- #1
spaghetti3451
- 1,344
- 33
The quantum states ##\psi(x)## of the infinite square well of width ##a## are given by
##\psi(x) = \sqrt{\frac{2}{a}}\sin\Big(\frac{n \pi x}{a}\Big),\ n= 1,2,3, \dots##
Now, I understand ##n \neq 0##, as otherwise ##\psi(x)## is non-normalisable.
But, can't we get additional states for ##n=-1,-2,-3,\dots##?
Of course, they have the same position probability densities and energy eigenvalues as the corresponding positive $n$ states, but still, don't they *exist* at all?
##\psi(x) = \sqrt{\frac{2}{a}}\sin\Big(\frac{n \pi x}{a}\Big),\ n= 1,2,3, \dots##
Now, I understand ##n \neq 0##, as otherwise ##\psi(x)## is non-normalisable.
But, can't we get additional states for ##n=-1,-2,-3,\dots##?
Of course, they have the same position probability densities and energy eigenvalues as the corresponding positive $n$ states, but still, don't they *exist* at all?