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 $(I)$?


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.
  • $\begingroup$ I get (I) and (II), but personally I don't get point (III). 1. Why $K$ presents that integral expression? 2. Why do we care to keep the energy finite in an infinitesimally small point? 3. We use the Dirac delta to define potentials all the time.. $\endgroup$
    – Noumeno
    Jun 21 at 16:26
  • $\begingroup$ Hi Noumeno. Thanks for the feedback. 1. We are here interested in the integrated kinetic energy $K$ integrated over 3-space, not in a local kinetic energy density per se. 2. If the kinetic energy $K$ is infinite, so is the total energy $E$. We are only interested in states with finite total energy $E$, since else we cannot produce them using a finite energy source in any realistic experiment. 3. Even though the delta potential blows up locally in a point, the integrated potential energy should be finite. $\endgroup$
    – Qmechanic
    Jun 21 at 16:43

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