What Causes Atomic Orbitals to Hybridise?
Hybridization is a mathematical concept developed to cope up the flaws of Valence Bond theory. It hinges solely on the linear combination of wavefunctions of atomic orbitals having more or less same energy. Hybridized state is a purely theoretical concept &amp; cannot be detected even spectroscopically. 

There is nothing like "collapse." The wavefunction of a hybridized orbital is "mixing" of the atomic orbitals only. The number of hybrid orbitals are same as the number of atomic orbitals participating in the "mixing up". The wavefunctions interfere constructively or destructively to give rise to equivalent orbitals known as hybrid orbitals.

The Methane Case:

Carbon has valence configuration: $2s^2 2p_x ^1  2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then promotion comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ &amp; $p$ orbitals are not  same. To deal with the problem, concept of hybridization was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$