# In order to solve for the states of a spherically symmetric parabolic potential do we need to use cartesian and cylindrical coordinates?

In the general case a spherically symmetric potential, the Time Independent Schrodinger Equation is separable in spherical coordinates but not in cartesian, or cylindrical coordinate as in general $$V(r)\neq{V(x)+V(y)+V(z)}$$, and $$V(r)\neq{V(\rho)+V(z)}$$. In the special case of a spherically symmetric parabolic potential, $$V(r)=ar^2=a\sqrt{x^2+y^2+z^2}^2=a\left(x^2+y^2+z^2\right)$$ and so $$V(r)=V(x)+V(y)+V(z)$$ meaning that the Time Independent Schrodinger Equation $$\nabla^2\Psi+(E-V)\Psi=0$$ is separable in cartesian coordinates. Also $$\rho=\sqrt{x^2+y^2}$$ meaning that $$\rho^2=\sqrt{x^2+y^2}^2=x^2+y^2$$, meaning that $$a\left(x^2+y^2+z^2\right)=a\left(\rho^2+z^2\right)$$, which means that $$V(r)=V(\rho)+V(z)$$, so the Time Independent Schrodinger Equation is also separable in cylindrical coordinates in the case of a spherically symmetric parabolic potential.

Does this mean that for a spherically symmetric parabolic potential, in addition to using separation of variables in spherical coordinates, we also need to use separation of variables for cartesian in order to find all the bound states? If not why don't we need to use separation of variables in cartesian and cylindrical coordinates in order to find all bound states for a spherically symmetric parabolic potential?

• You never need to use separation of variables. If it's available we generally always use it because it is by far more convenient than any alternatives, but it is never required as such. Commented Jun 11, 2022 at 23:29
• @EmilioPisanty well is separation of variables in spherical coordinates sufficient to find all the bound states for a spherically symmetric parabolic potential? Commented Jun 12, 2022 at 2:33

You don't need to, but you can.

Essentially, what you're doing is choosing between expressing your energy eigenstates as products of functions of spherical, cylindrical, or Cartesian coordinates. Any one of these options is equally good, while in other problems the calculations will get quite cumbersome in, e.g., Cartesian coordinates.

At the end of the day, you need to specify each point in space with three numbers. Which separation of variables you use is just a matter of whether you prefer $$(r, \theta, \phi)$$, $$(\rho, \phi, z)$$, or $$(x, y, z)$$.

To be a bit more specific, if you use spherical coordinates, the solution will be written in terms of, e.g., spherical harmonics and some radial function. If you use Cartesian, no spherical harmonics will appear. Instead, you'll get products of Hermite polynomials in each direction.

For an example a bit more explicit, I'll notice that the solution in spherical coordinates is given on Wikipedia (e.g., on this article). In there it is mentioned that the energy levels are given by $$E_n = \hbar \omega \left(n + \frac{3}{2}\right),$$ where $$n$$ is a non-negative integer and $$\frac{1}{2} \mu \omega^2 = a$$ ($$\mu$$ is the mass of the particle and I'm changing the choice of constants to that used in Wikipedia so the notation gets closer to the standard when dealing with quantum harmonic oscillators). Wikipedia also mentions that the degeneracy at energy level $$n$$ (i.e., the dimension of the eigenspace with energy $$E_n$$) is given by $$\frac{(n+1)(n+2)}{2}.$$

Let us reproduce this result in Cartesian coordinates. I'll just sketch the calculations, but they are done in more detail in Sec. 2.5 of Weinberg's Lectures on Quantum Mechanics. In Cartesian, the Hamiltonian can be written as \begin{align} H &= - \frac{\hbar^2}{2 \mu} \nabla^2 + \frac{1}{2}\mu \omega^2 r^2, \\ &= \left(- \frac{\hbar^2}{2 \mu} \frac{\partial^2}{\partial x^2} + \frac{1}{2}\mu \omega^2 x^2\right) + \left(- \frac{\hbar^2}{2 \mu} \frac{\partial^2}{\partial y^2} + \frac{1}{2}\mu \omega^2 y^2\right) + \left(- \frac{\hbar^2}{2 \mu} \frac{\partial^2}{\partial z^2} + \frac{1}{2}\mu \omega^2 z^2\right), \end{align} which is just three times the Hamiltonian to a one-dimensional QHO. Hence, the system can be treated as three independent QHOs, and hence the allowed energy levels are

\begin{align} E_{qrs} &= \hbar \omega \left(q + \frac{1}{2}\right) + \hbar \omega \left(r + \frac{1}{2}\right) + \hbar \omega \left(s + \frac{1}{2}\right), \\ &= \hbar \omega \left(q + r + s + \frac{3}{2}\right), \end{align} where the non-negative integers $$q$$, $$r$$, and $$s$$ are the quantum numbers for each of the independent 1D-QHOs. Notice then that the allowed energies are $$E_n = \hbar \omega \left(n + \frac{3}{2}\right),$$ just as we found in the spherical separation of coordinates. However, is the degeneracy right? Fix $$n$$. How many combinations of $$q$$, $$r$$, and $$s$$ lead to the name $$n$$? Following Weinberg, we notice we have $$q + r + s = n.$$ Since $$n$$ is fixed, this means that once we choose values for $$q$$ and $$r$$, $$s$$ will be given by $$s = n - q - r$$. The degeneracy then must be given by \begin{align} \sum_{q = 0}^{n} \sum_{r = 0}^{n - q} 1 &= \sum_{q = 0}^{n} (n - q + 1), \\ &= (n + 1)^2 - \frac{n (n+1)}{2}, \\ &= \frac{(n+1)(n+2)}{2}, \end{align} which is exactly the same result.

To show the different separations of variables only lead to different choices of basis will likely take a bit more calculation, but I hope this helps to illustrate the result.

• So should you find the same states from separation of variables in each coordinate system, but just expressed differently? Commented Jun 13, 2022 at 3:51
• @AndersGustafson Exactly. Commented Jun 13, 2022 at 12:37
• Does this just mean that I should find the same Energy Levels with each coordinate system or does it also mean that I should find the same number of configurations in using each coordinate system? Commented Jun 14, 2022 at 4:58
• @AndersGustafson I'm not sure I understand what you mean by "same number of configurations". About the energy levels, yes, you'll find the very same energy levels, although the energy eigenstates might be a bit different. For example, in spherical coordinates they are also angular momentum eigenstates, but in Cartesian coordinates they are not. This just means that, in each energy eingenspace, you're choosing a different basis in each coordinate system Commented Jun 14, 2022 at 8:07
• If by "number of configurations" you mean, e.g., the degeneracy of each energy eigenvalue, then yes, all coordinate systems will yield the very same degeneracies Commented Jun 14, 2022 at 8:08