quantum mechanics - Single particle operator in second quantization

I want to understand why we write in the formalism of second quantization for a single particle operator

\begin{equation} \hat H=\sum_i \varepsilon_i \hat a_i^{\dagger} \hat a_i \end{equation} where $\varepsilon_i$ is the eigenvalue of the solved Schroedinger equation. Is it just the fact that I know that $a_i^{\dagger} \hat a_i=\hat n_i$ and I associate the Hamiltonian with the energy of the system: $H=E=\sum_i n_i \varepsilon_i$, or how can one understand this?

Answer

In the second quantization (physics of many-body systems) language, the (physical) question is "How many particle in each state?". Suppose that there is $n_\alpha$ particles in state $\alpha$, each particle in this state has energy $\epsilon_\alpha$, then the total energy in this state is: $n_\alpha\epsilon_\alpha$. If we want to have the total energy of the system, we just simply add all possible state's energies: $$E = \sum_\alpha n_\alpha\epsilon_\alpha$$ We see that E and $n_\alpha$ are physical observables. In QM, each physical observable corresponds to a hermitian operator. Hence, naturally, E is corresponding to Hamiltonian, which is the energy operator; and $n_\alpha$ is corresponding to $\hat{n}_\alpha$, occupation number operator. It turns out that $n_\alpha$ and E are eigenvalues of $\hat{n}_\alpha$ and H, respectively. So we have: $$H = \sum_\alpha \hat{n}_\alpha\epsilon_\alpha$$

The question now is to find the number operator. We discuss here the bosonic case. In the many-body system, since the number of particle can be changed, we introduce the creation and annihilation operators: $a^\dagger$ and a. Their definitions are: $$a^\dagger(k)|0\rangle = |k\rangle$$ $$a^\dagger(k_{n+1})|k_1,k_2,...,k_n\rangle = \textrm{(constant)}_1|k_1,k_2,...,k_n,k_{n-1}\rangle$$

$$a|0\rangle = 0$$ $$a(k)|k_1,k_2,...,k_n\rangle = \textrm{(constant)}_2\sum_i\delta(k,k_i)|k_1,k_2,...,k_{i-1},k_{i+1},...,k_n\rangle$$ $\textrm{(constant)}_2$ and $\textrm{(constant)}_1$ can be obtained by permutation: $\textrm{(constant)}_1 = \sqrt{n+1}$, $\textrm{(constant)}_2 = 1/\sqrt{n}$. When acting $a^\dagger(k)a(k)$ on a n-particle state $|k_1,k_2,...,k_n\rangle$, we get: $$a^\dagger(k)a(k)|k_1,k_2,...,k_n\rangle = n|k_1,k_2,...,k_n\rangle$$ which is exactly the same as the action of the occupation number operator on the state $|k_1,k_2,...,k_n\rangle$. Then, $\hat{n} \equiv a^\dagger a$.

Is that what you want?