Derivation of Single Particle Operator in second quantization? See, I have looked through a bunch of scripts about second quantization on the internet, but everywhere at some point something weird is happening so I get stuck over and over and over again, which is a little bit depressing. So here is the thing: Let‘s assume some single particle operator $\mathcal{O}^{(1)}$. Now taken it is separable acting on a N-particle Hilbert space one can represent the operator by 
$$\mathcal{O}^{(1)}=\sum_{i}\mathcal{o}_{i}=\sum_{\alpha,\beta,i}\langle{\alpha}|\mathcal{o}_{i}|{\beta}\rangle|{\alpha}\rangle\langle{\beta}|.$$ 
Now $|\alpha\rangle={a^{\dagger}}_{\alpha}|0\rangle$ so I could write 
$$\mathcal{O}^{(1)}=\sum_{\alpha,\beta,i}\langle{\alpha}|\mathcal{o}_{i}|{\beta}\rangle{a^{\dagger}}_{\alpha}|{0}\rangle\langle{0}|a_{\beta}$$ 
which in the result should be equal to 
$$\sum_{\alpha,\beta,i}\langle\alpha|\mathcal{o}_{i}|\beta\rangle a^{\dagger}_{\alpha}a_{\beta}.$$ 
But I can‘t believe that these two expressions are equal... Can someone please explain to me how I get to the operator expression of second quantization or at least recommend some website or something where it is explained really well ? Thank you in advance!
 A: Okay, so I cleaned up the latex a bit (Use \langle and \rangle for the braket notation. Though I wish they'd install the braket package, which would make it even easier). 
First, you have a mistake in the first line, and it doesn't quite make sense when going to second quantization anyway: There should be no sum over individual particles any more! In your first line, what sort of basis do the states $|\alpha\rangle$ even refer to? If they are single particle basis states of the operator, we can't use them like that in the many-particle operator. 
And that's probably why you run into issues. I'd skip that step entirely. In "first" quantization, we have $\mathcal{O}^{(1)} = \sum_i o_i$ and in second quantization we have 
$$\mathcal{O}^{(1)} = \sum_{\alpha, \beta} \langle \alpha | o | \beta \rangle 
a^\dagger_\alpha a_\beta$$
To prove that they are indeed equivalent, we have to show that they have the same matrix elements in a given basis. We can pick any basis we want, so of course we pick the appropriate many-particle basis of, e.g., Slater determinants.
Some info on this can be found eg here: http://physics.gu.se/~tfkhj/OsloSecondQuant.pdf
