On further pondering, I realized the source of your confusion is that, if they "don't move", how could they have relativistic momentum? Well, they do move, once you accept QM motion as motion. I.e. as far as I'm concerned, and every quantum chemist I know, we talk about 'electronic motion' all the time, even though we clearly don't mean any sort of classical motion - Especially considering that most of the time we're talking about time-independent systems. The main reason for this is correlation. Since this question does pop up on a regular basis, I thought I'd give a thorough explanation for future reference. All about correlation:
Consider a two-electron system (e.g. helium). What if we assume that the electrons aren't 'really' moving? I.e. they "see" the electrostatic repulsion of the other electron's density cloud, and this determines their kinetic energy, but that there is no explicit interdependence between their kinetic energies. I.e. each electron only 'sees' the Coulomb potential of the other (Pauli repulsion need not be taken into account for the potential of ground-state Helium)
V_{ee} = \int \int \frac{\phi_1^*(\mathbf{r})\phi_2^*(\mathbf{r'})\phi_1(\mathbf{r})\phi_2(\mathbf{r'})}{|r-r'|}d\mathbf{r'}d\mathbf{r}
Since the nuclear-electronic potential of each electron is obviously independent of the other electron's location, and we're making the assumption that the respective "kinetic energy" terms of the Hamiltonian does not depend on both r and r', this is equivalent to assuming the total wave function is a product of single-electron wave functions.
Due to fermion antisymmetry, in singlet He this is the same spatial orbital with opposite spins:
\Psi(\mathbf{r},\mathbf{r'}) = \phi(\mathbf{r})\phi(\mathbf{r'})(\alpha(\sigma_1)\beta(\sigma_2) - \alpha(\sigma_2)\beta(\sigma_1))
(where alpha and beta are spin functions) However, the orthogonality of spin states leads to the pair probability distribution (probability of electron 2 being at r' with electron 1 at r) being:
|\Psi(\mathbf{r},\mathbf{r'})|^2 = \phi^2(\mathbf{r})\phi^2(\mathbf{r'})
So, if electron 1 is in a location, say, \mathbf{r} = (1,0,0) from the nucleus (at the origin), the probability density of electron 2 in a given spot (x,y,z) is \psi^2(1,0,0)\psi^2(x,y,z).
But - if electron 1 is at \mathbf{r} = (0,1,0) from the nucleus, the probability density of electron 2 in a given spot (x,y,z) is \psi^2(0,1,0)\psi^2(x,y,z).
This is the same function! Since \psi^2(1,0,0) = \psi^2(0,1,0) = constant, because the wave function is spherically symmetrical. (But regardless of the exact wave function is, there will be some contour where the wave function has the same constant value, and hence multiple positions of electron 1 which will give the same probability-distribution for electron 2.)
This is an obviously wrong consequence no matter what interpretation you use. If electron 1 is one one side of the nucleus, electron 2 should have a higher probability of being on the opposite side. But if you assume, as most do, that the kinetic-energy term of the Hamiltonian really does represent kinetic energy, the physical interpretation is straightforward: By assuming that the electrons interact merely as static "density clouds", you neglect the interdependence of the electronic kinetic energies. As a result of this, the electrons cannot 'avoid' each other by adapting their instantaneous motion to that of the other electron. The result is an uncorrelated wave function, and an overestimation of the total electronic energy as a result. This is dubbed the 'correlation energy', and determining it is the central problem of the quantum mechanics of atomic and molecular systems. Even though the motion is by no means classical, the correlation is directly analogous (in this interpretation) to correlation within the classical many-body problem.
So even though electronic motion is not classical, even though the resulting probability density is stationary, there is a definite and measurable effect on their energy (~5%) due to correlation, and correlation is difficult to rationalize in any other terms than as an effect due to the dynamics of electronic motion. Now for an alternate derivation, that assumes nothing about the wave function, the electronic Hamiltonian for Helium is:
\hat{H}_e = -\frac{1}{2}(\nabla^2_1 + \nabla^2_2) - \frac{Z}{|\mathbf{r_1}|} - \frac{Z}{|\mathbf{r_2}|} + \frac{1}{|\mathbf{r_{12}}|}
This is the form one usually sees, and it's deceptively simple, because it mixes coordinate systems by having a Cartesian kinetic-energy operator but spherical potential-energy operators. Now substitute Cartesian coordinates for the vectors r_1, r_2, r_{12} in the kinetic energy operator. After multiple applications of the chain rule, some tedious algebra, and some nice cancellations, you get:
\nabla_1^2 + \nabla_2^2 = \frac{\partial^2}{\partial r_1^2} + \frac{2}{r_1}\frac{\partial}{\partial r_1} + 2\hat{r_1}\cdot\hat{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} + \frac{\partial^2}{\partial r_2^2} + \frac{2}{r_2}\frac{\partial}{\partial r_2} + 2\hat{r_2}\cdot\hat{r_{12}}\frac{\partial^2}{\partial r_2 \partial r_{12}}<br />
+\frac{4}{r_{12}}\frac{\partial}{\partial r_{12}} + \frac{\partial^2}{\partial r_{12}^2}
Believe it or not, but this is a simplification. We've gone from six coordinates to five - the scalar lengths of r_1, r_2, r_{12} and the r1-r12 and r2-r12 angles (represented in the dot products). It's a less compact form, for sure, but it illustrates how totally interdependent the kinetic energies and coordinates of the two electrons are. So I've shown that the potential between the two electrons is dependent on r_{12}, and that the kinetic energy is indeed independent on r_{12} as well, then. Hence, the kinetic energies of the two electrons is totally interdependent, which is hard to rationalize in other terms than that they're 'moving' and in that motion, 'avoiding' each other.
In short: In quantum-mechanics, the time-independence of a state does not mean the particles are not exhibiting the usual many-body dynamics of motion.
