quantum mechanics - Free-particle solution to Schrödinger Equation

The free particle solution in stationary state (with definite energy) to the Schrödinger equation is

$$\psi(x,t) =Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)}$$

Since the energy is definite, and hence the momentum is definite, the uncertainty in position must be infinite. How is this reflected by the probability distribution function:

$$\Psi = |\psi(x,t)\psi^*(x,t)| $$

The book that I am using just look at the first term of the solution, and derive that the probability distribution function is $A^2$. However, I do not understand why we can do that?

Does it imply that if wave function is made up of n terms such that each individual term has a constant probability distribution function, the whole wave function also has a constant probability distribution function? If so, how can I prove it?

I know my question might be very vague but that is precisely the problem I am facing now, I don't even know how to ask about the things that I don't understand.

Answer

For any function of $x$ and $t$ that depends on the combination $x\pm vt$ (for constant $v$ represents a wave with a fixed shape that travels in the $\mp x$ direction with speed $v$. That is to say, $$ x\pm vt={\rm constant} $$ In your wave function, $$ \psi(x)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)}\tag{0}, $$ the first term represents a right-moving wave while the second term represents a left-moving wave. Since they differ only by the sign of $k$, we can simplify the wavefunction to $$ \psi(x)=Ae^{i(kx-\omega t)}\tag{1} $$ where $k>0$ means the right-moving wave (the first term of Eq (0) ) and $k<0$ means a left-moving wave (the second term of Eq (0)).

Using Equation (1) and multiplying by its complex conjugate, we get $$ \psi^*\psi(x)=\left(Ae^{-i(kx-\omega t)}\right)\left(Ae^{i(kx-\omega t)}\right)=A^2 $$