Pronounced duh broy His fundamental assumption was that particles's momentum and wavelength were inversely proportional.

λ=hp\lambda = \frac{h}{p}

i.e.,

⇔p=ℏk\Leftrightarrow p=\hbar k

We also experimentally found out that a momentum-p eigenstate has a single, Definite wavelength.

Let's write the momentum Wave function ψp(x)\psi_p(x) in polar form.

ψp(x)=A(x)eiθ(x)\psi_p(x)=A(x)e^{i\theta (x)}

Since we know that the wave number is Definite,

dθdx=k=const\frac{d\theta}{dx}=k=\text{const}

we integrate that to get

θ(x)=kx+θ0\theta(x) = kx + \theta_0

Because definite momentum forces the Magnitude of the wave to be constant i.e.,

∣ψp(x)∣2=A(x)2=const|\psi_p(x)|^2 = A(x)^2 = \text{const}

hence

A(x)=A0=constA(x) = A_0 = \text{const}

hence

ψp(x)=A0 ei(kx+θ0)=A0 eiθ0⏟= C eikx\psi_p(x) = A_0\,e^{i(kx+\theta_0)} = \underbrace{A_0\,e^{i\theta_0}}_{=\,C}\,e^{ikx}

In position space, a momentum eigenstate must look like a plane wave

ψp(x)=⟨x∣p⟩=Ceipx/ℏ\psi_p(x)=\braket{x|p}=Ce^{ipx/\hbar}