Contructive Proof of 2nd Quantization form of Operators
Is there a constructive proof for these forms of operators in second quantization
$$R= \sum \limits_a \sum \limits_b \langle a | R_1 | b \rangle C_a^\dagger C_b $$ using the general form $R = \sum \limits_i^N R_i$ where we sum over all $n$ single particle state operators?
* a more precise definition of these opertors on an $N$ particle Hilbert space would be $R|\Psi\rangle=\sum \limits_i^N |\Psi_1\rangle\otimes..\otimes R_i |\Psi_i\rangle\otimes...\otimes|\Psi_N\rangle$
All this is done with fermionic operators and states. $C_a^\dagger$ is the creation operator, where $a$ refers to a quantum number uniquely identifying a one particle state. The sums over the quantum numbers are just over all the possible states i suppose, as that would be sensible. I can not find a definition of the single particle operators in this book, which is of course not helping.
I was trying to understand <PERSON> proof for a general pair operator when this question occured to me. 669
It looks as if going directly should be a lot easyer to just apply to that case aswell.
I tried many things but i don't seem to have a firm enough grasp on the formalism to realize this on my own. I am not shure if that is total nonsense or if i just don't know enaugh to do the last step here. I just put this example in so you know as a rough direction what I mean by contructive proof. Somehow of course the vacuum states should vanish here and i would need to somehow discover a $ \delta_{i,1}$ in that equation but who knows...
$$R= \sum \limits_i R_i =\sum \limits_i \sum \limits_{a,b} |a\rangle\ \langle a|R_i|b\rangle\langle b|=\sum \limits_i \sum \limits_{a,b} C_a^\dagger |0\rangle\ \langle a|R_i|b\rangle\langle 0|C_b $$ $$ =\sum \limits_i \sum \limits_{a,b} \langle a|R_i|b\rangle\; C_a^\dagger |0\rangle\ \langle 0|C_b$$ I would really appriciate any Help on this and on the possibility of a construction of the general additive pair operator in a similar way. 66