Why is a nearby nucleus required for Pair Creation? I was recently studying Pair Production and Annihilation. The author mentions that a nearby nucleus is required when the photon materialises into a particle and an anti-particle. The explanation given is that the momentum and the energy must be conserved. However, there is no calculation given that shows the violation of energy. The reason is just blankly stated. Is there more to this concept. Please explain? As far as I know that from the knowledge of Particle Physics, virtual photons can violate the conservation laws if the time scales are very small due to the Heisenberg Uncertainty Principle. Then why can't we apply the same idea here?
PS: I've read the other answers but none of them include the contribution of Nucleus' energy/momentum to conserve momentum or energy. 
 A: You don't need a calculation. Let's go backwards: consider the electron-positron pair. There is an inertial frame of reference (the "centre of mass" frame) where the total momentum of the pair is zero, i.e. the centre of mass is still. Now, keeping this frame of reference, rewind the movie: before the creation of the pair there was only a photon. But a photon can never be at rest, no matter how you change your frames of reference, as a consequence of special relativity. So conservation of momentum was violated during pair creation.
In presence of a nucleus nearby, the matter is different, because, as you say, you can use the nucleus's momentum and energy to make up for the difference.
Another way of stating it is (but here you should seek confirmation from someone more expert than me) at least two interactions (two vertices) are necessary to make a physically meaningful Feynman diagram. As you say, virtual photons can temporarily violate the conservation law. A virtual photon must be sent out from a vertex of the diagram and reabsorbed in the other, otherwise it's not virtual at all, and the violation is actual.
A: The argument depends on a fact about photons that you may or may not have encountered yet: if you have a photon with energy $E$, that photon must carry momentum $p = E/c$ in some direction.  If you change your opinion about the photon's momentum by observing it from some other reference frame, you also must change your opinion about its energy.  For instance if you run away from your light source your photons will exchange less momentum when they interact with you than if you and the light source were relatively at rest; you'll see these photons as having less energy, or "red shifted."
Matter particles (or collections of matter particles, like electron-positron pairs) have a property that photons do not: a "rest frame" where the momentum is zero.
So suppose you have a single photon which transforms into an electron-positron pair with equal and opposite momentum --- that is, you're in the electron-positron rest frame.  In order to create these two particles, your single photon must have had a little more than 1 MeV of energy (more if the e-p pair have kinetic energy).  But in order to conserve momentum your photon must have had zero momentum, and therefore zero energy.  This contradiction is why a single photon cannot transform into an electron-positron pair.
The flaw goes away if the photon can steal momentum from its environment.  The inside of an atom has a strong electric field, which is made of (in quantum electrodynamics) "virtual" photons.  Your real photon can exchange momentum with the atom by scattering from one of these virtual photons; that's where the pair creation happens.  Theoretically it should be possible to generate e-p pairs by sending gamma rays into a very intense laboratory electric field or a microwave cavity, but I don't think those environments are experimentally accessible.
