Infinite Series from Perturbation Theory

[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!

 

[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.

 

[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??

 

[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...

 

[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?

 

[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.

 

[dextercioby]

Now I'm probably say something stupid, so dextercioby, feel free to slap me...

Nice wording...

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,

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"??

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.

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

Daniel.

 

[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

 

[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.

 

[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.

 

[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.

 

[dextercioby]

Okay,then,lemme give you an idea.Please make up your mind about this problem,will u??

State it in the original context.

Daniel.

 

[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?

 

[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?

 

[ALime88]

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

 

[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.

 

[ALime88]

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

 

[ALime88]

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

The answer is (as can be seen in Griffiths) proportional to $$\frac{1}{n^2}[\tex]$$

 

[ALime88]

Proportional to $$\frac{1}{n^2}$$ I meant... wrong slash