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I don't study physics, this is a layman question. From some online sources, some probability would be given by squaring the parameters of a wave function. Some sources also claim that wave function describes all the states of an entity.

Does wave function describe the probability of what new particles (or photons) would be generated after two particles annihilate each other? Does it describe the particle decay, or electron emitting and reabsorbing a virtual photon? Or anything involving new particles?

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  • $\begingroup$ wave functions are quantum mechanical mathematical models of particle interactions, and yes, the models can describe annihilation and predict probabilities for what new particles will appear.. The models are successful in describing interactions. Given the wave functions of the standard model, new particles can be predicted to be found in higher energies, as happened with the Higgs boson. $\endgroup$ – anna v Dec 12 '19 at 5:56
  • $\begingroup$ @annav Thank you! However, your answer seems to be a little opposite to the others, are you talking about different types of wavefunctions? $\endgroup$ – frt132 Dec 12 '19 at 7:19
  • $\begingroup$ I am talking of the calculations using field theory, feynman diagrams, of the crossections using the standard model. Crossections are connected with the probailty of interaction , therefore a wave function for the interacting systemg in your case particle scattering off antiparticle.. see my answer here physics.stackexchange.com/questions/367160/… $\endgroup$ – anna v Dec 12 '19 at 16:56
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When we use the term wave function we normally mean the function that is a solution to the Schrodinger equation, and particle creation and annihilation does not exist in the Schrodinger approach. So by definition the Schrodinger wave function cannot describe any process in which articles are created or destroyed.

To understand creation and annihilation we need to use quantum field theory. The difference in this approach is that particles are no longer fundamental objects. Instead they are excitations of a quantum field and they can be created by adding energy to the field and destroyed by removing energy from the field.

The idea of a wave function does not really exist in quantum field theory. The individual particles can be approximately described by a one particle wave function as long as the particles do not interact with any other particles. In this case the particles are described by functions called Fock states. However this is only an approximation since all real particles interact with each other and the interaction mixes up their states.

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  • $\begingroup$ Thank you, boss. If you could indulge me, in QFT, is excitation just a one/zero state or there is uncertainty too in the 'magnitude' of excitation? Are magnitudes quantized? The excitation also doesn't have an exact location? It seems there's infinitely many ways of fields interacting with each other too, what tools are used for approximating the results of an interaction, do you use something like Feynman diagram for that too? $\endgroup$ – frt132 Dec 12 '19 at 7:12
  • $\begingroup$ For bosonic fields (say, photons), excitations are to $n$-particle states. The free-particle state is usually taken to be a plane wave with definite four-momentum. It fills all space and is completely unlocalized. Feynman diagrams are the usual way to calculate interactions. Leaving kinematics aside, there are actually relatively few fundamental interactions. There are only 17 fundamental fields and not every one interacts with every other (or itself). $\endgroup$ – G. Smith Dec 12 '19 at 7:19
  • $\begingroup$ @frt132 the problem is that you're asking how QFT works, and that's not a short discussion. If you want to join the physics SE chat room I'd be happy to discuss it there. $\endgroup$ – John Rennie Dec 12 '19 at 7:25
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No. The Schrodinger-style wave functions used in non-relativistic quantum mechanics cannot describe particles being created and destroyed. (But the quantum fields used in quantum field theory can.)

An $N$-particle wave function in the position representation is a function of $3N$ position variables. $N$ simply has no way to change in this formalism. This limitation is the main reason that quantum field theory was invented.

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  • $\begingroup$ Thank you, but Feynman diagrams seem to be dealing with particles but not fields, and they could be used to calculate probability of particle creation too? $\endgroup$ – frt132 Dec 12 '19 at 7:15
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    $\begingroup$ Feynman diagrams may look like pictures of particles bouncing off each other but they are actually shorthand for complicated mathematical expressions (namely transition amplitudes) in quantum field theory. In QFT, particles are excitations (or “quanta”) of quantum fields. There are no particles without fields for them to be the quanta of. Feynman diagrams are used to calculate the probability of particle creation. $\endgroup$ – G. Smith Dec 12 '19 at 23:44

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