Question

show that linear combination of e^ikx and e^-ikx is a eigen function of operator d²/dx²

Answer 1

Let's make the combination:

$$f(x) = Ae^{ikx} + Be^{-ikx}$$

and apply the operator d²/dx² to it:

$$\dfrac{d^2}{dx^2}[f(x)] = \dfrac{d^2}{dx^2}[Ae^{ikx} + Be^{-ikx}] = \dfrac{d^2}{dx^2}[Ae^{ikx}] + \dfrac{d^2}{dx^2}[ Be^{-ikx}] =\\ =(ik)^2Ae^{ikx} + (-ik)^2Be^{-ikx} = -k^2Ae^{ikx}-k^2Be^{-ikx} =\\ = -k^2(Ae^{ikx} + Be^{-ikx}) = -k^2f(x)$$

Thus, obtain:

$$\dfrac{d^2}{dx^2}[f(x)] = -k^2f(x)$$

By definition, the function $f(x)$ is an egenfunction of some operator $A$ if $A[f(x)] = \lambda f(x)$, which is the case here. Thus, the linear combination of e^ikx and e^-ikx is a eigen function of operator d²/dx².

Q.E.D.