Additional quantum states of the infinite square well

[spaghetti3451]

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?

 

[Orodruin]

No, those are not additional states. In fact, they are the same states as they only differ by an arbitrary phase factor.

 

[spaghetti3451]

Well, do we not rely on the negative sign on the wavefunction to distinguish between bosons and fermions?

For bosons and fermions, we might as well call these the same particle as the wavefunction differs also by a phase factor ?

 

[Orodruin]

Well, do we not rely on the negative sign on the wavefunction to distinguish between bosons and fermions?

No we do not. We rely on the symmetry properties under exchange of particles in multi-particle states. Relative phase factors matter, overall phase factors do not.

 

[spaghetti3451]

Thanks! I get it now!