According to what I have read, the second quantisation originally came from the effort to quantise the many body wave function in the Schrodinger equation.
We could write down the commutation relation of $\psi$ and $\psi^{\dagger}$ by first applying
$$|\psi(t)\rangle = \sum_n \psi_n(t) |n\rangle $$
to the Schrodinger equation
$$H|\psi(t)\rangle=i\hbar \partial_t |\psi(t)\rangle$$
$$ \langle n|H| \psi(t) \rangle = \sum_m \psi_m(t) \langle n|H|m \rangle = i\hbar \dot \psi_n(t) $$
Then consider the Hamiltonian as a functional of $\psi$ and $\psi^{*}$, that is
$$H(\psi,\psi^*)=\langle H\rangle = \langle\psi|H|\psi\rangle = \sum_{m,n} \langle\psi| m \rangle \langle m|H|n \rangle \langle n|\psi \rangle = \sum_{m,n} \psi_m^* \psi_n \langle m|H|n \rangle$$
Hence, by doing partial differential on $H$ with respect to $\psi$ and $\psi^*$, we get
$$\dot{\psi_m} = \frac{\partial H}{\partial (i\hbar \psi_m^*)}$$
$$i\hbar\dot{\psi_m^*} = -\frac{\partial H}{\partial \psi_m}$$
This has the same form as the Hamilton equations. Therefore, regarding $\psi$ as an operator, we can identify $\hat\psi_n \rightarrow \hat q_n$ and $i\hbar \hat\psi_n^\dagger \rightarrow \hat p_n$.
The commutation relation between $\hat\psi$ and its conjugate is then
$$[\hat\psi,\hat\psi^\dagger] = 1$$
However, I still cannot really see what the operator $\hat\psi$ actually does.
I have seen in many textbooks and lecture notes that $\hat\psi$ is an operator that annihilates a particle from the many-body wave function. Why is that?