# Half-line vs radial Schrodinger equation

Assume that a quantum particle is constrained to move along a semi-infinite fixed rigid rod. This system could be described as a particle on a half-line, if we introduce the cartesian coordinates so that the $x$-axis runs along the rod. Alternatively, we could introduce polar coordinates so that the radial coordinate runs along the rod and the angle is fixed and defines the orientation of the rod.

For the cartesian approach, there is no $y$ dependence and the Schrodinger equation is $$-\frac{\hbar^2}{2m} \psi_c''(x) = E \psi_c(x).$$ On the other hand, in the polar approach, there is no $\phi$ dependence and the Schrodinger equation is $$-\frac{\hbar^2}{2m}\left(\psi_p''(r) + \frac{\psi_p'(r)}{r}\right) = E \psi_p(r).$$

I would (naively) expect that I would get the same Schrodinger equation as the underlying physics is the same and both coordinates, $x$ and $r$ denote the same position along the rod. However, this is obviously not the case. So, which equation is the right one and why?

OP's comparison has at least 2 issues:

1. The radial equation describes an $$s$$-wave in $$d=2$$ spatial dimensions $$\mathbb{R}^2\backslash\{(0,0)\}~\cong~ \mathbb{R}_+ \times \mathbb{S}^1,$$ i.e. a wavefunction independent of (/uniform in) the angular variable $$\phi$$; not a wavefunction constrained to a fixed angle $$\phi$$.

This issue may be resolved (at least partially) by instead going to $$d=1$$ spatial dimension $$\mathbb{R}\backslash\{0\}~\cong~ \mathbb{R}_+ \times \mathbb{S}^0,$$ i.e. view the real line as two half-lines. Then the halfline and the radial Schrödinger equations become the same.

2. The half-line wavefunction has a Dirichlet boundary condition $$\psi(r\!=\!0)~=~0$$ at its finite endpoint $$r=0$$ (because of an infinite wall potential); it is more delicate what boundary condition at the center $$r=0$$ should be imposed for the radial equation, cf. e.g. this Phys.SE post.

• Would it be correct to say that the Schrodinger equation for the polar wave function $\psi_p(r,\varphi) = \psi_c(r)\delta(\varphi-\varphi_0)$ reduces to the half-line wave function for $\psi_c(r)$? – Fizikus Dec 30 '17 at 18:44
• $\psi_p(r,\varphi) = \psi_c(r)\delta(\varphi-\varphi_0)$ would not satisfy the 2D TISE. – Qmechanic Dec 30 '17 at 19:52
• OK, why not? What's missing? The Jacobian? – Fizikus Dec 30 '17 at 19:54
• Because of the $\delta^{\prime\prime}(\varphi-\varphi_0)$ term. – Qmechanic Dec 30 '17 at 19:58
• So how would a 2D wave function constrained to a half-line look like? – Fizikus Dec 30 '17 at 23:18