# Usage of $\hat{\psi}^{\dagger}$ operator

When I was reading the original paper about Runge-Gross theorem (Phys. Rev. Lett. 52, 997 (1984)), I saw $\hat{\psi}$ operator notation.

I've never seen these notations from my QM text, and I'm confused now.

In this paper, each operator consisting of the Hamiltonian $\hat{H}(t) = \hat{T} + \hat{V}(t) + \hat{W}$ is assummed to have the form of

kinetic energy: $$\hat{T} = \sum_{s} \int d^{3}r \hat{\psi}_{s}^{\dagger}(\vec{r}) \left( -\frac{1}{2} {\nabla}^{2} \right) \hat {\psi}_{s}(\vec{r})$$

one-particle potential: $$\hat{V} = \sum_{s} \int d^{3}r \hat{\psi}_{s}^{\dagger}(\vec{r}) v(\vec{r} \; t) \hat {\psi}_{s}(\vec{r})$$

two-particle potential: $$\hat{W} = \frac{1}{2}\sum_{s}\sum_{s'} \int d^{3}r \int d^{3}r' \hat{\psi}_{s}^{\dagger}(\vec{r}) \hat{\psi}_{s}^{\dagger}(\vec{r}') w(\vec{r'}, \vec{r}) \hat {\psi}_{s}(\vec{r}') \hat {\psi}_{s}(\vec{r})$$

At first I thought $\hat{\psi}_{s} \equiv \langle \mathbf{r}_{s}|$, but it seemed wrong when I plugged in above formula, and calculated $\langle \psi|\hat{T} | \psi \rangle$.

Actually, I found a similar question posted earlier (What exactly is $\hat{\psi}^\dagger(x)$? An operator or a function?), but it doesn't give a clear answer for that.

I want to know textbooks or articles about the operator $\hat{\psi}_{s}$, and how to operate it manually when it applied to a physical state $|\psi \rangle$.

You have encountered a formalism, that for better or worse (probably for worse) has come to be associated with the name second quantization. This formalism is primarily designed for situations where you can have one-particle wavefunctions $\psi(x)$, but also two-, three-, and even $n$-particle states with wavefunctions $\psi(x_1,x_2)$, $\psi(x_1,x_2,x_3)$ or $\psi(x_1,x_2,\ldots,x_n)$, and so on, and it enables you to treat the cases of arbitrary $n$ (and even superpositions of different $n$) within one unified setting.
(The name 'second quantization' is misleading and unfortunate. Nothing is getting "quantized again", and the appearance of hats on $\psi$s is purely cosmetic. The 'first-quantization' and 'second-quantization' formalisms are essentially equivalent and the former fits in unmodified as the special case $n=1$ of the latter.)
Returning to $\hat \psi^\dagger(x)$, though: in short, this describes an operator that takes an $n$-particle position eigenstate $| x_1,\ldots,x_n\rangle$, and it transforms it into an $(n+1)$-particle position eigenstate by inserting an additional particle at position $x$, i.e., at the argument of $\hat\psi(x)$: $$\hat \psi^\dagger(x)| x_1,\ldots,x_n\rangle = | x_1,\ldots,x_n,x\rangle.$$ Thus, $\hat\psi^\dagger(x)$ "creates" a particle at $x$. Similarly, and following from the above, $\hat \psi(x)$, "destroys" a particle at $x$, in some suitable sense.
You should be careful, however, with normalization, and with the bosonic or fermionic symmetry of the particles - the state $| x_1,\ldots,x_n\rangle$ is not equal to $| x_1\rangle \otimes \ldots \otimes |x_n\rangle$, but rather to its symmetrized or antisymmetrized version. This should make it clearer that the formula above is a huge simplification of what would otherwise be a mess of symmetrizers, Slater determinants, and whatnot.