Contructive Proof of 2nd Quantization form of Operators

Question

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 Ballentines proof for a general pair operator when this question occured to me. 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.

Answer 1

It is posslible to do this proof in a direct way using a slightly different and in my opinion a lot more sensible approach to the creation/annihilation operators. I will write it up here for the benefit of any interested.

The inner product of $N$ particle states can be written as: $$ \langle\chi_1,...,\chi_N|\Psi_1,...,\Psi_N\rangle= \det {\begin{pmatrix}\langle\chi_1|\Psi_1\rangle &...& \langle\chi_1|\Psi_N\rangle \\.\\.\\\langle\chi_N|\Psi_1\rangle& ...& \langle\chi_N|\Psi_N\rangle\end{pmatrix}}$$ Using th Slater determinant for fermion- and the permanent for Boson-states.

With this we get a new representation of the annihilation operator $C_a$. We are now looking at $N-1$ particle states:

$$ \langle\chi_1,...,\chi_{N-1}|C_a |\Psi_1,...,\Psi_N\rangle = \langle\Psi_1,...,\Psi_N|C_a^\dagger |\chi_1,...,\chi_{N-1}\rangle^* \\ \langle\Psi_1,...,\Psi_N| a,\chi_1,...,\chi_{N-1}\rangle^*=\det {\begin{pmatrix}\langle \Psi_1 |a \rangle &\langle\Psi_1|\chi_1\rangle &...& \langle\Psi_1|\chi_{N-1}\rangle \\.\\.\\ \langle \Psi_1|a \rangle & \langle\Psi_N|\chi_1\rangle& ...& \langle\ \Psi_N|\chi_{N-1}\rangle\end{pmatrix}}^*$$ from the definition of the determinant we know we can write this equally as: $$=\sum \limits_{k=1}^N \xi^{k-1}\langle a|\Psi_k \rangle \langle\chi_1,...,\chi_{N-1}|\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$ whre $\xi=-1$ for fermions and $+1$ for bosons and $\sim\Psi_k$ means $\Psi_k$ is not part of that state. This gives a new representation of the annihilatinon operator $$C_a|\Psi_1,...,\Psi_N\rangle=\sum \limits_{k=1}^N \xi^{k-1}\langle a|\Psi_k \rangle |\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$

We now need to look at a one particle Operator $R_{ab}=|a\rangle \langle b|$ definied as in the question above: $$R_{ab}|\Psi_1,...,\Psi_N\rangle=\sum \limits_i^N |\Psi_1\rangle\otimes..\otimes \underbrace{| a \rangle}_{\text{i-th ket}} \otimes...\otimes|\Psi_N\rangle \langle b | \Psi_i\rangle $$ with this and the newfound representation of the operators we see that the operator $R_{ab}$ can be writtten as $R_{ab}=C_a^\dagger C_b$. How can we see this?

First $$C_a^\dagger C_b |\Psi_1,...,\Psi_N\rangle= \sum \limits_{k=1}^N \xi^{k-1}\langle b|\Psi_k \rangle |a,\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$ second using the symmetry of the states we get: $$|a,\Psi_1,...,\Psi_{k-1},\Psi_{k+1},...,\Psi_N\rangle=\xi^{k-1}|\Psi_1,...,\Psi_{k-1},a,\Psi_{k+1},...,\Psi_N\rangle$$

This results in the desired equality: $R_{ab}=C_a^\dagger C_b$. With this we now have finished the proof since i allows us to write any one particle operator A as: $$A=\sum \limits_{k=1}^N \sum_{a,b} |a\rangle \langle a| A |b\rangle \langle b|= \sum_{a,b} C_a^\dagger C_b \langle a| A |b\rangle $$ the sum over $k$ went into the operator $C_b$, as it should according to the introduced representation. This is what we wanted. Have a nice day. Please feel free to critique.

EDIT: this method can be used for the two particle interaction as well.