Why are electron wavefunctions standing waves? How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?
Is it a consequence of the fact that electrons don't drift away from the molecule?
In other words, can one prove from the Schrödinger equation that, unless $\psi(x,t)$ can be represented as $\phi(x)\theta(t)$, then $\lim_{t \to \infty}\int_U |\psi(\bar x,t)|^2d\bar x=0$ for any bounded set $U\subset \mathbb R^3$ (or something along those lines)?
Or are there physical considerations that explain the standing waves?
Update. Apparently «standing wave» is an ambiguous/controversial term here, so let me reformulate my question in a more mathematical and unambiguous way without referring to standing waves.
Let a wavefunction $\psi$ correspond to a stationary state, i.e. $|\psi(x,t)|=\mathrm{const}(t)$. We can conclude, then, that $\psi(x,t)=\phi(x)\theta(x,t)$, where $|\theta(x,t)|=1$. In order to separate the variables and move on to the time-independent Schrödinger equation, we also need to establish that $\theta(x,t)$ doesn't depend on $x$. Where does this assumption follow from?
 A: 
How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?

Actually, it's better not to. In modern Quantum Physics the idea of electrons as standing waves is increasingly seen as no more than an analogy and not a very good one either. In some cases like this system it's a rather compelling one but even there it's not necessary to think of bound particles as standing waves.
Instead look at the wave function $\psi$ as a mathematical function that contains all the information about the particle and with these properties.
Wave functions of bound particles are the eigenvalues of the Time Independent Schrödinger Equation, $\hat{H}\psi=E\psi$. $\psi$ contains information like the probability density distribution of the particles, so that orbital 'shapes' can be determined as iso-probability surfaces.
Looking at the electrons in orbitals as standing waves adds nothing to this approach.
As regards bound states and scattered states, I recommend this part of the Feynman Lectures on it.
A: 
How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?

It seems to me there is a confusion between a Bohr type model of atoms and molecules and the quantum mechanical framework with the orbitals. 
One can design an orbit of an electron as a standing wave classical solution and then one has to postulate the stability, i.e. that only quantized states can exist.
The orbitals around atoms and molecules are not standing waves in space in the same sense. The wavefunction is sinusoidal, but the wave nature appears in the probability distribution, which is the complex square of the wavefunction, and can be verified over many measurements. In quantum mechanics one does not have a trajectory for the electron around the atom or molecule but an orbital, as you state.

Here is a measurement of the hydrogen orbitals. Each dot is an individual measurement of a different electron, not a path for the same electron. It can only be seen as a probability distribution.

The figure at the top of this article shows the team's main result – the raw camera data for four measurements, where the hydrogen atoms were excited to states with zero, one, two and three nodes in the wavefunction for one of the parabolic coordinates. "If you look at the measured projections on the detector, you can easily recognize the nodes, and see their radial, ring-like structure," says Vrakking.

The wavefunction since it is sinusoidal, will have nodes and peaks , but it is a probability that is varying at the nodes and peaks , the electron itself is within the bounds of the uncertainty principle.
