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I'd like to understand crossing symmetry in QFT better. I've only found somewhat detailed treatments of the scalar case, somewhat contradictory comments for the spin-1/2 case, and nothing so far for the spin-1 case.

Broadly speaking, does anyone have some good references for a rigorous treatment of the crossing symmetry of particles with non-zero spin?

Speaking of the spin-1/2 case, Peskin writes somewhat cryptically below Eq. 5.68 that when crossing spin-1/2 particles one gets an extra minus, but that "[t]he minus sign can be compensated by changing our phase convention for $v(k)$."

Weinberg, on the other hand, suggests that the minus sign comes from Fermi statistics and continues that "crossing symmetry is not an ordinary symmetry (it involves an analytic continuation in kinematic variables) and it is difficult to use it with any precision for general processes." (!)

So, in particular, what would the explicit spinors $u(p)$ and $v(p)$ be for Peskin under the different sign convention?

Further, is crossing symmetry trivial for spin-1 particles like it is for scalars?

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General concepts

Crossing symmetry is basically the CPT theorem applied in the context of the LSZ formula, using microcausality to re-order the field operators. The role of the CPT theorem is to relate particles to their antiparticles. The CPT transform is not unique: in particular, we can compose it with any proper Poincaré transform to get another equally-good CPT transform. The CPT theorem defines a conjugate relation between two sets of single-particle states (like between the set of all single-electron states and the set of all single-positron states), but it doesn't define a unique one-to-one relationship between individual single-particle states. So questions like "does the antiparticle of a spin-up electron have spin up or spin down?" don't have unique answers. The answer is convention-dependent.

Crossing symmetry involves replacing an incoming particle with an outgoing antiparticle (or conversely), and since the relationship between individual single-particle and single-antiparticle states is convention-dependent, we can compensate for a minus sign that comes from Fermi statistics by switching conventions, as Peskin & Schroeder wrote.

Crossing symmetry is not an "ordinary symmetry" that relates physical states to other physical states, and maybe this limits its utility as Weinberg suggested, but neither of these points contradict what Peskin & Schroeder wrote.

To make the relationship between crossing symmetry and CPT more explicit, consider a time-ordered correlation function $$ \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\dpsi}{\psi^\dagger} \la 0|T\,X_A(x) \psi_a(y)|0\ra $$ where $\psi_a(y)$ is an individual field operator with Lorentz index $a$ and at the spacetime point $y$, and where $X_A(x)$ is an abbreviation for some product of field operators with indices collectively denoted $A$ and spacetime points collectively denoted $x$. If $\psi$ is a fermion field, then the overall sign of the correlation function (and hence of the scattering amplitude) is affected by how the field-operator factors are ordered.

Starting with this correlation function, we can use the LSZ reduction formula to construct a scattering amplitude in which the particle associated with $\psi$ is either in the initial state or in the final state. CPT says that the single-particle part of the state $\psi_a(y)|0\ra$ is an antiparticle of the single-particle part of the state $\la 0|\psi_a(y)$, or equivalently of the state $\dpsi_a(y)|0\ra$. The idea behind LSZ is that we can isolate the desired single-particle contributions to the in/out states by isolating the associated poles. The field operator $\psi_a$ can be written as the sum of its positive- and negative-frequency parts, $\psi_a(y)=\psi_a^+(y)+\psi_a^-(y)$, which act on a state-vector (ket) to their right as annihilation and creation operators, respectively, and conversely when acting on a state-vecctor (bra) to their left. The LSZ formula uses this to select one of the two poles, either incoming or outgoing. The identitities $$ \big(\psi_a^+\big)^\dagger = \big(\dpsi_a\big)^- \hskip2cm \big(\psi_a^-\big)^\dagger = \big(\dpsi_a\big)^+ $$ say that the particles corresponding to these two poles are antiparticles of each other. Crossing symmetry amounts to a relationship between the formulas that LSZ uses to select either of these two poles. So in general, what crossing symmetry does to the crossed particle's spin-state is determined by the relationship between the single-particle parts of $\psi_a|0\ra$ and $\la 0|\psi_a$.

References for special cases

is crossing symmetry trivial for spin-1 particles like it is for scalars?

Crossing symmetry for spin-1 particles (like photons) doesn't have any minus signs from Fermi statistics, but the amplitudes still involve specific components of the field operators (photon polarizations). Equations (13.5.1)-(13.5.9) in Weinberg give a photon example.

Section 2.1 in https://arxiv.org/abs/1605.06111 gives some convention-dependent details for the case of a "vector, Dirac, and left- or right-handed (massless) Weyl representation respectively" with a footnote that says "the overall sign that relates $u^\sigma$ with $v^{-\sigma}$ ... is conventional since it depends on the choice of the CPT phase." Section 3 in the same paper shows some detailed examples for various spins, for both massive and massless particles.

Itzykson and Zuber's book Quantum Field Theory also works out an example involving crossing symmetry in a process involving electrons and photons (section 5-2-2). They also show a detailed derivation of the LSZ formula for Dirac fermions (section 5-1-6) from which the details of crossing symmetry can be inferred, and it illustrates the general concepts outlined above.

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  • $\begingroup$ Thank you very much for this answer! I especially appreciate the linked paper, which was very helpful. In terms of "is crossing symmetry trivial for spin-1 particles like it is for scalars?" I meant: are there any minus signs that show up when crossing spin 1 particles? My understanding is that no, there aren't any extra minus signs (beyond possibly coming from evaluating the polarization vectors at minus the momentum). In terms of Weinberg, my sense is he actually missed the boat here; i.e. that one can use crossing symmetry with precision $\endgroup$ Commented May 4, 2020 at 13:04
  • $\begingroup$ I believe you've sufficiently answered the question, so I'll accept it and award the bounty. Thank you very, very much! Briefly, do you by any chance know an example of a convention for fermions such that the minus sign is compensated (or know of somewhere where this is done)? $\endgroup$ Commented May 4, 2020 at 13:05
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    $\begingroup$ @WAH You are very welcome. I think P&S meant that we can compensate for a Fermi-statistics minus sign by switching conventions -- I mean, by using two different conventions on the two different sides of the crossing-symmetry equation. I don't know of a reference that shows an explicit example of that, but I did another reference to the answer (Itzykson and Zuber's book), because sometimes having multiple perspectives from multiple references can be helpful. $\endgroup$ Commented May 4, 2020 at 14:16
  • $\begingroup$ Very nice additional reference. I'd looked through I&Z before, but didn't catch their nice photon/fermion example. My guess is that P&S mean that one could choose to use a different explicit realization of spinors such that, e.g., $u^\sigma(-p) = -v^{-\sigma}(p)$. I'm not clever enough at the moment to cook up such a representation, though $\endgroup$ Commented May 4, 2020 at 18:17

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