# Is it okay to put singularities into the wave function to test behavior around unstable potentials? [closed]

$$\psi(r)=\sqrt[4]{\frac{ a}{8\pi^3 }}\frac{ \exp (-a r)}{r^{1.25}}$$

The wave function above is an example of a function that is normalizable in 3D space and $r=\sqrt{x^2+y^2+z^2}$.

$$-\psi ''(r)-\frac{2 \psi '(r)}{r} + \left( \frac{a }{2 r}+\frac{5 }{16 r^2} \right) \psi (r) = -a^2 \psi (r)$$

It appears to be a solution of Schrodinger's equation with a potential of the form $V(r)=V_1/r+V_2/r^2$ which is classically unstable. The potential's radial force is always outward.

Another example of a normalizable wave function with a singularity is $$\psi(r) = \sqrt[4]{\frac{2 a^3}{\pi ^3}} \frac{ \exp (-a r)}{r^{0.75}}$$ satisfying $$-\psi ''(r)-\frac{2 \psi '(r)}{r} + \left( -\frac{a}{2 r}-\frac{3}{16 r^2} \right)\psi(r) = -a^2 \psi(r)$$ In general 3D space, the condition of normalizability of a wavefunction with one singularity $$\int_0^\epsilon (r^2dr) \left(\dfrac{f(r)}{r^b}\right)^2$$ is satisfied when $b<1.5$ and $f(r)$ has no singularity and approaches zero sufficiently fast when $r$ approaches infinity.

Is it physically invalid to test the behavior of particles around unstable potentials by putting singularities into the wave function? What does QFT have to say about this form of testing?

## closed as unclear what you're asking by ACuriousMind♦, Prahar, user36790, Kyle Kanos, yuggibOct 5 '15 at 15:26

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• I don't understand at all what you are trying to do. What do you mean by "punching singularities into the wave function"? – ACuriousMind Oct 3 '15 at 14:49
• @ACuriousMind Punching singularities means forcing the wavefunction to have singularities, but still maintain the normalizability condition. – linuxfreebird Oct 3 '15 at 14:51
• @ACuriousMind If you flagged my question to be closed, you might as well tell me the reason why so I can fix it. Thanks. – linuxfreebird Oct 3 '15 at 14:56
• It doesn't look normalizable. Normalizability is unrelated to which coordinate system you use (it's in e.g. $L^2(\mathbb R^{3n})$ or it isn't). The phrase "be belong" is not standard. And there is no choice of $a$ and $b$ that puts it in the form you mention, and it's a wave function anyway, not a potential. Is it supposed to be an energy eigenstate? Can you rewrite to be more clear? – Timaeus Oct 3 '15 at 15:02
• @Timaeus I implemented your suggestions. – linuxfreebird Oct 3 '15 at 15:36

## 1 Answer

One thing that goes wrong here is that the Hamiltonian is no longer a self-adjoint operator acting on these singular wave-functions. Let's focus on the kinetic energy portion of the Hamiltonian: $$\hat K = - \partial_r^2 - {2 \over r} \partial_r = -{1 \over r^2} \partial_r r^2 \partial_r$$ To prove self-adjointness, consider the natural inner product on two wave functions where we act with $\hat K$ on one of them $$\langle \phi, \hat K \psi \rangle = -\int_0^\infty \phi(r) \left({1 \over r^2} \partial_r r^2 \partial_r \psi(r) \right)r^2 dr$$ We need to integrate by parts which introduces boundary terms at $r \to \infty$ and $r=0$. The boundary term at $r \to \infty$ will vanish because the wave functions in question are exponentially damped. Let's look at a boundary term at $r=0$: $$\langle \phi, \hat K \psi \rangle = \int_0^\infty (\partial_r \phi) (\partial_r \psi) r^2 dr + \lim_{r \to 0} r^2 \phi(r) \partial_r\psi(r)$$ While the bulk integral over $r$ is nice and symmetric as it needs to be for self-adjointness, the boundary term is problematic for the wave functions in question. If we let $\phi$ and $\psi$ scale as $1/r^b$ as $r \to 0$, the boundary term will scale as $r^{1 - 2b}$ as $r \to 0$, which is singular for $b>1/2$, or in particular for the values $b=3/4$ and $b=5/4$ used in the question.

One moral here is that self-adjointness (Hermiticity) is not just a local property of the operator. It depends also sensitively on boundary conditions. Here, we need non-singular boundary conditions at $r=0$ for the Hamiltonian to be a Hermitian operator.

• Excellent answer. I did not consider the Hermiticity/B.C. of the problem. – linuxfreebird Oct 5 '15 at 13:51
• Must we consider the possibility that the operator may still have real eigenvalues, but is not necessarily hermitian? – linuxfreebird Oct 5 '15 at 13:58
• By assumption, observables in QM should be Hermitian operators. If the operator is not Hermitian, it's no longer clear what you are computing. I think you might be able to make your Hamiltonian Hermitian by adding some delta function like boundary term at $r=0$, but now physically it would become clearer what's going on -- you have bound states because of some additional local physics at $r=0$. – user2309840 Oct 5 '15 at 15:33
• Wait local physics, is that when QFT comes into play? – linuxfreebird Oct 5 '15 at 16:19
• This question was put on hold, does that mean everything will be deleted? – linuxfreebird Oct 5 '15 at 16:32