- #1

- 10

- 0

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.

- Thread starter relativist
- Start date

- #1

- 10

- 0

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.

- #2

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

- #3

- 10

- 0

Can anybody please answer my question

- #4

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

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

- #5

Matterwave

Science Advisor

Gold Member

- 3,965

- 326

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).

- #6

- 10

- 0

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.

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 again,

Relativist.