# Why does the finite square well in 3D only admit odd radial solutions?

In these MIT Lecture notes (section 4.1.2, Page 7(53)) it says:

"We must then choose the odd-parity solution in order to obtain a finite wavefunction at $$r= 0$$. Thus in 3D, only the odd solutions are possible and we need a minimum potential well depth in order to find a bound state."

Could you explain to me why that's the case? Why are even functions not admissible solutions? Precluding these solutions doesn't make any physical sense, because an even distribution at the center does not seem to break any symmetry of the equation.

Is there a mistake in the lecture? or did I miss something?

This is an attempt to export the 1d result to the 3d spherical well.

In particular, for $$\ell=0$$, the radial part of the Schrodinger equation is identical to the 1d problem when solving $$\chi(r)=rR(r)$$ so the solutions of the 1d problem which satisfy the correct boundary conditions can be taken as solutions of the 3d problem.

Since the radial coordinate $$r\ge 0$$, the function $$\chi(r)$$ must be $$0$$ at $$r=0$$ by continuity since $$\chi$$ must be $$0$$ in the unphysical region where $$r<0$$. In other words, the condition $$r\ge 0$$ is equivalent to a hard wall at $$r=0$$ so $$\chi(0)$$ must be $$0$$ there.

Only the odd solutions to the 1d problem satisfy this condition of a node at the origin, so only those can be exported to the 3d problem.

Note that, for $$\ell\ne 0$$, there is an extra centrifugal term in the effective potential and the solutions are spherical Bessel functions; there is no analogue of this term and this situation in the 1d problem infinite well problem.

• Why couldn't it use $-L$ to $L$ as boundary? That's the correct one right? So basically they say odd function can be directly transform into 3D, but it did not actually prove that even function could not exists by it self in 3D, right? – ShoutOutAndCalculate Oct 29 at 5:38
• The $r<0$ values make no sense since the distance to origin is strictly non-negative so you need $\chi(o)=0$. – ZeroTheHero Oct 29 at 8:28
• No. I mean not using spherical harmonic, then the boundary condition was actually $-L$ to $L$ in $(x,0,0)$. – ShoutOutAndCalculate Oct 29 at 12:46
• We're talking at cross purposes. You want to take $\psi(x)$ to the radial function $\chi(r)$. Yes $x$ goes from $-L$ to $L$ but $r$ only goes from $0$ to $L$. If you are using a Cartesian well then there is no parity restriction. – ZeroTheHero Oct 29 at 13:05

That quote refers to the finite spherical well, not the square well. And it's referring the transformed radial coordinate function $$u(r) = r R(r)$$ with $$r > 0$$. So it's really only formally even or odd, since $$r > 0$$. But morally, odd $$u(r)$$ corresponds to even $$R(r)$$, so the radial function is basically even.