I I cannot normalise this: F= Cexp(-r/a)

[jqmhelios]

TL;DR

I cannot normalise this, in fact, I cannot normalise any integral in QM at all unless it is a 1D infinite square well (and struggle even then)

I need to normalise F= Cexp(-r/a)
To do this, I squared the integrand to get C^2exp(-2r/a).
Then I integrated with infinite limits (from 0 to infinity) and equated to 1. The answer to the integral (confirmed by symbolab) is -a/2exp(-2r/a). When I set the limits I get sqrt(2/a). The book says the answer should be (pi*a^3)^1/2. I do not understand what I am doing wrong- I checked everything.

 

[PeroK]

At this level, it's going to be difficult for you without Latex. Also, you need to show us what you are doing. I guess you are doing a 3D integral in spherical coordinates.

 

[Vanadium 50]

In addition, when you post what you do in LaTeX, please be clear what is equal to what. "I get..." can mean several different things. Be explicit.

 

[Nugatory]

Do you have a specific textbook exercise on mind here? If so you will get better and more helpful answers by posting in the homework forums and using the homework template that appears when you start a thread there.

 

[anuttarasammyak]

Isn’t it 3D with
r^2 =x^2+y^2+z^2?

 

[Charles Link]

Use $dv=4 \pi r^2 \, dr$. If your calculus is good enough that you can do inegration by parts a couple times, it should get you there. I get the answer for $C$ is the inverse of the book's answer that you gave.

 

[Vanadium 50]

Isn’t it

Quite possibly. But isn't it the OP's responsibility to post a clear question and not ours to figure out what they must have meant? And isn't it the OP's responsibility to look in on his thread and answer clarifying questions rather than to leave us to guess?

 

[anuttarasammyak]

Constat C has physical dimension of L^(-d/2) where d is space dimension. The "right" answer suggests d=3. But Sherlock Holmes seems not be welcomed here.

 

[Vanadium 50]

Sherlock Holmes should not be required. A question that requires detective work is not a good question, and the OP should clarify. Also, our detective work is not always perfect and we run the substantial risk of being told "No, that's not what I meant at all". The forum has plenty of examples.

 

[Demystifier]

I cannot normalise any integral in QM at all unless it is a 1D infinite square well (and struggle even then)

Your problem is not QM. Your problem is calculus, which you should learn before trying to learn QM.

As others told you, in this case you need to know that $r$ is a radial coordinate in 3-dimensional space, so you should not integrate just over $dr$. You should integrate over $4\pi r^2 dr$. If you have no idea why, you should spend some time with learning calculus first.

 

[Frabjous]

The OP has not been back since @PeroK’s comment (#2). I am guessing that was sufficient.