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Playing around with summations...

by Hazzattack
Tags: playing, summations
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Hazzattack
#1
Oct30-13, 12:46 PM
P: 69
Hi guys, i'm hoping someone could explain this for me... and was also hoping that some decent literature could be suggested to help me get more comfortable with series expansions of things etc...

So i was playing around with the Q function (its used in quantum optics) and in the book they say that;

For the sum from 0 to infinity Ʃn^n = 1

(the bold case n's represent 'average' values - not sure if that makes a difference?)

Why is this the case ?

Thanks to anyone who can shed some light on this.
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Office_Shredder
#2
Oct30-13, 12:49 PM
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Can you explain what you mean by average values? Is n totally independent of n here?
Hazzattack
#3
Oct30-13, 12:59 PM
P: 69
I think the index of the sum only applies to the power and not to the average value. (which i think is what you mean by them being independent?

In this case the sum represents the photon distribution where n represents the average photon number, so i'm assuming it doesn't make sense to sum over that?

Sorry if that isn't very clear...

Hazzattack
#4
Oct30-13, 01:00 PM
P: 69
Playing around with summations...

The page i'm looking at if it helps at all is; (the very end of page17)

http://books.google.co.uk/books?id=7...xample&f=false
Hazzattack
#5
Oct30-13, 04:16 PM
P: 69
Never mind i've realised why. Thanks for anyone who took a look.


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