- #1
I_laff
- 41
- 2
If you have a particle in a 1-d box with a finite potential when ##0 < x < L ## and an infinite potential outside this region, then the normalised wavefunction used to describe said particle is ## \psi (x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})##.
However, if you had say instead a finite potential when ##\frac{-L}{2} < x < \frac{L}{2}## and an infinite potential outside this region, then wouldn't the wavefunction now change?
However, if you had say instead a finite potential when ##\frac{-L}{2} < x < \frac{L}{2}## and an infinite potential outside this region, then wouldn't the wavefunction now change?