# Homework Help: Infinite Series from Perturbation Theory

1. Jan 16, 2005

### ALime88

Hey there, I'm working on a perturbation theory problem, and I have no clue where to start in solving an infinite series.

It's an infinite square well with a delta function potential in the centre and I'm trying to find the 2nd order energy correction to Energy En. Anyway, what I've got is

SIGMA [sin(m*Pi/2)sin(n*Pi/2)]^2/(n^2-m^2)

where that sum is over m, from 1 to Infinity but not equalling n (the denominator would be 0 then). It can be solved explicitly, but again, I have no clue where to start. Thanks for any pointers!

2. Jan 16, 2005

### dextercioby

Did u mean:

$$\sum_{m=1,m \neq n}^{+\infty} \frac{\sin\frac{m\pi}{2}\sin\frac{n\pi}{2}}{n^{2}-m^{2}}$$
??

Daniel.

3. Jan 16, 2005

### ALime88

Yeah that's exactly it! Sorry for not posting it in such a clear form... Is there a built-in equation editor in the posting process??

4. Jan 16, 2005

### dextercioby

Can u compute this sum???

$$\sum_{n=a}^{+\infty} \frac{\sin n}{n}$$

Daniel.

PS.If u know Tex,then u can type your formulas and a built-in compiler will show them...

5. Jan 16, 2005

### ALime88

I thought about it and scanned some textbooks, and no I don't know how to solve that... I'm sure I could figure it out relatively easily, so how could that help with the initially stated question?

6. Jan 17, 2005

### wizzart

Now I'm probably say something stupid, so dextercioby, feel free to slap me...If it's the TEX-sum in your post, that ALime is talking about then it's fairly easy: those sinusoids are always equal to one, so the term to be summed is just $$\frac{1}{n^2-m^2}$$. It's hellish arithmetic, and I can't really reproduce it now, but the sum over such a term equals $$\frac{1}{n^2}$$, if I'm not mistaking. If you have Griffiths, check problem 6.4.

Last edited: Jan 17, 2005
7. Jan 17, 2005

### dextercioby

:tongue2: :rofl: Nice wording... :tongue2:

Not really.Those synusoides are either plus or minus (remember that "n" is natural,arbitrary and fixed,the summing is after "m"),so
$$\sin\frac{n\pi}{2}=\pm 1$$

On the other side,"m" takes all natural values for which $m\neq n$,which means that this "sine" can also be "+1" or "-1".
$$\sin\frac{m\pi}{2}=\pm 1$$

Now consider their product.Are u sure that for every "m" and "n" possible,their product is "+1"??

I'm not at the library,but i'll check it.
Is it in:
David J.Griffiths:"Introduction to Quantum Mechanics" ??

Daniel.

8. Jan 17, 2005

### vincentchan

I checked the griffith book already.. the sum should look like this
$$\sum_{m=1,n=1m \neq n}^{m=+\infty,n=+\infty} \frac{\sin\frac{m\pi}{2}\sin\frac{n\pi}{2}}{n^{2}-m^{2}}$$
the answer is trivial and the reason is obvious

9. Jan 17, 2005

### dextercioby

That changes things.I didn't know the sum would go over "n" as well...

Using that
$$\sin u \sin v =\frac{1}{2}[\cos(u-v)-\cos(u+v)]$$

Then
$$\sin\frac{n\pi}{2}\cos\frac{m\pi}{2}=\frac{1}{2}\{\cos[\frac{\pi}{2}(n-m)]-\cos[\frac{\pi}{2}(n+m)]\}$$

which is identically zero for "obvious reasons"...

Daniel.

P.S.Ty Vincentchan,u spared me a visit to the library.

Last edited: Jan 17, 2005
10. Jan 17, 2005

### wizzart

stupid me...ow well, still, if there weren't sinusoids, my post would be true ;), except when n is also a changing variable...Don't think it should be, but if you say so.

11. Jan 17, 2005

### ALime88

Hey guys, the sum should NOT be over n. The problem I'm doing actually is Griffith's 6.4, and I'm looking for the correction to the nth energy level, so the n stays in as the index variable. Also, the sinusoids are multiplied, then squared on top, so they will only ever be 0 or 1. So I guess I could still use some help. The prof said that it's rather involved to do explicitly.

12. Jan 17, 2005

### dextercioby

State it in the original context.

Daniel.

13. Jan 17, 2005

### vincentchan

so, now you are doing
$$\sum_{m=1,m \neq n}^{+\infty} \frac{ (\sin\frac{m\pi}{2}\sin\frac{n\pi}{2})^2}{n^{2}-m^{2}}$$
right?

14. Jan 17, 2005

### ALime88

Ok...

$$E^2_{n}=CONSTANT\sum_{m=1,m\neq n}\frac{(\sin(\frac{m\pi}{2})\sin(\frac{n\pi}{2}))^2}{n^2-m^2}$$

This is it. And it's the right side of that equation that I can't solve. Hey VincentChan, why did you think it was summed over n as well?

15. Jan 17, 2005

### ALime88

Sorry VC, didn't mean for that to sound rude.. just wondering if I might be doing it wrong.

16. Jan 17, 2005

### dextercioby

Well,even in this case,why would anything be different??My decomposition into difference of cosines would not be affected whatsover,"n" and "m" are still natural numbers and the cosines would still be zero.Zero squared is still zero...Sum of zeros is still zero.

Daniel.

17. Jan 17, 2005

### ALime88

They are sines, not cosines. $$(sin(\frac{m\pi}{2}))^2=1$$ for odd m.

18. Jan 17, 2005

### ALime88

Oh I what you're talking about with the cosines. Anyway the answer is not zero.