How is reaction possible? The verticies do not conserve charge. Also, why is the arrow for the positron pointing downwards when as time increases, the positron should move towards its vertex?

Sorry, I just a bit all confused about this diagram and how it's formed enter image description here

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    $\begingroup$ Why is charge not conserved at the vertices? At each vertex one elementary charge enters and leaves. Furthermore, labelling Feynman diagrams with time and space axes is a dangerous idea (as it will easily lead to false reasoning). The arrow does not depict the direction the particle is moving, but the direction the fermion number is flowing. See the answers at <physics.stackexchange.com/questions/183648/…>. $\endgroup$ – Sebastian Riese May 24 '15 at 15:00
  • $\begingroup$ Adding to what Sebastian already said, Feynman diagrams are a mathematical series expansion of a physical process. They are not physical processes in themselves, only the expectation value of their complex sum is (and keep in mind that each diagram is actually a multidimensional integral!), and only in cases where the problem can be described with a convergent series of diagrams. This ain't new. You can also describe planetary motion with a perturbation series similar to old style epicycles. None of the terms in that series would be a physical process in itself, even if they converge. $\endgroup$ – CuriousOne May 24 '15 at 16:14
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    $\begingroup$ Hi.The total charge is zero at the beginning $q_{e^-} + q_{e_+}=0 $ and at the end(two photons). $\endgroup$ – Constantine Black May 25 '15 at 8:07
  • $\begingroup$ Thank You very much. I don't feel my lecturer really explained this properly as he always draws the arrows representing particle direction (I guess for our understanding as undergraduates). It was only looking online that I found the arrows seemingly to be in a confusing direction. $\endgroup$ – SomePhysicsStudent May 27 '15 at 19:40

It is the way one reads/writes Feynman diagrams, a particle going backwards in time is the antiparticle. The electron radiates a gamma, and continues to meet the positron , annihilating charge with another photon. Two real particles are needed for momentum conservation in the center of mass, and two photon vertices are the simplest case within the standard model.

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    $\begingroup$ And the two vertices are also needed because there is no fundamental term coupling a charge to two photons in QED (though and effective one arises as a loop correction). $\endgroup$ – Sebastian Riese May 24 '15 at 15:30

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