Mean number of particles in BCS theory This is my first time using the stack, so I apologize if this question does not abide fully to the rules. Also English is not my native language, and I apologize for any mistakes I make.
I would like to evaluate the mean square fluctuation in the number of particles in the BCS theory of superconductivity. To do this I know that I must compute:
$\left<N^2\right>-\left<N\right>^2$
I did find some helpful lecture notes:
http://nptel.ac.in/courses/115101012/downloads/M6/lec8.pdf
But while working out the maths on my own I got stuck at equation (95) of the notes and do not understand why "$l=l'$ are the only surviving terms. 
I would like to see a clearer argument on why that is true.
 A: Take one of the terms
$$
\left<0\right|\prod_{l>0}(u_l^*+v_l C_{-l} C_l)C_\alpha^\dagger C_\alpha \prod_{l'>0}(u_{l'}^*+v_{l'} C_{l'}^\dagger C_{-l'}^\dagger) \left|0\right>
$$
and use anti-commutation on $C_\alpha^\dagger C_\alpha$ to put all annihilation operators to the left of creation operators. 
If you look closely, expanding the products will result now in terms that each contain as many annihilation ops as creation ops, in order from left to right, something like
$$
\langle 0 | \left[ (C_{-l'1}C_{l'1})(C_{-l'2}C_{l'2})\dots C_\alpha \right] \left[ C_\alpha^\dagger \dots (C_{l2}^\dagger C_{-l2}^\dagger) (C_{l1}^\dagger C_{-l1}^\dagger)\right]|0 \rangle 
$$
But the creation ops acting on the ket vacuum generate one multi-electron state, say 
$$
\left[ C_\alpha^\dagger \dots (C_{l2}^\dagger C_{-l2}^\dagger) (C_{l1}^\dagger C_{-l1}^\dagger)\right] |0 \rangle \sim |\alpha \dots (l2)(-l2)(l1)(-l1)|0\rangle  
$$
while the annihilation ops acting on the bra vacuum generate another, say 
$$
\langle 0 |\left[ C_\alpha^\dagger \dots (C_{l'2}^\dagger C_{-l'2}^\dagger) (C_{l'1}^\dagger C_{-l'1}^\dagger)\right]  \sim \langle \alpha \dots (l'2)(-l'2)(l'1)(-l'1)|    
$$ 
(in bra form). So each term would read $\langle \alpha \dots (l'2)(-l'2)(l'1)(-l'1)\; |\;\alpha \dots (l2)(-l2)(l1)(-l1)\rangle$. 
But the two states are in each case either orthogonal to each other (different quantum numbers) or identical (same quantum numbers). Obviously only the identical states survive, in which case $l'1 = l1$, $l'2 = l'2$, etc, and overall these correspond to keeping factors with $l=l'$:
$$
\left<0\right|\prod_{l>0}(u_l^*+v_l C_{-l} C_l)C_\alpha^\dagger C_\alpha (u_{l}^*+v_{l} C_{l}^\dagger C_{-l}^\dagger) \left|0\right>
$$
If you use again the anti-commutation relations to set aside factors in $\alpha$ and $l \neq \alpha$, you can calculate separately the vacuum average for each mode, which is what the text you link to shows how to do. In essence it all comes down to the fact that only terms containing the exact same type of creation and annihilation operators survive, while all others generate mutually orthogonal states and vanish. Equivalently, unpaired annihilation ops annihilate the ket vacuum and unpaired creation ops annihilate the bra vacuum. 
Same logic applies to the calculation of $\langle N^2\rangle$, just somewhat more involved.
