# Quantum free particle in spherical coordinate

I am trying to understand free particle in both cartesian and spherical coordinate. So a free particle going in, say $$x$$ direction with some energy $$E$$. We know the wavefunction of such particle is:

$$\psi(x)=Ae^{ikx} + Be^{-ikx}.\tag{1}$$

Now lets do the same calculation in spherical coordinate and derive the wave function. In Spherical equations take the following form with $$\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$$:

Now the angular part, $$Y$$, I can take it as constant, and $$l=0$$ as there is no angular momentum. Now if I solve now the radial part using the substitution $$u=rR(r)$$, and $$V=0$$, I get $$u=Ce^{ikr}+De^{-ikr}$$, and hence $$R=\frac {1}{r}(Ce^{ikr}+De^{-ikr}).$$

Now I know that $$\theta=\pi/2, \phi=\pi/2$$, and hence $$x=rsin(\theta)sin(\phi)=r$$, Now clearly just by substituting $$r$$ with $$x$$, I am unable to recover my cartesian solution as described above (eq. 1). Moreover at $$r\to0$$, the solution blows up which was not happening in the cartesian solution.

I am unable to understand this dilemma, that solution should be identical in both coordinates but they are giving me different results!

1. Technically, a spherical coordinate system is defined in 3-space $$\mathbb{R}^3\backslash\{0\}$$ except the origin $$r=0$$. Therefore, spherical coordinates are a poor description of the system at the origin $$r=0$$.

2. The free particle itself has no beef with the origin. There is no actual physical boundary conditions at the origin. The moral is that we shouldn't use a spherical coordinate system to describe a translation invariant system, since it artificially distinguishes a point of the system.

3. Now say that we nevertheless choose to use spherical coordinates. So we have to give up the description at $$r=0$$. Hence we are effectively studying the free particle on $$\mathbb{R}^3\backslash\{0\}$$ instead. For each $$\ell\in\mathbb{N}_0$$, the radial TISE has 2 modes: One of them diverges as $$r\to 0$$, the other is regular. That's fine, because as far as our new description goes, $$r=0$$ no longer exists. Also note that the divergent modes are needed if we try to translate them into the rectangular modes of a rectangular coordinate system.

4. Up until now we have discussed scattering states of the free particle. The situation is different for bound states. There additional physical boundary conditions may arise at the origin, cf. e.g. this & this Phys.SE posts.

The solutions for the radial part are in terms of spherical Bessel functions: \begin{align} R_\ell(\rho)=j_\ell(\rho) \end{align} This is done in details in Landau&Lifshitz's QM text.

Your radial equation for $$u$$ should come out to be \begin{align} u''-\frac{\ell(\ell+1)}{r^2}+k^2 u(r)=0\, . \end{align} There's a boundary condition at $$r=0$$ that should guarantee that your $$u$$ doesn't diverge there since $$u(0)=0$$. Thus, for $$\ell=0$$ we have \begin{align} j_0(\rho)&=\frac{1}{\rho}\sin(\rho)\, ,\\ u_0(\rho)=\rho R_0(\rho)&=\sin(\rho)\, . \end{align} which goes to $$0$$ as $$\rho\to 0$$ as expected.

Note that it is $$\rho R(\rho)$$ that must go to $$0$$, even if $$R(\rho)$$ behave differently. This is because the condition is on the probability density $$\vert u(\rho)\vert^2 = \vert \rho R(\rho)\vert^2$$. The same type of behaviour happens in the hydrogen atom.

• I got it, mistake I did was in that equation I set $l=0$, and then solved it, but I should set $l=0$ only after solving with $l$ intact. Thanks a lot. Commented Jun 11, 2020 at 14:00

Your formula for R is correct.

However, it requires the origin of coordinates to be at the center of the wave. That won’t work if for, among other things, the double slit experiment, where there are two radial waves with offset centers.

I think (an ansatz) that the more general eigenfunctions, of the unconstrained, free-particle, spherical Hamiltonian, are $$A\lvert \vec r -\vec r_a \vert^{-1} e^{-i(k (\lvert \vec r -\vec r_a \vert) } + B\lvert \vec r -\vec r_a \vert^{-1} e^{i(k (\lvert \vec r -\vec r_a \vert) }\tag1$$ where $$\vec r_a$$ is the position vector of the center of the radial wave. However, I have not performed the tedious calculation to prove that. Nevertheless, I think any wave function of an object with only radial momentum can be built from a weighted superposition of $$(1)$$ over the variable $$k$$. It would have its center at $$\vec r_a$$. I also think a valid solution for something like a plane wave (eigenfunctions of the cartesian free-particle Hamiltonian) is a superposition of eigenfunctions $$(1)$$, where one would integrate over $$\vec r_a$$, along a line that would be parallel to the wavefront of the plane wave.

To keep it simple, let’s consider the space being a 2-D plane. I think the $$r^{-1}$$ in your formula has to do with the requirement that the probability density over a length of arc (defined by angle $$\theta$$ and at a radial distance r) times $$\Delta r$$ should be the same as the probability density over a different arc length arc (also defined by that same angle $$\theta$$ but at a different radial distance) times $$\Delta r$$. In the cartesian case, that is not the true. A mathematical consequence is a singularity at $$r=0$$. Nevertheless, spherical coordinates are much easier to work with in cases of angular momentum or when tbe iso-surfaces of probability density are spherical, rather than planar.