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  Does a real electron possess a certain spin?

+ 2 like - 2 dislike

A real electron is permanently coupled to the electromagnetic degrees of freedom. Experimentally we observe a certain spin in the electron. In theory we assign the spin to a free (non interacting) electron. When we turn the coupling on (in theory), the latter can make the resulting spin state uncertain - because the spin of a constituent particle is not exactly conserved quantity. Is there any proof that the coupling "conserves" the spin of the constituent electron?

asked Nov 4, 2014 in Theoretical Physics by Vladimir Kalitvianski (102 points) [ no revision ]
recategorized Nov 10, 2014 by Dilaton

When we turn the coupling on (in theory), the latter can make the resulting spin state uncertain - because the spin of a constituent particle is not exactly conserved quantity.

Who says or claims that, in particular in the context of elementary particles such as an electron which is not a composite of anything else?

The only thing I could imagine is some supersymmetric interections that could for example do conversions between the electron and the selectron. The spin is one of the unique quantum numbers of a quantum state, it can (apart from anyons and other exotic things?) only take half or integer values. So there is no uncertainty in the spin of a elementary particle (such as the electron) at all.

If the spin of the electron (or any other elementary particle) changes, it is no longer called an electron!

This question looks rather confused about basic things in quantum mechanics and particle physics.


Also, apart from being confused about basic facts, the question and the follow up comments apply a lot of non-standard or incorrect terminology in stranges ways (for example the ongoing talking about "real electrons", " real photons", etc. So even though the answer nicely clarifies the confusions, I think the question itself is not a god fit for a graduate-level upward physics site and should therefore be moved to chat or closed.

1 Answer

+ 5 like - 0 dislike

Let me first define the context in which the question can be answered.

In any quantum field theory, real particles appear as 1-particle states of asymptotic generalized free field operators that propagate freely long after a scattering event is over (and long before the next scattering event begins); in all other situations, the notion of a particle (and of particle number) is ill-defined, as the interacting Hilbert space does not have a Fock space structure. Thus one can point to individual particles only when this asymptotic situation is valid.

In a quantum field theory without massless particles, the asymptotic field operators have definite masses, one for each bound state of the theory. In this case, the physical particles are precisely the bound states of the theory. (In QCD, for example, the physical particles are the mesons and baryons, not the quarks.) This is the situation described by the LSZ theorem. In a quantum field theory with massless particles, the asymptotic field operators cannot have definite masses since (in a vague but useful imagery) asymptotic particles are always dressed by a cloud of arbitrarily many soft massless particles - soft meaning with very low energy. However, this imagery is again ill-defined because of the lack of an interacting Fock space. On a better defined level, one has states composed of a coherent state of the massless field and a 1-particle state of an asymptotic generalized free field operator corresponding to a so-called infraparticle. The Fourier transform of the 2-point function of the infraparticle field is an integral $$\int_{m_0}^\infty dm \rho(m) (p^2+m^2)^{-1}$$ where $m_0$ is the physical rest mass of the infraparticle. (In contrast, the Fourier transform of the 2-point function of an ordinary particle is just $(p^2+m^2)^{-1}$.)

Since QED contains a massless electromagnetic field, a physical electron is an infraparticle, associated with an asymptotic generalized free field with the electron charge $e$. This infraparticle is multiplied with (aka ''dressed by'') a coherent state of the electromagnetic field. This dressed infraparticle is the physical electron including its e/m field, a boosted Coulomb field. It behaves correctly under scattering, and (unlike in the conventional, LSZ-based scattering theory, which is valid only in the massive case) no infrared divergences appear. See the references in this answer for the nonrelativistic case, and work by Kulish and Faddeev in the relativistic case.

Since the electron is characterized by a generalized free field operator, there is an associated representation of the Poincare group. However, in the center of mass frame of the electron, we still have an irreducible representation of the Lorentz group, and this fixes the spin. To 1/2, as we know from experiment, and as it is built in into QED.

Note that spin is not a conserved quantity, only total angular momentum. However, the word spin has a different meaning in the sentence ''spin is not conserved'' and in the sentence ''the electron has spin 1/2''.

When we say that the electron has spin 1/2 we mean that the rotation group is represented by a 2-dimensional projective representation, nothing more. But when we measure the spin of an electron (e.g., in a Spin-Gerlach like experiment), we measure the state of the projection of the spin operator in the direction of flight - giving spin up or spin down. In this latter situation, if spin is (not) conserved, the state of the spin remains (does not remain) constant, but the electron always remains a particle of spin 1/2 - meaning just having two (nonrelativistic) components.

answered Nov 5, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

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