Why is $x^n \psi^{(m)}(x)$ zero at infinity?

Question

Prove: Product of any polynomial of $x$ with $\psi(x)$ or any of its derivatives goes to zero in the limit $x\to\pm\infty$.

This comes from a footnote written by the professor in his quantum mechanics problem set and I am curious to know why. I have unsuccessfully attempted L'Hôpital, Taylor expansion, and many other weird methods. Any help is appreciated.

$\Psi(x)$ is a part of the normalized wave function representing the state of a quantum system. It is the part that is only dependent on position and is a solution to the time independent Schrodinger equation. — XYSquared, Jun 28 '19 at 23:34
Then what you have to prove is wrong. There are normalizable wavefunctions whose product with a polinomial in x go to infinity, when x goes to infinity. — DanielC, Jun 28 '19 at 23:42
Would you please show me a counterexample to prove your point? — XYSquared, Jun 28 '19 at 23:47
@Zachary Any function is a solution to the Schrödinger equation, for some suitable potential. For a given $\psi$, just solve for $V$ in the Sch. eq. — AccidentalFourierTransform, Jun 28 '19 at 23:59
You've either dropped some of the crucial context, or your professor did by accident when writing the question. — knzhou, Jun 29 '19 at 01:44
Related: https://physics.stackexchange.com/q/92517/2451 , https://physics.stackexchange.com/q/331976/2451 and links therein. — Qmechanic, Jun 29 '19 at 03:05

tparker · Accepted Answer · 2019-06-29T18:19:04.273

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In general, not all states in the Hilbert space satisfy your professor's requirement. One guess for what your professor had in mind is that they wanted the wave function to be "nice" enough that the $\hat{X}$ and $\hat{P}$ operators can act on the state arbitrarily many times while still outputting a physically valid state in the Hilbert space. In general, the $\hat{X}$ and $\hat{P}$ operators are not closed on the Hilbert space: there exist wave functions that are square-integrable, but if you act $\hat{X}$ on them enough times (i.e. multiply by them by a high enough power of $x$), then the result is no longer square-integrable. So technically, the $\hat{X}$ and $\hat{P}$ operators are not actually linear operators on the entire Hilbert space $\mathcal{L}_2(\mathbb{R}^n)$. But they are closed on a (non-complete) subspace of the Hilbert space called the Schwartz space. It's pretty straightforward to show that the Schwartz space respects your professor's claim. Your professor may have been considering this special subset of "nice" wave functions, but a general state doesn't meet the stated requirement.

edited Jun 29 '19 at 18:19

answered Jun 29 '19 at 03:31

tparker

47,418

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Hilbert space means complete complex vector space with respect to the norm induced by the scalar product.for a quantum system it is unique. No other definitions exist. X and P are never defined on the whole Hilbert space, so " as they should be" does not mean anything. This answer is misleading in my view. – Valter Moretti Jun 29 '19 at 06:28
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Observables, described by selfadjoint operators, are used in physics because they posses an associated decomposition in terms of eigevectors (also generalized, more properly in terms of a projection valued measure). The existence of such a decomposition is one of the most important facts for the physical interpretation of the general formalism of Quantum Theory. Completeness of the space is necessary to assure the existence of this decomposition for every selfadjoint operator. – Valter Moretti Jun 29 '19 at 06:44
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So, this requirement cannot be relaxed dealing for instance with spaces of differentiable functions instead of the whole Hilbert space where most elements are not even continuous. – Valter Moretti Jun 29 '19 at 06:44
@ValterMoretti You're right, I fixed my answer. – tparker Jun 29 '19 at 11:04

MannyC · Answer 2 · 2019-06-29T16:05:03.070

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A counterexample where one can provide an explicit Hamiltonian is the case of a particle under a constant force (linear potential).

$$ \frac{\mathrm{d}^2}{\mathrm{d}x^2} \psi(x) = F\cdot \,(x-x_0)\,\psi(x)\,. $$ The solution is famously given by the Airy function. $$ \psi(x) = c_A \mathrm{Ai}(\sqrt[3]{F}\cdot(x-x_0)) + c_B \mathrm{Bi}(\sqrt[3]{F}\cdot(x-x_0))\,. $$ If $x < x_0$ and $|x-x_0|\gg1$, then, letting $z= - \sqrt[3]{F}\cdot(x-x_0)$, $$ \mathrm{Ai}(-z) \sim \frac{\sin \left(\frac23z^{\frac{3}{2}}+\frac{\pi}{4} \right)}{\sqrt\pi\,z^{\frac{1}{4}}} \left(1 + O(z^{-3/2})\right)\,. $$ This clearly does not go to zero when multiplied by a polynomial.

edited Jun 29 '19 at 16:05

answered Jun 29 '19 at 02:43

MannyC

6,706

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I agree. But this is a physical system that exists and so any general claim about QM must admit the possibility that a Hilbert state containing these functions is a valid choice. – MannyC Jun 29 '19 at 03:44
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A free particle under a constant force. – MannyC Jun 29 '19 at 03:49
Parenthetically, I'd say that "free particle under a constant force" is a contraction in terms; if the particle experiences a linear potential then it isn't free. – tparker Jun 29 '19 at 03:50
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I borrowed the terminology "free" from QFT, where interactions are those from cubic degree onwards. Anyway physics is no stranger to idealized model to describe nature. It's not necessary for the force to be constant out to spatial infinity. It's sufficient that it's constant on a scale much larger than the particle at the point that no difference can be detected experimentally. – MannyC Jun 29 '19 at 04:07
If there are linear and quadratic terms, then you can always complete the square and eliminate the linear term via a constant shift in the field or $\hat{X}$ operator, so the linear term is indeed qualitatively unimportant. But if there's only a linear term then this trick doesn't work, and the space of solutions clearly changes qualitatively from the free case. – tparker Jun 29 '19 at 12:00
Also, I'm curious, do you consider a single nonrelativistic quantum harmonic oscillator to be a "free particle"? The reason it's called "free" in QFT is that its collective excitations don't interact, but a single harmonic oscillator doesn't have any collective excitations. – tparker Jun 29 '19 at 12:02

Why is $x^n \psi^{(m)}(x)$ zero at infinity?

2 Answers2