Quantum mechanics uses notions of systems, states, observables, and dynamics. These notions pervade all textbook physical theories.

A system is any object we want to study the properties of. We say that a system occupies a state, which, operationally, completely specifies information about the system (as much as the physical theory allows). An observable is (usually) a physical quantity of interest. Given the system in some initial state, a dynamic relation (usually a differential equation) tells us how the state changes as time evolves.

In (pure) quantum mechanics, systems are Hilbert spaces, or complex vector spaces. We denote such structures by $\mathcal{H}$. States of the system are unit vectors of the space, denoted $\lvert \psi \rangle \in \mathcal{H}$. Observables are Hermitian operators over Hilbert space $$\hat{O}: \mathcal{H}\to \mathcal{H} \text{ such that } \hat{O}^\dagger = \hat{O}.$$ The dynamic relation is the Schrodinger equation $$i\hbar \frac{\partial}{\partial t} \lvert \psi\rangle = \hat{H}\lvert \psi \rangle.$$ Finally, we must posit a rule that connects all these mathematics to something in the real world. This is called Born's rule $$\text{Pr}(\lvert \psi \rangle \to \lvert a \rangle)= \lvert \langle a \lvert \psi \rangle \lvert^2,$$ which gives the probability of obtaining the measurement outcome $a$ having made a measurement of the observable $\hat{A}$ on initial state $\lvert \psi \rangle$. I have stated these facts in a basis independent manner. Almost all of the quantities in your question are written in the position basis, which may be contributing to your confusion.

Now, to get to your question. In classical mechanics, observables always take on definite values. For example, I can solve for the trajectory of a point-like object using Newtonian mechanics. In doing so, I will obtain equations that exactly describe the position of the point-like object $\{x(t), y(t)\}$.

In quantum mechanics, observables do not always take on definite values. In particular, we are forced to consider the expectation values of observables with respect to a state of the system $$\langle \psi \lvert \hat{O} \lvert \psi \rangle.$$

Now, let $\hat{p}$ be the momentum operator. The quantity $\hat{p}\lvert \psi \rangle$ is indeed another state (recall that observables are simply maps over the space of states). It, to my understanding, doesn't really have a generally useful physical interpretation.

Quantum mechanics uses notions of systems, states, observables, and dynamics. These notions pervade all textbook physical theories.

A system is any object we want to study the properties of. We say that a system occupies a state, which, operationally, completely specifies information about the system (as much as the physical theory allows). An observable is (usually) a physical quantity of interest. Given the system in some initial state, a dynamic relation (usually a differential equation) tells us how the state changes as time evolves.

In (pure) quantum mechanics, systems are Hilbert spaces, or complex vector spaces. We denote such structures by $\mathcal{H}$. States of the system are unit vectors of the space, denoted $\lvert \psi \rangle \in \mathcal{H}$. Observables are Hermitian operators over Hilbert space $$\hat{O}: \mathcal{H}\to \mathcal{H} \text{ such that } \hat{O}^\dagger = \hat{O}.$$ The dynamic relation is the Schrodinger equation $$i\hbar \frac{\partial}{\partial t} \lvert \psi\rangle = \hat{H}\lvert \psi \rangle.$$ Finally, we must posit a rule that connects all these mathematics to something in the real world. This is called Born's rule $$\text{Pr}(\lvert \psi \rangle \to \lvert a \rangle)= \lvert \langle a \lvert \psi \rangle \lvert^2,$$ which gives the probability of obtaining the measurement outcome $a$ having made a measurement of the observable $\hat{A}$ on initial state $\lvert \psi \rangle$. I have stated these facts in a basis independent manner. Almost all of the quantities in your question are written in the position basis, which may be contributing to your confusion.

Now, to get to your question. In classical mechanics, observables always take on definite values. For example, I can solve for the trajectory of a point-like object using Newtonian mechanics. In doing so, I will obtain equations that exactly describe the position of the point-like object $\{x(t), y(t)\}$.

In quantum mechanics, observables do not always take on definite values. In particular, we are forced to consider the expectation values (an average) of observables with respect to a state of the system. Expectation values are computed as $$\langle \psi \lvert \hat{O} \lvert \psi \rangle.$$

Now, let $\hat{p}$ be the momentum operator. The quantity $\hat{p}\lvert \psi \rangle$ is indeed another state (recall that observables are simply maps over the space of states). It, to my understanding, doesn't really have a generally useful physical interpretation.