I was playing around with a 3-D potential $V$ such that $V_{(r)} = 0$ for $r<a$, and $V_{(r)} = V_0>0$ otherwise. By using the Schrödinger Equation, I showed that: $$\frac{-\hbar}{2m}\frac{1}{r^2}\frac{d}{dr}\bigl( r^2\frac{d}{dr}\bigr)\psi = E\psi$$

I then used the substitution $\psi_{(r)}=f_{(r)}/r$ and $k=\sqrt{2mE}/\hbar$ to get: $$\frac{1}{r}\frac{d^2f_{(r)}}{dr^2}=-\frac{k^2}{r}f_{(r)} \tag{I}$$

which describes the wavefunction $\psi_{(r)}=f_{(r)}/r$ inside the sphere. Hence, the differential equation has the domain $0\leq r<a$, and I cannot multiply both sides by $r$. This is unfortunate, because there is a similar equation for the outside of the sphere: $$\frac{1}{r}\frac{d^2f_{(r)}}{dr^2}=\frac{k'^2}{r}f_{(r)}$$ As this is outside the sphere, I can multiply both sides by $r$ to get a familiar differential equation that can be solved easily: $$\frac{d^2f_{(r)}}{dr^2}=k'^2f_{(r)}$$

If I do the same thing to $(I)$, I obtain the equation for simple harmonic motion, but substituting the solution back into $(I)$ as a sanity check gives a division by zero when evalutating for $r=0$. After that, I tried a number of substitutions to make $(I)$ have a more recognisable form - to no avail. Then I had the idea of multiplying my trial solution by some other function of $r$ so that upon substitution into $(I)$, the evaluation of $r=0$ doesn't give an infinity... but I don't know quite how to do that...

Long story short..... my question is: what trick do I need to get a meaningful solution to $(I)$?

  • $\begingroup$ Perhaps I'm missing something or forgetting the reasoning, but why can't you multiply by $r$ in $(I)$? $\endgroup$ – Kyle Kanos Sep 9 '14 at 14:25
  • $\begingroup$ $(I)$ applies to the wavefunction inside the sphere, so you shouldn't multiply of divide by a variable that could be equal to zero. $\endgroup$ – Matt Elliot Sep 9 '14 at 14:28
  • $\begingroup$ I don't buy that argument because as $(I)$ stands, you are dividing by a term that could be equal to zero. $\endgroup$ – Kyle Kanos Sep 9 '14 at 14:40
  • $\begingroup$ Multiplying by $r$ gives: $$\frac{d^2f_{(r)}}{dr^2}=-{k^2}f_{(r)} $$. Solving for the odd solution (I'm finding the ground state wavefunction) gives $$f_{(r)}=A\cos(kr)$$ Using this as a trial solution, substitute it back into (I) and evaluate for r=0. I end up with infinities... $\endgroup$ – Matt Elliot Sep 9 '14 at 15:05
  • $\begingroup$ Related: physics.stackexchange.com/q/90987/2451 $\endgroup$ – Qmechanic Nov 5 '14 at 20:44

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1.

II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical coordinates.

III) However in order to keep the kinetic energy $K=\frac{\hbar^2}{2m} \int d^3x~ |\nabla \psi|^2$ finite, a $1/r$ singularity of $\psi(r)$ at $r=0$ is unacceptable, i.e., we must discard the cosine solution and only keep the sine solution. This corresponds to imposing that the wavefunction $\psi(r)$ should be bounded.


  1. D. Griffiths, Intro to QM, Section 4.1.3.

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