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Not sure what "trivial" means here. 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.

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.

General concepts

References for special cases

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.

Crossing symmetry doesn't require choosing a specific CPT transform, but 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.

For perspective, 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 the states $\psi_a|0\ra$ and $\la 0|\psi_a$. We don't need the LSZ context for this, and we don't need to choose a specific convention for this, either.

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 (as in the spin-1/2 case), which can be handled according to the general principle described above. Equations (13.5.1)-(13.5.9) in Weinberg give a photon example.

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.

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 $\la 0|\psi_a$.

Not sure what "trivial" means here. 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.