Probability density of time-dependent wave functions

Why is it so that probability density of eigenfunctions of time-dependent schrodinger equation are time independent while that of general wave functions (which are a combination of the eigenfunctions) are not?

It is important to understand how the eigenfunctions are defined. In particular, here we talk about the eigenfunctions of the hamiltionian, i. e. functions that satisfy

$$\hat{H} \psi(x) = E \psi(x) ,$$

where $$\psi(x)$$ is an eigenfunction (function of space only) and $$E$$ is its corresponding eigenvalue. In case of the hamiltonian the eigenvalue is also the energy of its associated quantum state.

Now, the total eigenfunction is a function of space AND time. The form of these full eigenfunctions is:

$$\psi(x, t) = \psi(x) \exp{(-iEt)},$$

and the probability density is $$P(x, t) =\psi(x, t) \psi^* (x, t)$$.

Now you can see that if one takes a single eigenfunction, then the time-dependence (the exponential part) cancels with its complex conjugate in the probability density.

If one takes a linear combination, e.g. $$\psi_1 + \psi_2$$, then the probability density is:

$$P(x, t) = \left( \psi_1(x)\exp{(-iE_1 t) } + \psi_2(x)\exp{(-iE_2 t) } \right) \left( \psi_1(x)\exp{(-iE_1 t) } + \psi_2(x)\exp{(-iE_2 t)} \right)^*.$$ Taking the complex conjugate one obtains

$$P(x, t) = \left( \psi_1(x)\exp{(-iE_1 t)} + \psi_2(x)\exp{(-iE_2 t)} \right) \left( \psi_1(x)\exp{(iE_1 t)} + \psi_2(x)\exp{(iE_2 t)} \right).$$

Now there is no reason why the above should be time independent. The exponentials with $$E_1$$ do not cancel exponentials with $$E_2$$, therefore one expects mixing terms such as $$\exp{i(E_1-E_2)t}$$ in the final probability density, and thus it will no longer be time independent.