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In section 4.1.1 of quantum field theory book by M. Schwartz, the author wants to calculate electron scattering by photons and writes the following interaction: $$ V= \frac{1}{2}e\int dx \psi_e(x)\phi(x)\psi_e(x)$$ Where, obviously, $\psi_e$ is the electron field and $\phi$ is the photon field. The only difference between the real QED interaction term is that he treats the photon and electron fields as scalars. In fact he writes:

"the factor of 1/2 comes from ignoring spin and treating all fields as representing real scalar particles"

It seems to me that the author says that this formula can be derived from QED in a particular limit. I tried to work out such calulation, but I failed to derive the 1/2 factor.

  1. What does ignoring spin mean?

  2. There is a particular limit in wich it makes sense to derive this result from QED?

  3. If this is the case, how can we obtain such result starting from real QED?

More generally,

  1. What is the relation between spin in QFT, non-relativistic QM and classical physics?
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closed as too broad by Ryan Unger, ACuriousMind, Gert, HDE 226868, Daniel Griscom Jan 9 '16 at 19:29

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ note that when you have $n$ equivalent fields in an interaction, you want to include a factor $\frac{1}{n!}$ to account for overcounting in the diagrams. In this case, if the electron field is real, $\psi_e^2$ has $n=2$ equivalent fields, so you want to write the interaction with a factor $1/2$ (and this has nothing to do with spin). As to "So, what is spin really?", see, eg, What is spin as it relates to subatomic particles? $\endgroup$ – AccidentalFourierTransform Jan 9 '16 at 12:48
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    $\begingroup$ This is far too broad, and many of the individual questions have been answered on this site already in one form or another. $\endgroup$ – ACuriousMind Jan 9 '16 at 15:01
  • $\begingroup$ ok, it is a nice and siple answer!So, do you think that interaction can be derived from QED? maybe I misunderstanding schwartz words. I'll take a look to your link $\endgroup$ – Alessandro Iannone Jan 9 '16 at 15:06
  • $\begingroup$ @AlessandroIannone if I were you, I wouldnt worry too much about the 1/2: its there to simplify the Feynman rules for the theory. If you dont write the interaction with the 1/2, the Feynman rule for the vertices is to include a factor of $-2ie$; if you write the interaction with a 1/2 in front, the rules are $-ie$ for each vertex, which is simpler. You probably dont know what a Feynman rule is yet, so keep on reading the book and when you fully understand how Feynman diagrams work, come back to this post and re-read it in detail. $\endgroup$ – AccidentalFourierTransform Jan 9 '16 at 16:01
  • $\begingroup$ @AlessandroIannone For now: this interaction cannot be derived from QED, because QED and your theory are inherently different and unrrelated. The theory Schwartz is developing is just a toy model to explain the gross details of QFT and interactions, without taking into account the subtleties of QED. $\endgroup$ – AccidentalFourierTransform Jan 9 '16 at 16:03
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You shouldn't ignore spin even in the nonrelativistic limit. Schwartz is only ignoring spin for pedagogical reasons. Calculating scattering amplitudes with scalars is easier to do, so he wants you to learn how to do that first before getting into all the complications of spin 1/2 and spin 1 particles (where you need to worry about fermions and gauge invariance, respectively).

As an example of where spin shows up in a nonrelativistic problem, consider filling the electron orbitals of, say, helium. By the Pauli exclusion principle, if the electron had no spin you'd expect to have to fill the lowest two orbitals. But electrons do have spin, so in addition to the $n,l,m$ quantum numbers there are two internal polarization states associated with spin. So in fact we can fit two electrons into the lowest orbital, one with spin up and one with spin down. If you ignored the electron spin this would not be possible and chemistry would be completely different.

I wouldn't worry too much about where the 1/2 comes from for now, it's a detail and really the only way to fully understand where it comes from is to understand the scattering calculation in the first place. But, just as a side note, different place where you see a similar factor of 1/2 is in comparing the kinetic term for a real scalar field: \begin{equation} -\frac{1}{2} \partial _\mu \phi \partial^\mu \phi, \end{equation} to a complex scalar field \begin{equation} -\partial_\mu \Phi^* \partial_\mu \Phi. \end{equation} You can see you need this for example by constructing the propagator. You will find in both cases the normalization guarantees that the propagator goes like $Z/p^2-i\epsilon$ with $Z=1$ (any other normalization would give you $Z\neq 1$). There's a similar normalization requirement underlying the $1/2$ in your cubic vertex.

Spin is "really" about representations of the poincaire group. There is a classical analogue of spin at least for bosons (integer spin), it more or less amounts to dealing with tensor valued fields. from the perspective of quantum theory, a "classical" field is a coherent state with many quanta. For fermions there isn't really a classical analogue because the Pauli exclusion principle forbids you from making coherent states of fermions.

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  • $\begingroup$ Thanks for the answer. Now I realize why the 1/2 factor has to be there, but I still not know if and how we can derive it from QED. Of course I know that we cannot eliminate the concept of spin in QM (your argument is a clear example). What I would ask is when (and if) we can "ignore" it in studying the dynamics of the theory (e.g. spin orbit coupling is a relativistic correction to the shrodinger eqn). Is there any limit in wich it makes sense to approximate the QED interaction to the one I write in the question? Finally, if spin is poincaire rep, what's about NQRM and galilean rep? $\endgroup$ – Alessandro Iannone Jan 9 '16 at 14:58
  • $\begingroup$ No, there's no sense that interaction is an approximation to qed. It is purely a pedagogical toy example. So you can't derive the 1/2 from qed. The Galilean group is a limit (Wigner-Innonu contraction) of the poincaire group. If you go through the representation theory for the poincaire group you see you need to worry about the rotation group--basically for a massive particle you can go into its rest frame and then the states of the particle on that frame are reps of the rotation group. $\endgroup$ – Andrew Jan 9 '16 at 15:05
  • $\begingroup$ I don't know of any case where you can simply ignore spin. I think there are cases where you could focus on the physics of a single polarization state (for example if you have a massless fermion and you work in the infinite momentum frame), but even in those cases the direction of the polarization is still a dof you have to keep track of. It can also be that there are certain special questions you can ask where the spin doesn't matter to some approximation (energy levels of hydrogen ignoring spin-orbit coupling), but it's not as simple as taking a nonrelativistic limit, you need extra luck. $\endgroup$ – Andrew Jan 9 '16 at 15:13
  • $\begingroup$ Ok, now it is all clear. Anyway, do you know any nice reference where I can study Poincare and galilean groups and their rep? $\endgroup$ – Alessandro Iannone Jan 9 '16 at 15:19
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    $\begingroup$ @AccidentalFourierTransform no worries! So long as everyone's learning something (especially me) that's the goal! $\endgroup$ – Andrew Jan 10 '16 at 0:50

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