# Single particle operator in second quantization

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

$$\hat H=\sum_i \varepsilon_i \hat a_i^{\dagger} \hat a_i$$ 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?

• This is quite a broad question. Are you simply asking what does the expression for $H$ mean, which should be covered in any textbook covering second quantization, or are you asking about why do we choose to write it this way rather than some "more obvious" way that is more recognisable from elementary QM? – By Symmetry Oct 30 '17 at 11:03
• It's not that I want to know what are advantages of this formulation. I want to understand why this formulation follows from the elementary QM. In addition why can we write for a two particle operator than $$V(t) = \frac{1}{2} \sum_{klmn} v(kl; nm) a^{\dagger}_k(t) a_l^{\dagger}(t) a_m(t) a_n(t)$$ with $v(kl; nm) = v(\mathbf{k}-\mathbf{n}) \delta_{k+l,m+n} \delta_{\sigma_k} \delta_{\sigma_n} \delta_{\sigma_m\sigma_l}$ The question is: Where is the connection between elementary and second quantization? – Leviathan Oct 30 '17 at 11:08
• You can look at first two chapters of "Quantum Field Theory of Non-equilibrium States" by Rammer to get a good grasp of second quantization formalism for non-relativistic systems. – Sunyam Oct 30 '17 at 13:40
• Have a look into Fetter, Walecka "Quantum theory of many-particle systems". – WoofDoggy Oct 30 '17 at 14:37
• Thanks for the books, I will have a look on them and ask again, when I'm deeper in the thematic. – Leviathan Oct 30 '17 at 18:34

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, $\hat{n} \equiv a^\dagger a$.
• Yes, exactly but there is some mistake: $$a^{\dagger} |0\rangle = 0$$ I think or ? – Leviathan Nov 2 '17 at 9:56