Time evolution of eigenstates superposition If a system is in a state $\psi$ which is superposition of, let's say two, energy eigenfunction, namely $\psi_1$ and $\psi_2$, so that $$\psi(t)=\psi_1e^{-i\omega_1t}+\psi_2e^{-i\omega_2t}$$ (I am omitting normalization constants for simplicity) then $$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2+\psi_1^* \psi_2 e^{i(\omega_1-\omega_2)t}+\psi_2^* \psi_1 e^{i(-\omega_1+\omega_2)t}$$
right?
But isn't it true that since $\psi_1$ and $\psi_2$ are orthogonal eigenfunctions then $$\psi_1^* \psi_2$$ and $$\psi_2^* \psi_1$$ must be zero and so one would have:
$$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2$$
How does it work? thanks
 A: Cross-terms like $\psi_1^*(x)\psi_2(x)$ integrate to $0$ but their product is not $0$ everywhere.  As a result, $\psi_1^*(x)\psi_2(x) e^{i(\omega_1-\omega_2)t}\ne 0$ in general.
Maybe two simple examples will illustrate the point:


*

*Infinite-well solutions 
For infinite-well wavefunction with $\psi_1(x)\sim\sin(\pi x/L)$ and $\psi_2(x)\sim \sin(2\pi x/L)$.  Certainly then $\psi_1^*(x)\psi_2(x)\ne 0$ in general although $\int_0^L dx\,\psi_1^*(x)\psi_2(x)=0$.

*Harmonic oscillator solutions
Take $\psi_0\sim e^{-\lambda x^2/2}$ and $\psi_1\sim x e^{-\lambda x^2/2}$.  Then again $\int_{-\infty}^\infty \psi_0^*(x)\psi_1(x)=0$ but $\psi_1(x)^*\psi_0(x)\sim x e^{-\lambda x^2}\ne 0$ except at $x=0$ and $x=\pm \infty$. 
A: While it's true that
$$\langle\psi_1|\psi_2\rangle = 0$$
it isn't necessarily true that
$$\psi^*_1(x)\psi_2(x) = 0$$
Why?  Because we can insert the identity
$$1 = \int\,\mathrm{d}x\,|x\rangle\langle x|$$
in the first equation to get
$$\int\,\mathrm{d}x\,\langle \psi_1|x\rangle\langle x|\psi_2\rangle = \int\,\mathrm{d}x\,\psi^*_1(x)\psi_2(x) = 0 $$
But the fact that the integral is zero doesn't imply that the integrand is zero, correct?
A: Your notation is ambiguous so it's not quite clear what you're actually asking. If by $\psi_1$ you mean the position basis-wavefunction $\psi_1(x)$, then as the other answers point out, only the integrals of the cross-terms with respect to $x$ vanish, not the pointwise products. If by $\psi_1$ you mean the basis-independent ket $|\psi_1\rangle$, and by $\psi_1^*$ the corresponding bra $\langle \psi_1|$, then your last equation is correct as written (with no need for any integration), and simply says that the magnitude of $|\psi\rangle$ is preserved over time, so if it's correctly normalized at one time then it remains correctly normalized at all other times.
