But can I apply this reasoning? This result has been obtained from a classical phenomenon(Electrostatics).

Any time you take a classical problem and quantize it, you are making an educated guess as to what the Hamiltonian will look like. In general, there are many different quantum mechanical systems which have the same classical limit, so there's no unique way of translating a classical problem into the "correct" underlying quantum mechanical problem.

Your reasoning makes sense, and in accordance with Ehrenfest's theorem, it respects the correspondence principle. I don't think you can say (or ask for) any more validation than that.

Secondly why in this case $E>V(x)$

You could answer this a few different ways. From a physical standpoint, one could argue that the free particle Hamiltonian $H_{free}=-\frac{d^2}{dx^2}$ is just the kinetic energy operator, and for a that state with definite kinetic energy, that kinetic energy should be positive.

You could also observe that the eigenstates corresponding to negative energies take the form $e^{\pm\alpha x}$ with $\alpha >0$, and "blow up" either at large positive or large negative values.

Of course, you could also argue that neither of these arguments is satisfactory. The first, for example, assumes something about kinetic energy (whatever that is, in the context of quantum mechanics) - maybe in quantum mechanical systems, kinetic energy can be negative. The second argument is also problematic because while it's true that functions of the form $e^{\pm i\alpha x}$ don't diverge to infinity, they are also not normalizable, so by a similar argument we should throw them away as well.

The correct answer to this question is a bit technical. Roughly speaking, operators with continuous spectra (such as the momentum and energy operators in this case) do not actually have eigenvectors and eigenvalues, because if $\psi$ obeys the equation

$$P \psi = \lambda \psi$$

then it can be shown that $\lambda \notin L^2(\mathbb R)$. In physics, we still hold on to those states, but simply note that the are not physical (i.e. in real life, you cannot truly have a free-particle state of definite momentum or energy). In such cases, we replace the requirement that such states be square-integrable with the requirement that their inner products with physical states be well-defined. In other words, if $\psi_P$ is an unphysical state of definite momentum and $\phi$ is an actually physical state, then we demand that $$\langle \psi_P , \phi\rangle < \infty$$

This is what separates functions like $e^{i\alpha x}$ from functions like $e^{\alpha x}$. The former satisfy the above requirement, while the latter do not. Therefore, even though neither are proper, normalizable states, we keep the former (and call them "generalized eigenstates") but must throw away the latter.

From this, it follows that the only allowed states of definite energy take the from $e^{\pm i\alpha x}$, which correspond to strictly positive kinetic energy terms when you plug them into $H_{free}$.

But can I apply this reasoning? This result has been obtained from a classical phenomenon(Electrostatics).

Any time you take a classical problem and quantize it, you are making an educated guess as to what the Hamiltonian will look like. In general, there are many different quantum mechanical systems which have the same classical limit, so there's no unique way of translating a classical problem into the "correct" underlying quantum mechanical problem.

Your reasoning makes sense, and in accordance with Ehrenfest's theorem, it respects the correspondence principle. I don't think you can say (or ask for) any more validation than that.

Secondly why in this case $E>V(x)$

You could answer this a few different ways. From a physical standpoint, one could argue that the free particle Hamiltonian $H_{free}=-\frac{d^2}{dx^2}$ is just the kinetic energy operator, and for a that state with definite kinetic energy, that kinetic energy should be positive.

You could also observe that the eigenstates corresponding to negative energies take the form $e^{\pm\alpha x}$ with $\alpha >0$, and "blow up" either at large positive or large negative values.

Of course, you could also argue that neither of these arguments is satisfactory. The first, for example, assumes something about kinetic energy (whatever that is, in the context of quantum mechanics) - maybe in quantum mechanical systems, kinetic energy can be negative. The second argument is also problematic because while it's true that functions of the form $e^{\pm i\alpha x}$ don't diverge to infinity, they are also not normalizable, so by a similar argument we should throw them away as well.

The correct answer to this question is a bit technical. Roughly speaking, operators with continuous spectra (such as the momentum and energy operators in this case) do not actually have eigenvectors and eigenvalues, because if $\psi$ obeys the equation

$$P \psi = \lambda \psi$$

then it can be shown that $\psi \notin L^2(\mathbb R)$. In physics, we still hold on to those states, but simply note that the are not physical (i.e. in real life, you cannot truly have a free-particle state of definite momentum or energy). In such cases, we replace the requirement that such states be square-integrable with the requirement that their inner products with physical states be well-defined. In other words, if $\psi_P$ is an unphysical state of definite momentum and $\phi$ is an actually physical state, then we demand that $$\langle \psi_P , \phi\rangle < \infty$$

This is what separates functions like $e^{i\alpha x}$ from functions like $e^{\alpha x}$. The former satisfy the above requirement, while the latter do not. Therefore, even though neither are proper, normalizable states, we keep the former (and call them "generalized eigenstates") but must throw away the latter.

From this, it follows that the only allowed states of definite energy take the from $e^{\pm i\alpha x}$, which correspond to strictly positive kinetic energy terms when you plug them into $H_{free}$.