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We say that the wave function must go to zero at infinity faster than $1/x^{0.5}$ in order for it to be normalizable.
What about other quantities like the probability current? What is the general rule in computing how fast must something go to zero at infinity?

Qmechanic
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  • Probability current is not supposed to be normalizable, so this question does not make sense. As it stands the question also is to broad, because there is no general rule, but rather a lot of special cases, it always depends on what you actually want to compute, in which fashion, how and why a quantity should vanish at infinity, and so on. – Sebastian Riese Mar 07 '16 at 16:30

2 Answers2

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The statement that the wavefunction has to go to zero in some prescribed manner to be normalizable is wrong.

The space of all normalizable wavefunctions is the space of square-integrable functions $L^2(\mathbb{R}^3)$, and there are square-integrable functions that do not fall off towards infinity, like \begin{align} f(x) & = \sum_n f_n(x) \\ f_n(x) &= \left(\theta(x-n+\frac{1}{2n^2}) - \theta(x-n - \frac{1}{2n^2})\right) \end{align} where $\theta$ is the Heaviside function. $f_n$ is a rectangle of height $1$ and width $\frac{1}{n^2}$ centered at $n$, so the integral of $f_n$ is $\frac{1}{n^2}$ (and also the square-integral - the values of $f$ are 0 and 1, those do not change when squaring) and since $\sum_n \frac{1}{n^2}$ converges, $f$ is square-integrable. But $f$ does not fall off towards infinity, there are arbitarily large values of $x$ for which $f$ is $1$.

However, many operators, like multiplication by $x$ and differentiation, are not defined on the whole of $L^2$, and one often wants to work with invariant domains of definitions for these operators. For that, one often chooses the Schwartz functions.

ACuriousMind
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There aren't any general rules for the probability current. The closest you get are some conventions, so it's more like a notation for some things that already exist when the wavefunction already exists.

If you took a probability current and added the curl of a vector field (field in configuration space, and the vectors are valued in configuration space, so take the curl in three coordinates of the domain and three components of the codomain at a time) then you'd get another vector field (field in configuration space, and configuration vector valued).

And if someone handed you both, how would you know which to use? Both satisfy a continuity equation for the probability density. It's more a convention to pick one possible probability flux and call it the current.

In fact the natural current for a relativistic first quantization has a low speed limit that differs from the popular current in nonrelativistic quantum mechanics.

Timaeus
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