- #1

Irishdoug

- 102

- 16

- Homework Statement
- See below

- Relevant Equations
- See below

A particle of mass m is in the ground state on the infinite square well. Suddenly the well expends to twice it's original size (x going from 0 to a, to 0 to 2a) leaving the wave function monetarily undisturbed.

On answering, for ##\Psi_{n}## I got ##\Psi_{n}## = ##\sqrt{\frac{1}{a}} sin(\frac{n\pi}{2a})## however in the answer booklet ##\Psi_{n}## = ##\sqrt{\frac{2}{2a}} sin(\frac{n\pi}{2a})##. I am unsure why this is. I employed the boundary conditions that ##\Psi(0) = \Psi(2a)##. As such ##k = \frac{n\pi}{2a}##; ##E = \frac{\pi^{2}\hbar^{2} n^{2}}{8ma^{2}}##

My integral was ## A^{2}\int_{0}^{2a} sin^{2}(\frac{n\pi}{2a}) dx = 1##. Thus, ##A^{2}a=1## ; ##A## = ##\frac{1}{a}##.

##\Psi(x,0) = \sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})## as always.

I cannot figure where I have gone wrong? I can post a link to the answer if need be. Any help appreciated.

On answering, for ##\Psi_{n}## I got ##\Psi_{n}## = ##\sqrt{\frac{1}{a}} sin(\frac{n\pi}{2a})## however in the answer booklet ##\Psi_{n}## = ##\sqrt{\frac{2}{2a}} sin(\frac{n\pi}{2a})##. I am unsure why this is. I employed the boundary conditions that ##\Psi(0) = \Psi(2a)##. As such ##k = \frac{n\pi}{2a}##; ##E = \frac{\pi^{2}\hbar^{2} n^{2}}{8ma^{2}}##

My integral was ## A^{2}\int_{0}^{2a} sin^{2}(\frac{n\pi}{2a}) dx = 1##. Thus, ##A^{2}a=1## ; ##A## = ##\frac{1}{a}##.

##\Psi(x,0) = \sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})## as always.

I cannot figure where I have gone wrong? I can post a link to the answer if need be. Any help appreciated.