# How fast does a wavefunction vanish at infinity?

When solving one-dimensional quantum mechanical systems, I find myself very confused about the behavior of wavefunctions at infinity. Let's first impose three reasonable constraints:

1. The potential energy $V(x)$ is bounded and has a finite number of discontinuities.
2. The wavefunction is normalized in position space (so Dirac Delta distributions are not allowed).
3. The momentum operator is Hermitian in position space(this condition demands that the wavefunction vanishes at infinity so that the surface term vanishes).

What I find interesting is that, in simple examples, the wavefunctions always vanish exponentially fast at infinity. For example, the free particle wave packet falls off as Gaussians, the harmonic oscillator wavefunctions are Gaussians multiplied by Hermite polynomials, and the finite square well wavefunctions fall off as $e^{-x}$. Given such exponential falloffs, it is obvious that the position operator is Hermitian and its expectation value is well defined. However, we can cook up wavefunctions that vanish at the rate of inverse polynomials and the expectation value of position operator can become ill defined in these contexts. For example, consider a wavefunction that falls off as $1/x^2$. Then it is obvious that the expectation value of $X^3$ is not well defined: $$\int \psi^*(x) x^3 \psi(x) dx \propto \int \frac{1}{x} dx \rightarrow \infty$$ So do these wavefunctions appear in real Hamiltonians? Given the three constraints above, can we find a potential $V(x)$ such that the wavefunction falls off as an inverse polynomial instead of an exponential?

• Your question is related to the regularity and decay of functions and their Fourier transform (position and momentum if you prefer). I suggest you take a look at the theory of tempered distribution which involve in particular the Schwartz space $\mathcal S$ of rapidly falling function, see article. – user130529 Feb 5 '17 at 8:38
• @claudechuber I was just thinking about making an answer along these ideas when I saw your comment. But I'm not sure either of us would be quite answering the question even though it is definitely relevant information. The question I think boils down to why or when only tempered distributions / Schwartz space elements arise as the solutions of the Schrödinger equation. – WetSavannaAnimal Feb 5 '17 at 9:13
• well, you can always make up your own eigenfunction $\psi(x)$, such as for example $\psi(x)=\frac{1}{x^{2n}+1}$, and plug this into the Schrödinger equation and solve for $V(x)$. In any case, for power law potentials the wave-function always vanishes exponentially fast at spatial infinity (see e.g., arxiv.org/abs/quant-ph/9902081 equation 2.10). – AccidentalFourierTransform Feb 5 '17 at 11:16
• Possible duplicates: physics.stackexchange.com/q/331976/2451 and links therein. – Qmechanic Feb 5 '17 at 13:37

## 1 Answer

Very interesting question! I'll start by outlining some of the mathematical basics for this question:

• You're looking for a bounded state of some potential $V$, i.e. a non-scattering state. Mathemematically, this translates to an $L^2$-integrable eigenfunction of the Schrödinger operator $-\Delta+V$.

• By elliptic regularity, for these functions you instantly get what I understand as your condition (3) (The more precise statement would be that $\psi$ lies in the domain of the self-adjoint version of $\hat{p}$). Basically, the argument here is, that $\psi$ has to be two-times differentiable, by rearranging the Schrödinger equation, for your class of potentials you will have $\Delta \psi\in L^2$, by Fourier transform you then get that the first derivative will also be $L^2$. Hence, $\hat{p}$ is well-defined for $\psi$.

The basic idea why most bound states you will come across are exponentially decaying comes from the following idea: Assume that far away from the origin, $V$ is monotone i.e. it doesn't oscillate. This allows us toestimate $V$ from below by a box potential, which implies that a bounded state of $V$ will be dominated by a bounded state of the box potential. Bounded states of box potentials do exponentially decay, hence the state will decay exponentially. This argument can be made explicit using maximum principles for elliptic PDEs, you may up the mathematical details in e.g.

Berezin and Shubin, The Schrödinger equation (Springer 1991).

So from this argumentation, the answer to your question is almost no for potentials which are monotone far outside. By "almost", I mean that there may be such functions at distinguished, but physically irrelevant values of $E$, for example, consider the potential $$V(x)=\frac{2-6x^2}{(1+x^2)^2}$$which looks like this:

You may now check that $\psi(x)=\frac{2}{\sqrt{\pi}}\frac{1}{1+x^2}$ is a normalized eigenfunction to this potential with eigenvalue 0. The momentum operator is well-defined for this $\psi$ and $\psi$ obviously decays only polynomially for $x\rightarrow\infty$. So, what happened here? If you try to use the "box" argument, you would compare to a box which is completely negative away from the origin (remember, the box estimates the potential from below), so 0 is already a scattering state for the box! However, looking at the potential, you see that this can only be the case for this exact value of $E$ - for even an $\epsilon$ more energy, you will obtain a scattering state since $V\rightarrow 0$ for $x\rightarrow \infty$; and for an $\epsilon$ less, you will get a bound state you can again estimate by a box, hence it decays exponentially. Since you can't prepare a state with an exact energy, this is not physically relevant. Generally speaking, this phenomenon should only occur at $E=\limsup_{|x|\rightarrow\infty}V(x)$, since this will correspond to the lowest possible energy for scattering states.

So, what can happen if we drop the "monotonicity-far-outside"-condition? I think that in this case, it should be possible to obtain the kind of states you are searching for. My attempt on the construction goes as follows: Let $V$ be a collection of box potentials where the boxes have constant height and grow thinner as $x$ gets bigger, e.g. something that looks vaguely like this:

If the infinitely many discontinuities bother you, the behavior I'll describe should be exactly the same for a smoothed version of this potential.

Now, a bound state of this potential would oscillate around $0$ where $V=-0.5$ and decay where $V=+0.5$. The (exponential) decay rate where $V=0.5$ is always the same, by controlling the width of the boxes, you can exactly control how fast your bound state decays, e.g. you can achieve, that each time you pass the positive part of the box, your amplitude goes down at the rate $1/x^2$. The details are probably very technical and fishy, but I think in principle this should work.

• Thank you so much for this brilliant answer! Now assuming that your construction works rigorously, do you know of a physical situation in which such a potential arises? If so, how do we cope with the failure of high powers of position operator to be Hermitian? If not, do you think there is a reason why physical potentials will not create these mildly decaying wavefunctions? – Zhengyan Shi Feb 7 '17 at 0:34
• I'm not sure, but I don't really think that the potential arises in any real situations - It's not periodic, so it can't appear in any lattice setting. In order to model it, you would have to put many particles at very special distances to each other, which I assume is almost impossible since there is no crystal-like setting where you can do this. Also, in every physical situation you can just cut off your space at some very high distance, i.e. you end up in some $L^2(\Omega)$ for bounded $\Omega$, so all the powers of your operators should remain well-defined. – Daniel Feb 7 '17 at 22:42
• I also think that a nontrivial physical reason (i.e. apart from "nature tends to be well-defined") is probably hard to come up with. In nonrelativistic QM, all this stuff is closely related to the structure of the Hamiltonian. Mathematically speaking, the nonrelativistic Hamiltonians generally have a very regular behaviour, which results in all the physically important stuff being well-defined. Hence, I'd say you have to ask why the Hamiltonian has this particular structure, which is a very, very deep question nobody really has an answer for as far as I know. – Daniel Feb 7 '17 at 22:50
• Maybe I can add this thought: there is no physical reason that the expectation value of x^3 or x^2 or even x exist. The prediction of QM is, if you do a measurement you look at the spectral decomposition of the operator and the expectation value of this outcome is given by the ex pectation value of the projector from this decomposition. So basically in this example you get a probability for each intervall of the real line. This probability has to be summable of course. But putting an arbitrary value (like 'x' or 'x^2') to it can easily break that. – lalala Mar 6 '17 at 10:38