# Why wave function should goto 0 f faster than 1/sqrt(x) at infinity

• relativist
In summary, the conversation discusses the reason why a normalizable wave function decays faster than 1/sqrt(x) as x approaches infinity. The concept of quadratic integrability is also mentioned, with the understanding that this does not necessarily imply the faster decay rate. The conversation also touches on the difference between physicists and mathematicians when it comes to math, and concludes with a thank you note for the helpful explanation.

#### relativist

Can anybody please explain the reason why a normalizable wave function ψ(x) → 0 faster than 1/√x as x → ∞.

I can understand the reason why ∫ψψ*dx < ∞ But do not understand how quadratic integrability implies that.

I would be very thankful to anybody who can give me some idea.

It doesn't. Check out example 2, starting on page 4 in this article. In particular, you should look at figure 2.1 (page 5).

I need a better explanation than that. Books like Introduction to Quiantum Mechanics by Griffiths state that see footnote 11 in Griffiths under section 1.4.

Your question was answered, so I don't know why you're acting like it wasn't. The statement that you want to prove ("if $\psi$ is square integrable, it goes to zero faster than $1/\sqrt x$ ") is false. The article I linked to explains why, and contains counterexamples.

I took a quick look at Griffiths. It's footnote 8 on page 11 in the copy I could get hold of quickly, so maybe the claim is different in your edition. In the text I'm looking at, he's talking about square integrable solutions of Schrödinger's equation. Maybe it's possible to show that there's no choice of the potential V that allows solutions whose first derivatives with respect to x aren't bounded. Since all the counterexamples I've seen have unbounded derivatives, I expect that it's possible to prove that square-integrability and bounded first derivatives implies that the function goes to zero.

If memory serves me right, Griffiths does explicitly mention that there are pathological functions which do not obey the 1/sqrt(x) rule but are square integrable over the whole real line. I believe he goes on to say that if this bothers you, you should become a mathematician. =]

Physicists are usually sloppy when it comes to math as compared with mathematicians. Suffice it to say that the 1/sqrt(x) rule is at the very least a good rule of thumb for square integrability.

1/x, when integrated over the part of the real line where x>1 (giving you a natural log function) diverges due to it not "decaying fast enough". That's why functions must decay FASTER than 1/x in order that this integral does not diverge. This corresponds to a wave-function which must decay faster than 1/sqrt(x) (since it will be squared).

Matterwave said:
If memory serves me right, Griffiths does explicitly mention that there are pathological functions which do not obey the 1/sqrt(x) rule but are square integrable over the whole real line. I believe he goes on to say that if this bothers you, you should become a mathematician. =]

Physicists are usually sloppy when it comes to math as compared with mathematicians. Suffice it to say that the 1/sqrt(x) rule is at the very least a good rule of thumb for square integrability.

1/x, when integrated over the part of the real line where x>1 (giving you a natural log function) diverges due to it not "decaying fast enough". That's why functions must decay FASTER than 1/x in order that this integral does not diverge. This corresponds to a wave-function which must decay faster than 1/sqrt(x) (since it will be squared).

Thanks for a nice answer. I fact I stumbled upon a problem in chapter 3 (problem 3.2) which got me thinking in the right direction after working on it. Your reply has helped to gain a good deal of confidence that I am making some progress in learning qm.

Thanks again,

Relativist.