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What is the correct separable Schrödinger equation in spherical coordinates?

Some articles use this formula: $$ \Psi(r,\theta,\phi) = R(r)\cdot\Theta(\theta)\cdot \Phi(\phi), $$ and some of them use: $$ \Psi(r,\theta,\phi) =\frac{R(r)\cdot \Theta(\theta)\cdot \Phi(\phi) }{r}. $$

So which is true?

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2 Answers 2

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Both formulas are correct. And it is essentially a matter of taste which one you prefer:

  • If you use the separation approach $$\Psi(r,\theta,\phi) = R(r) \cdot \Theta(\theta) \cdot \Phi(\phi)$$ then you get the radial differential equation $${\frac {1}{r^{2}}}{\frac {d}{dr}}\left(r^{2}{\frac {dR}{dr}}\right)-{\frac {\ell (\ell +1)}{r^{2}}R+{\frac {2m}{\hbar ^{2}}}\left[E-V(r)\right]R=0}$$ (like in Wikipedia - Particle in a spherically symmetric potential - Derivation of the radial equation) with the normalization condition $$\int_0^\infty dr\ r^2 |R(r)|^2=1.$$
  • If you use the separation approach $$\Psi(r,\theta,\phi) = \frac{u(r)}{r} \cdot \Theta(\theta) \cdot \Phi(\phi)$$ then you get the radial differential equation $$\frac {d^2u}{dr^2}-{\frac {\ell (\ell +1)}{r^{2}}u+{\frac {2m}{\hbar ^{2}}}\left[E-V(r)\right]u=0}$$ with the normalization condition $$\int_0^\infty dr\ |u(r)|^2=1.$$

At the end you get the same solutions $\Psi(r,\theta,\phi)$ in both cases because $R(r)=\frac{u(r)}{r}$. However, the approach with $u(r)$ leads to simpler math and has more similarity with a 1-dimensional Schrödinger equation.

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If I am not wrong, then the second one is used for (I guess) when we solve the radial part. We have to put $R(r)=u(r)/r$, so maybe in second one they have written $R(r)/r$ instead of $R(r)$ is because they put directly $R(r)/r$. I am not sure though.

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  • $\begingroup$ Firstly I need to find radial, angular and azimuthal parts, then solve them. $\endgroup$ Jun 21 at 16:09
  • $\begingroup$ Generally the angular and azimuthal parts are solved first and then radial but u can do either. $\endgroup$
    – Souvik
    Jun 21 at 16:23
  • $\begingroup$ You will get a different equation because you have to derive $R/r$ twice. you can take a look at a laplacian equation. $\endgroup$ Jun 21 at 16:30
  • $\begingroup$ U can solve radial eqn using the following method, Near r=0 the differential equation for radial part is d^2u/dx^2=l(l+1)/r^2, now let u=r^s,then putting the value of u in the diiferential eqn, we gets(s-1)=l(l+1), therefore s=l+1, and the solution is u(r)=r^(l+1) or R(r)=r^l $\endgroup$
    – Souvik
    Jun 21 at 17:30

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