Second Quantization-Kinetic operator

[ChrisVer]

I am feeling a little stupid tonight... So let me build the problem...
For a single particle operator O, we have in the basis |i> we have that:
O= \sum_{ij} o_{ij} |i><j| with o_{ij}=<i|O|j>
Then for N particles we have that:
T=\sum_{a}O_{a}= \sum_{ij} o_{ij} \sum_{a} |i>_{a}<j|_{a} with a=1,2...,N

How can we write this afterwards (for bosons) in respect to the creation and annihilation operators as:
T= \sum_{ij} o_{ij} c_{i}^{t} c_{j}

From what I suspect, \sum_{a} |i>_{a}<j|_{a} must be equal to the number operator... But for some reason I'm unable to see it...Also I tried taking this:
O= \sum_{ij} o_{ij} |i><j|
O= \sum_{ij} o_{ij} c_{i}^{t}|0><0|c_{j}
And then summing over the particles:
O= \sum_{ij} o_{ij} c_{i}^{t}c_{j} \sum_{a}|0>_{a}<0|_{a}
but I see no reason why the sum must be equal to unity \sum_{a}|0>_{a}<0|_{a} =1

Please give hints, not answers

 

[DrClaude]

Consider a $N$-particle operator $O$ as a sum of one-particle operators
$$
O = \sum_{i=1}^{N} O_i
$$
First assume that you can write it as
$$
O = \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta
$$
If you change the basis, you can write
$$
O' = \sum_{\gamma, \delta} \langle \gamma | O_1 | \delta \rangle b_\gamma^\dagger b_\delta
$$
The creation and annihiliation operators satisfy the linear transformations
$$
\begin{align}
b_\gamma^\dagger &= \sum_{\alpha} c_\alpha^\dagger \langle \alpha | \gamma \rangle \\
b_\gamma &= \sum_{\alpha} \langle \gamma | \alpha \rangle c_\alpha
\end{align}
$$
Therefore,
$$
\begin{align}
O' &= \sum_{\gamma, \delta} \sum_{\alpha, \beta} c_\alpha^\dagger \langle \alpha | \gamma \rangle \langle \gamma | O_1 | \delta \rangle \langle \delta | \beta \rangle c_\beta \\
&= \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta \\
&= O
\end{align}
$$
This means that the assumed representation of $O$ is independent of the basis. If you choose a basis $| \gamma \rangle$ that diagonalizes $O_1$, then
$$
O = \sum_\gamma o_\gamma b_\gamma^\dagger b_\gamma
$$
where $o_\gamma$ is the eigenvalue associated with the ket $| \gamma \rangle$, which is what we expected from the first formula above. Therefore, the assumption for the form of $O$ in any basis was correct.