Self-annihilating photons and Z bosons?

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(If my question seems irrational, it's ok - I'm an undergrad!)

So I was reading up on the standard model and read that Z bosons are their own antiparticles. Does this mean that when two Z bosons are in physical contact, they annihilate? I ask this for photons too since I read this on the "Photon" wiki article, but I'd like some more assurance.
 
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Yes, but we need to discuss what the word "annihilate" means. Originally it meant "turns into a pair of photons", but it can be used more generally. In that sense an electron and positron can annihilate into a neutrino-antineutrino pair. Or given sufficient energy, a neutrino-antineutrino pair can annihilate back into positron and electron. When a proton and antiproton "annihilate", multiple particles will be the typical outcome.

Two photons can interact, with the aid of a closed electron loop, to form another pair of photons. This is a very small effect that has been predicted but never observed.

Z0's can interact with each other, but it is called vector boson fusion. They may combine to form a Higgs boson, which then decays into something else. One way that has been proposed to search for the Higgs boson is to look for a resonance in Z-Z or W-W scattering.
 
I am undergrad and it was my question too for some month( intraction of two photons),
for example what is the photonic explanation of static electromagnetic waves?
or two destroyer electromagnetic wave?
 
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I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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