Basis vectors in quantum mechanics In quantum mechanics, sometimes we take discrete basis and other time we take continuous basis. I know the mathematical difference between both but what is the physical significance of both the basis.
Is it true to say that the discrete basis are used for bounded systems like particle in potential well and continuous basis are used for free particles?
 A: The basis is related to an observable; if the observable is discrete, then the basis is discrete and vice versa.

Is it true to say that the discrete basis are used for bounded systems like particle in potential well and continuous basis are used for free particles?

No. As a counterexample, consider an infinite square well. It has discrete energies, but the position is continuous. The same can hold for other bounded systems such as a finite square well or harmonic oscillator.
A: Yes, you are basically correct.  The basis set are the solution to the Schrödinger equation for a known / simpler system - generally in which there is an analytical / closed form solution.  If that system contains a bound particle, these will be a  discrete set of solutions to the Schrödinger equation; if it is an unbound or free particle there will be a continuous set of solutions.
Also important to note this is due to imposing the boundary conditions (bound vs. free) on the mathematical functions that are solutions to the Schrödinger equation.
