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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?

frt132
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    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. – anna v Dec 12 '19 at 05:56
  • @annav Thank you! However, your answer seems to be a little opposite to the others, are you talking about different types of wavefunctions? – frt132 Dec 12 '19 at 07:19
  • 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 https://physics.stackexchange.com/q/367160/ – anna v Dec 12 '19 at 16:56
  • Yes, it does, - in the formalism of occupation numbers it is evident ("second quantization representation"). – Vladimir Kalitvianski Jul 14 '21 at 16:35

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I can see what the other answers are trying to say, but they're not quite right. There IS something very much like a wave function in QFT, and it does describe those things. There also is a Schrodinger equation.

Since you are a layman, I have to ask, to what extent do you know the basics of QM? Are you aware of the Schrodinger equation and the bra-ket formalism? If yes, then I can explain to you very simply what QFT is: exactly the same thing as QM, except that where in QM you typically have observable operators of, say, position $\hat{x}$, and a wave function that depends on time, in QFT position is no longer an operator but a parameter just like time, and the role played by $\hat{x}$ in QM is played by fields $\hat{\phi}(x)$.

Notice how the field depends on the parameter $x$ (which is NOT an operator, like in QM), while in QM the position operator $\hat{x}$ doesn't depend on anything. This is what people mean when they QFT has infinite degrees of freedom, to specify a field completely you need to specify it in every point of space. Conversely, notice that if you got rid of all spacial parameters in QFT and retained only time, you would get something that looks very much like ordinary QM, with $\hat{x}$ being a "field". That's why people say QM is QFT in 1 dimension (that dimension being time).

What is the wavefunction then? Well, it's kinda different from ordinary QM because it's an abstract vector in Hilbert space, you can't just write its value as a number in some place, like you can with QM. It's a bit more complicated than that because it carries a lot more data (remember the infinite degrees of freedom). But what is relatively simple is to produce the wavefunction that corresponds to, say, an electron and a positron as excitations of a quantum EM field at some distant time in the past, produce the wavefunction corresponding to a photon in the distant future, and then find the probability that one state evolves into the other, ie that the electron-positron pair annihilates into a photon (this way of treating problems is called scattering theory). Performing that calculation is how you get the Feynman diagrams.

Bare in mind in QFT people typically use the Heisenberg picture, in which the states ("wavefunctions") remain the same and the fields (or position/momentum operators in QM) evolve with time. I've explained it with the Schrodinger picture because it is probably what you are familiar with. But really both are exactly the same, one transforms into the other by a simple mathematical transformation.

Andreas Christophilopoulos
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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.

G. Smith
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  • 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? – frt132 Dec 12 '19 at 07:15
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    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. – G. Smith Dec 12 '19 at 23:44
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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.

John Rennie
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  • 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? – frt132 Dec 12 '19 at 07:12
  • 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). – G. Smith Dec 12 '19 at 07:19
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    @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. – John Rennie Dec 12 '19 at 07:25