The time independent wave function for a bound state given some potential function $V(r)$ is given by the time independent Schrödinger Equation


One example of a wavefunction that is in a bound state would be the wavefunction of an electron in a hydrogen atom. For the Hydrogen Atom when $l$ and $m$ are both $0$ the wavefunction is spherically symmetric, and for a spherically symmetric wavefunction in a bound state the time independent Schrödinger Equation reduces to


and for a second order differential equation the initial value of a function, and the initial value of the functions derivative are needed for a unique solution to the differential equation. In the case of a spherically symmetric wave function an additional requirement to following the Schrödinger Equation is that the integral of the square of the wavefunction from $0$ to $\infty$ must be finite, non zero, and converge. This puts restrictions on the initial values for $\frac{\partial\Psi}{\partial{r}}$ as not all initial values will satisfy the second condition given the initial value for $\Psi$.

In the case of the electron in a hydrogen atom $V\propto\frac{1}{r}$ and there are analytical solutions to the wavefunction for an electron in a hydrogen atom.

For most potential functions $V(r)$ there are no analytical solutions to the wavefunction, and also no analytical solutions for finding the energy levels. This means that in general the wavefunction must be modeled numerically, and the Energy levels must also be approximated numerically.

I understand that in the case of a hydrogen atom $\frac{\partial\Psi}{\partial{r}}$ is not $0$, and for the ground state there is no location, in which it would be $0$, but for $n>1$ there are points, there are values of $r$, for which $\frac{\partial\Psi}{\partial{r}}=0$.

When the wavefunction for a bound state cannot be found analytically can the value for $\frac{\partial\Psi}{\partial{r}}$ at $r=0$, or the values for $r$ in which $\frac{\partial\Psi}{\partial{r}}=0$ be found analytically? If not would approximating these values be similar to approximating the values for the Energy Levels?

  • $\begingroup$ The Laplacian of $\Psi(r)$ in spherical coordinates is not $d^2\Psi/dr^2$. $\endgroup$
    – G. Smith
    Sep 19, 2020 at 22:45
  • $\begingroup$ @G.Smith Thanks for pointing that out. I figured figured out that in cartesian coordinates, for the part of the wavefunction along the x axis the second derivates of the wavefunction in terms of y and z are not $0$, meaning that my laplacian was wrong. I switched to cartesian coordinates took the second derivative of the equation of a cone with respect to x, set y and z to $0$, and switched back to spherical coordinates to derive the other part of the laplacian. I edited the second equation, so it should now have the correct form of the laplacian for the special case of spherical symmetry. $\endgroup$ Sep 20, 2020 at 22:39
  • $\begingroup$ Yes, that looks correct now. You can compare it with Wikipedia. $\endgroup$
    – G. Smith
    Sep 20, 2020 at 22:58

1 Answer 1


Usually the BC are not on the derivative of $\psi$ but on $\psi$ itself. For hydrogen, $\lim_{r\to 0}r^2 \psi^2(r)\to 0$ and $\lim_{r\to\infty}\psi(r)\to 0$. The prob. density must have a node at $r=0$ by continuity since $r<0$ is not physical.

In practice the condition $\psi(r)\to 0$ as $r\to \infty$ is very difficult to implement numerically because of (unavoidable) roundoff errors: quantization occurs because the eigenvalue is exact, else the series for the differential equation does not exactly truncate and eventually diverges. Thus, the solutions are extremely sensitive to the guess energy and the accuracy of the integration scheme: even guess energy within 0.1% of the actual value will eventually produce divergences. In practice one chooses some “reasonably far” value of $r$ and looks for non-diverging solutions up to that point. It’s a bit of an art.

  • $\begingroup$ If I have one guess energy slightly greater than the real energy, and one slightly less than the real energy, is the way the wavefunctions for the two guess energies always opposite? I mean it it always the case that in one instance it diverges to $+\infty$ and in the other $-\infty$? $\endgroup$ Sep 21, 2020 at 15:50
  • $\begingroup$ That’s an interesting hypothesis but I’ve never investigated this systematically. It may depend on the step size as well. $\endgroup$ Sep 21, 2020 at 16:50
  • $\begingroup$ I was also wondering if it takes a long time for the simulated wavefunction to diverge is that necessarily a sign of being close to the actual energy? $\endgroup$ Sep 22, 2020 at 3:33
  • $\begingroup$ This is certainly true if by “long time” you mean larger values of $r$. $\endgroup$ Sep 22, 2020 at 12:06

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