neutrino said:
That's right. But what I want to know is how does one come up with that particular expression for the momentum operator - that combination of i, h-bar etc. I hope I'm clear.
Each physical observable has associated with it an operator, which you know. These operators that correspond to physical entities have a geometric interpretation. In quantum mechanis the relations (i.e. commutation rules) between operators are derived by considering the behaviour and the nature of the generators of infinitesimal symmetry transformations, and then using these generators we can produce unitary representations of finite transformations of the symmetry group that act on wave-functions. From these we can then identify the generators of said objects with dynamical quantities, whose eigenvalues give observables.
An infinitesimal transformation by a parameter s can be expanded as a taylor series to first order in s, assuming U(0) is the indentity transformation,
U(s)=1+\left.\frac{dU}{ds}\right|_{s=0}s+\mathcal{O}(s^2)
Assuming unitarity UU^{\dagger}=1
UU^{\dagger}=1+s\left.\left[\frac{dU}{ds}+\frac{dU^{\dagger}}{ds}\right]\right|_{s=0}+\mathcal{O}(s^2)
To satisfy unitarity we require \left.\frac{dU}{ds}\right|_{s=0}=iK, with K self-adjoint, this K is the generator of infinitesimal transformations. One can also generate finite transformations using this generator.
In non-relativistic quantum mechanics this group of symmetry operations is the Galilei group, consisting of 10 generators of the following symmetry operations on the space-time of classical physics: spatial and temporal translations and non-relativistic rotations and boosts. Respectively, these generators are: momentum, total energy (Hamilontian), angular momentum and some operator that I recall be expressable as a multiple of some other operator (cannot remember precisely). In relativistic quantum mechanics this symmetry group is the Poincare or lorentz group (the lorentz group only consists of rotations and boosts, not translations as well, which Poincare does).
Momentum is responsible for a spatial translations, thus some finite coordinate shift by a vector a of a scalar wave-function is
\langle\mathbf{x}|\exp\left[\tfrac{i\mathbf{a}\cdot\mathbf{P}}{\hbar }}\right]|\psi \rangle=\psi (\mathbf{x-a})
Expanding the exponential on the left gives one a series of ascending powers of the momentum operator. Expanding the right as a taylor series about \mathbf{x}=\mathbf{a} gives one a series with ascending derivatives. Identifying the terms in expansions of the left and right hand side allows one to conclude the form of the momentum operator. This is true in both relativistic and non-relativistic quantum mechanics, as this derivation does not rely on any kind of motion. The same procedure gives one an energy operator.
. I know there aren't any derivations for an operator associated with an observable and they are usually defined in a certain form. So why do they have those particular forms. Was it trial and error method? Or any other reason(s)? I'm still a bit rusty with my understanding of basic QM, so please be gentle. 
May be a few months of intense study will help me understand what you're saying. 
One really nice book is by Nouredine Zettili. The great thing about this book (besides being a generally clear treatment) is that Zettili provides numerous nontrivial worked example problems at the end of each chapter. There are then lots more problems afterwards for you try your hand at. It's really quite helpful, especially for self study. I don't know if the book is in your price range, but I would recommend taking a look.
