Is it possible for an electron in the atom to spontaneously change the direction of its spin component on the z-axis? By "spontaneously" I mean without interacting with any external field. 
More specifically, can the $S_z$ component of an electron's spin collapse into one of its two possible states ($S_{z\pm} = \pm\frac{\hbar}{2}$) spontaneously within the atom?  
I suppose that normally the state of $S_z$ for an electron in the atom is a linear combination of its two possible states, e.g. $S_z = \frac{1}{\sqrt{2}} S_{z+} + \frac{1}{\sqrt{2}} S_{z-}. $ 
Can $S_z$ collapse into one state and then back to the superposition state, then collapse again etc., without any external interference? 
If so, what would be the conditions for this to happen? Could it be related to spin-orbit coupling or other phenomena inside the atom? 
 A: We are in the quantum mechanical regime when talking of electrons and atoms. This means the electrons are in bound states, i.e. with fixed energy levels, and cannot "jump" levels because of energy consrvation. Generally to be found with a different spin orientation would require energy input or output, i.e. an interaction:
Look at the fine structure of the hydrogen atom.:

When the familiar red spectral line of the hydrogen spectrum is examined at very high resolution, it is found to be a closely-spaced doublet. This splitting is called fine structure and was one of the first experimental evidences for electron spin.



The small splitting of the spectral line is attributed to an interaction between the electron spin S and the orbital angular momentum L. It is called the spin-orbit interaction.

So there can be no "collapse" as you envisage it. After all "collapse" is another vocabulary for "measurement", which means "interaction".
A: Without any external field and in the absence of orbital momentum the spin states are degenerate. The system is then in a mixed state described by a diagonal density matrix $$\begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}~.$$
