\,dk$$ $$\Psi(x,t)=\sqrt{\frac{1}{4L}}\left[e^{i\frac{\pi}{2L}[x-\frac{\pi}{4mL}t]}+e^{-i\frac{\pi}{2L}[x-\frac{\pi}{4mL}t]}\right]$$ Am I correct to this point? It's a superposition of two plane wave with same energy traveling to opposite direction. This matches the intuition in my mind where a wavepacket spread out
then the momentum are $\pm\frac{\hbar\pi}{2L}$, and this should be normalised since we start from a normalized initial wave function?
$\textbf{Update}$
To match $t=0$ the first integral should be integrated over $[-L,L]$ $$\varphi(k)=\sqrt{\frac{1}{2\pi L}}\int^{L}_{-L}\cos\frac{\pi x}{2L}e^{-ikx}\,dx=\sqrt{\frac{1}{2\pi L}}\frac {4 L\pi\cos (kL)} {-4 k^2 L^2 + \pi^2}=\frac{\sqrt{8\pi L} \cos(kL)}{-4 k^2 L^2 + \pi^2}$$ and the second integral, $t>0$ immediately extend over all space. So it would be over all space $$\Psi(x,t)=\sqrt{4L}\int^{\infty}_{-\infty}\frac{\cos(kL)}{-4k^2L^2+\pi^2}e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}\,dk$$ and I think this is the correct time evolution
